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Electronic
Applications
An Online Text
Bob Zulinski
Associate Professor
of Electrical Engineering
Michigan Technological University
Version 2.0
Electronic Applications ii
Dedication
Human beings are a delightful and complex amalgam of
the spiritual, the emotional, the intellectual, and the physical.
This is dedicated to all of them, especially to those
who honor and nurture me with their friendship and love.
Electronic Applications iii
Table of Contents
Preface x
Philosophy of an Online Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x
Notes for Printing This Document . . . . . . . . . . . . . . . . . . . . . . . . . xii
Copyright Notice and Information . . . . . . . . . . . . . . . . . . . . . . . . . xii
Thermal Considerations in Amplifiers 1
Junction Temperature, T
J
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Thermal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Specifying Device Ratings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Power Derating Curves 3
Device Maximum Ratings 4
Thermal Resistance, Junction-to-Case . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Thermal Resistance, Case-to-Sink . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Thermal Resistance, Sink-to-Ambient . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Power Devices 6
Power BJTs vs. Small-Signal BJTs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Effect of Temperature on BJTs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Thermal Runaway 6
BJT Safe Operating Area (SOA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Power MOSFETs (Enhancement-Mode Only) . . . . . . . . . . . . . . . . . . . . . 8
Power MOSFET Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Power Amplifiers 10
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Calculating Power Dissipation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Amplifier Performance Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Efficiency 12
Power Capability 12
Class A Integrated-Circuit Power Amplifier . . . . . . . . . . . . . . . . . . . . . . 13
Class A Analysis 13
Class A Waveforms 15
Class A Calculations 16
Electronic Applications iv
Class B Complementary Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Class B Analysis 18
Class B Calculations 19
Crossover Distortion 22
Single-Ended Power Amplifiers 23
Class A Single-Ended Power Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 23
Circuit Operation 23
Calculations 25
Class B Single-Ended Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Circuit Operation 27
Calculations 31
Introduction to Feedback 33
Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Positive Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Closed-Loop Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Gain Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Reduction of Nonlinear Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Reduction of Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Noise Sources 37
Noise Models 37
Signal-to-Noise Ratio, SNR 37
Reduction of Noise Using Feedback 38
Types of Feedback 39
On the Input Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
On the Output Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
The Amplifier Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
The Feedback Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Gain Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Effect on Input Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Series Mixing 43
Parallel Mixing 44
Effect on Output Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Voltage Sensing 45
Current Sensing 46
Summary of Feedback Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Electronic Applications v
Practical Feedback Networks 49
Identifying Feedback Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
At the Input 49
At the Output 49
Identifying Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Estimating the Feedback Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Design of Feedback Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Transient and Frequency Response in Feedback Amplifiers 52
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Transient Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Transient Response and Pole Location . . . . . . . . . . . . . . . . . . . . . . . . . 53
Frequency Response and Pole Location . . . . . . . . . . . . . . . . . . . . . . . . 54
Graphical Details of Complex Conjugate Poles . . . . . . . . . . . . . . . . . . . 55
Frequency Characteristics of Feedback Amplifiers 56
Dominant-Pole Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Gain-Bandwidth Product 57
Dominant-Pole Gain-Bandwidth Example 58
Feedback Factor and Pole Location 59
Summary 59
Two-Pole Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Amplifiers With 3 or More Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
Gain Margin and Phase Margin 66
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Gain Margin, Phase Margin, and Bode Plots . . . . . . . . . . . . . . . . . . . . . 69
Example #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Example #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Example #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Compensation of Amplifiers 76
Compensation by Adding a Dominant Pole . . . . . . . . . . . . . . . . . . . . . . 77
Compensation by Moving an Existing Pole . . . . . . . . . . . . . . . . . . . . . . 78
Compensation by Adding a Pole and a Zero . . . . . . . . . . . . . . . . . . . . . 80
Compensation in the LM741 Operational Amplifier . . . . . . . . . . . . . . . . 81
Electronic Applications vi
Oscillators 82
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
RC Oscillator Example - Wien-Bridge Oscillator . . . . . . . . . . . . . . . . . 85
Practical Wien-Bridge Oscillator #1 89
Practical Wien-Bridge Oscillator #2 90
Practical Wien-Bridge Oscillator #3 91
LC Oscillators - The Hartley Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 92
LC Oscillators - The Colpitts Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 97
Hartley - Colpitts Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
Crystal Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
The Piezoelectric Effect 99
Quartz Crystal Characteristics 99
Pierce Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Series-Resonant Crystal Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Comparators and Schmitt Triggers 102
Comparators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
Ideal Transfer Characteristic 102
Actual Transfer Characteristic 102
Speed vs. Overdrive 103
Open-Loop Comparators 104
Totem-Pole vs. Open Collector Outputs 105
Schmitt Triggers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
Inverting Schmitt Trigger 106
Noninverting Schmitt Trigger 107
Schmitt Trigger Design 108
Schmitt Trigger Example 109
Single-Pole Circuits 111
General Response to Piecewise-Constant Inputs . . . . . . . . . . . . . . . . 112
Time Interval Between Known Values of Exponential Response . . . . . 114
Example 115
Pulse Response of Single-Pole High-Pass . . . . . . . . . . . . . . . . . . . . . 116
Pulse Response of Single-Pole Low-Pass . . . . . . . . . . . . . . . . . . . . . . 118
Steady-State Rectangular-Wave Response of Single-Pole High-Pass 120
Special Cases 124
Steady-State Rectangular-Wave Response of Single-Pole Low-Pass . 125
Electronic Applications vii
555 IC Precision Timer 129
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
Monostable Multivibrator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Operation 132
Calculating the Output Pulse Width 133
Leakage, Bias, and Discharge Currents 135
Triggering in Monostable Mode 136
Astable Multivibrator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
Operation 137
Varying the Control Voltage 140
Energy Storage Elements 143
Capacitor in Series with Arbitrary Impedance . . . . . . . . . . . . . . . . . . . 143
Inductor in Parallel with Arbitrary Impedance . . . . . . . . . . . . . . . . . . . . 144
Capacitive Voltage Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
Miscellany . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
RC Attenuators 146
The Compensated Attenuator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
The Generalized RC Attenuator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
Poles and Zeros 150
Undercompensated Attenuator Rectangular-Wave Response 152
Overcompensated Attenuator Rectangular-Wave Response 152
Practical RC Attenuators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
An Oscilloscope Probe Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
Diode Static Characteristics 156
Measuring Reverse Saturation Current . . . . . . . . . . . . . . . . . . . . . . . . 156
Diode Cut-in Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Temperature Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Depletion Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Diffusion Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
Zener Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
IC Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
Electronic Applications viii
BJT Static Characteristics 162
Output Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
BJT Saturation-Region Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
BJT Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Rule of Thumb Voltages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Power Dissipation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Inverse Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Diode Switching Characteristics 167
Zero Bias . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
Reverse Bias . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
Forward Bias . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
Diode Switching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
Schottkv Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
BJT and FET Switching Characteristics 175
BJT Operation in the Active Region . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
BJT Operation in the Saturation Region . . . . . . . . . . . . . . . . . . . . . . . 177
Typical Switching Waveforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
Schottkv Transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
Increasing BJT Switching Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
FET Switching Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Introduction to DC-DC Conversion 181
Converter Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
Capacitor-Based Converters 183
Peak Rectifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Clamp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
Building a Capacitor-Based Converter . . . . . . . . . . . . . . . . . . . . . . . . . 185
Nonideal Converters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
Simulation of a Four-Stage Converter . . . . . . . . . . . . . . . . . . . . . . . . . 189
Electronic Applications ix
Inductor-Based Converters 192
Techniques and Assumptions for Inductive-Converter Analysis . . . . . 192
Buck Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
Switching Details 193
Inductor Current 194
Output Current 195
Output Voltage 195
Input Current 196
Input and Output Power 197
Real Buck Converters 197
Estimating Peak-to-Peak Output Voltage Ripple 199
Buck Converter Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
Boost Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
Inductor Current 206
Output Voltage 207
Input Current 207
Current Waveforms 208
Estimating Peak-to-Peak Output Voltage Ripple 209
Output Voltage Sensitivity 210
Boost Converter Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
Buck-Boost Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
Inductor Current 214
Output Voltage 215
Current Waveforms 216
Estimating Peak-to-Peak Output Voltage Ripple 217
Output Voltage Sensitivity 218
(End of contents)
Electronic Applications x
1
I use the word supposedly because, in my view, the official rewards for textbook
authoring fall far short of what is appropriate and what is achievable through an equivalent
research effort, despite all the administrative lip service to the contrary. These arguments,
though, are more appropriately left to a different soapbox.
Preface
Philosophy of an Online Text
I think of myself as an educator rather than an engineer. And it has
long seemed to me that, as educators, we should endeavor to bring
to the student not only as much information as possible, but we
should strive to make that information as accessible as possible,
and as inexpensive as possible.
The technology of the Internet and the World Wide Web now allows
us to virtually give away knowledge! Yet, we dont, choosing
instead to write another conventional text book, and print, sell, and
use it in the conventional manner. The whys are undoubtedly
intricate and many; I offer only a few observations:
G Any change is difficult and resisted. This is true in the habits
we form, the tasks we perform, the relationships we engage.
It is simply easier not to change than it is to change. Though
change is inevitable, it is not well-suited to the behavior of any
organism.
G The proper reward structure is not in place. Faculty are
supposedly rewarded for writing textbooks, thereby bringing
fame and immortality to the institution of their employ.
1
The
recognition and reward structure are simply not there for a text
that is simply posted on the web.
G No economic incentive exists to create and maintain a
Electronic Applications xi
structure that allows all authors to publish in this manner; that
allows students easy access to all such material, and that
rigorously ensures the material will exceed a minimum
acceptable quality.
If I were to do this the way I think it ought to be done, I would have
prepared the course material in two formats. The first would be a
text, identical to the textbooks with which you are familiar, but
available online, and intended to be used in printed form. The
second would be a slide presentation, la Corel

Presentations
or Microsoft

PowerPoint

, intended for use in the classroom or in

an independent study.
But, alas, I am still on that journey, so what I offer you is a hybrid of
these two concepts: an online text somewhat less verbose than a
conventional text, but one that can also serve as classroom
overhead transparencies.
Other compromises have been made. It would be advantageous to
produce two online versions - one intended for use in printed form,
and a second optimized for viewing on a computer screen. The two
would carry identical information, but would be formatted with
different page and font sizes. Also, to minimize file size, and
therefore download times, font selection and variations are
somewhat limited when compared to those normally encountered
in a conventional textbook.
You may also note that exercise problems are not included with this
text. By their very nature problems quickly can become worn out.
I believe it is best to include problems in a separate document.
Until all of these enhancements exist, I hope you will find this a
suitable and worthwhile compromise.
Enough of this; lets get on with it...
Electronic Applications xii
Notes for Printing This Document
This document can be printed directly from the Acrobat

Reader -
see the Acrobat

Reader help files for details.

If you wish to print the entire document, you may wish to do so in
two sections, as most printer drivers will only spool a maximum of
255 pages at one time.
Copyright Notice and Information
This entire document is 1999 by Bob Zulinski. All rights reserved.
I copyrighted this online text because it required a lot of work, and
because I hold a faint hope that I may use it to acquire
immeasurable wealth, thereby supporting the insatiable, salacious
lifestyle that Ive always dreamed of.
Thus, you will need my permission to print it. You may obtain that
permission simply by asking: tell me who you are and what you
want it for. Route your requests via email to rzulinsk@mtu.edu, or
by USPS mail to Bob Zulinski, Dept. of Electrical Engineering,
Michigan Technological University, Houghton MI 49931-1295.
Generous monetary donations included with your request will be
looked upon with great favor.
Electronic Applications 1 Thermal Considerations in Amplifiers
Fig. 1. Examples of heat sinks. Clockwise from
top: power supply heat sink with two TO-3 case-
style BJTs; pressure-fit heat sink over TO-5 case-
style BJT; heat sink with TO-220 case-style BJT.
Thermal Considerations in Amplifiers
In every amplifier, some power is dissipated as heat, thus, some
form of heat removal is required.
In power amplifiers heat is significant, and heat sinks are required:
Junction Temperature, T
J
Most of the device dissipation occurs at the collector-base junction.
The device/chip must be operated below a specified T
J max
.
Typically,
G T
J max
= 200 C (silicon devices in metal cases)
G T
J max
= 150 C (silicon devices in plastic cases)
Electronic Applications 2 Thermal Considerations in Amplifiers
X Y D XY
T T P
(1)
Thermal Resistance
Heat flows between any two points X and Y if a temperature
difference exists between them.
The steady-state temperature difference is proportional to the
thermal power.
Our immediate concern is with electronic devices:
G Heat flow: Junction Case Heat Sink Ambient Air
G The power dissipated in the device is the thermal power.
In steady-state, we have the generic equation:
where: T
X
and T
Y
are the temperatures at points X and Y,
P
D
is the power dissipated in the device, and

XY
is the thermal resistance between X and Y.
Note that the units of
XY
are C/W or K/W.
Also note the similarity of eq. (1) to Ohms Law: V = IR . . .
We can construct a thermal circuit that is analogous to an
electrical circuit!!!
Electronic Applications 3 Thermal Considerations in Amplifiers
P
D
T
J
T
C
T
S
T
A

JC

CS

SA
Fig. 2. Heat flow circuit analogy.
P
D
(W)
T
C
( C)
25 200
40
0
Slope =
-1/
JC
Fig. 3. Power derating curve example.
JA JC CS SA
+ + (2)
200 C 25 C
40 W
C
4.38
W
J C
JC
D
T T
P

(3)
J A
JA
D
T T
P

(4)
In constructing a thermal circuit,
we use subscripts to indicate:
Junction temperature,
Case temperature,
Sink temperature, and
Ambient temperature.
As with electrical circuits:
Specifying Device Ratings
Various manufacturers use various methods.
Power Derating Curves:
This is a plot of allowable power
dissipation vs. case temperature. In
this example T
J max
= 200 C, and
Sometimes T
A
is used rather than T
C
.
Then:
Electronic Applications 4 Thermal Considerations in Amplifiers
150 C 25 C C
8.33
15 W W
J C
JC
D
T T
P

(5)
Device Maximum Ratings:
Some manufacturers just give T
J max
and P
D max
, e.g.,
T
J max
= 150 C and P
D max
= 15 W at T
C
= 25 C
For this example, then:
Thermal Resistance, Junction-to-Case
For lowest
JC
the collector (or drain) is often in direct electrical and
thermal contact with the metal case.
The circuit designers only influence on
JC
is in choosing an
appropriate device.
Thermal Resistance, Case-to-Sink
Designers influence on
CS
:
G Choosing device with the desired type of case.
G Applying a thermally-conductive compound between case and
sink.
G Because the collector (or drain) is usually electrically
connected to the case, and the collector is not usually at
ground potential:
Electronic Applications 5 Thermal Considerations in Amplifiers
The case must be electrically insulated from the sink (usually
with a washer of mica or similar material), or
The sink must be electrically insulated from the chassis
(usually not done - can be very dangerous),
G Without additional information, it is usually assumed that

CS
1 C/W.
Thermal Resistance, Sink-to-Ambient
The designers influence on
SA
is by selection of heat sink.
Heat sinks range . . .
. . . from none (device case is the heat sink),
. . . to clip-on units,
. . . to large extruded aluminum units,
. . . to forced-air cooled arrangements,
. . . to water-cooled copper units.
Reference
For an excellent and thorough technical description of extruded,
fabricated, and water-cooled heat sinks, download the catalogs [sic]
from R-Theta

Corporation of Mississauga, Ont.

They are available via links at:
http://www.r-theta.com/products/info.html
Electronic Applications 6 Power Devices
Power Devices
Power BJTs vs. Small-Signal BJTs
Power devices have lower , e.g., 10 vs. 100.
Power devices require larger collector area due to the higher P
D
,
thus:
G The internal capacitances are larger.
G The transition frequency (f
T
) is lower.
G The reverse leakage current (I
CBO
) is higher.
Effect of Temperature on BJTs
As temperature increases:
G increases, approximately 3 times from -55 C to 150 C.
G I
CBO
increases, approximately 2 times per 10 C rise.
G V
BE
(for fixed I
C
) decreases, -2.5 mV / C.
As a result of these three effects, P
D
increases, which further
increases the junction temperature.
Thermal Runaway:
This describes the positive feedback process of higher T
J
, higher
P
D
, higher T
J
, etc., which eventually destroys the device.
Electronic Applications 7 Power Devices
I
C
V
CE
I
C
V
CE
= P
D max
Fig. 4. BJT constant-power
parabola.
max
log log log
C CE D
I V P + (6)
I
C
(log scale)
V
CE
(log scale)
I
C max
(bonding-wire limit)
Safe
Operating
Area
P
D max
(power-dissipation limit)
Second Breakdown limit
V
CE max
(voltage limit)
Fig. 5. BJT Safe Operating Area
(usually specified at T
C
= 25 C).
BJT Safe Operating Area (SOA)
The maximum power parabola, from
I
C
V
CE
= P
D max
, is plotted at left.
This becomes a straight line on a log-log
plot, as we see from:
Using this and other restrictions on BJT operation gives rise to the
BJT Safe Operating Area:
Note that second breakdown results from localized hot spots at
the collector-base junction, which then cause failure due to thermal
runaway.
Electronic Applications 8 Power Devices
n
+
n
+
n
+
n
-
p
+
p
+
channel channel
body
SiO
2
metal
D
S
G
Fig. 6. Illustration of power MOSFET cell construction. Many cells
can be paralleled for higher power devices.
Fig. 7. Power
MOSFET
model.
Power MOSFETs (Enhancement-Mode Only)
G Power MOSFETs are constructed vertically
(low-power MOSFETs are constructed horizontally).
G Positive v
GS
enhances the channel between source
metalizations.
G Positive v
DS
reverse biases the body-drain parasitic diode.
G Low channel resistance results from a short channel length,
and a heavily doped n
+
substrate.
G High breakdown voltage results from a lightly doped n

region.
Electronic Applications 9 Power Devices
i
D
v
GS
T
C
= 125 C
T
C
= 25 C
T
C
= -55 C
Fig. 8. Power MOSFET transfer characteristics.
Note point of zero temperature coefficient.
Power MOSFET Characteristics
G Can be biased with zero temperature coefficient.
G Very low static drive requirements (can be logic driven).
G Lower switching times than BJT (no minority carriers, no
stored charge).
G At large I
D
, I
D
decreases with temperature - no thermal
runaway.
G No second breakdown.
Electronic Applications 10 Power Amplifiers
Power Amplifiers
Introduction
Basically, power amplifiers are used to amplify two types of signals:
G Baseband f
H
/f
L
large audio, video
G Bandpass f
H
/f
L
small communication systems
Power amplifiers also fall into many different classes of operation:
G Class A devices in active region always (
cond
= 360)
suitable for baseband or bandpass signals
G Class B devices active for half cycle (
cond
= 180)
suitable for baseband or bandpass signals,
depending on the amplifier configuration
G Class C devices active < half cycle (
cond
< 180)
suitable for bandpass signals only
G Class D devices act as switches
suitable for baseband or bandpass signals,
depending on the amplifier configuration
G Class E devices act as switches
suitable for bandpass signals only
G Other, less-common, classes with specialized uses, e.g.,
Class F, Class G, Class H, Class S.
Electronic Applications 11 Power Amplifiers
Any element
or circuit
i
A
(t)
v
A
(t)
+
-
Fig. 9. Calculating power.
( ) ( ) ( )
A A A
p t v t i t
(7)
0 0
( ) ( ) ( )
T T
A A A A
E p t dt v t i t dt

(8)
0
1
( ) ( )
T
A
A A A
E
P v t i t dt
T T

(9)
0 0
1 1
( ) ( )
T T
CC CC CC CC CC CC CC
P V i t dt V i t dt V I
T T


(10)
Calculating Power Dissipation
Instantaneous power:
We usually assume that the amplifier
signals are periodic to simplify
calculations.
The energy delivered per cycle for a periodic signal is:
And the average power delivered is:
For a constant voltage source, the average power delivered is:
Note that I
CC
= i
CC average
, which is the dc component of i
CC
(t)!!!
For a constant current source we also obtain P
CC
= V
CC
I
CC
, except
in this case V
CC
is the average, or dc component, of v
CC
(t).
Electronic Applications 12 Power Amplifiers
signal power delivered to load
total power delivered by dc sources
O
IN
P
P
(11)
max
max max
O
P
CE C
P
c
v i
(12)
Amplifier Performance Measures
Efficiency:
Efficiency is almost always expressed as a percentage:
Power Capability:
Power capability is a measure of how much signal power I can
deliver to a load with a specific device:
where, P
o max
= maximum possible output power,
v
CE max
= maximum device voltage throughout cycle
i
C max
= maximum device current throughout cycle
Electronic Applications 13 Power Amplifiers
V
CC
-V
EE
I
BIAS
v
IN
v
O
+
-
i
O
R
L
Q
1
Fig. 11. Generic class A amplifier,
equivalent to Fig. 10.
V
CC
-V
EE
Amplifier
Current
Mirror
Fig. 10. IC Class A amplifier.
0 and 0
sat fwd
CE BE
v v (13)
max max
and
CC
O CC O
L
V
v V i
R
(14)
max
1
CC
C BIAS
L
V
i I
R
+ (15)
Class A Integrated-Circuit Power Amplifier

Note that Q
1
is an emitter follower. Thus, it has low Z
out
, high A
i
,
high G, and unity A
v
.
Class A Analysis:
We use two assumptions to simplify the analysis:
Now, when v
IN
swings positive (and keeping Q
1
from saturation):
Thus:
Electronic Applications 14 Power Amplifiers
V
CC
-V
EE
I
BIAS
v
IN
v
O
+
-
i
O
R
L
Q
1
Fig. 12. Class A amplifier (Fig. 11
repeated).
min min
1
0
C O BIAS O BIAS L
i i I v I R (16)
min min
1
EE EE
O EE O C BIAS
L L
V V
v V i i I
R R
(17)
EE
BIAS
L
V
I
R
(18)
Now, when v
IN
swings negative,
v
O min
depends on whether:
Q
1
reaches cutoff first, or
Q
3
reaches saturation first.
(Q
3
is the current mirror output
transistor.)
If Q
1
reaches cutoff first:
For Q
1
to reach cutoff before Q
3
reaches saturation, v
O min
> -V
EE
.
If Q
3
reaches saturation first:
For Q
3
to reach saturation before Q
1
reaches cutoff, I
BIAS
> V
EE
/R
L
.
Q. How can obtain maximum negative swing, without excessive
I
BIAS
?
A. If we design the circuit so that Q
1
reaches the edge of cutoff
and Q
3
reaches the edge of saturation at the same instant !!!
We can do this if we choose, by design:
Electronic Applications 15 Power Amplifiers
V
CC
-V
EE
I
BIAS
v
IN
v
O
+
-
i
O
R
L
Q
1
Fig. 13. Class A amplifier (Fig. 11
repeated.
t
v
O
V
CC
-V
CC
Fig. 14. Maximum output voltage magnitude in class A amplifier. Source
v
IN
must be capable of producing this amplitude.
t
i
C1
2V
CC
/R
L
V
CC
/R
L
I
BIAS
Fig. 15. Collector current in Class A amplifier with maximum output
voltage.
CC EE
V V
(19)
Class A Waveforms:
In most class A circuits
which allows these maximum-
amplitude waveforms:
Electronic Applications 16 Power Amplifiers
max
2 2
and
2 2
CC m
o o
L L
V V
P P
R R

(20)
2
CC CC
CC CC CC CC
L L
V V
P V I V
R R
j \

, (
( ,
(21)
2
CC CC
EE EE BIAS CC
L L
V V
P V I V
R R
j \

, (
( ,
(22)
2
2
CC
in CC EE
L
V
P P P
R
+
(23)
2
2
2
/ 2
0.25
2 /
o m L m
in CC CC L
P V R V
P V V R

j \

, (
( ,
(24)
max
25% (25)
Class A Calculations:
1. Output Power
Recall that v
O
v
IN
. Usually, we assume that v
O
= V
m
sin t.
Because V
m max
= V
CC
, we have:
2. DC Supply Power
This is more typically called input power, P
in
. From eqs. (10)
and (18) we have for the positive supply, V
CC
:
Similarly, for the negative supply, V
EE
:
Thus:
3. Efficiency
Thus:
Electronic Applications 17 Power Amplifiers
( )( )
max
max max
2
/ 2
0.125
2 2 /
o
CC L
P
CE C CC CC L
P
V R
c
v i V V R
(26)
2
CC
mirror BIAS BIAS EE BIAS
L
V
P V I V I
R

(27)
1
2
2
2 2 2
2
2
2
m
CC
CC CC m
Q in mirror o
L L L L
V
V
V V V
P P P P
R R R R

(28)
1max
2
when 0
CC
Q m
L
V
P V
R

(29)
1min
2
when
2
CC
Q m CC
L
V
P V V
R

(30)
4. Power Capability
5. Transistor Power Dissipation
Because the voltage across it is small, the power dissipated in
the current mirror reference transistor 0.
The power dissipated in the current mirror output transistor:
where V
BIAS
is the average voltage across I
BIAS
.
Finally, the power dissipated in amplifier transistor is:
from which:
and
Electronic Applications 18 Power Amplifiers
t
i
o
I
m
-I
m
T
Fig. 17. Class B output current.
t
i
C1
I
m
T
Fig. 18. Q
1
collector current.
V
CC
-V
EE
= -V
CC
v
in
v
o
+
-
i
o
R
L
Q
1
Q
2
Fig. 16. Class B complementary amplifier.
t
i
o
I
m
T
Fig. 19. Q
2
collector current.
0
0
sat
fwd
CE
BE
C E
v
v
i i

(31)
( ) sin sin
( ) sin sin
O m m
m
O m
L
v t V t V
V
i t t I
R




(32)
Class B Complementary Amplifier
Complementary means that we
have npn and pnp devices with
identical characteristics.
It is not critical that these devices
be truly complementary.
Q
1
is active on positive half-
cycles; Q
2
is active on negative
half-cycles.
Class B Analysis:
We begin the analysis with the
idealizing assumptions:
We further assume input and
output are sinusoidal, and use
some notational convenience:
Electronic Applications 19 Power Amplifiers
max
2 2
and
2 2
CC m
o o
L L
V V
P P
R R

(33)
CC m m
CC CC CC CC
L
V V I
P V I V
R
j \

, (
( ,
(34)
CC m m
EE EE EE CC
L
V V I
P V I V
R
j \

, (
( ,
(35)
2
CC m
in CC EE
L
V V
P P P
R
+ (36)
Class B Calculations:
1. Output Power
This is the familiar result obtained with a sinusoidal voltage
across a purely resistive load:
2. Input Power
For the power delivered by the positive supply we need the
average current through Q
1
. This is just the average of a half-
wave-rectified waveform:
We have a similar result for the power delivered by the
negative supply:
And the total input power is the sum of these two:
Electronic Applications 20 Power Amplifiers
2
/ 2
2 / 4
o m L m
in CC m L CC
P V R V
P V V R V

j \

, (
( ,
(37)
max
78.5%
4

(38)
( )( )
max
max max
2
/ 2
0.25
2 /
o
CC L
P
CE C CC CC L
P
V R
c
v i V V R
(39)
1 2
2
2
2
Q Q
CC m m
D D in o
L L
V V V
P P P P
R R
+
(40)
3. Efficiency
Because V
m
cannot be greater than V
CC
(Q
1
cannot become
saturated):
4. Power Capability
Because there are two devices, c
p
= 0.125 per device . . . the
same as in the Class A amplifier.
5. Transistor Power Dissipation
The difference between input and output power must be
dissipated in the devices:
From symmetry we can conclude P
D Q1
= P
D Q2
, but how do we
determine the maximum dissipation, P
DQ max
???
Electronic Applications 21 Power Amplifiers
Fig. 20. Plots of efficiency, output power, and device dissipation
for a class B amplifier with normalized values of V
CC
= 1 V and
R
L
= 1 .
( )
1 2
0
Q Q
D D
m
P P
V
+

(41)
1max 2 max max
2
2 2
2
Q Q
CC
D D o
L
V
P P P
R

(42)
2
CC
m
V
V

(43)
To find P
DQ max
, we set the derivative of (40) to zero:
from which
which occurs for
Various class B parameters are plotted in the figure below:
Electronic Applications 22 Power Amplifiers
V
CC
-V
EE
= -V
CC
v
in
v
o
+
-
i
o
R
L
Q
1
Q
2
Fig. 21. Class B amp (Fig. 16
repeated).
V
CC
-V
EE
= -V
CC
v
in
v
o
+
-
i
o
R
L
Q
1
Q
2
+
+
V
BIAS
V
BIAS
Fig. 23. Class AB amplifier.
t
v
in
v
o
Fig. 22. Illustration of crossover distortion.
Crossover Distortion:
In a real amplifier v
BE fwd
= 0.7 V (not
zero)!!!
So v
o
= 0 for |v
in
| < 0.7 V!!! The
effect on the output voltage is called
crossover distortion:
Eliminating crossover distortion
requires biasing slightly into the
active region.
One approach is to add a small bias
voltage, V
BIAS
0.7 V.
The result is a Class AB amplifier.
Details of various approaches are
left to your curiosity.
Electronic Applications 23 Single-Ended Power Amplifiers
v
sig
V
GG
V
DD
C
blocking
R
L
v
o
v
DS
i
D
Choke
+
+
+
-
Fig. 24. Class A single-ended amplifier.
I
M
I
DQ
V
DSQ
=V
DD V
M
|v
o
| |v
o
|
i
D
v
DS
Q
dc load line
ac load line
Fig. 25. FET output characteristics.
Single-Ended Power Amplifiers
The term single-ended refers to a single active device, usually
referenced to ground.
Class A Single-Ended Power Amplifier
Circuit Operation:
Gate voltage V
GG
is large enough so the FET is always in the pinch-
off region (class A). For maximum output swing, the Q-point should
be placed in the middle of the ac load line, as shown. (We
determine the details of placing the Q-point later.)
The choke is a very large inductance, with essentially infinite
impedance at the operating frequency, and zero impedance at dc.
Thus only dc current can flow through choke!!!
The blocking capacitor is also very large, with infinite impedance at
dc, and nearly zero impedance at the operating frequency. Thus
only signal current (zero average) can flow through load!!!
These two form a frequency-sensitive current divider!!!
Electronic Applications 24 Single-Ended Power Amplifiers
v
sig
V
GG
V
DD
C
blocking
R
L
v
o
v
DS
i
D
Choke
+
+
+
-
Fig. 26. Class A single-ended amplifier
(Fig. 24 repeated).
I
M
I
DQ
V
DSQ
=V
DD V
M
|v
o
| |v
o
|
i
D
v
DS
Q
dc load line
ac load line
Fig. 27. FET output characteristics (Fig. 25
repeated).
( )sin sin
O M DD m
v V V t V
(44)
0 thus 2
M DD DD M DD
V V V V V
(45)
and
2
M M
M DQ
L L
V V
I I
R R
(46)
From a study of Fig. 27, note that the maximum drain-source
voltage is noted as V
M
.
For a sinusoidal output (which we typically assume):
And, for maximum output swing:
Eq. (45) mathematically shows that Q must bisect the ac load line
for maximum output swing to be possible. We would still need v
sig
to be large enough to achieve maximum output swing.
Because R
L
determines the slope of the ac load line, we can write:
Electronic Applications 25 Single-Ended Power Amplifiers
v
sig
V
GG
V
DD
C
blocking
R
L
v
o
v
DS
i
D
Choke
+
+
+
-
Fig. 28. Class A single-ended amplifier
(Fig. 24 repeated).
I
M
I
DQ
V
DSQ
=V
DD V
M
|v
o
| |v
o
|
i
D
v
DS
Q
dc load line
ac load line
Fig. 29. FET output characteristics (Fig. 25
repeated).
DD
DQ
L
V
I
R
(47)
max
2 2
and
2 2
m DD
o o
L L
V V
P P
R R

(48)
2
DD
in DD DQ
L
V
P V I
R

To allow for maximum output amplitude, and thus maximum P
o
:
The bias voltage V
GG
must be adjusted to achieve this I
DQ
. A
higher bias voltage would create a larger I
DQ
, but could not provide
for a larger output voltage. Thus it would only waste power.
Calculations:
1. Average Output Power
2. Average Input Power:
Electronic Applications 26 Single-Ended Power Amplifiers
v
sig
V
GG
V
DD
C
blocking
R
L
v
o
v
DS
i
D
Choke
+
+
+
-
Fig. 30. Class A single-ended amplifier
(Fig. 24 repeated).
I
M
I
DQ
V
DSQ
=V
DD V
M
|v
o
| |v
o
|
i
D
v
DS
Q
dc load line
ac load line
Fig. 31. FET output characteristics (Fig. 25
repeated).
2 2
max
2 2
/ 2 1
thus 50%
2 /
o m L m
in DD L DD
P V R V
P V R V

j \

, (
( ,
(50)
( )( )
max
max max
2
/ 2 1
0.125
2 2 / 8
o
DD L
P
DS D DD DD L
P
V R
c
v i V V R
(51)
2 2
2
DD m
Q in o
L L
V V
P P P
R R

(52)
min max max
2 2
at and at 0
2
DD DD
Q o o Q o
L L
V V
P P P P P
R R

(53)
3. Efficiency
4. Power Capability
5. FET Power Dissipation
Electronic Applications 27 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 32. Class B single-ended amplifier.
Class B Single-Ended Amplifier
Circuit Operation:
This circuit is biased at the threshold voltage, so only positive half-
cycles of the signal can move the device into the pinch-off region.
As we have done before, we let t = for convenience.
Also as before, the choke has infinite impedance at the operating
frequency, and zero impedance at dc. Only dc current can flow
through the choke !!!
The blocking capacitor has infinite impedance at dc, and zero
impedance at operating frequency. Only signal current (zero
average) can flow toward load !!!
In addition, L
o
- C
o
is a parallel resonant circuit. It has infinite
impedance at the operating frequency, , and zero impedance (we
assume) at all harmonics (2, 3, etc.). Thus:
G All harmonic signal current flows through L
o
-C
o
, not R
L
!!!
G Only fundamental signal current (at ) flows through R
L
!!!
Electronic Applications 28 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 33. Class B single-ended amplifier (Fig. 32
repeated).
i
D

I
p
2
Fig. 34. Drain current in class B amplifier. The
magnitude of the current pulses is yet
undetermined.
0 1 2 3
sin sin2 sin3
D
i a a a a + + + + (55)
2 2
0
0 0
1 1
sin
2 2
p
DC D p
I
I a i d I d





(56)
sin sin
sig
v K t K (54)
To repeat, the FET is
biased at V
T
.
Thus it is in the cutoff
region for v
sig
< 0,
and in the pinch-off region
for v
sig
> 0.
We make the standard
assumption: a sinusoidal
input at :
Drain current flows only for
positive half-cycles of v
sig
,
as shown at left.
This is a periodic function,
and therefore it has Fourier
components:
The dc component a
0
can flow only through the choke:
Note that I
DC
is simply the average value of i
D
.
Electronic Applications 29 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 35. Class B single-ended amplifier
(Fig. 32 repeated).
i
D

I
p
2
Fig. 36. Drain current in class B amplifier.
(Fig. 34 repeated)
D DC x
i I i + (57)
1 2 3
sin sin2 sin3
x
i a a a + + + (58)
x res o
i i i + (59)
1
sin
o
i a (60)
2
2
1
0 0
1 1
sin sin
2
p
D p
I
a i d I d





(61)
A KCL equation at the drain node gives:
so we know that all of the sinusoidal components of i
D
comprise i
x
:
Further, a KCL equation at the output node gives:
We know that the fundamental component at cannot flow through
L
o
-C
o
, while the harmonic components at 2, 3, etc., can only flow
through L
o
-C
o
. Thus:
The magnitude a
1
of this fundamental component is found from a
Fourier integral:
Electronic Applications 30 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 37. Class B single-ended amplifier
(Fig. 32 repeated).
i
D

I
p
2
Fig. 38. Drain current in class B amplifier
(Fig. 34 repeated).
sin sin
2
p L
o m
I R
v V
(62)
2
m
p
L
V
I
R
(63)
sin
blocking
DS C o DD m
v v v V V +
(64)
0 thus
DS m DD
v V V (65)
The output voltage is produced by i
o
flowing through R
L
:
which gives us the relationship between I
p
and V
m
:
Now, we write a KVL equation around the drain-output loop:
and recognize that v
DS
cant be negative, which limits the magnitude
of v
o
(before clipping of the output voltage occurs):
It has taken us some effort (!!!), but we may finally proceed to the
calculations.
Electronic Applications 31 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 39. Class B single-ended amplifier
(Fig. 32 repeated).
i
D

I
p
2
Fig. 40. Drain current in class B amplifier
(Fig. 34 repeated).
max
2 2
and
2 2
m DD
o o
L L
V V
P P
R R

(66)
max
2
2 2
and
DD P DD m DD
in DD DC in
L L
V I V V V
P V I P
R R

(67)
2
max
/ 2
and 78.5%
2 / 4 4
o m L m
in DD m L DD
P V R V
P V V R V

j \

, (
( ,
(68)
( )
( )
( )( )
max
max max
max
2 2
/ 2 / 2 1
2 2 / 8
2
o
DD L DD L
P
DS D DD DD L
DD p
P
V R V R
c
v i V V R
V I

(69)
Calculations:
1. Output Power
2. Input Power
3. Efficiency
4, Power Capability
Electronic Applications 32 Single-Ended Power Amplifiers
V
GG
=V
T
V
DD
L
o
R
L
v
o
v
DS
i
D
Choke
+
+
-
v
sig
+
C
blocking
C
o
+
-
I
DC
i
x
i
o
i
res
Fig. 41. Class B single-ended amplifier
(Fig. 32 repeated).
i
D

I
p
2
Fig. 42. Drain current in class B amplifier
(Fig. 34 repeated).
2
2
2
DD m m
Q in o
L L
V V V
P P P
R R

(70)
2
0
Q DD m
m L L
P V V
V R R

(71)
max max
2
2 2
2 2 4
at
DD DD
Q o m
L
V V
P P V
R

(72)
5. FET Power Dissipation
To find P
Q max
we need to set the derivative to zero:
from which
Notice that all of these results are identical to those obtained earlier
for the class B complementary amplifier!!!
Electronic Applications 33 Introduction to Feedback
Introduction to Feedback
Feedback describes a technique used in amplifiers in which a
fraction of the output signal is returned (fed back) to the input.
Negative Feedback
In negative feedback the returned portion subtracts from (opposes)
the original input.
Benefits of negative feedback include:
G Stabilized gain (with variation of circuit parameters)
G Reduced nonlinear distortion
G Reduced noise (only some types of noise are reduced)
G Controlled Z
in
and Z
out
G Increased bandwidth
The price we pay for these benefits, increased circuit complexity
and reduced overall gain, are readily accepted given the capability
of todays circuits.
Positive Feedback
In positive feedback the returned portion adds to (aids) the original
input.
At one time (early 20
th
century) positive feedback was used to
increase the gain over that obtainable with conventional circuits.
The price was a degradation of all the items in the bulleted list,
above.
No longer useful in amplifiers, it is required in oscillator circuits,
Schmitt triggers, etc. We leave a discussion of positive feedback
until later.
Electronic Applications 34 Introduction to Feedback
Source
Amplifier
Load
F.b. network
Gain = A

Gain =
+
-
x
s
x
s
-x
f
=x
i
Ax
i
=x
o
x
f
=x
o
Fig. 43. Block diagram of generic feedback system. For electrical systems the
variable x represents a voltage or a current.
i s f s o
x x x x x (73)
( )
o i s o
x Ax A x x
(74)
o o s
x A x Ax + (75)
1
o
f
s
x A
A
x A

+
(76)
Closed-Loop Transfer Function
Consider the system represented by the block diagram, below. The
negative sign at the summing block reflects our emphasis on
negative feedback at this time.
If the path through the feedback network is opened, the gain from
source to load is simply A, which is called the open-loop gain.
With the path through the feedback network closed, as shown, the
gain changes to its closed-loop value.
Calculating the closed-loop gain is a matter of writing simple circuit
equations for the above figure:
A
f
is called the closed-loop gain; the product A is the loop gain.
Electronic Applications 35 Introduction to Feedback
1
o
f
s
x A
A
x A

+
(77)
1
f
A

(78)
( )( ) ( )( )
( ) ( )
2 2
1 1
1
1 1
f
A A
dA
dA
A A


+

+ +
(79)
( ) ( )
2 2
1
1
1 1
f
f
A dA A dA
dA dA
A A A
A A



+
+ +
(80)
1
1
f
f
dA dA
A A A

+
(81)
Repeating eq. (44):
Note that:
G For A > 0 A
f
< A negative feedback
G For A < 0 (and |A| < 1) A
f
> A positive feedback
G If A = -1 oscillator
G In amplifiers, usually A >> 1 (A very large, small). Thus:
Gain Stability
We can show that, for a typical negative feedback amplifier with
A >> 1, the closed loop gain A
f
is much more stable than the open-
loop gain A. From eq. (77)
Thus, an incremental change in A results in a much smaller change
in A
f
.
Electronic Applications 36 Introduction to Feedback
v
o
v
i
A
v
=10
A
v
=5
Fig. 44. Amplifier with exaggerated nonlinear
characteristic.
t
v
o
Fig. 45. Nonlinear amplifier output.
Source
Amplifier
Load
Nonlinear

= 0.1
+
-
Preamp
A
v
=1000
Fig. 46. Improving overall linearity with negative feedback.
Reduction of Nonlinear Distortion
It is relatively easy to produce a small-signal amplifier that is
reasonably linear. It is much more difficult to produce a linear
power amplifier. Suppose we have a power amplifier with a
nonlinear characteristic. The output would be very distorted, as
shown below for an assumed sinusoidal input:
But if we embed this amplifier in a feedback system with a linear high-
gain preamplifier:
For positive v
i
, A = (1000)(10) = 10,000, A
f
= A / (1 + A) = 9.99
For negative v
i
, A = (1000)(5) = 5,000, A
f
= A / (1 + A) = 9.98
Note that the closed-loop gain is very nearly linear!!!
Electronic Applications 37 Introduction to Feedback
Amplifier
Load
Gain = A
1

+
+
x
sig
x
noise
x
o
Fig. 47. Noise referred to input.
Amplifier
Load
Gain = A
1

+
+
x
sig
x
o
A
1
x
noise
Fig. 48. Noise referred to output.
signal power delivered to load
noise power delivered to load
signal
noise
P
SNR
P
(82)
( ) ( )
2
2
1
1
and
sig
noise
signal noise
L L
A x
A x
P P
R R

(83)
2
2
sig
noise
x
SNR
X

(84)
Reduction of Noise
If we can construct a single low-noise amplifier stage, we can
reduce considerably the overall noise in an amplifier system.
Noise Sources:
Johnson noise - thermally generated in resistors.
Shot noise - due to discrete nature of current flow.
Microphonic noise - due to vibration of components.
Miscellaneous noise - power supply hum, stray coupling.
Noise Models:
Signal-to-Noise Ratio, SNR:
where:
Thus:
Electronic Applications 38 Introduction to Feedback
Source
Amplifier
F.b. network
Gain=A
2

Gain =
+
-
x
s
x
s
-x
f
=x
2
x
f
=x
o
x
noise
Load
Amplifier
Gain = A
1
+
x
o

x
1
Fig. 49. The noise model of Fig. 47 embedded in an otherwise noiseless feedback network.
10log 10log 20log
signal sig
dB
noise noise
P x
SNR SNR
P x
(85)
( ) ( )
1 1 1 2 2 1 2 o noise s o noise
x A x A A x x A A x x x + + , ]
]
(86)
( )
1 2 1 2 1
1
o s noise
A A x A A x A x + +
(87)
1 2 1
1 2 1 2
1 1
o s noise
A A A
x x x
A A A A
+
+ +
(88)
( )
( )
2
st
2
2
2
2 2
nd
1 term
2 term
s
noise
x
SNR A
x

(89)
Signal-to-Noise Ratio is usually expressed in decibels:
Reduction of Noise Using Feedback:
Beginning at the output and working backwards gives the equation:
Now, solving eq. (86) for x
o
:
The signal-to-noise ratio is greatly improved (by A
2
2
):
Electronic Applications 39 Types of Feedback
Source
Amplifier
Load
F.b. network
Gain = A

Gain =
+
-
x
s
x
s
-x
f
=x
i
Ax
i
=x
o
x
f
=x
o
Fig. 50. Block diagram of generic feedback system. For electrical
systems the variable x represents a voltage or a current (Fig. 43
repeated).
+
-
+
-
v
i
+
-
v
f
+
-
v
o
+
-
v
o
+
-
v
s
+
-
R
L
Gain = A
v
(Voltage)
Fig. 51. Series-voltage feedback.
i
i
i
o
i
s
R
L
i
f
i
o
Gain = A
i
(Current)
Fig. 54. Parallel-current feedback.
+
-
+
-
v
i
+
-
v
f
+
-
i
o
i
o
+
-
R
L
Gain = G
m
(Transconductance)
v
s
+
-
Fig. 52. Series-current feedback.
i
i
v
o
+
-
v
o
i
s
R
L
i
f
Gain = R
m
(Transresistance)
Fig. 53. Parallel-voltage feedback.
Types of Feedback
As described in the figure above, either voltage or current can be
sampled (measured) at the output, and either voltage or current
can be fed back to the summing block. Thus, there are four
possible feedback configurations. The descriptions follow the form
input - output:
Electronic Applications 40 Types of Feedback
+
-
+
-
v
i
+
-
v
f
+
-
v
o
+
-
v
o
+
-
v
s
+
-
R
L
Gain = A
v
(Voltage)
Fig. 55. Series-voltage feedback.
i
i
i
o
i
s
R
L
i
f
i
o
Gain = A
i
(Current)
Fig. 58. Parallel-current feedback.
+
-
+
-
v
i
+
-
v
f
+
-
i
o
i
o
+
-
R
L
Gain = G
m
(Transconductance)
v
s
+
-
Fig. 56. Series-current feedback.
i
i
v
o
+
-
v
o
i
s
R
L
i
f
Gain = R
m
(Transresistance)
Fig. 57. Parallel-voltage feedback.
On the Input Side
We use the term series-mixing with voltages, because voltages add
or subtract in series, and we model the signal as a voltage source.
We use the term parallel-mixing (or shunt-mixing) with currents,
because currents add or subtract in parallel, and we model the
signal as a current source.
On the Output Side
We use the term voltage-sensing when voltage is the sampled
output variable, which must be done with a parallel connection.
We use the term current-sensing when current is the sampled
output variable, which must be done with a series connection.
Electronic Applications 41 Types of Feedback
+
-
+
-
v
i
+
-
v
f
+
-
v
o
+
-
v
o
+
-
v
s
+
-
R
L
Gain = A
v
(Voltage)
Fig. 59. Series-voltage feedback.
i
i
i
o
i
s
R
L
i
f
i
o
Gain = A
i
(Current)
Fig. 62. Parallel-current feedback.
+
-
+
-
v
i
+
-
v
f
+
-
i
o
i
o
+
-
R
L
Gain = G
m
(Transconductance)
v
s
+
-
Fig. 60. Series-current feedback.
i
i
v
o
+
-
v
o
i
s
R
L
i
f
Gain = R
m
(Transresistance)
Fig. 61. Parallel-voltage feedback.
The Amplifier Block
Any real amplifier can be modeled as voltage, current,
transconductance, or transresistance.
We (naturally) choose the model appropriate for the type of
feedback connection.
The Feedback Block
Units of the feedback factor are the inverse of the units of open-
loop gain A, i.e., the loop gain A is always unitless!!!
In our models the feedback network is assumed to be ideal, i.e.,
either an ideal voltage or current source at the input, and either a
short-circuit or open-circuit at the load.
Electronic Applications 42 Types of Feedback
+
-
+
-
v
i
+
-
v
f
+
-
v
o
+
-
v
o
+
-
v
s
+
-
R
L
Gain = A
v
(Voltage)
Fig. 63. Series-voltage feedback.
1
f
o m
m
s m
v R
R
i R

+
(92)
i
i
i
o
i
s
R
L
i
f
i
o
Gain = A
i
(Current)
Fig. 66. Parallel-current feedback.
1
f
o i
i
s i
i A
A
i A

+
(93)
+
-
+
-
v
i
+
-
v
f
+
-
i
o
i
o
+
-
R
L
Gain = G
m
(Transconductance)
v
s
+
-
Fig. 64. Series-current feedback.
i
i
v
o
+
-
v
o
i
s
R
L
i
f
Gain = R
m
(Transresistance)
Fig. 65. Parallel-voltage feedback.
1
f
o v
v
s v
v A
A
v A

+
(90)
1
f
o m
m
s m
i G
G
v G

+
(91)
Gain Equations
The form of our generic gain equation (44) can be changed to more
appropriately reflect the type of feedback:
Series-voltage:
stabilized voltage gain
Series-current:
stabilized transconductance gain
Parallel-voltage:
stabilized transresistance gain
Parallel-current:
stabilized current gain
Electronic Applications 43 Types of Feedback
+
-
+
-
v
i
+
-
v
f
+
-
i
s
x
o
+
-
R
i
v
s
+
-
Load
x
o
=Av
i
R
if
Fig. 67. Feedback system with series-mixing input and
generic output.
( ) ( )
( )
1
s i f
s i o s i i s i s i
s i
v v v
i R x i R Av i R Ai R
A i R

+
+ + +
+
(94)
( )
( )
1
1
f
s i
s
i i
s s
A i R
v
R A R
i i

+
+
(95)
Effect on Input Resistance
The type of feedback effects the input resistance of the system, as
we see in the following pages.
Series Mixing:
The input resistance of the amplifier is R
i
, though the signal source
sees an input resistance of R
if
.
From KVL:
Thus,
Series-mixing increases input resistance!!!
Electronic Applications 44 Types of Feedback
i
i
x
o
i
s
i
f
Load
x
o
=Av
i
v
s
+
-
R
i
R
if
Fig. 68. Feedback system with parallel-mixing input and generic
output.
( )
( )
1
s i f
s s s s
o i
i i i i
s
i
i i i
v v v v
x Av A
R R R R
v
A
R

+
j \
+ + +
, (
( ,
+
(96)
( )
1
1
f
s s i
i
s
s
i
v v R
R
v
i A
A
R

+
+
(97)
Parallel Mixing:
As in the previous example, the input resistance of the amplifier is
R
i
, though the signal source sees an input resistance of R
if
.
From KCL:
Thus,
Parallel-mixing decreases input resistance!!!
Electronic Applications 45 Types of Feedback
+
-
i
test
v
test
+
-
R
o +
-
x
s
=0

x
i
A
oc
x
i
R
of
+
-
v
test
Fig. 69. Feedback system with generic input and voltage sensing at the
output.
( )
test test o oc i
test o oc test
v i R A x
i R A v
+
+
(98)
( )
1
oc test test o
A v i R +
(99)
1
f
test o
o
test oc
v R
R
i A

+
(100)
Effect on Output Resistance
The type of feedback also effects the output resistance of the
system, as we see in the following pages.
Voltage Sensing:
The output resistance of the amplifier is R
o
, though the load (or test
source) sees an output resistance of R
of
.
From KCL:
Rearranging:
Thus:
Voltage sensing decreases output resistance!!!
Electronic Applications 46 Types of Feedback
+
-
i
test
v
test
+
-
R
o
x
s
=0

x
i
A
sc
x
i
R
of
+
-
i
o
=-i
test
i
o
Fig. 70. Feedback system with generic input and current sensing at the
output.
( )
test test
test sc i sc test
o o
v v
i A x A i
R R
(101)
( )
1
test
sc test
o
v
A i
R
+ (102)
( )
1
f
test
o sc o
test
v
R A R
i
+ (103)
Current Sensing:
As in the previous example, the output resistance of the amplifier is
R
o
, though the load (or test source) sees an output resistance of
R
of
.
From KCL:
Rearranging:
Thus:
Current sensing increases output resistance!!!
Electronic Applications 47 Types of Feedback
+
-
+
-
v
i
+
-
v
f
+
-
v
o
+
-
v
o
+
-
v
s
+
-
R
L
Gain = A
v
(Voltage)
Fig. 71. Series-voltage feedback.
+
-
+
-
v
i
+
-
v
f
+
-
i
o
i
o
+
-
R
L
Gain = G
m
(Transconductance)
v
s
+
-
Fig. 72. Series-current feedback.
Summary of Feedback Types
Input: voltage source
Output variable: voltage
R
if
and R
of
tend toward: Ideal
Voltage Amplifier
1
f
v
v
v
A
A
A

+
( ) 1
f
i v i
R A R +
1
f
oc
o
o
v
R
R
A

+
Input: voltage source
Output variable: current
R
if
and R
of
tend toward: Ideal
Transconductance Amplifier
1
f
m
m
m
G
G
G

+
( ) 1
f
i m i
R G R +
( )
1
f sc
o m o
R G R +
Electronic Applications 48 Types of Feedback
i
i
i
o
i
s
R
L
i
f
i
o
Gain = A
i
(Current)
Fig. 74. Parallel-current feedback.
i
i
v
o
+
-
v
o
i
s
R
L
i
f
Gain = R
m
(Transresistance)
Fig. 73. Parallel-voltage feedback.
Input: current source
Output variable: voltage
R
if
and R
of
tend toward: Ideal
Transresistance Amplifier
1
f
m
m
m
R
R
R

+
1
f
i
i
m
R
R
R

+
1
f
oc
o
o
m
R
R
R

+
Input: current source
Output variable: current
R
if
and R
of
tend toward: Ideal
Current Amplifier
1
f
i
i
i
A
A
A

+
1
f
i
i
i
R
R
A

+
( )
1
f sc
o i o
R A R +
Electronic Applications 49 Practical Feedback Networks
R
1
R
L
R
2
R
S
v
s
+
-
v
o
+
-
v
f
+
-
v
i
+
-
+
-
i
o
Fig. 75. Series-voltage feedback.
R
L
R
S
v
s
+
-
v
o
+
-
R
f
v
f
+
-
i
o
-
+
v
i
+
-
Fig. 76. Series-current feedback.
R
L
v
o
+
-
i
o
R
f
R
S
i
s
i
i
i
f
-
+
Fig. 77. Parallel-voltage feedback.
R
L
v
o
+
-
R
1
R
2
i
o
R
S
i
s
i
i
i
f
-
+
Fig. 78. Parallel-current feedback.
Practical Feedback Networks
Its impossible, of course, to show all possible practical feedback
networks. The four shown below are very typical examples:
Identifying Feedback Configuration
At the Input:
Look for series or parallel connection. Remember that the signal
source, the amplifier input port, and the feedback network output
port each have two terminals.
At the Output:
Look for series or parallel connection here, also. Alternatively, if a
short-circuited load causes x
f
to be zero, the connection is voltage
sensing. Conversely, if an open-circuited load causes x
f
to be zero,
the connection is current sensing.
Electronic Applications 50 Practical Feedback Networks
R
1
R
L
R
2
R
S
v
s
+
-
v
o
+
-
v
f
+
-
v
i
+
-
+
-
i
o
Fig. 79. Series-voltage feedback.
R
L
R
S
v
s
+
-
v
o
+
-
R
f
v
f
+
-
i
o
-
+
v
i
+
-
Fig. 80. Series-current feedback.
R
L
v
o
+
-
i
o
R
f
R
S
i
s
i
i
i
f
-
+
Fig. 81. Parallel-voltage feedback.
R
L
v
o
+
-
R
1
R
2
i
o
R
S
i
s
i
i
i
f
-
+
Fig. 82. Parallel-current feedback.
Identifying Negative Feedback
The feedback is negative if:
G a change in x
s
that tends to increase x
i
. . .
G causes a change in x
f
that tends to decrease x
i
.
Estimating the Feedback Factor
We estimate the feedback factor, , by idealizing the circuit. Note
that the amplifier input acts as a load to the feedback network.
To estimate , we assume the amplifier input is ideal, i.e.:
G an pen-circuit for series mixing,
G a short-circuit for parallel mixing.
Electronic Applications 51 Practical Feedback Networks
Design of Feedback Amplifiers
1. Determine type of feedback and value of required.
2. Choose appropriate feedback network (basic versions in Figs.
75 - 78).
3. Choose appropriate resistor values
(a) often dependent only on ratio of resistors.
(b) Real feedback networks have nonideal input and output
resistances - they are not the ideal-dependent-source
models of Figs. 71 - 74.
(c) These real feedback networks load the amplifier output
and insert unwanted resistance into the amplifier input.
(d) To minimize loading effects (must often compromise):
(i) Series Mixing - small Rs to minimize series voltage
drop
(ii) Parallel Mixing - large Rs to minimize load on signal
source
(iii) Voltage Sensing - large Rs to minimize load on
amplifier output
(iv) Current Sensing - small Rs to minimize series
voltage drop
4. Analyze/simulate/test to insure design specifications are met.
Electronic Applications 52 Transient and Frequency Response in Feedback Amplifiers
( )
( )
1 ( ) ( )
f
A s
A s
A s s

+
(104)
( )
( )
cos sin
j t st t
e e e A t jB t


+
+
(105)
Transient and Frequency Response in Feedback Amplifiers
Introduction
As a practical matter, in feedback systems both the gain, A, and the
feedback factor, , are functions of frequency:
Recall:
G Zeros are those values of s for which A
f
(s) = 0.
G Poles are those values of s for which A
f
(s) = ,

i.e., the roots of 1 + A(s)(s) = 0.

G Poles and zeros may be complex (real ; imaginary j).
G Poles and zeros of a transfer function may be illustrated
graphically on the complex plane (the s-plane).
G For physically realizable circuits, complex poles always occur
in conjugate pairs.
Transient Response
Transfer function poles lead to transient response terms of the form:
Electronic Applications 53 Transient and Frequency Response in Feedback Amplifiers
j

x
x
x
x x x
Fig. 83. Illustration of various transient responses with s-plane pole
locations.
Transient Response and Pole Location
For poles on the negative real axis, the transient response has the
form e
-t
, where 1/ = .
For poles on the positive real axis, the transient response has the
form e
t
.
For complex conjugate poles at s = j
G Transient response form: e
t
(A cos t + jB sin t)
G Transfer function form: (s + + j)(s + - j) = s
2
+ 2
n
+
n
2
G Natural frequency
2 2
n
+
G Damping ratio = /
n
(Quality factor, Q = 1/2)
Electronic Applications 54 Transient and Frequency Response in Feedback Amplifiers
j

x
x
x x
Fig. 84. Illustration of various frequency
(magnitude) responses with s-plane pole locations.
Frequency Response and Pole Location
Bode Plots use poles and zeros to determine the asymptotes of the
magnitude and phase responses. The Bode asymptotes are shown
above in red.
As the poles move off of the negative real axis, the asymptotes do
not change, but the actual magnitude response exhibits gain peaks.
These gain peaks increase as the poles approach the j axis.
Electronic Applications 55 Transient and Frequency Response in Feedback Amplifiers

n
x
s = - + j

cos

= /
n
=
Fig. 85. One of two complex conjugate
poles showing the relationship among
transfer function parameters
45
Acceptable
Pole
Locations
Excessive
ringing/gain peaks
Unstable
transient
response

n
s = - + j
x
Fig. 86. Illustration of acceptable pole
locations using 45 rule of thumb.
Graphical Details of Complex Conjugate Poles
These are graphical details, so lets let the figures tell the story:
In amplifiers we normally choose, somewhat arbitrarily, a region of
pole locations where transient response ringing and frequency
response gain peaks are considered to be within acceptable limits.
The typical rule of thumb allows poles to be located within a 45
angles from the negative real axis.
Note that the 45 rule of thumb illustrated in Fig. 86 requires
(which means 0.707 and Q 0.707) for acceptable stability.
Electronic Applications 56 Frequency Characteristics of Feedback Amplifiers
f
b
A
v
, dB
f
20

log

A
0
Fig. 87. Single-pole lowpass Bode
magnitude response.
0
( )
1
2
b
A
A s
s
f

+
(106)
0
0
0 0
0 0
( ) 1
( )
1 ( )
1 1
1
1
1 1 1
1
f
b
b
b b b
b
A A s
A s
s A
A s
s
A A
s s s
A A
s

+
+ +
+

+ + + + +
+
(107)
Frequency Characteristics of Feedback Amplifiers
Dominant-Pole Amplifiers
A dominant-pole amplifier may have many poles, but one pole is
much lower in frequency than the others, and dominates the
responses as if there were only one pole.
All real amplifiers have one or more zeros at s = . In this sense
they are lowpass. So consider a single-pole (dominant-pole)
lowpass amplifier:
Now lets add negative feedback to this amplifier using a purely
resistive, and thus frequency independent, feedback network:
Electronic Applications 57 Frequency Characteristics of Feedback Amplifiers
( )
0
0
0 0
0
0
1
( )
1 1 1
1 2
f
f
f
b b b
A
A
A A
A s
s s s
A
A f

+

+ + + +
+
(108)
( )
0
0 0
0
and 1
1
f f
b b
A
A f f A
A

+
+
(109)
( )
0
0 0 0
0
1
1
f f
b b b
A
A f f A A f
A

+
+
(110)
0 0
f f
b b
GBW A f A f (111)
Continuing with eq.(107) and solving:
where
Thus, the feedback amplifier also has a single-pole lowpass
response, but:
1. DC gain is reduced by the factor (1 + A
0
) as usual.
2. Bandwidth is increased by a factor of (1 + A
0
)!!!
Gain-Bandwidth Product:
The notation GB or GBW is normally used to denote this measure.
We obtain it by noting that:
i.e.,
This latter equation says that the gain-bandwidth product of a
feedback system using a dominant pole amplifier is independent of
the amount of feedback, , applied!!!
It is also equal to the GBW of the original dominant-pole amplifier!!!
Electronic Applications 58 Frequency Characteristics of Feedback Amplifiers
R
1
R
L
R
2
R
S
v
s
+
-
v
o
+
-
v
f
+
-
v
i
+
-
+
-
i
o
A
0
, f
b
Fig. 88. Dominant-pole example.
A
v
, dB
f, Hz
100
80
60
40
20
0
10 1 10
2
10
3
10
4
10
5
10
6
Fig. 89. Bode plots showing constant GBW.
Dominant-Pole Gain-Bandwidth Example:
An operational amplifier
has a dominant (single)
pole response with:
A
0
= 10
5
100 dB
f
b
= 10 Hz
Feedback factors to be
used with this amplifier
are = 0.01, 0.1, and 1
A
0f
= A
0
/(1+A
0
) A
0f
1/ f
bf
= f
b
(1+A
0
)
0.01 99.90 100 10.01 kHz
0.1 9.999 10 100.0 kHz
1 1.000 1 1.000 MHz
Electronic Applications 59 Frequency Characteristics of Feedback Amplifiers
x
j

-2f
b
(1+A
0
)
x
-2f
b
Fig. 90. Movement of the dominant pole with
feedback.
Feedback Factor and Pole Location:
The pole moves to the
left with feedback, as
shown.
From the form of the
transient response, e
-t
,
we can see that the
response decays faster
as pole moves left.
Summary:
G Many multiple-pole amplifiers have a dominant pole, i.e., a
pole much closer to the origin than all other poles.
G The dominant pole transient response lasts much longer than
the transient response of the other poles.
G A dominant-pole amplifier may be treated like a single-pole
amplifier (usually).
G Dominant-pole amplifiers have a constant gain-bandwidth
product.
Electronic Applications 60 Frequency Characteristics of Feedback Amplifiers
+
-
v
in
A
vo
v
in
C
in
v
s
R
S
+
-
R
L
C
L
R
in
R
out
+
-
Fig. 91. Model of amplifier with two lowpass poles
0
1 2
( )
1 1
2 2
A
A s
s s
f f

j \j \
+ +
, (, (
( ,( ,
(112)
( )
( )
1 ( )
f
A s
A s
A s

+
(113)
0
1 2
0
2
1 2 1 2
1 ( ) 1
1 1
1
1
A
A s
s s
A
s s s

+ +
j \j \
+ +
, (, (
( ,( ,
+
+ + +
(114)
Two-Pole Amplifiers
Lets presume we have a two-pole amplifier with both poles on the
negative real axis:
Now lets add negative feedback to this amplifier using a purely
resistive (thus frequency independent) feedback network:
The closed-loop poles (the poles of the resulting feedback amplifier)
are given by the roots of 1 + A(s) :
Electronic Applications 61 Frequency Characteristics of Feedback Amplifiers
( )
( )
( ) ( )
0 1 2
2
1 2 1 2
2
1 2 1 2
0 1 2
2 2
1 2 1 2 1 2 1 2
0 1
A
s s
s s
A
s s s s





+
+ + +
+ + +
+
+ + + + + +
(115)
( ) ( )
2
1 2 0 1 2
1 0 s s A + + + +
(116)
( ) ( )
2 2
1 2 0 1 2
2 2 1 4 0 s s f f A f f + + + +
(117)
2
4
2
b b ac
s
a

(118)
( ) ( ) ( )
2
2
1 2 1 2 1 2 0
1
2 2 2 2 16 1
2
s f f f f f f A + + +
(119)
Setting eq. (114) to zero and continuing to work the right-hand side:
Now we have a common denominator, thus, the sum of the
numerators must equal zero:
Note that eq. (117) has the form as
2
+ bs + c = 0, and the roots can
be found with the quadratic formula:
i.e.,
Eq. (119) gives two roots:
G The roots are a function of the feedback factor, .
G Therefore, the roots will move as increases from zero.
G We can plot a root locus!!!
Electronic Applications 62 Frequency Characteristics of Feedback Amplifiers
-2f
2
-2f
1
=0 =0
1_
2
(2f
1
+

2f
2
)
j

Fig. 92. Root locus of poles in a two-pole amplifier with negative

feedback.
One can see that, as increases from zero, the amplifier eventually
exhibits excessive ringing (excessive gain peaking).
However, the poles never move into the right half-plane, i.e., the
amplifier is always stable.
Electronic Applications 63 Frequency Characteristics of Feedback Amplifiers
3
1000
( )
1
2
b
A s
s
f

j \
+
, (
( ,
(120)
3
1000
1
1
2
b
s
f

j \
+
, (
( ,
(121)
3
3
10
1
1
2
b
s
f

j \
+
, (
( ,
(122)
3
3
10
2 1
1
b
s f

j \

, (

( ,
(123)
Amplifiers With 3 or More Poles
Well illustrate the behavior of these amplifiers by example.
Consider an amplifier with an open-loop transfer function that has
three identical poles on the negative real axis at -2f
b
:
To find the closed-loop poles we need the roots of 1 + A(s) = 0,
from which A(s) = -1, i.e.:
Taking the cube root of both sides of eq. (121):
And solving for s to find the closed-loop poles:
But , or , or .
( )
3
1 1
( )
3
1 1 60
( )
3
1 1 60
So there are three closed-loop poles.
Electronic Applications 64 Frequency Characteristics of Feedback Amplifiers
( )
( )
3
1
3
2
3
3
2 10 1 60 1
2 10 1 60 1
and 2 10 1
b
b
b
s f
s f
s f



, ]

]
, ]

]
, ]
+
]
(124)
j

-2f
b
60
Fig. 93. Root locus of closed-loop poles in
our 3-pole amplifier.
The three closed-loop poles are:
The root-locus of these three poles as a function of the feedback
factor, , is shown below. Notice that, because poles move into the
right half-plane, the amplifier eventually becomes unstable.
Electronic Applications 65 Frequency Characteristics of Feedback Amplifiers
4 4
4 4
1 1 45 1 1 45
1 1 135 1 1 135


(125)
j

-2f
b
Fig. 94. Root locus of closed-loop poles in a 4-
pole amplifier.
We will have a similar result for an amplifier with an open-loop
transfer function that has four identical poles on the negative real
axis at -2f
b
, because:
Notice that with sufficient feedback, this amplifier becomes unstable
also. We could extend this result to amplifiers with any larger
number of poles.
Electronic Applications 66 Gain Margin and Phase Margin
( ) ( )
( ) ( )
1 ( ) 1 ( )
f f
A s A f
A s A f
A s A f

+ +
(126)
( )
1
1 1 180 A f
(127)
Gain Margin and Phase Margin
Introduction
Consider the closed-loop gain equation evaluated at s = j2f :
where we have assumed to be independent of frequency.
Now, suppose there is some particular frequency f
1
for which
For this case, we may conclude:
G The closed-loop gain A
f
(f
1
) will be infinite.
G Poles will exist on the j axis.
G The transient response will have a constant-amplitude
sinusoid!!!
In fact, what we have just described is an oscillator!!!
We will study oscillators in a subsequent section where this
situation is desirable, but in an amplifier it is to be avoided!!!
Electronic Applications 67 Gain Margin and Phase Margin
A(f)

+
-
v
s
=0
Assume
1
2
3
Fig. 95. Examining conditions for oscillation in a feedback amplifier.
1
cos2
m
V f t
(128)
1 1 1
( ) cos2 cos2
m m
A f V f t V f t i
(129)
( )
1 1
0 cos2 cos2
m m
V f t V f t
(130)
Lets examine this situation more carefully:
1. Presume that a signal exists at the amplifier input ( 1 ):
2. The output of the feedback network ( 2 ) is:
3. Then the signal applied to amplifier input ( 3 ) is:
4. Thus, the input signal can maintain itself, and we have a
constant-amplitude sinusoid at the frequency f
1
!!!
Again, this is certainly a desirable condition in oscillators, but in an
amplifier it is to be avoided!!!
Electronic Applications 68 Gain Margin and Phase Margin
Lets review the key point:
G Placing poles on the j axis requires
1
( ) 1 180 A f
More generally, we can observe what happens at the frequency for
which :
1
( ) 180 A f
G If the magnitude of A(f) is greater than 1, poles are in the
right half-plane and the amplifier will be unstable.
G If the magnitude of A(f) is less than 1, poles are in the left
half-plane and the amplifier will be stable.
G The lower the magnitude of A(f), the farther the poles are
from the j axis, and the more stable the amplifier
Obviously, in an amplifier, we want to avoid placing poles on, or
even near, the j axis!!!
Definitions
The value of the magnitude of A(f), when the phase angle of
A(f) = 180 would be a good measure of amplifier stability:
Gain Margin: the amount that 20 log |A(f)| is below 0 dB, at
the frequency where the angle of A(f) = -180.
Another good measure of amplifier stability would be how close the
angle of A(f) is to -180 when the magnitude of A(f) = 1:
Phase Margin: the phase difference between the angle of A(f)
and -180, at the frequency where 20 log |A(f)| = 0 dB.
Electronic Applications 69 Gain Margin and Phase Margin
20 log |A(f)|
0
A(f)
-180
G.M.
P.M.

Fig. 96. Illustration of gain margin and phase margin.

20log ( ) 20log ( ) 20log A f A f +
(131)
Gain Margin, Phase Margin, and Bode Plots
Gain Margin and Phase Margin can be determined from the a Bode
magnitude and phase plots. Note that:
Thus, magnitude plot of A(f)
is just the magnitude plot of
open-loop gain A(f), shifted
downward by 20 log .
A(f) = A(f) because is
independent of frequency by
assumption.
Thus, the phase plot of A(f)
is same as the phase plot of
A(f).
To avoid excessive ringing in the transient response, and excessive
gain peaking in the frequency response, we generally require
G.M. 10 dB and P.M. 45.
Electronic Applications 70 Gain Margin and Phase Margin
20log ( ) 20log 20log ( ) 0 A f A f +
(132)
1
20log ( ) 20log 20log A f

(133)
To repeat:
The magnitude plot of A(f) is just the magnitude plot of open-loop
gain A(f), shifted downward by 20 log .
This would normally require us to take the 20 log A(f) plot and
redraw it on a new set of axes. This approach is tedious at best.
But there is an easier way:
G Rather than shift the magnitude plot of A(f) down by 20 log ,
we can move the 0 dB axis up by 20 log , i.e.,
G We can draw a new axis at 20 log 1/.
G This is usually easy, because 1/ A
f
.
We can show this mathematically. To determine phase margin, the
point we want in Fig. 96 is where:
or, where
This is illustrated on the following page.
Electronic Applications 71 Gain Margin and Phase Margin
20 log |A(f)|
0
-180
G.M.
P.M.
20 log 1/
A(f)
Fig. 97. Typical method of determining gain margin
and phase margin.
Electronic Applications 72 Gain Margin and Phase Margin
20 log |A|
60 dB
40 dB
20 dB
0 dB f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
39 dB

Fig. 98. G.M. and P.M. example #1.

Example #1
Given: An op amp with an open-loop dc gain of 1000, and three
low-pass poles, all at 100 kHz.
Find: Maximum allowable feedback factor, , for stability.
Solution: 1. Find frequency for which Phase Margin is 0.
2. Find open loop gain at this frequency (39 dB).
3. Thus 20 log 1/ = 39 dB for P.M. = 0.
4. Solving gives 0.0112.
Electronic Applications 73 Gain Margin and Phase Margin
20 log |A|
60 dB
40 dB
20 dB
0 dB f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
P.M.
G.M.

Fig. 99. G.M. and P.M. example #2.

Example #2
Given: An op amp with an open-loop dc gain of 1000, and three
low-pass poles, all at 100 kHz.
Find: Gain margin and phase margin for = 0.002.
Solution: 1. Draw 20 log 1/ line (at 54 dB) - note the frequency.
2. Find phase shift at that frequency: P.M. = 36.
3. Find frequency for which phase shift is -180.
4. Find gain at that frequency: G.M. = 25 dB.
Electronic Applications 74 Gain Margin and Phase Margin
20 log |A|
60 dB
40 dB
20 dB
0 dB
f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
80 dB
100 dB
100 Hz 10 Hz 1 Hz
P.M.=45
20 log 1/

Fig. 100. G.M. and P.M. example #3.

Example #3
Given: An op amp with an open-loop dc gain of 100 dB, and
poles at 1 kHz, 100 kHz., and 10 MHz.
Find: Maximum allowable for P.M. 45.
Solution: 1. Find frequency for which phase shift is -135.
2. Find gain at that frequency: 20 log 1/ = 60 dB.
3. Solving gives = 0.001.
Electronic Applications 75 Gain Margin and Phase Margin
Note in the previous example that the 20 log 1/ line goes through
2
nd
pole on the magnitude plot. This is not just a coincidence!!!
If the poles are sufficiently spaced, the 20 log 1/ line will always go
thru the second pole for P.M. of 45:
G This is because a P.M. of 45 corresponds to a phase shift of
-135.
G If were way past the first pole we have -90 shift from that
pole . . .
G . . . plus an additional -45 shift at the second pole, for a total
phase shift of -135.
This is why we determined that a P.M. of 45 produced acceptable
stability in the first place !!!
Electronic Applications 76 Compensation of Amplifiers
20 log |A|
60 dB
40 dB
20 dB
0 dB
f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
80 dB
100 dB
100 Hz 10 Hz 1 Hz
20 log 1/
P.M.=21

Fig. 101. Insufficiently stable amplifier.

Compensation of Amplifiers
Suppose we must use the open-loop amplifier from example #3 of
the previous section . . .
A
0
= 100 dB; Poles at 1 kHz, 100 kHz., and 10 MHz
. . . in a closed-loop amplifier with a gain of 40 dB.
We also want the resulting feedback amplifier to be stable
(P.M. 45), but unfortunately, P.M. is only 21:
Electronic Applications 77 Compensation of Amplifiers
20 log |A|
60 dB
40 dB
20 dB
0 dB
f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
80 dB
100 dB
100 Hz 10 Hz 1 Hz
P.M.=45
20 log 1/

Fig. 102. Compensation by adding a dominant pole.

Compensation by Adding a Dominant Pole
However, we can force P.M. to be 45 by adding another pole - a
dominant pole.
To be dominant, the additional pole must be at a frequency much
less than the current lowest pole (1 kHz), and the 1 kHz pole will
become the second pole!!!
The 20 log 1/ line will intersect the new magnitude plot at the
second pole, like it always does for P.M. = 45. Then we simply
construct the new magnitude plot!!! In this case we need to add
the dominant pole at 1 Hz:
Electronic Applications 78 Compensation of Amplifiers
r

+
-
v

g
m
v

r
o
C

B C
E E
B
C
2
C
1
Fig. 103. A single capacitor, C
1
or C
2
, can be added to the open-loop amplifier to produce a
dominant pole.
Adding a dominant pole is usually very simple:
The drawback of this method is that we sacrifice a tremendous
amount of bandwidth to achieve stability.
Compensation by Moving an Existing Pole
Instead of adding a dominant pole, we may be able to move an
existing pole to a much lower frequency so that it becomes
dominant.
The reduction in bandwidth with this approach is not as severe.
The least reduction in bandwidth results if the lowest existing pole
is the one that can be moved - see the figure on the next page.
When we move the lowest pole, the second pole determines the
frequency where the new P.M. = 45, and we simply construct the
new magnitude plot. The construction is shown on the following
page.
To move a pole we need to identify (at least approximately) a
capacitance that causes the pole we wish to move. We then move
the pole by adding additional capacitance in parallel.
See the Bode diagram on the following page for details.
Electronic Applications 79 Compensation of Amplifiers
20 log |A|
60 dB
40 dB
20 dB
0 dB
f, kHz
10 kHz 100 kHz 1 kHz 1 Mhz 10 Mhz
A
f, kHz
-30
-60
-90
-120
-150
-180
-210
-240
-270
80 dB
100 dB
100 Hz 10 Hz 1 Hz
P.M.=45
20 log 1/

Fig. 104. Compensation by moving an existing pole. Note the less severe reduction in
bandwidth compared to introducing an additional, dominant pole (here shown in dotted
green for comparison).
Electronic Applications 80 Compensation of Amplifiers
V
in
V
out
R
A
R
B
C
+ +
- -
Fig. 105. Simple circuit with a pole and a zero in the
transfer function.
1
( ) 2
( )
1
2
out z
in
p
s
V s f
s
V s
f

+
(134)
( )
1 1
and
2 2
z p
B A B
f f
R C R R C

+
(135)
Compensation by Adding a Pole and a Zero
Sometimes we cant move a pole directly, but we can cancel it with
a zero, and add a new pole at the desired location.
The result is the same as moving that particular pole!!!
The transfer function of the circuit in the figure above is:
where
Electronic Applications 81 Compensation of Amplifiers
Fig. 106. Schematic diagram of the LM741 operational amplifier, obtained from National
Semiconductors web site at http://www.national.com.
Compensation in the LM741 Operational Amplifier
The 30 pF capacitor in the LM741 introduces a dominant pole in the
op amp (open loop) transfer function at approximately 5 Hz.
The dominant pole assures unconditional stability for feedback
factors as large as 1, i.e., even for a voltage follower.
The required capacitance is minimized by placing the capacitor at
a point of large Thevenin resistance, and by using the Miller Effect
to increase the effective capacitance.
Electronic Applications 82 Oscillators
( ) f
(136)
Oscillators
Introduction
G Oscillators generate periodic signals - sine, square, etc.
G They convert dc power to ac power spontaneously.
G Examples are:
Local Oscillator - generates a sinusoidal signal used to select
the proper frequency in a radio, TV, etc.
Clock - computer-generated rectangular waves for timing
G In switching oscillators, active devices are used as switches.
G In linear oscillators, active devices are used in a linear
feedback amplifier.
The feedback path around the amplifier is usually intentionally
frequency selective :
We have just studied how to avoid oscillation in feedback amplifiers,
but by doing so, we have also already studied how to produce
oscillation.
Electronic Applications 83 Oscillators
A(f)

+
-
v
s
=0
Assume
1
2
3
Fig. 107. Examining conditions for oscillation in a feedback
amplifier (Fig. 95 repeated).
1
cos2
m
V f t
(137)
1 1 1 1
( ) ( ) cos2 cos2
m m
f A f V f t V f t i
(138)
( )
1 1
0 cos2 cos2
m m
V f t V f t
(139)
Lets re-visit a part of our discussion of gain margin and phase
margin from p. 67:
1. Presume that a signal exists at the amplifier input ( 1 ):
2. The output of the feedback network ( 2 ) is:
3. Then the signal applied to amplifier input ( 3 ) is:
4. Thus, the input signal can maintain itself, and we have a
constant-amplitude sinusoid at the frequency f
1
!!!
We can build an oscillator if the gain around the loop is unity!!!
We usually dont need to identify the amplifier block, the feedback
block, or the summing block. We need only concern ourselves with
the gain around the loop.
(Dont confuse gain around the loop with loop gain which we
already defined as the product A(f)(f) )
Electronic Applications 84 Oscillators
The requirement of unity gain around the loop means:
G The magnitude of the gain around the loop equals 1.
G The total phase shift around the loop is 0 or 360.
This is known as the Barkhausen Criterion.
Practical Linear Oscillators:
G Have a gain magnitude greater than unity so that oscillations
will build.
G Eventually, amplifier nonlinearity (clipping) changes the gain
back to unity.
G This assures oscillation will always begin despite component
aging, temperature and humidity changes, etc.
Electronic Applications 85 Oscillators
R
R
C
C
Ideal Voltage Amplifier
R
in
= Gain = A
v
Z
p Z
s

Fig. 108. Wien-Bridge Oscillator

R
R
C
C
Ideal Voltage Amplifier
R
in
= Gain = A
v

Z
p Z
s
V
in V
o
+
+
-
-
V
o
+
-
Fig. 109. Modified oscillator circuit, ready for gain
calculation.
( )
( )
1/
1/
1/
1/
1/
p
in
o s p
R j C
Z
V R j C
R j C
V Z Z
R j C
R j C

+

+
+ +
+
(140)
RC Oscillator Example - Wien-Bridge Oscillator
Here, we will put the principles
of the previous section to work.
To analyze this circuit, we:
1. Break the loop at any
convenient place.
2. Insert a test source.
3. Calculate gain all the way
around the loop.
Well start with the passive network & note that R
in
= , so:
Electronic Applications 86 Oscillators
( )
( )
1/
1/
1/
1/
1/
p
in
o s p
R j C
Z
V R j C
R j C
V Z Z
R j C
R j C

+

+
+ +
+
(141)
2
2
2 2
2
2
2 2
2
1
1 1
1
1 1 1
2
1
1
1 1
3
3
1
3
in
o
R
j C V
V
R R
j C j C
R
j C
R R R
j C C j C
R
j C R
j CR j R
R R
C
C j C
R
R j CR
C

j \
, (
( ,

j \ j \
+ +
, ( , (
( , ( ,
j \
, (
( ,

j \ j \
+ +
, ( , (
( , ( ,
j \
, (
( ,

j \
j \
+
+
, (
, (
( ,
( ,

j \
+
, (
( ,
(142)
Now, repeating eq. (?):
and multiplying the numerator and the denominator by R + 1/jC :
Note that we have manipulated V
in
/V
o
so that all the frequency
terms are in the denominator!!! This is a standard approach!!!
Electronic Applications 87 Oscillators
R
R
C
C
Ideal Voltage Amplifier
R
in
= Gain = A
v

Z
p Z
s
V
in V
o
+
+
-
-
V
o
+
-
Fig. 110. Modified oscillator circuit, ready for gain
calculation (Fig. 109 repeated).
2
1
3
in
o
V R
V
R j CR
C

j \
+
, (
( ,
(143)
2
1 0
1
3
o v
o
V A R
V
R j CR
C

j \
+
, (
( ,
(144)
Lets catch our breath and review . . .
Were working our way
around the loop . . .
So far weve obtained
the transfer function
from our test source to
the input of the ideal
amplifier.
That transfer function is
repeated here for
convenience:
To complete the loop, we need only account for the amplifier gain,
so our transfer function around the loop is:
where we have already applied the Barkhausen Criterion by
choosing the voltage V
o
for both sides of the break in the loop.
Be sure to take note of more than this particular equation. The
method used here is quite standard in most oscillator problems!!!
Electronic Applications 88 Oscillators
2
1 0
1
3
o v
o
V A R
V
R j CR
C

j \
+
, (
( ,
(145)
2
2 2 2
1
0 1 0
1 1
2
osc osc
osc
osc osc
CR C R
C
f
RC RC

(146)
1 3
3 0
v
v
A R
A
R j

+
(147)
To complete the analysis, we first repeat eq. (144):
Notice that transfer function must have zero phase shift, which
means the imaginary term in the denominator must be zero. This
allows us to determine the frequency of oscillation, which well
denote as
osc
:
And, at f
osc
, the magnitude must be unity (or greater) to insure
oscillation:
Electronic Applications 89 Oscillators
R
R
C
C
Z
p Z
s
R
1
R
2
v
o
+
-
-
+
Fig. 111. Implementing the Wien-Bridge
oscillator with an op amp.
R
R
C
C
Z
p Z
s
Lamp
R
2
v
o
+
-
-
+
Fig. 112. A practical Wien-Bridge oscillator with
AGC.
Practical Wien-Bridge Oscillator #1:
We can implement the Wien-
Bridge oscillator by using an
op amp with the gain set to be
> 3 to insure oscillation will
always begin (e.g. R
2
= 22 k
and R
1
= 10 k).
However, there is a problem:
The oscillations grow until the
op amp saturates.
This causes clipping of the
output peaks at V
OH
& V
OL
.
The solution for this is to
implement an automatic gain
control (AGC).
In Fig. 112, the lamp filament
has a low resistance when
cold, and A
v
> 3.
As oscillations begin to build,
increasing current will flow
through the lamp. The lamp
resistance increases, and the
gain is reduced automatically.
The drawback is that a very
low value is required for R
2
,
so the op amp is very heavily
loaded.
Electronic Applications 90 Oscillators
R
R
C
C
Z
p Z
s
R
2A
v
o
+
-
-
+
R
2B
R
1
Fig. 113. A second Wien-Bridge oscillator with
AGC.
Practical Wien-Bridge Oscillator #2:
We can implement the desired AGC in a different manner, as
shown in Fig. 113.
In this circuit, at low output
levels, A
v
= R
2A
/R
1
> 3. The
diodes are off while the
oscillation builds.
As oscillation grows, the
diodes conduct, putting R
2B
in
parallel with R
2A
.
Gain is reduced, and the
output level stabilizes without
clipping.
However, theres a drawback
here, too. The nonlinear
diodes introduce some
distortion of their own.
Electronic Applications 91 Oscillators
R
R
C
C
Z
p Z
s
R
2
v
o
+
-
-
+
R
1A
R
1B
R
filter
C
filter
Fig. 114. Our last, and fanciest, Wien-
Bridge oscillator with AGC.
( )
2
1 1
1 3
||
v
B A FET
R
A
R R R
+ >
+
(148)
2
1
1 3
v
B
R
A
R
+ < (149)
Practical Wien-Bridge Oscillator #3:
The implementation of a fancier AGC circuit is our final Wien-Bridge
oscillator.
The JFET is used in its voltage
controlled resistance region:
R
FET
is low for V
GS
0.
R
FET
is high for V
GS
< 0.
The diodes, C
filter
, and R
filter
form
a peak rectifier circuit:
The diodes conduct for
v
o
< -(V
Z
+ 0.7 V).
C
filter
stores negative
voltage.
R
filter
discharges C
filter
slowly.
For low-amplitude oscillation, V
GS
is near zero, and R
FET
is small:
For high-amplitude oscillation, V
GS
is negative, and R
FET
is large:
Electronic Applications 92 Oscillators
V
CC
L
1
L
2
L
3
C
R
L
Fig. 115. Hartley oscillator.
L
1 L
2
R
C
g
m
v
gs
v
gs
+
-
Fig. 116. Hartley oscillator small-signal
equivalent.
L
1
L
2
R
C
g
m
v
gs
v
gs
+
-
v
gs
Z
2
Z
1
i
x
Fig. 117. Hartley equivalent redrawn.
LC Oscillators - The Hartley Oscillator
Because inductances are less
practical at low frequencies, LC
oscillators are generally considered
to be high-frequency (rf) oscillators.
For our analysis we note that:
R
L
is reflected into the L
2
primary as R.
All elements are assumed to
be ideal.
All device capacitances are
ignored.
We draw the small-signal
model and recall that we
want gain around the loop
to be unity.
But its difficult to see a loop,
so we redraw Fig. 116.
For the analysis to proceed,
note that:
Z
1
comprises L
1
and C in
series.
Z
2
comprises L
2
and R in
parallel.
Z
1
, Z
2
, and g
m
v
gs
form a
current divider.
Electronic Applications 93 Oscillators
L
1
L
2
R
C
g
m
v
gs
v
gs
+
-
v
gs
Z
2
Z
1
i
x
Fig. 118. Hartley equivalent redrawn (Fig. 117
repeated).
1
1
Z j L
C

j \

, (
( ,
(150)
2
2
2
j L R
Z
R j L

+
(151)
( )
2
2 2
2
2 1
1
2
2
2 2 2
2
2 1 1 2
2
2
2
1 2 1 2
1
1
j L R
Z R j L
j L R
Z Z
j L
R j L C
j L R
L R
j L R j L R j j L L j
C C
j L R
L
L L jR L L
C C

+ j \
+
, (
+
( ,

+ +

j \ , ]
+ +
, (
, ]
] ( ,
(153)
2
2 1
x m gs
Z
i g v
Z Z

+
(152)
The impedances are:
and
And the current divider equation is:
where
Electronic Applications 94 Oscillators
L
1
L
2
R
C
g
m
v
gs
v
gs
+
-
v
gs
Z
2
Z
1
i
x
Fig. 119. Hartley equivalent redrawn (Fig. 117
repeated).
( )
2
2
2
1 2 1 2
1
m gs
x
j L Rg v
i
L
L L jR L L
C C

j \ , ]
+ +
, (
, ]
] ( ,
(154)
( )
( )
1 2
2
2
1 2 1 2
1
m gs
gs
j L j L Rg v
v
L
L L jR L L
C C

j \ , ]
+ +
, (
, ]
] ( ,
(155)
( )
2
1 2
2
2
1 2 1 2
1 0
1
m
L L Rg
L
L L jR L L
C C

j \ , ]
+ +
, (
, ]
] ( ,
(156)
Now, repeating the current divider equation for convenience:
Now, v
gs
is the voltage across L
1
. It is simply i
x
(jL
1
):
Multiplying the numerator terms together, and dividing both sides by
v
gs
, gives the familiar form:
Electronic Applications 95 Oscillators
L
1
L
2
R
C
g
m
v
gs
v
gs
+
-
v
gs
Z
2
Z
1
i
x
Fig. 120. Hartley equivalent redrawn (Fig. 117
repeated).
( ) ( )
( )
1 2 1 2
1 2
1 1
0
1
osc
L L L L
C C
L L C

+ +

+
(157)
2
1 2
2
2
1 2
1
osc m
osc
L L Rg
L
L L
C

(158)
2
2
1 2
2 2
1 2 1
1 1
osc
m
osc osc
L
L L
C
g
R L L R LCR

(159)
For 0 phase shift, the imaginary term in the denominator of eq.
(156) must be zero:
And, for eq. (156) to have a magnitude of unity at
osc
:
from which:
Electronic Applications 96 Oscillators
L
1
L
2
R
C
g
m
v
gs
v
gs
+
-
v
gs
Z
2
Z
1
i
x
Fig. 121. Hartley equivalent redrawn (Fig. 117
repeated).
V
CC
L
1
L
2
L
3
C
R
L
Fig. 122. Hartley oscillator (Fig. 115
repeated.
( )
1 2 1 2
1
1 1 1
1 2
1 1
m
L L L L
g
LCR
R L R L R L R
L L C
+

+
(160)
( )
1 2
1
osc
L L C

+
(161)
2
1
m
L
g
L R
(162)
Finally, we substitute eq. (157) for
osc
2
in eq. (159):
Weve done a lot of algebra!!! Lets review the original circuit, and
the two important results:
Electronic Applications 97 Oscillators
R
D
R
G
C
1
C
2
C
B
L
Fig. 123. Colpitts oscillator.
L
R
C
1
g
m
v
gs
v
gs
+
-
C
2
Fig. 124. Colpitts oscillator small-signal equivalent.
1 2 1
1 2 2
and
osc m
C C C
g
LCC C R

+

(163)
LC Oscillators - The Colpitts Oscillator
The Colpitts Oscillator is very
similar topologically to the Hartley:
R
G
provides a dc ground for
gate.
C
B
blocks dc drain voltage
from the gate circuit, and
provides an ac short circuit.
C
1
and C
2
can include device
capacitances.
The small-signal equivalent circuit
is shown below:
Again, note the similarity to Hartley oscillator. The analysis
proceeds in a similar fashion. We show only the results here:
Electronic Applications 98 Oscillators
R
D
R
G
C
1
C
2
C
B
L
Fig. 127. Colpitts oscillator (Fig. 123
repeated).
V
CC
L
1
L
2
L
3
C
R
L
Fig. 125. Hartley oscillator (Fig. 115
repeated.
L
R
C
1
g
m
v
gs
v
gs
+
-
C
2
Fig. 128. Colpitts oscillator small-signal
equivalent (Fig. 124 repeated).
L
1 L
2
R
C
g
m
v
gs
v
gs
+
-
Fig. 126. Hartley oscillator small-signal equivalent
(Fig. 116 repeated).
( )
2
1
1 2
1
osc m
L
g
L R
L L C

+
(164)
1 2 1
1 2 2
osc m
C C C
g
LCC C R

+

(165)
Hartley - Colpitts Comparison
Hartley Oscillator
Colpitts Oscillator
Electronic Applications 99 Oscillators
Crystal Oscillators
The Piezoelectric Effect:
The piezoelectric effect describes the phenomenon whereby a
mechanical stress applied to certain crystals produces a voltage
proportional to the applied stress - the phenomenon is reciprocal.
The phenomenon was proven to exist in the early 1880's by the
Curie brothers, who performed experiments with tourmaline,
Rochelle salt, and cane sugar in addition to the now-well-known
quartz crystals.
Piezoelectric crystals are used in various transducers, e.g.,
microphones, strain gauges, and phonograph cartridges (does
anyone remember these???)
Well look only at the use of piezoelectric crystals in oscillators.
Quartz Crystal Characteristics:
The electro-mechanical resonance characteristics of quartz crystals
are extremely stable.
RC and LC oscillators are stable to within 100 - 1000 ppm
Crystal oscillators are stable to within 1 ppm !!!
Depending on the mode of mechanical vibration employed, crystal
oscillators can be obtained with resonant frequencies from a few
kHz to several hundred MHz.
Electronic Applications 100 Oscillators
L
C
S
R
C
P

Fig. 129. Quartz crystal schematic

symbol and equivalent ckt.
f
P
f
S
f
x
crystal
Fig. 130. Reactance of
a quartz crystal.
In the quartz crystal equivalent
circuit at left, L, C
S
, and R
represent the electro-mechanical
properties of the crystal.
L as high as to 100's of H
C
S
as low as a fraction of a F
R usually only a few 10's of
C
P
capacitance between
electrodes, usually 1 - 10 pF
The quality factor, Q, is defined for a series circuit as L/R.
Conventional LC circuits at resonance have a Q of 200 to 300, but
for a crystal at resonance, Q can be several 100,000 !!!
The reactance of a quartz crystal is shown
at left.
The series-resonant frequency, f
S
, is
where L and C
S
resonate.
Above f
S
, the series branch becomes
inductive.
The parallel-resonant frequency, f
P
, is
where this inductance resonates with C
P
.
Because of the high Q, f
S
and f
P
are within
a few tenths of a %.
Electronic Applications 101 Oscillators
R
D
R
G C
1
C
2
Fig. 131. Pierce oscillator.
V
CC
V
EE
L
C
2
C
1 R
E
R
B
Fig. 133. Series-resonant crystal
oscillator.
V
CC
V
EE
L
C
2
C
1 R
E
Fig. 132. Common-base
Colpitts oscillator.
Pierce Oscillator
The Pierce Oscillator is an
example of an antiresonant
oscillator - the crystal is used as
inductance.
The circuit is simply a Colpitts
oscillator with L replaced by the
crystal.
Zero dc gate bias is provided by
the resistance R
G
.
The frequency of oscillation is
between f
S
and f
P
.
Series-Resonant Crystal Oscillator
This is based on the Common-Base Colpitts Oscillator. The base
is grounded, resulting in sufficient gain for oscillation, only at the
crystal resonant frequency.
Electronic Applications 102 Comparators and Schmitt Triggers
+
-
V
DC+
V
DC-
V
OH
or
V
OL
v
+
v
-
Fig. 134. Comparator.
V
OH
=V
DC+
V
OL
=V
DC-
v
O
v
d
=v
+
-v
-
Fig. 135. Ideal transfer characteristic.
v
O
v
d
=v
+
-v
-
V
OH
V
OL
V
DC+
V
DC-
V
IOS
Fig. 136. Actual transfer
characteristic.
Comparators and Schmitt Triggers
Comparators
A comparator is a linear amplifier,
like an op amp, but intended for
switching rather than for use in the
linear region with negative feedback.
The output switches from V
OH
to V
OL
,
i.e., from one saturated state to the
other.
Ideal Transfer Characteristic:
Infinite gain
V
IOS
, input offset voltage, = 0
|v
d
| for switching = 0
Infinite input impedance
Infinite speed:
t
d
= t
r
= t
s
= t
f
= 0
Actual Transfer Characteristic:
Finite gain (60 dB typical)
V
IOS
usually a few mV
|v
d
| for switching < 1 mV
Finite speed
Electronic Applications 103 Comparators and Schmitt Triggers
Fig. 137. Speed vs. overdrive, negative output
transition, National LM393 comparator.
Fig. 138. Speed vs. overdrive, positive output
transition, National LM393 comparator.
Speed vs. Overdrive:
Overdrive is defined as the amount
by which a step change in the input
goes past zero.
The larger the overdrive, the faster
the output transition.
The effect of overdrive on the
speed of the output transition is
different for a negative-going output
than for a positive-going output.
Figs. 137 and 138 are taken from
the LM393 data sheets, available
from National Semiconductor, Inc.,
at http://www.national.com .
Electronic Applications 104 Comparators and Schmitt Triggers
V
REF
+
-
v
I +
-
v
O
Fig. 139. Noninverting open-loop
comparator
V
OH
V
OL
v
O
v
I min
v
I max
v
I
V
REF
V
IOS
Fig. 140. Noninverting transfer function.
V
REF
+
-
v
I
v
O
+
-
Fig. 141. Inverting open-loop compartator.
V
OH
V
OL
v
O
v
I min
v
I max
v
I
V
REF
V
IOS
Fig. 142.
Inverting transfer function.
Open-Loop Comparators:
The two possible configurations and their idealized output
characteristics are shown.
Comparators are not compensated. When operated open-loop a
slowly varying input will likely result in an oscillating output, because
the device is in the linear region long enough for oscillations to
build.
Electronic Applications 105 Comparators and Schmitt Triggers
V
DC+
V
DC-
v
O
Fig. 143. Totem-pole output.
V
DC+
V
DC-
v
O
R
pull-up
Fig. 144. Open-collector output.
Totem-Pole vs. Open Collector Outputs:
IC comparators have one of two output stages.
A totem-pole output, also called an
active pull-up, is shown at left.
Advantages are:
Fast switching
V
OH
0.7 V to 1 V below V
CC

V
OL
V
CE sat

The major disadvantage is that it

consumes power. Because
transistors turn off more slowly than
they turn on, a large current spike
occurs at each transition.
An open-collector output uses a pull-
up resistor instead of an active
device.
Advantages are::
User-selected R, V
DC
, and V
OH
V
OL
V
CE sat

Hot element off chip

Wired NOR gate possible
But V
OH
varies with load, and the
circuit will be slow if = R
L
C
L
is large.
Electronic Applications 106 Comparators and Schmitt Triggers
+
-
v
I
V
R
R
F
R
P
V
OH
or
V
OL
Fig. 146. Inverting Schmitt trigger.
v
O
v
I
V
2
V
1
V
OH
V
OL
v
I min
v
I max
Fig. 145. Transfer function for inverting
Schmitt trigger.
2
R F OH P
F P
V R V R
v V
R R
+
+

+
(166)
1
R F OL P
F P
V R V R
v V
R R
+
+

+
(167)
( )
2 1
P
H OH OL
F P
R
V V V V V
R R

+
(168)
Schmitt Triggers
A Schmitt Trigger is a comparator with positive feedback, which
provides hysteresis in the transfer characteristic. This solves the
oscillation problem and provides noise immunity.
Inverting Schmitt Trigger:

This is just a comparator with V
REF
= v
+
, the voltage at the
noninverting input. Note that V
REF
depends on V
R
and v
O
.
Switching occurs when v
I
= v
+
. For v
O
= V
OH
:
and, for v
O
= V
OL
:
From eqs. (166) and (167), we define the hysteresis voltage, V
H
:
Electronic Applications 107 Comparators and Schmitt Triggers
+
-
v
I
V
R
R
F
R
P
V
OH
or
V
OL
Fig. 147. Noninverting Schmitt trigger.
v
O
v
I
V
2
V
1
V
OH
V
OL
v
I min
v
I max
Fig. 148. Transfer function for noninverting
Schmitt trigger.
( )
1
1
R F P OH P
F OH P
R
F P F
V R R V R
VR V R
v V V
R R R
+
+
+

+
(169)
( )
2
2
R F P OL P
F OL P
R
F P F
V R R V R
V R V R
v V V
R R R
+
+
+

+
(170)
( )
2 1
P
H OH OL
F
R
V V V V V
R
(171)
Noninverting Schmitt Trigger:
Again, we have a comparator, but now v
+
depends on v
I
and v
O
.
For v
O
= V
OH
, let v
I
= V
1
at the state change (i.e., when v
+
= V
R
):
And, for v
O
= V
OL
, let v
I
= V
2
at the state change (when v
+
= V
R
):
In this case, the hysteresis voltage, V
H
, is
Electronic Applications 108 Comparators and Schmitt Triggers
( )
P
H OH OL
F
R
V V V
R
(172)
Schmitt Trigger Design:
To avoid oscillation for slowly changing inputs, only a few mV of
hysteresis voltage is usually required.
This means that R
P
<< R
F
. A very approximate typical ratio for
these resistors is 1:100.
For R
P
<< R
F
eq. (168) becomes nearly equal to (171), and
hysteresis voltage for either the noninverting or the inverting Schmitt
trigger is given by:
To aid your understanding of Schmitt triggers, remember these two
key points:
1. Switching level depends on the state of the output
2. The feedback is always positive, regardless of the
connection used.
Electronic Applications 109 Comparators and Schmitt Triggers
+
-
v
I
V
R
R
F
R
P
V
OH
or
V
OL
Fig. 149. Inverting Schmitt trigger of
example.
v
O
v
I
10 V
2 V 6 V
Fig. 150. Transfer function
for example.
1
2 V
F
R
F P
R
V V
R R

+
(173)
2
10
6 V
R F P
F P
V R R
V
R R
+

+
(174)
10 10
4 V
R F P R F P
H
F P F P F P
V R R V R R
V
R R R R R R
+

+ + +
(175)
Schmitt Trigger Example:
A comparator with a totem-pole output operates from a single 12 V
supply with V
OH
= 10 V and V
OL
= 0.
Design an inverting Schmitt trigger that switches at 2V and 6V.
When v
O
= 0 we want switching at v
I
= 2 V:
When v
O
= 10 we want switching at v
I
= 6 V:
Now, V
2
- V
1
= V
H
:
Note we have only one equation, but two unknowns . . . we get to
arbitrarily choose one of the unknowns!!!
Electronic Applications 110 Comparators and Schmitt Triggers
40 k and 60 k
P F
R R
(176)
+12 V
V
TH
=

3.33 V
R
TH
=

40 k
Fig. 151. Using a voltage divider to obtain V
R
and R
P
.
2 V 3.33 V
F P
R
F
R R
V
R
+
(177)
We may select R
F
, R
P
, or R
F
+ R
P
. . . well let R
F
+ R
P
= 100 k.
Then, from eq. (175), 10R
P
= 400 k, thus:
Finally, we can use either (173) or (174) to solve for V
R
.
From (173) (its the easier one!):
We already have a 12 V supply, so we obtain V
R
and R
P
with a
voltage divider from that supply:
The details of the calculations implied by Fig. 151 are left for you to
work out . . .
Electronic Applications 111 Single-Pole Circuits
C
R
v
1
v
2
+ +
- -
Fig. 152. High Pass
C
R
v
1
v
2
+ +
- -
Fig. 153. Low Pass
s
P
s
Z
Fig. 154. HP pole-zeros.
s
P
Fig. 155. LP pole-zeros.
f
P
0 dB
20 dB
dec
Fig. 156. HP Bode plot.
f
P
0 dB
-20 dB
dec
Fig. 157. LP Bode plot.
1

exp(-
t

) u(t)
1

-
(t) -
Fig. 158. HP impulse resp
1

exp(-
t

) u(t)
1

Fig. 159. LP impulse resp

Single-Pole Circuits
Example
Time Constant = RC or = L/R
Transfer Function
2
1
( )
( )
( )
V s
T s
V s

( )
1
s
T s
s

+
1
( )
1
T s
s

j \
+
, (
( ,
Pole-Zero Pattern
s
P
= -1/
Bode Magnitude Plot

P
= 1/
Impulse Response
v
2
(t)
Natural Response v
2
(t) = (1/) exp(-t/) u(t)
Differential Eq. v
2
+ (1/)v
2
= v
1
v
2
+ (1/)v
2
= v
1
/
Electronic Applications 112 Single-Pole Circuits
0
t
start
t
end
r( ) = A
r(0)
r(t
0
)
t
0
r(t)
t

Fig. 160. General exponential (single-pole) response.

( ) exp ,
start end
t
r t A B t t t

j \
+ < <
, (
( ,
(178)
( ) A r
(179)
General Response to Piecewise-Constant Inputs
In any single-pole circuit, within any interval in which the input is
constant, the response will always have the form:
The time constant is usually known or can be determined from the
circuit. The remaining unknowns in eq. (178) are the coefficients A
and B, so we need to know the value of r(t) at two points.
Usually r() can be determined by inspection . . . and from eq. (178)
with t = , we have:
This is already noted on the r(t) axis in Fig. 160.
Electronic Applications 113 Single-Pole Circuits
0
t
start
t
end
r( ) = A
r(0)
r(t
0
)
t
0
r(t)
t

Fig. 161. General exponential response (Fig. 160 repeated).

(0) exp(0) r A B A B + +
(180)
(0) (0) ( ) B r A r r
(181)
[ ]
( ) ( ) (0) ( ) exp
t
r t r r r

j \
+
, (
( ,
(182)
[ ]
( )
0
0
( ) ( ) ( ) ( ) exp
t t
r t r r t r

, ]
+
, ]
]
(183)
We can also write:
which is mathematically valid even if r(0) is not within the interval.
Often, though r(0) is not only within the interval, but it is the start of
the interval. In either case:
So we can write the general response expression in the form:
Similarly, if the response r(t) is known at t = t
0
rather than at t = 0,
eq. (182) becomes:
Electronic Applications 114 Single-Pole Circuits
r( )
r(t
1
)
r(t
2
)
t
1
t
2
t

Fig. 162. General exponential response.

[ ]
( )
1
1
( ) ( ) ( ) ( ) exp
t t
r t r r t r

, ]
+
, ]
]
(184)
[ ]
( )
2 1
2 1
( ) ( ) ( ) ( ) exp
t t
r t r r t r

, ]
+
, ]
]
(185)
( )
2 1
2
1
( ) ( )
exp exp
( ) ( )
t t
r t r t
r t r
, ]
j \

, (
, ]

( ,
]
(186)
2 2
1 1
( ) ( ) total change in after
( ) ( ) total change in after
r t r r t
p
r t r r t



(187)
Time Interval Between Known Values of Exponential Response
Given:
, r(t
1
), r(t
2
), r()
i.e., the function is known at
three points.
We wish to determine an
expression for the interval t.
We know r(t
1
), so well let it serve as the time reference point for the
general response equation (183):
Then, at t
2
:
Now we solve eq. (185) for t
2
- t
1
, which is the interval t of interest:
For notational convenience, we let
Electronic Applications 115 Single-Pole Circuits
r( )
r(t
1
)
r(t
2
)
t
1
t
2
t

Fig. 163. General exponential response (Fig. 162

repeated).
1
exp exp
t t
p
p

j \ j \

, ( , (
( , ( ,
(188)
1
ln t
p

j \

, (
( ,
(189)
7.5 V 3
0.5 s
10 V 4
4
0.5ln 144 ms
3
RC p
t

j \

, (
( ,
(190)
Continuing with our derivation, we have:
From which:
Example:
A power supply filter capacitor of 10,000 F is in parallel with a 50
load in a 10 V system. How long will it take for the load voltage to
drop to 7.5 V after the supply is turned off?
Solution:
Electronic Applications 116 Single-Pole Circuits
v
2
(t)
T
t
V
X
=

V
1
V
Y
V
Z
v
H
(t)
v
L
(t)
Fig. 165. Pulse response of HPF.
v
1
(t)
V
1
T 0
t
C
R
v
1
v
2
+ +
- -
Fig. 164. High-pass filter and input
pulse.
1 2 1 1 X
v v V V V
(191)
[ ]
0
0
( ) ( ) ( ) ( ) exp
H H H H
t t
v t v v t v

, ]
+
, ]
]
(192)
0 0 1
( ) 0, 0, and ( )
H H
v t v t V (193)
2 1
( ) exp , 0
t
v t V t T

j \
< <
, (
( ,
(194)
Pulse Response of Single-Pole High-Pass

Assume v
1
(t) = 0 for t < 0 . . . then v
2
(t) = 0 for t < 0 also.
For t = 0:
Capacitor voltage cannot change instantaneously without a current
impulse, which would require infinite source voltage, thus:
For 0 < t < T:
We have a piecewise-constant input, thus:
where
from which:
Electronic Applications 117 Single-Pole Circuits
v
2
(t)
T
t
V
X
=

V
1
V
Y
V
Z
v
H
(t)
v
L
(t)
Fig. 167. Pulse response of HPF (Fig. 165,
171 repeated).
v
1
(t)
V
1
T 0
t
C
R
v
1
v
2
+ +
- -
Fig. 166. High-pass filter and input
pulse (Fig. 164 repeated).
2 1
( ) exp
Y
T
v T V V

j \

, (
( ,
(195)
1 1 1 1
exp exp 1
Z Y
T T
V V V V V V

, ]
j \ j \

, ( , (
, ]
( , ( ,
]
(196)
[ ]
0
0
( ) ( ) ( ) ( ) exp
L L L L
t t
v t v v t v

, ]
+
, ]
]
(197)
0
( ) 0, , and ( )
L L Z
v t T v T V (198)
2
( ) exp ,
Z
t T
v t V t T

j \
>
, (
( ,
(199)
Note from eq. (194) that at t = T
-
:
For t = T:
Again v
1
= v
2
= V
1
, and:
For T < t < :
We again have piecewise-constant input, thus:
where
from which
Electronic Applications 118 Single-Pole Circuits
v
2
(t)
T 0
t
V
1
V
A
Fig. 169. Pulse response of LPF.
v
1
(t)
V
1
T 0
t
C
R
v
1
v
2
+ +
- -
Fig. 168. Low-pass filter and input
pulse.
2
(0 ) 0 v
+

(200)
[ ]
0
0
( ) ( ) ( ) ( ) exp
H H H H
t t
v t v v t v

, ]
+
, ]
]
(201)
1 0 0
( ) , 0, and ( ) 0
H H
v V t v t (202)
2 1
( ) 1 exp , 0
t
v t V t T

, ]
j \
< <
, (
, ]
( ,
]
(203)
Pulse Response of Single-Pole Low-Pass

Assume v
1
(t) = 0 for t < 0 . . . then v
2
(t) = 0 for t < 0 also.
At t = 0:
Capacitor voltage cannot change instantaneously without a current
impulse, which would require infinite source voltage, thus:
For 0 < t < T:
We have a piecewise-constant input, thus:
where:
from which:
Electronic Applications 119 Single-Pole Circuits
v
2
(t)
T 0
t
V
1
V
A
Fig. 171. Pulse response of LPF (Fig. 169 repeated).
v
1
(t)
V
1
T 0
t
C
R
v
1
v
2
+ +
- -
Fig. 170. Low-pass filter and input
pulse (Fig. 168 repeated).
2 1
( ) 1 exp
A
T
v T V V

, ]
j \

, (
, ]
( ,
]
(204)
[ ]
0
0
( ) ( ) ( ) ( ) exp
L L L L
t t
v t v v t v

, ]
+
, ]
]
(206)
2 2
( ) ( )
A
v T v T V
+

(205)
0
( ) 0, , and ( )
L L A
v t T v T V (207)
2 1
( ) exp 1 exp exp ,
A
t T T t T
v t V V t T

, ]
j \ j \ j \
>
, ( , ( , (
, ]
( , ( , ( ,
]
(208)
Note from eq. (203) that at t = T
-
:
At t = T
+
:
Again, because v
capacitor
= 0:
For T < t < :
We again have piecewise-constant input, thus:
where:
from which:
Electronic Applications 120 Single-Pole Circuits
Fig. 172. High pass
v
1
(t)
V
1H
T
0
t
V
1L
V
1 P-P
t
H
t
L
Fig. 173. Rectangular wave input.
v
2
(t)
V
A
T
0
t
V
1 P-P
t
H
t
L
V
B
V
A
V
C
V
D
V
1 P-P
v
H
(t)
v
L
(t)
Fig. 174. Steady-state HP response.
Steady-State Rectangular-Wave Response
of Single-Pole High-Pass
Key Ideas: 1. v
2
has zero dc component.
2. Capacitor voltage cant change instantly
V
A
- V
D
= V
1 P-P
= V
B
- V
C
.
3. v
H
() = 0 = v
L
().
4. Steady-State Natural response has died.
5. Each cycle is identical to the preceding one.
Approach: 1. Assume arbitrary initial value V
A
.
2. Calculate response waveform through a full cycle.
3. Set value at end of period to initial value V
A.
Electronic Applications 121 Single-Pole Circuits
Fig. 175. High pass
v
1
(t)
V
1H
T
0
t
V
1L
V
1 P-P
t
H
t
L
Fig. 176. Rectangular wave input (Fig. 173
repeated).
v
2
(t)
V
A
T
0
t
V
1 P-P
t
H
t
L
V
B
V
A
V
C
V
D
V
1 P-P
v
H
(t)
v
L
(t)
Fig. 177. Steady-state HP response (Fig.
174 repeated).
[ ]
( )
0
0
( ) ( ) ( ) ( ) exp
t t
r t r r t r

, ]
+
, ]
]
(209)
( ) exp
H A
t
v t V

j \

, (
( ,
(210)
( ) exp
H
B H H A
t
V v t V

j \

, (
( ,
(211)
1 1
exp
H
C B P P A P P
t
V V V V V

j \

, (
( ,
(212)
To calculate the response waveform through a full cycle, well begin
with the general response to a piecewise-constant input:
For 0 < t < t
H
:
For t
H
-
< t < t
H
+
:
and
Electronic Applications 122 Single-Pole Circuits
Fig. 178. High pass
v
1
(t)
V
1H
T
0
t
V
1L
V
1 P-P
t
H
t
L
Fig. 179. Rectangular wave input (Fig. 173
repeated).
v
2
(t)
V
A
T
0
t
V
1 P-P
t
H
t
L
V
B
V
A
V
C
V
D
V
1 P-P
v
H
(t)
v
L
(t)
Fig. 180. Steady-state HP response (Fig.
174 repeated).
( ) exp
H
L C
t t
v t V

j \

, (
( ,
(213)
1
( ) exp
exp exp
H
D L C
H L
A P P
T t
V v T V
t t
V V

j \

, (
( ,
, ]
j \ j \

, ( , (
, ]
( , ( ,
]
(214)
2 1
( )
A D P P
v T V V V
+

+
(215)
For t
H
< t < T:
For T
-
< t < T
+
:
And, finally:
Electronic Applications 123 Single-Pole Circuits
Fig. 181. High pass
v
1
(t)
V
1H
T
0
t
V
1L
V
1 P-P
t
H
t
L
Fig. 182. Rectangular wave input (Fig. 173
repeated).
v
2
(t)
V
A
T
0
t
V
1 P-P
t
H
t
L
V
B
V
A
V
C
V
D
V
1 P-P
v
H
(t)
v
L
(t)
Fig. 183. Steady-state HP response (Fig.
174 repeated).
1 1
exp exp
H L
A A P P P P
t t
V V V V


, ]
j \ j \
+
, ( , (
, ]
( , ( ,
]
(216)
1 1
exp exp exp
H L L
A A P P P P
t t t
V V V V


j \ j \ j \

, ( , ( , (
( , ( , ( ,
(217)
1
1 exp
1 exp
L
P P
A
t
V
V
T

, ]
j \

, (
, ]
( ,
]

, ]
j \

, (
, ]
( ,
]
(218)
Now, we re-write eq. (215) using the expression for V
D
from eq.
(214):
Rearranging:
And finally solving for V
A
:
Electronic Applications 124 Single-Pole Circuits
v
2
(t)
t

Fig. 185. HP Steady-state

rectangular-wave
response, for large .
v
2
(t)
t

Fig. 184. HP Steady-state

rectangular-wave
response, for small .
With V
A
determined we can find V
B
, V
C
, and V
D
by going back
through the derived equations.
Special Cases:
1. For << t
H
, t
L
:
We will have V
B
0, V
D
0,
and V
2 P-P
2 V
1 P-P
.
In this case the circuit is known as a
differentiator.
2. For >> t
H
, t
L
,
We will have V
B
V
A
and V
D
V
C
,
and V
2 P-P
V
1 P-P

In this case the output is just input with dc

removed!!!
3. For any square wave input, t
H
= t
L
= T/2,
The output must still have zero average, so waveform must be
symmetrical around the t axis, i.e., V
C
= -V
A
and V
D
= -V
B
Electronic Applications 125 Single-Pole Circuits
C
R
v
1
v
2
+ +
- -
Fig. 186. Low pass.
v
1
(t)
V
1H
T
0
t
V
1L
t
H
t
L
Fig. 187. Rectangular wave
input.
v
2
(t)
T
0
t
t
H
V
B
V
A

v
H
(t)
v
L
(t)
Fig. 188. Steady-state LP response.
[ ]
( )
0
0
( ) ( ) ( ) ( ) exp
t t
r t r r t r

, ]
+
, ]
]
(219)
Steady-State Rectangular-Wave Response
of Single-Pole Low-Pass
Key Ideas: 1. Average of v
2
= average of v
1
(same dc component)
2. v
H
() = V
1H
and v
L
() = V
1L
.
3. Steady-State Natural response has died.
4. Each cycle is identical to the preceding one.
Approach: 1. Assume arbitrary initial value.
2. Calculate response waveform through a full cycle.
3. Set value at end of period to initial value.
We begin with the general response to a piecewise-constant input:
Electronic Applications 126 Single-Pole Circuits
C
R
v
1
v
2
+ +
- -
Fig. 189. Low pass.
v
1
(t)
V
1H
T
0
t
V
1L
t
H
t
L
Fig. 190. Rectangular wave input
(Fig. 187 repeated).
v
2
(t)
T
0
t
t
H
V
B
V
A

v
H
(t)
v
L
(t)
Fig. 191. Steady-state LP response (Fig. 188 repeated).
[ ]
1 1
( ) exp
H H B H
t
v t V V V

j \
+
, (
( ,
(220)
[ ]
1 1
1
exp
1 exp exp
H
A H B H
H H
H B
t
V V V V
t t
V V

j \
+
, (
( ,
, ]
j \ j \
+
, ( , (
, ]
( , ( ,
]
(221)
[ ]
1 1
( ) exp
H
L L A L
t t
v t V V V

j \
+
, (
( ,
(222)
[ ]
1 1
exp
H
B L A L
T t
V V V V

j \
+
, (
( ,
(223)
For 0 t t
H
:
At t = t
H
:
For t
H
t T:
At t = T:
Electronic Applications 127 Single-Pole Circuits
C
R
v
1
v
2
+ +
- -
Fig. 192. Low pass.
v
1
(t)
V
1H
T
0
t
V
1L
t
H
t
L
Fig. 193. Rectangular wave input
(Fig. 187 repeated).
v
2
(t)
T
0
t
t
H
V
B
V
A

v
H
(t)
v
L
(t)
Fig. 194. Steady-state LP response (Fig. 188 repeated).
1
1 exp exp
L L
B L A
t t
V V V

, ]
j \ j \
+
, ( , (
, ]
( , ( ,
]
(224)
1 1
1 exp 1 exp exp
exp exp
H L H
A H L
L H
A
t t t
V V V
t t
V


, ] , ]
j \ j \ j \
+
, ( , ( , (
, ] , ]
( , ( , ( ,
] ]
j \ j \
+
, ( , (
( , ( ,

(225)
1 1
1 exp
1 exp exp 1 exp
A
H H L
H L
T
V
t t t
V V

, ]
j \

, (
, ]
( ,
]
, ] , ]
j \ j \ j \
+
, ( , ( , (
, ] , ]
( , ( , ( ,
] ]

(226)
Continuing at t = T, we re-write eq.(223):
Now we use the above equation to substitute for V
B
in eq. (221):
Rearranging:
Electronic Applications 128 Single-Pole Circuits
C
R
v
1
v
2
+ +
- -
Fig. 195. Low pass.
v
1
(t)
V
1H
T
0
t
V
1L
t
H
t
L
Fig. 196. Rectangular wave input
(Fig. 187 repeated).
v
2
(t)
T
0
t
t
H
V
B
V
A

v
H
(t)
v
L
(t)
Fig. 197. Steady-state LP response (Fig. 188 repeated).
1 1
1 exp exp 1 exp
1 exp
H H L
H L
A
t t t
V V
V
T

, ] , ]
j \ j \ j \
+
, ( , ( , (
, ] , ]
( , ( , ( ,
] ]

j \

, (
( ,
(227)
1 1
1 exp exp 1 exp
1 exp
L L H
L H
B
t t t
V V
V
T

, ] , ]
j \ j \ j \
+
, ( , ( , (
, ] , ]
( , ( , ( ,
] ]

j \

, (
( ,
(228)
Finally, we solve eq. (226) for V
A
:
In similar fashion, we could have used eq. (221) to substitute for V
A
in eq. (224), and solved for V
B
:
Electronic Applications 129 555 IC Precision Timer
555 IC Precision Timer
Introduction
The 555 is an integrated circuit that performs the basic function of
a precision timer. Other functions and features are:
G Astable multivibrator, monostable multivibrator, pulse width
modulator, pulse position modulator, many other applications.
G Timing from microseconds to hours!!!
G Timing determined by resistor ratios - independent of V
CC
.
G Operating frequency up to 500 kHz.
G Wide supply voltage range - 4.5 V to 18 V.
G Compatible with TTL and CMOS logic.
G Totem-pole output.
G Output can source or sink 200 mA.
G Can be reset (i.e., gated, or disabled).
Electronic Applications 130 555 IC Precision Timer
Fig. 198. Function diagram of 555 timer, from Motorolas MC1455 data sheet,
available via www.motorola.com .
The functional diagram of a 555 timer is shown above. Note the
following characteristics from the diagram.
Inputs:
V
TH
, threshold voltage. V
CV
, control voltage.
V
TR
, trigger voltage. V
R
, reset voltage.
RS flip-flop:
When reset (R input is high), Q output is low.
When set (S input is high), Q output is high.
Comparator reference voltages:
Threshold comparator (Comp A): V
CC
or V
CV
2
3
Trigger Comparator (Comp B): V
CC
or V
CV
1
3
1
2
Electronic Applications 131 555 IC Precision Timer
Fig. 199. Function diagram of 555 timer (Fig. 198 repeated).
Features of the threshold comparator:
Active high when V
TH
> V
CV

When active, the flip-flop is reset, v

O
= V
OL
, and the discharge
transistor is on.
Features of the trigger comparator:
Active low when V
TR
< V
CV
/2
When active, the flip-flop is set, v
O
= V
OH
, and the discharge
transistor is .
Simultaneous active threshold and trigger comparators not allowed.
Features of the reset input:
Active low when V
R
<< V
BE
( 0.4 V)
When active, the flip-flop is reset. The reset input overrides
the flip-flop R and S inputs.
Electronic Applications 132 555 IC Precision Timer
Fig. 200. Monostable MV using 555 timer.
v
TR
v
O
v
TH
V
CC
/3
2V
CC
/3
V
OH
Fig. 201. Monostable MV waveforms.
Monostable Multivibrator
The 555 monostable multivibrator, shown above, generates a single
output pulse when the input is triggered.
Operation:
1. The circuit rests in its stable,
reset state:
v
O
= V
OL
, Q
discharge
on, V
TH
= 0
2. When the circuit receives a
trigger pulse V
TR
< V
CC
/3:
f-f is set, v
O
= V
OH
Q
discharge
off
C charges toward V
CC
3. When V
TH
reaches 2V
CC
/3:
f-f resets to stable state
v
O
= V
OL
, Q
discharge
on, V
TH
= 0
Electronic Applications 133 555 IC Precision Timer
Fig. 202. Monostable MV using 555 timer (Fig. 200
repeated).
v
TR
v
O
v
TH
V
CC
/3
2V
CC
/3
V
OH
T
Fig. 203. Monostable waveforms.
A
R C
(229)
1
ln T
p
(230)
2
1
3
3
CC CC
CC
V V
p
V

(231)
ln3 1.10 T
(232)
Calculating the Output Pulse Width:
Once the circuit is triggered, C charges through R
A
, with;
Thus, the output pulse width will be:
where
Thus:
Electronic Applications 134 555 IC Precision Timer
Fig. 204. Monostable MV using 555 timer (Fig. 200
repeated).
v
TR
v
O
v
TH
V
CC
/3
2V
CC
/3
V
OH
T
Fig. 205. Monostable waveforms.
CC CV
CC
V V
p
V

(233)
Other comments . . .
Output pulse may be terminated
early if reset pin brought low.
Trigger pulse must remain below
V
CC
/3 for 50-100 ns.
Control voltage input (V
CV
) can be
used to alter timing, in which case:
Electronic Applications 135 555 IC Precision Timer
Fig. 206. Monostable MV using 555 timer (Fig.
200 repeated).
V
CC
R
A
C
I
R
I
LEAKAGE
I
TH
Fig. 207. Bias and leakage
currents.
R LEAKAGE TH
I I I >> + (234)
15 V 10 V
1M
5
A
R
A

(235)
Leakage, Bias, and Discharge Currents:
The input bias currents are:
For the threshold comparator:
approximately 0.25 A
For the trigger comparator:
approximately 1.5 A
The capacitor is not ideal, but
has leakage current. The
amount of leakage current
depends on the type of C
For accurate timing, we want:
Note that I
R
is smallest for V
TH
= 2V
CC
/3.
We can estimate the largest practical R
A
:
Assume V
CC
= 15 V, I
LEAKAGE
= 0.25 A
For I
R
>> I
LEAKAGE
+ I
TH
, we let:
I
R
5 A at V
TH
= 10 V
Then
We must also keep R
A
5 k (or so), to limit Q
discharge
current.
Electronic Applications 136 555 IC Precision Timer
V
CC
R
C
TR
V
TR
to
trigger
Fig. 208. RC-coupled trigger pulse.
V
TR
to
trigger
to
reset
V
BE
V
CV
/2
Fig. 209. Positive trigger pulse.
Triggering in Monostable Mode:
1. Negative Trigger Pulse - If the trigger pulse is shorter than the
output pulse, a direct connection is OK.
But if the trigger pulse is longer than the output pulse, a direct
connection would result in simultaneous trigger and threshold
commands. The state of v
O
would be unknown.
RC-coupling allows V
TR
to be well
above V
CV
/2 by end of timing cycle.
Typically, should be > 10 ns.
2. Positive Trigger Pulse - In this case we can apply the pulse to
both the trigger input and the reset input.
On leading edge of the trigger pulse, reset input goes high
before trigger input does, allowing an active timer to be
triggered.
However, the trigger pulse must be
longer than output pulse.
Otherwise an additional inverting
circuit must to be used to provide a
negative trigger.
Electronic Applications 137 555 IC Precision Timer
Fig. 210. Astable MV using 555 timer.
v
TR
v
O
v
TH
,
V
CC
/3
2V
CC
/3
V
OH

1

2
t
H
t
L
Fig. 211. Astable waveforms.
1
1
ln
H
H
t
p
(236)
( )
2
3
1
1
3
1
and
2
CC CC
A B H
CC CC
V V
R R C p
V V

(237)
( ) ln2
H A B
t R R C +
(238)
Astable Multivibrator
The astable multivibrator generates a free-running rectangular
waveform, i.e., it is an oscillator.
Operation:
Well assume steady-state, with:
v
O
= V
OH
, Q
discharge
off
The capacitor charges toward V
CC
through R
A
and R
B
until v
C
= 2V
CC
/3
Thus:
where:
Thus:
Electronic Applications 138 555 IC Precision Timer
Fig. 212. Astable MV using 555 timer (Fig. 210
repeated).
v
TR
v
O
v
TH
,
V
CC
/3
2V
CC
/3
V
OH

1

2
t
H
t
L
Fig. 213. Astable waveforms (Fig. 211
repeated).
2
1
ln
L
L
t
p
(239)
1
3
2
2
3
0 1
and
0 2
CC
B L
CC
V
R C p
V

(240)
ln2
L B
t R C
(241)
Now v
C
has reached 2V
CC
/3, and
the timer changes state. The flip-
flop resets, with:
v
O
= V
OL
0, Q
discharge
on
Now, C charges only thru R
B
toward
zero: Thus:
where
Thus:
Electronic Applications 139 555 IC Precision Timer
Fig. 214. Astable MV using 555 timer (Fig. 210
repeated).
v
TR
v
O
v
TH
,
V
CC
/3
2V
CC
/3
V
OH

1

2
t
H
t
L
Fig. 215. Astable waveforms (Fig. 211
repeated).
( )
1 1 1
2 ln2
H L A B
f
T t t R R C

+ +
(242)
2
H A B
A B
t R R
D
T R R
+

+
(243)
Finally, we combine eq. (238)
and eq. (241) to calculate
frequency and duty ratio:
And
Notice that D must be between 0.5 and 1. It approaches 0.5 for
R
A
<< R
B
, and 1 for R
A
>> R
B
. We can achieve duty ratios less than
0.5 only by varying the control voltage.
Electronic Applications 140 555 IC Precision Timer
Fig. 216. Function diagram of 555 timer (Fig. 198
repeated).
v
TR
v
O
v
TH
,
V
OH

1

2
t
H
t
L
V
CV
V
CV
/2
Fig. 217. Astable waveforms with change in
notation for switching levels.
Varying the Control Voltage:
Recall that the CV input
is the reference to the
threshold comparator.
It is also connected to
upper junction of the
voltage divider.
Note also that the
reference to the trigger
comparator is always
half of the threshold
comparator reference.
Thus, t
L
wont change with changes
in V
CV
! But:
If V
CV
increases, t
H
increases,
If V
CV
decreases, t
H
decreases
So we can change f and D directly,
by varying V
CV
!!!
The two most common techniques used to vary the control voltage
are shown on the following pages.
Electronic Applications 141 555 IC Precision Timer
Fig. 218. Function diagram of 555 timer (Fig. 198
repeated).
+
-
2V
CC
/3
R
TH
=

3.33 k
V
CV
Fig. 219. Thevenin Eq. of
internal divider.
There are 2 basic ways to change the control voltage . . .
1. The simple (but less accurate) way:
Note that the internal voltage divider can
be represented by its Thevenin
Equivalent.
We can add a pull-up resistor from V
CV
to V
CC
to increase the control
voltage, or
we can add a pull-down resistor from V
CV
to ground to decrease the
control voltage.
Problem: the internal resistors are not known accurately.
Solution: attach a voltage source to the control voltage pin.
Electronic Applications 142 555 IC Precision Timer
Fig. 220. Function diagram of 555 timer (Fig. 198
repeated).
+
-
V
CC
to
pin 5
Fig. 221. Setting V
CV
with a
voltage source.
2. Controlling V
CV
with a voltage source:
The op amp voltage follower serves as
a voltage source.
The resistive divider keeps V
CV
a fixed
fraction of V
CC
so that circuit operation
remains independent of V
CC
.
We might even choose resistors with
equal temperature coefficients.
Electronic Applications 143 Energy Storage Elements
i
C
v
C
v
O
v
I
+
+
+
-
- -
Z
Fig. 222. Capacitor in series with arbitrary Z.
C
C
dv
i C
dt

(244)
Energy Storage Elements
Here we review some interesting behaviors in capacitive and
inductive circuits . . .
Capacitor in Series with Arbitrary Impedance
Recall the basic current-voltage
relationship:
We make the following interpretations from the equation:
G Current i
C
must be finite i.e., there are no current impulses
(presuming v
I
contains no impulses). Thus:
The voltage v
C
is continuous, i.e., there are no step changes,
i.e., v
C
= 0.
Any step change in v
I
must equal the step change in v
O
, i.e.,
v
I
= v
O
.
G If i
C
is constant, the capacitor voltage is a ramp function.
Electronic Applications 144 Energy Storage Elements
i
L
v
L i
I
+
-
i
O
Z
Fig. 223. Inductor in parallel with arbitrary Z.
L
L
di
v L
dt

(245)
Inductor in Parallel with Arbitrary Impedance
Recall the basic current-voltage
relationship:
We make the following interpretations from the equation:
G Voltage v
L
must be finite, i.e., there are no voltage impulses
(presuming i
L
contains no impulses). Thus:
The current i
L
is continuous, i.e., there are no step changes,
i.e., i
L
= 0.
Any step change in i
I
must equal the step change in i
O
, i.e.,
i
I
= i
O
.
G If v
L
is constant, the inductor current is a ramp function.
G This is merely the dual of the previous example.
Electronic Applications 145 Energy Storage Elements
C
1
C
2
v
1
v
2
v
I
+
+
+
-
- -
Fig. 224. Capacitive voltage divider.
1
C C
v i dt
C

(246)
2 1
1 2
1 2 1 2
and
I I
C C
v v v v
C C C C

+ +
(247)
Capacitive Voltage Divider
Recall the basic current-voltage
relationship:
We make the following interpretations from the equation, assuming
an input voltage step, as shown above:
G The input voltage step divides in inverse proportion to C.
G The current is an impulse, and
The dual of this circuit is the inductive current divider, however, it is
not discussed here
Miscellany
G Any elements in parallel with a voltage source do not affect the
remaining circuit.
G Any elements in series with a current source do not affect the
remaining circuit.
Electronic Applications 146 RC Attenuators
R
1
R
2
C
1
C
2
+ +
- -
v
1
(t) v
2
(t)
Fig. 225. The compensated attenuator.
1 1 1
RC
(248)
2 2 2
R C
(249)
1
1
1
( )
1 1
1
R
sC
C
Z s
R s
C s
sC RC

j \
, (
( ,

j \
+ +
+
, (
( ,
(250)
1 1 1 2 2 2
1 2
1 2
1 1
( ) || , ( ) ||
1 1
Z s R C Z s R C
C s C s


j \ j \
+ +
, ( , (
( , ( ,
(251)
RC Attenuators
RC attenuators are important circuits for use with general
waveforms, especially pulse and rectangular waveforms. Of these,
the compensated attenuator is most important and the easiest to
study, so it is where we begin.
The Compensated Attenuator
We begin by defining:
and
For a compensated attenuator we require that
1
=
2
.
To help with the mathematical development, lets recall that for any
parallel RC combination we have:
Thus, we may write:
Electronic Applications 147 RC Attenuators
R
1
R
2
C
1
C
2
+ +
- -
v
1
(t) v
2
(t)
Fig. 226. The compensated attenuator (Fig.
225 repeated).
2
2
2 2
1 1 2
1 2
1 2
1
1
( ) ( )
( )
1 1
( ) ( ) ( )
1 1
C s
V s Z s
H s
V s Z s Z s
C s C s

j \
+
, (
( ,

+
+
j \ j \
+ +
, ( , (
( , ( ,
(252)
2 1 1 2
1 2 1 2
1 2 1 2
1
( )
1 1
C C R R
H s
C C R R
C C R R

+ +
+ +
(253)
The transfer function is:
Because
1
=
2
, the (s + 1/) terms in the above equation will
cancel, leaving:
Note that H(s) is independent of s, i.e., constant for all frequency,
with the ratio of the capacitive voltage divider or the resistive
voltage divider.
This means v
2
(t) will always be an attenuated replica of v
1
(t),
regardless of the type of waveform of v
1
(t)!!!
Electronic Applications 148 RC Attenuators
R
1
R
2
C
1
C
2
+ +
- -
v
1
(t) v
2
(t)
Fig. 227. A generalized RC attenuator.
1 1 1
RC
(254)
2 2 2
R C
(255)
( )( )
1 2 1 2
||
eq eq
R C R R C C +
(256)
1 1 1 2 2 2
1 2
1 2
1 1
( ) || , ( ) ||
1 1
Z s R C Z s R C
C s C s


j \ j \
+ +
, ( , (
( , ( ,
(257)
2 1
2 1
1 2
1 2
1 2
1 2
1
1 1
( )
1 1
1 1
1 1
C s C s
H s
C s C s
C s C s



j \ j \
+ +
, ( , (
( , ( ,

j \ j \
+
+ + +
, ( , (
j \ j \
( , ( ,
+ +
, ( , (
( , ( ,
(258)
The Generalized RC Attenuator
With a little more effort, we can generalize the previous result.
Again we let
and
and, for convenience, we add the additional expression:
Also, as before, we have
Now, the transfer function is:
Electronic Applications 149 RC Attenuators
( ) ( )
( ) ( )
( )
1 2
1 2 1 2
1 2 1 2
1 2 1 2
1 2 1 2
1 2
1 2 1 2
1 2
1 1
1 1 1
||
1
eq
C C
C s C s Cs C s
C C s C C s
R R R R
C
C C
C C s C C s
C C s

j \ j \
+ + + + + +
, ( , (
( , ( ,
+ + + + +
+
+ + + +
j \
+ +
, (
( ,
(259)
( )
1
1
1 2
1
( )
1
C s
H s
C C s

j \
+
, (
( ,

j \
+ +
, (
( ,
(260)
( )
1
1 1 1 2
1 2
1 2
1 2
1 2 1 2
1 1
(0)
1
||
R
C
R R R
H K
R R
C C
R R
R R RR

+
+
+
(261)
Working with just the denominator of H(s):
Thus, the transfer function becomes:
Now, at dc, s = 0, and:
Thus, for dc, the transfer function is merely a resistive voltage
divider, as expected.
Electronic Applications 150 RC Attenuators
( )
1 1
1 2 1 2
lim ( )
C
s
C s C
H s K
C C s C C

+ +
(262)
( )
1
1
1 2
1
( )
1
C s
H s
C C s

j \
+
, (
( ,

j \
+ +
, (
( ,
(263)
( )( )
( )
( )
1 2 1 2
1 1
1 2
1 2
1 1 1 1 2
1
2
1 2
1 2
1
1
||
1
1 1
1
P
Z
C
R
R R C C
s RC
RR
s
C C
RC R R
K C
R
K
C C
R R

+

+
+

+
+
i
(264)
Now, as s becomes infinite:
Thus for infinite frequency (like the edge of a step function), the
transfer function is merely a capacitive voltage divider, as expected.
The significance of K
R
and K
C
becomes more apparent when we
look at the poles and zeros of H(s).
Poles and Zeros:
Recall H(s):
This has a zero at s
Z
= -1/
1
and a pole at s
P
= -1/. Now we look at
the ratio of s
P
to s
Z
:
Electronic Applications 151 RC Attenuators
log f
dB
f
P
f
Z
20 log K
R
20 log K
C
0
Fig. 228. Undercompensated RC attenuator,
K
C
< K
R
, s
P
< s
Z
.
log f
dB
f
P
f
Z
20 log K
R
20 log K
C
0
Fig. 229. Overcompensated RC attenuator,
K
C
> K
R
, s
P
> s
Z
.
Eq. (264) shows that:
G For K
C
< K
R
(called undercompensated), s
P
< s
Z
G For K
C
> K
R
(called overcompensated), s
P
> s
Z
We can illustrate this with Bode magnitude plots:
Note that:
G The undercompensated attenuator has a lowpass effect
G The overcompensated attenuator has a highpass effect
G For a compensated attenuator, K
C
= K
R
, s
P
= s
Z
, and the Bode
magnitude plot will merely be a horizontal line at
20 log K
C
= 20 log K
R
.
Thus it will not alter the frequency components of any
waveform, which is the same as saying v
2
(t) will always be an
attenuated replica of v
1
(t), regardless of the type of waveform
of v
1
(t)!!!
We can see the lowpass and highpass effects by looking at the
general response of an RC attenuator to a rectangular wave input.
Electronic Applications 152 RC Attenuators
V
1H
V
1L
K
R
V
1H
K
R
V
1L
K
R
V
1PP
V
1PP
K
C
V
1PP
K
C
V
1PP

Fig. 230. Undercompensated attenuator rectangular-wave response.

K
C
V
1PP
K
C
V
1PP
V
1PP

Fig. 231. Overcompensated attenuator rectangular-wave response.

Undercompensated Attenuator Rectangular-Wave Response:
K
C
< K
R
. Thus the step sizes of the output waveform are
smaller than (K
R
V
1H
- K
R
V
1L
) = K
R
V
1PP
:
Overcompensated Attenuator Rectangular-Wave Response:
K
C
> K
R
. Thus the step sizes of the output waveform are larger
than (K
R
V
1H
- K
R
V
1L
) = K
R
V
1PP
:
Electronic Applications 153 RC Attenuators
R
1
R
2
C
1
C
2
+ +
-
-
v
1
(t)
v
2
(t)
R
S
v
S
(t)
+
-
Fig. 232. Practical RC attenuator circuit.
R
1
R
2
C
1
C
2
+
-
v
1
(t)
R
S
v
S
(t)
+
-
v
2A
v
2B
Fig. 233. Practical RC attenuator circuit re-drawn.
Practical RC Attenuators
A practical RC attenuator is not driven by an ideal voltage source
but, rather, by a source which includes its internal resistance:
In general this is not a single-pole circuit. But we can show that it
becomes a single-pole circuit if compensated, by re-drawing the
above figure:
If the attenuator is precisely compensated we will have
1
=
2
(i.e.,
R
1
C
1
= R
2
C
2
). Then we will also have K
C
= K
R
and voltage v
2A
will
equal voltage v
2B
even with the connection between them removed.
With the connection removed, we can see that there is a single
equivalent capacitance. Therefore we have a single-pole circuit!!!
Electronic Applications 154 RC Attenuators
R
2
C
2
+
-
v
2
(t)
R
S
v
S
(t)
+
-
100
1 M 30 pF
Fig. 234. Step voltage source connected to oscilloscope.
R
2
and C
2
represent the scope input impedance. Typical
values are shown.
( ) ( ) 30pF 100 || 1M 30pF 100 3ns
(265)
1
ln ln9 6.59ns
r
t
p
(266)
An Oscilloscope Probe Example
An RC attenuator has many uses, the most common of which is the
oscilloscope / oscilloscope probe combination.
Lets first look at a step voltage source connected directly to an
oscilloscope input:
The time constant of this circuit is:
and the resulting rise time of v
2
is:
Thus, the step input will appear on the oscilloscope as an
exponential, single-pole response.
Electronic Applications 155 RC Attenuators
R
1
R
2
C
1
C
2
+
-
v
1
(t)
R
S
v
S
(t)
+
-
v
2A
v
2B
100
1 M
30 pF
9 M
3.33 pF
Fig. 235. 10X oscilloscope probe inserted between voltage
source and oscilloscope.
( )( )
3.33pF 30pF
3pF
3.33pF 30pF
eq
C
+
(267)
( ) ( ) 3pF 100 || 10M 3pF 100 300ps
(268)
ln9 659ps
r
t
(269)
Now we measure the same step input, v
S
, with a 10X oscilloscope
probe. This is nothing more than a 10:1 compensated attenuator
which produces K
C
= K
R
= 10:
This is a perfectly compensated attenuator, so we can remove the
short-circuit (shown in red). This leaves a single-pole circuit with:
The time constant of this circuit is:
Because v
2
is an attenuated replica of v
1
, both voltages will have
the same rise time:
The oscilloscope response is ten times faster using the 10X probe!!!
The price we pay for this improvement is a smaller amplitude signal
at v
2
.
Electronic Applications 156 Diode Static Characteristics
exp 1
D
D S
T
v
i I
nV
, ]
j \

, ] , (
( ,
]
(270)
25mV at 300K
T
kT
V T
q
(271)
exp
D
D S
T
v
i I
nV
j \

, (
( ,
(272)
ln ln
D
D S
T
v
i I
nV
+ (273)
2.3log 2.3log
1
log log
2.3
D
D S
T
D D S
T
v
i I
nV
i v I
nV
+
j \
+
, (
( ,
(274)
Diode Static Characteristics
Recall that the diode i-v characteristic is described by the Shockley
equation:
where I
S
is the reverse saturation current, and
Measuring Reverse Saturation Current
Attempts to measure this directly result in values much larger than
I
S
, due to surface leakage current, but Is can be obtained from
forward bias measurements in the following manner.
Beginning with the forward bias approximation:
We then find the natural logarithm of both sides:
And change to common (i.e., base 10) logarithms:
Electronic Applications 157 Diode Static Characteristics
1
log log
2.3
D D S
T
i v I
nV
j \
+
, (
( ,
(275)
Fig. 236. PSpice-simulated semi-log i-v characteristic of 1N4148
diode.
y mx b +
(276)
Comparing the final form of eq. (274):
to the standard notation describing the equation of a straight line:
we see that the diode forward i-v characteristic should be a straight
line when i
D
is plotted on the log axis of a semi-log graph, and the
y-axis intercept is the common logarithm of I
S
!!!
With laboratory data, the straight line we use should be a least-
squared-error fit at low and moderate currents, to avoid the
influence of the semiconductor IR drop at higher currents.
Typical I
S
values are within two or three decades of 10 fA.
Electronic Applications 158 Diode Static Characteristics
exp
NOMINAL
NOMINAL S
T
v
i I
V
j \

, (
( ,
(277)
0.01 exp
CUT IN
NOMINAL S
T
v
i I
V

j \

, (
( ,
(278)
100 exp
ln100 115mV
NOMINAL CUT IN
T
NOMINAL CUT IN T
v v
V
v v V

j \

, (
( ,

(279)
Diode Cut-in Voltage
From observation, for small-signal diodes, a nominal diode voltage
of 0.75 V results in a nominal diode current of 10 mA. Assuming
n = 1 we could write:
For diodes to be used as a switch, let us define the off state to be
when i
D
is less than 1% of the nominal value. We define the
corresponding voltage as the cut-in voltage:
Dividing eq. (277) by eq. (278) and solving for v
CUT-IN
:
Thus v
CUT-IN
0.75 V - 0.115 V = 0.635 V. For the base-emitter
diode of BJTs, we round this value to arrive at these typical
assumptions:
v
BE
< 0.65 V cutoff
v
BE
= 0.70 V active
v
BE
> 0.75 V saturation
Electronic Applications 159 Diode Static Characteristics
mV
2
K
D
v
T

(280)
0
0
1
j
j
m
DQ
C
C
V

j \

, (

( ,
(281)
Temperature Dependence
For a constant diode current:
Depletion Capacitance
With the symbol C
j
(presumably for junction capacitance, by which
this is also known), this capacitance arises from the bound charges
that exist in the depletion region under reverse bias (and low
forward bias). It is given by:
where C
j0
is the zero-bias depletion capacitance ( 4 pF)
V
DQ
is the diode quiescent voltage

0
is the built-in barrier potential ( 1 V)
and m is the grading coefficient ( 0.5)
(Values in parentheses are typical for the 1N4148 diode)
In addition to the term junction capacitance, depletion capacitance
is also called transition capacitance, and barrier capacitance.
A plot of eq. (281), normalized to C
j0
for the values given above, is
shown on the following page.
Electronic Applications 160 Diode Static Characteristics
Fig. 237. Typical normalized depletion
capacitance vs. diode voltage.
Diffusion Capacitance
This arises from the charges which diffuse across the junction under
forward bias. It is very important in the base-emitter diode of BJT
amplifiers (C

), but of little importance in diodes used for switching.

It will not be considered here.
Zener Diodes
Recall that there are two different mechanisms for breakdown:
G Avalanche breakdown (V
Z
> 6 V), due to kinetic energy of
carriers, produces curves with sharper knees.
G Zener breakdown (V
Z
< 6 V), due to high E field.
G The temperature coefficient of zener diodes is approximately
(V
Z
- 5) mV/K.
Electronic Applications 161 Diode Static Characteristics
Fig. 238.
IC diode.
IC Diodes
It is not economical to use a different masking for
diodes than that already used for transistors, so
transistors are simply wired to function as diodes.
Of all possible connections, the connection shown
at left has:
G lowest forward drop
G fastest switching,
G lowest C
j
G BV
EB
from 6 V to 9 V
Electronic Applications 162 BJT Static Characteristics
collector
n-type
n-type
emitter
p-type base
C
B
E
v
CE
v
BE
+
+
-
-
i
C
i
E
i
B
C
B
E
collector
p-type
p-type
emitter
n-type base
C
B
E
v
EC
v
EB
+
+
-
-
i
C
i
E
i
B
C
B
E
Fig. 239. Npn and pnp (left and right, respectively) bipolar junction transistors.
and
C C c
FE fe
B B b
i i i
h h
i i i

(282)
BJT Static Characteristics
Lets review the basics of BJTs:
Operating Region EBJ CBJ Feature
cutoff rev. rev. i
C
= i
E
= i
B
= 0
active fwd. rev. i
C
= h
FE
i
B

saturation fwd. fwd. i
C
< h
FE
i
B

inverse rev. fwd. i
C
< 0, very small
Notes:
G At the active-saturation boundary, v
CB
= 0 and i
C
= h
FE
i
B
.
G
G h
FE
= h
fe
in the normal temperature range for silicon devices.
G h
FE
variation of 10:1 possible . . . 3:1 typical.
Electronic Applications 163 BJT Static Characteristics
Fig. 240. BJT output characteristics in the saturation and near-saturation
regions.
Output Characteristics
G With i
B
fixed, i
C
is a function of v
CE
, and with v
CE
fixed, i
C
is a
function of i
B
.
G The curves do not go through the origin, rather, i
C
= 0 for v
CE
= few mV.
G There is no mathematical model for the saturation region that
is simple and accurate.
For high accuracy we should use PSpice, or an equivalent.
For easy, but more approximate, analysis we can use a simple
piecewise-linear model.
Electronic Applications 164 BJT Static Characteristics
i
C
v
CE
I
CM
=h
FE
I
B
V
CR
V
BE
I
B
active saturation
1/R
S
Fig. 241. Piecewise-linear approximation to BJT output
characteristics
CR
S
FE B
V
R
h I
(283)
, 0
,
CE FE B
C CE CE CR
S CR
FE B CR CE CEO
v h I
i v v V
R V
h I V v BV


(284)
BJT Saturation-Region Model
For the saturation-region piecewise-linear model, we define the
corner voltage, V
CR
, which is typically 0.2 V to 0.4 V.
Our default value for the corner voltage is V
CR
= 0.3V.
We also define the saturation resistance:
thus
Note that the Early effect is ignored. It is important in analog
circuits, but not in switching circuits.
Electronic Applications 165 BJT Static Characteristics
R
C
= 1 k
R
B
V
BB
+10 V
h
FE
= 100
Fig. 242. BJT example.
i
C
v
CE
I
B
= 200 A
100 A
50 A
20 mA
10 mA
5 mA
0.3 V 0.7 V
load line
Fig. 243. Output characteristics.
( ) 10V 1k
C CE
i v
(285)
30
10V 10V 0.291V
1k 30
S
CE
S C
R
v
R R

+ +
(286)
BJT Example
We let V
CR
= 0.3 V and V
BE
= 0.7 V for the BJT in the circuit below,
and calculate the collector-emitter voltage for various base currents.
Note that the load line equation for this circuit is:
I
B
I
CM
v
CE
R
S
Comments:
50 A 5 mA 5 V 60 active region
93 A 9.3 mA 0.7 V 32.3 edge of saturation
97 A 9.7 mA 0.3 V 30.1 at corner
100 A 10 mA 0.291 V 30 left of corner (see below)
200 A 20 mA 0.148 V 15 left of corner
Example calculation for I
B
= 100 A:
Electronic Applications 166 BJT Static Characteristics
i
C
i
E
Fig. 244. Inverse BJT.
, where 0.1
E FC B FC
i h i h (287)
Rule of Thumb Voltages
Recall the typical assumptions detailed in the section on static diode
characteristics:
v
BE
< 0.65 V cutoff
v
BE
= 0.70 V active
v
BE
> 0.75 V saturation
Power Dissipation
The instantaneous collector power dissipation in a BJT is given by
p
C
= v
CE
i
C
. We note the following:
G Power dissipation is essentially zero in cutoff.
G Power dissipation is very small in saturation.
G Power dissipation is maximum at v
CE
= V
CC
, i
C
= V
CC
/R
C
.
Inverse Mode
The collector and emitter terminals are not
interchangeable, because the BJT is not
constructed symmetrically. If we reverse
these connections, we have:
If i
B
is large enough then the BJT operating point moves into the
inverse saturation region, for which:
v
EC sat - inverse mode
<< v
CE sat - normal mode
(typically 1 mV)
Switching into and out of inverse saturation is slower, thus this
mode is used only in special applications (e.g., TTL input circuit).
Electronic Applications 167 Diode Switching Characteristics
Anode Cathode
p-type n-type
+
-
i
D
v
D
+
+
+
+
+
+
+
-
-
-
-
-
-
-
free
electrons
free
holes
Fig. 245. Simple (abrupt) p-n junction.
+ -
I
D
v
D
=

0
p n -
-
-
+
+
+
I
S
Fig. 246. p-n junction under zero bias,
showing bound charges in the depletion
region.
Diode Switching Characteristics
Recall that a diode is a p-n junction.
In the p-type material:
G Holes, the majority carriers, are generated primarily from
doping.
G Electrons, the minority carriers, are generated thermally.
In the n-type material:
G Electrons, the majority carriers, are generated primarily from
doping.
G Holes, the minority carriers, are generated thermally.
Zero Bias
Holes diffuse from the p-type
material to the n-type material,
where the hole density is lower.
Electrons diffuse from the n-type
material to the p-type material,
where the electron density is lower.
Electronic Applications 168 Diode Switching Characteristics
+ -
I
D
v
D
=

0
p n -
-
-
+
+
+
I
S
Fig. 247. P-n junction under zero bias,
showing bound charges in the depletion
region (Fig. 246 repeated).
p
p
n0
n
n
p0
x
Fig. 248. Minority carrier distribution of an abrupt p-n
junction under zero bias. The relative densities of n
p0
and p
n0
were chosen arbitrarily.
The diffusion of the majority
carriers gives rise to a diffusion
current, I
D
.
The diffusion current exposes
bound charges in the vicinity of
the junction. The region of bound
charges is called the depletion
region (because it is depleted of
carriers.
The E field created by the bound
charges opposes the flow of the
diffusion current.
Thermally generated minority carriers will diffuse randomly to the
depletion region where they are swept across the junction by the E
field created by the bound charges, giving rise to a drift current, I
S
At equilibrium, I
D
=I
S
, and net current is zero. The steady-state
minority carrier distribution at equilibrium is shown below:
Electronic Applications 169 Diode Switching Characteristics
+ -
I
D
V
R
p n -
-
-
+
+
+
I
S
Fig. 249. Abrupt p-n junction under
reverse bias.
p
p
n0
n
n
p0
x
p
n
n
p
Fig. 250. Minority carrier distribution of reverse-
biased abrupt p-n junction.
Reverse Bias
The potential across the junction
increases due to the applied reverse
bias.
This causes more of the minority carriers
to be swept across the junction.
In a sense the depletion region grows,
but more insight is gained by looking at
the minority carrier distribution:
Because of this minority carrier distribution, minority carriers
continue to diffuse toward the junction, thus the current I
S
continues
to flow.
However, the current I
D
is essentially reduced to zero, because the
increase in junction potential prevents diffusion of the majority
carriers.
The resulting net current is just I
S
, the reverse saturation current.
This current is quite small because the minority carrier density is
very low (remember, the minority carriers arise only from thermal
generation).
Electronic Applications 170 Diode Switching Characteristics
+ -
I
D
V
F
p n -
-
-
+
+
+
I
S
Fig. 251. Abrupt p-n junction under
forward bias.
p
p
n0
n
n
p0
x
p
n
n
p
n
p
-

n
p0
p
n
-

p
n0
Fig. 252. Minority-carrier distribution of forward-
biased abrupt p-n junction
Forward Bias
Applied forward bias decreases the
junction potential from its value at
zero bias.
This allows the majority carriers to
diffuse across the junction,
increasing the current I
D
. This
current can be quite large because
of the number of majority carriers.
Note, however, that when
majority carriers cross the
junction they become minority
carriers.
The large number of carriers
crossing the junction causes a
tremendous increase in the
minority carrier density, as
shown.
Note also that the higher the forward current, the higher the minority
carrier density at the junction. The excess minority carriers at the
junction, p
n
- p
n0
, or n
p
- n
p0
, is called the stored charge.
Electronic Applications 171 Diode Switching Characteristics
v
I
v
D
v
R
+ +
+ -
-
-
i
D
Fig. 253. Simple diode switching
circuit.
v
I
t
V
F
V
R
t
p
n
-

p
n0
t
i
D
V
F
/R
V
R
/R
t
s
t
t
t
rr
I
S
Fig. 254. Diode switching waveforms.
Diode Switching
Consider this simple diode circuit with
the input voltage shown below. It is
assumed that V
F
and |V
R
| are much
larger than the forward drop of the diode.
In the forward biased interval,
the diode has a minority carrier
density as shown in Fig. 252.
The stored charge and the
diode current as functions of
time are shown at left.
When the voltage switches
from V
F
to V
R
, the diode cannot
perform as a reverse-biased
diode until the minority carrier
distribution returns to that
shown for the reverse-biased
case (Fig. 250).
Thi s cannot happen
instantaneously.
Electronic Applications 172 Diode Switching Characteristics
v
I
t
V
F
V
R
t
p
n
-

p
n0
t
i
D
V
F
/R
V
R
/R
t
s
t
t
t
rr
I
S
Fig. 255. Diode switching waveforms (Fig.
254 repeated).
ln 1
F
s d
R
I
t
I

j \
+
, (
( ,
(288)
In fact, because the forward
biased minority carrier density
at the junction is quite large,
there can be a large reverse
diode current, as shown.
The larger reverse current
continues to flow until the
stored charge is removed and
the minority carrier density at
the junction drops to the
zero-bias level.
The time required for this to
occur is called the storage
ti me, t
s
; i t i s di rectl y
proportional to the forward
current, I
F
, and inversely
proportional to the reverse
current, I
R
.
Storage ti me can be
empirically described:
where
d
is a constant, characteristic of the particular diode.
Electronic Applications 173 Diode Switching Characteristics
v
I
t
V
F
V
R
t
p
n
-

p
n0
t
i
D
V
F
/R
V
R
/R
t
s
t
t
t
rr
I
S
Fig. 256. Diode switching waveforms (Fig.
254 repeated).
rr s t
t t t + (289)
After the stored charge has
been removed, reverse current
continues to fl ow, but
decreases in magnitude until
the minority carrier density at
the junction drops to zero and
the junction capacitance is
charged through R to the
voltage V
R
.
This additional interval is
called the transition time, t
t
.
The total reverse recovery
time is the sum of the storage
time and the transition time:
t
rr
can be less than 1 ns in
switching diodes, to as high as
several s in high-current
diodes.
Electronic Applications 174 Diode Switching Characteristics
Fig. 257. Schottky diode.
Schottkv Diodes
These diodes use a metal - n-type
semiconductor junction.
The metal functions as p-type material. However, with applied
forward bias, electrons flow into the metal where they are not
minority carriers.
Thus, there is no stored charge and t
s
= 0.
Typical total recovery times are 50 ps, and, for aluminum-on-silicon,
the forward voltage is 0.35 V.
Drawbacks of Schottky diodes include increased reverse saturation
current and low breakdown voltage.
Electronic Applications 175 BJT and FET Switching Characteristics
n p
C
B
E
n
Fig. 258. Sandwich construction
model of an npn BJT.
EBJ CBJ
minority carrier density
Fig. 259. Base-region minority-carrier
density of BJT in active region.
BJT and FET Switching Characteristics
We first tackle switching characteristics
of BJTs. An npn BJT is assumed to be
the device in the descriptions that follow.
Recall the sandwich construction of the
conceptual BJT:
Emitter heavily doped n-type
Base lightly doped p-type
Collector lightly doped n-type
Actual devices are not constructed in this sandwich fashion, but this
model serves us very well for the illustration of switching principles.
BJT Operation in the Active Region
The emitter-base junction (EBJ) is
forward biased.
Electrons are injected from the emitter
into the base, where they become
minority carriers and diffuse toward the
collector.
The minority carrier density in the base
region is shown at left.
At the reverse-biased collector-base junction (CBJ) the electrons
are swept into the collector producing collector current.
Electronic Applications 176 BJT and FET Switching Characteristics
EBJ CBJ
minority carrier density
Fig. 260. Base-region minority-carrier
density of BJT in active region (Fig. 259
repeated).
Because the base is lightly doped, the
holes that are injected from the base into
the emitter comprise a negligible part of
the emitter current.
In addition, light base doping produces
little recombination of the minority
carriers in the base.
As a result, most of the injected electrons are allowed to reach the
CBJ, resulting in 1 and large .
If we account for recombination in the base region, the minority
carrier density is slightly bowed, as indicated by the dashed orange
line in the figure.
It takes some time to set up this profile after forward bias is applied
to the EBJ.
Upon removal of the forward bias, collector current will continue
until the stored charge (the area under the curve) is removed.
Electronic Applications 177 BJT and FET Switching Characteristics
EBJ CBJ
minority carrier density
EBJ CBJ
minority carrier density
excess
stored
charge

Fig. 261. Base-region minority-carrier density of a BJT in the saturation region.

v
i
V
CC
Fig. 262. Typical BJT
switching circuit.
BJT Operation in the Saturation Region
In saturation, the CBJ is also forward biased and additional stored
charge is present in the base, as shown in the figure below.
This excess stored charge must be removed before collector
current can cease. Thus the switching speed of the device is
significantly reduced.
Typical Switching Waveforms
A simple BJT circuit driven with a pulse input yields the typical
switching waveforms and definitions of intervals shown on the
following page.
Study these waveforms carefully until you are familiar with all of the
intervals that are defined.
Electronic Applications 178 BJT and FET Switching Characteristics
v
I
t
t
i
B
90%
50%
10%
i
C
t
t
d
t
r
t
s
t
f
t
on
t
off
I
B2
I
B1
V
2
V
1
I
C
Fig. 263. Typical BJT switching waveforms.
t
on
, turn-on time:
t from the + step change in i
B
to the rising 50% I
C
level.
t
off
, turn-off time:
t from the - step change in i
B
to the falling 50% I
C
level.
t
d
, delay time: t from the + step change in i
B
to the 10% I
C
level.
t
r
, rise time: t from 10% I
C
to 90% I
C
.
t
s
, storage time: t from the - step change in i
B
to the 90% I
C
level.
t
f
, fall time: t from 90% I
C
to 10% I
C
.
Electronic Applications 179 BJT and FET Switching Characteristics

Fig. 264. Schottky transistor.

v
i
V
CC
Fig. 265. Circuit to increase BJT
switching speed.
Schottkv Transistor
This bipolar junction transistor uses a
Schottky diode between base and
collector to prevent the collector from
falling more than 0.35 V below the base.
Thus the transistor can never become
heavily saturated.
Additional base current that would serve only to increase the excess
stored charge is shunted through the diode to the collector.
The Schottky diode is easy to manufacture - the base metalization
is merely extended over the collector region.
Increasing BJT Switching Speed
Because of the need to remove stored charge, the BJT behaves
somewhat as a capacitor does. The addition of a speed-up
capacitor as shown produces an RC attenuator circuit and
increases switching speed:
Electronic Applications 180 BJT and FET Switching Characteristics
FET Switching Characteristics
A field-effect transistor is a majority carrier device.
There are no minority carriers, no stored charge, and therefore, no
storage time.
Speed limitations result from RC delays.
Low-power FETs have a high R
on
and are generally slower than
BJTs.
High-power FETs have a lower R
on
and are generally faster than
BJTs.
For a given FET, we can increase switching speed by reducing the
load capacitance at the drain.
Electronic Applications 181 Introduction to DC-DC Conversion
I
S
I
O
V
S
V
O
+ +
- -
converter
Fig. 266. Representative dc-dc
converter system.
S S S
P V I (290)
O O O
P V I (291)
O O O
S S S
P V I
P V I
(292)
Introduction to DC-DC Conversion
The purpose of a dc-dc converter is to convert dc voltage from one
level to another, at high efficiency, for the purpose of supplying
power to a load.
For the representative system at
left, we can define input and output
power, respectively:
and
We can also define the converter efficiency:
Converter efficiency is usually a function of I
O
. In an ideal converter,
input and output power are equal, i.e., the efficiency is 100%.
It is normally assumed that V
S
and V
O
are constant. In a well-
designed converter this is almost true, though V
O
usually contains
a small ripple component.
Converter Basics
All converters use active devices or diodes as switches, and
capacitors or inductors to store energy in electric or magnetic fields,
respectively.
Capacitor-based converters use switches to transfer charge from
one capacitor to another (and another, and another, etc.), and
eventually to a load.
Electronic Applications 182 Introduction to DC-DC Conversion
B, Wb/m
2
H, A-t
saturation
Fig. 267. Magnetization curve.
max max L
v f fAB (293)
Salient features of capacitor-based converters include:
G Usually useful only at lower power levels.
G V
O
depends on C and Q (remember Q = CV ?).
G Higher switching frequencies reduce C required and also
reduce output voltage ripple.
But switching losses (mostly active region dissipation during
on-to-off and off-to-on transitions) are fixed per cycle, and
increase with frequency.
G Typical switching frequencies are currently in the range of
several hundred kHz to several MHz.
Inductor-based converters typically store energy in the magnetic
field of one or more inductors, and use switches to transfer this
energy to a capacitor (and a load) at the output. Features are:
G More efficient at higher power levels.
G Large energy storage requires high L values which, in turn,
requires magnetic material for inductor cores.
To reduce core losses and distortion of
the internal switching waveforms,
magnetic saturation of the core is
avoided.
High output voltages require high
inductor voltages, and from
small sizes (areas) imply higher switching frequencies, with the
same increase in switching losses noted previously.
Electronic Applications 183 Capacitor-Based Converters
C R
L
v
D
v
O
+ -
Fig. 268. Peak rectifier circuit.
V
P
v
O
t
Fig. 269. Peak rectifier waveforms.
Blue indicates when the diode carries
forward current.
V
P
v
O
t
Fig. 270. Peak rectifier waveforms with
square-wave input.
Capacitor-Based Converters
These are usually based on the peak rectifier and the diode clamp
circuits.
Peak Rectifier
Recall peak rectifier operation with a
sinusoid:
G C charges to maximum input
voltage, V
P
, on positive half-cycle.
G C discharges through R
L
throughout
remainder of cycle.
G Diode forward current consists of
one pulse per cycle.
G Output ripple is reduced with higher
f and higher R
L
.
Operating with a square-wave input:
G Increases the diode conduction
angle.
G Reduces output ripple.
G Provides higher average output
voltage.
As you might expect, we will operate our converter with square-
wave (or rectangular-wave) inputs.
Electronic Applications 184 Capacitor-Based Converters
C
v
D
v
O
+
-
+ -
V
P
Fig. 271. Diode clamp circuit.
For convenience the clamp is left
unloaded.
V
P
v
O
t
2V
P
Fig. 272. Clamp circuit output voltage.
Clamp
Interchanging the positions of the diode and the capacitor in the
peak rectifier circuit gives us a diode clamp:
Circuit operation:
G During the first negative half-cycle,
the diode conducts.
G C charges to a maximum input
voltage, V
P
, with polarity shown.
G On all subsequent cycles, v
O
equals
the source voltage plus V
P
.
G As a result, the output is clamped
above zero.
Also note:
G Connecting the anode of the diode to a voltage other than
ground, clamps the output above that voltage.
G With a load attached, C discharges (slightly, if C is large)
during each cycle, and is recharged at each negative peak of
the input with a current pulse through the diode.
G Driving the clamp circuit with a square wave increases the
diode conduction angle, and reduces the discharge of C with
an attached load.
Electronic Applications 185 Capacitor-Based Converters
v
I
t
V
P
-V
P
Fig. 273. Square-wave input to clamp +
peak detector circuit.
+ -
V
P
2V
P
+
-
v
I
Fig. 274. Clamp followed by a peak
detector. Steady-state capacitor
voltages are noted in the figure.
v
I
t
V
P
0
Fig. 275. Single-polarity square wave
driving the doubler at right.
+ -
V
P
2V
P
+
-
v
I
V
P
Fig. 276. Alternative voltage doubler
circuit.
Building a Capacitor-Based Converter
We begin with a clamp, driven by a square wave, and follow that
circuit with a peak rectifier:
Note that the dc output voltage is twice the input voltage magnitude.
This circuit is called a voltage doubler.
We obtain the same output if we begin with a zero-to-V
P
square
wave, and clamp it above V
P
volts:
Normally, the square wave is derived from a dc supply (of V
P
volts)
using switches.
The single polarity square-wave is much easier to obtain, so the
doubler circuit of Fig. 276 is preferred.
Electronic Applications 186 Capacitor-Based Converters
+ -
V
P
2V
P
+
-
v
I
V
P
+ -
2V
P
Fig. 277. Additional clamp added to the
circuit of Fig. 276.
+ -
V
P
2V
P
+
-
v
I
V
P
+ -
2V
P
3V
P
+
-
Fig. 278. Additional peak rectifier added to the circuit
of Fig. 277.
We continue to build our capacitor-based converter by adding
another clamp circuit to the doubler of Fig. 276:
G The voltage reference for the
second clamp is 2V
P
.
G The input square wave varies
from zero to V
P
.
G The capacitor of the second
clamp charges to a dc voltage
of 2V
P
.
Now we add another peak rectifier circuit:
G Note that now we
have a dc output
voltage of 3V
P
!!!
G We can continue
with this process
i ndef i ni t el y ( i n
theory).
We can redraw Fig. 278 in a more straightforward manner (on the
following page . . .)
Electronic Applications 187 Capacitor-Based Converters
V
P
V
O
0
V
P
2V
P
+
-
2V
P
+
-
3V
P
+
-
V
P
+
-
Fig. 279. Basic capacitor-based dc-dc converter. Capacitor
voltages are steady-state values.
V
P
V
O
0
V
P
3V
P
+
-
2V
P
+
-
4V
P
+
-
V
P
+
-
0
V
P
Fig. 280. Improved capacitor-based dc-dc converter. Capacitor
voltages are steady-state values.
With the topology presented as above, the bucket-brigade nature
of the circuit operation is more obvious.
We can reduce the number of stages needed by pulling the
intermediate capacitors up to V
P
when their neighbors are pulled to
zero . . . this requires a second square wave, 180
o
out of phase with
the first:
The generation of the square waveforms from the dc supply (V
P
) is
relatively easy. The reduced losses resulting from fewer stages
usually makes this latter approach more attractive than the versions
shown earlier in this section.
Electronic Applications 188 Capacitor-Based Converters
V
P
V
O
0
V
P
3V
P
+
-
2V
P
+
-
4V
P
+
-
V
P
+
-
0
V
P
5V
P
+
-
Fig. 281. Five-stage converter.
0
V
P
2V
P
+
-
2V
P
+
-
V
P
+
-
0
V
P
V
O
Fig. 282. Negative converter.
We can add more stages by alternating capacitor connections to the
square wave sources:
We can also generate negative voltages by reversing the diodes
(and polarized capacitors such as electrolytics). Negative
converters require one additional stage for an output magnitude
equal to that of a corresponding positive converter.
Electronic Applications 189 Capacitor-Based Converters
V
O
0
0
1N4148 1N4148 1N4148 1N4148
10 F 10 F 10 F 10 F
5 V
5 V
5 V
10 kHz
Fig. 283. The 4-stage converter circuit used for simulation.
Piecewise-linear sources were used to simulate the square waves.
Nonideal Converters
G The dc output voltage is less than an integer multiple of the dc
input voltage, due to the diode drop within each stage.
Schottky diodes reduce the effect of multiple diode drops.
G Additional voltage drops occur across the internal resistance
of the square wave generators. These resistances should be
kept as low as possible.
G Output ripple results from the discharge of the final capacitor
through the load.
For a given load current, output ripple decreases as switching
frequency (i.e., square-wave frequency) increases.
For a given frequency, output ripple decreases as load current
decreases.
Simulation of a Four-Stage Converter
PSpice was used to simulate the four-stage converter shown below.
Output voltages for various loads are shown on the following pages.
Electronic Applications 190 Capacitor-Based Converters
Fig. 284. Simulated no-load output voltage for four-stage
converter. Note that the steady-state output voltage is less than
the theoretical value, but has very little output ripple.
Fig. 285. Simulated output voltage, 1 k load, for the four-stage
converter. The steady-state output voltage has decreased, and
output ripple has increased.
Electronic Applications 191 Capacitor-Based Converters
Fig. 286. Simulated output voltage, 100 load, for the four-stage
converter. With this heavy load, the steady-state output voltage
has decreased significantly, and ripple has become quite large.
Electronic Applications 192 Inductor-Based Converters
from
L L
L
V I di
v L
dt L t

(294)
Inductor-Based Converters
There are three basic configurations of inductive converters:
G Buck Converter:
Output voltage is less than input voltage.
G Boost Converter:
Output voltage is greater than input voltage.
G Buck-Boost Converter:
Output voltage can be either greater or less than input voltage
(in magnitude only, because as a consequence, the output
voltage is negative).
Techniques and Assumptions for Inductive-Converter Analysis
G All elements are lossless (except for load resistance).
G Diode forward voltage is zero.
G In steady state (i.e., all waveforms are periodic)
G Output voltage is constant (negligible ripple voltage).
We will see that this is equivalent to the requirement that the
output filter capacitor is infinitely large.
We will also see many linear inductor current waveforms, because
we will often assume inductor voltages are constant. Note that:
The latter equality merely describes the slope of the inductor current
waveform.
Electronic Applications 193 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 287. Buck converter.
C
t
D
T

(295)
C
t DT (296)
( ) 1
O
t D T
(297)
Buck Converter
G S is implemented with an active device.
G The freewheeling diode provides a path for inductor current
when the switch is open.
Switching Details:
We define:
t
C
and t
O
, the switch-closed and switch-open intervals,
T, the switching period,
and D, the switch duty ratio:
Thus, we can write:
and, because t
C
+ t
O
= T :
Electronic Applications 194 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 288. Buck converter (Fig 287 repeated).
L S O
v V V (298)
t
i
L
I
2
I
1
i
L ave
=

I
O
t
C
t
O
I = I
2
-

I
1
Fig. 289. Inductor current in the buck converter.
S O L
C
V V di I
dt L t

(299)
L O
v V (300)
O L
O
V di I
dt L t

(301)
Inductor Current:
When the switch is closed:
So the inductor current increases linearly in this interval:
When the switch is open:
And the inductor current decreases linearly:
Electronic Applications 195 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 290. Buck converter (Fig 287 repeated).
1 2
2
ave
L O
I I
i I
+

(302)
( ) ( )
S O C S O
V V t V V DT
I
L L


(303)
( )
1
O
O O
V D T
V t
I
L L

(304)
Output Current:
G The capacitor voltage can neither increase nor decrease.
G Therefore, the average capacitor current is zero!!!
G Therefore, the average inductor current must equal the
average output current:
The variables in this equation are illustrated in Fig. 289.
Output Voltage:
First, we solve eq. (299) for I:
Now, we solving eq. (301) for I:
Electronic Applications 196 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 291. Buck converter (Fig 287 repeated).
( ) ( )
( ) ( )
1
1
S O O
S O O
V V DT V D T
V V D V D
L L


(305)
t
i
S
I
2
I
1
I
O
t
C
t
O
Fig. 292. Input current in the buck converter.
S O
DV V (306)
Recognizing the equality of eq. (303) and eq. (304) allows us to
determine the output voltage:
from which:
In a buck regulator, D is varied using feedback to regulate V
O
.
Input Current:
Input current can be nonzero only when the switch is closed. It
equals the inductor current in the switch-closed interval:
Electronic Applications 197 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 293. Buck converter (Fig 287 repeated).
1 2
area
2
ave
C
O C
S S O
I I
t
I t
i I DI
T T T
+

(307)
S
O
I
I
D

(308)
( )
S
O O O S S S S
I
P V I DV V I P
D
j \

, (
( ,
(309)
We can calculate the average input current:
which can be written as
Input and Output Power:
We can determine average power with the average current values
calculated previously:
This is consistent with our lossless assumption.
Real Buck Converters:
G Switch and diode have nonzero voltage drops when on.
G Switch, diode, inductor, and capacitor all have resistive losses.
The equivalent series resistance (ESR) of the capacitor is very
important, especially at high frequencies.
Electronic Applications 198 Inductor-Based Converters
t
i
L
I
2
I
O
t
C
t
O
Fig. 295. Discontinuous-conduction mode inductor
current in the buck converter.
t
i
L
I
2
I
1
i
L ave
=

I
O
t
C
t
O
I = I
2
-

I
1
Fig. 294. Inductor current in the buck converter (Fig. 289
repeated).
G Diode stored charge results in energy loss per cycle. As a
result of these losses we have:
V
O
< DV
S
I
S
> DI
O
and P
S
> P
O
G Another phenomenon arises at light loads. First, lets recall
the inductor current waveform:
G At low values of I
O
, the minimum inductor current, I
1
, can fall
to zero. Should I
O
decrease further (i.e., less than I
2
/2), the
inductor current becomes discontinuous, and the model that
we used to derive the converter equations becomes invalid.
G Operation of the buck converter with I
1
0 is called the
continuous conduction mode.
I n t he di scont i nuous
conduction mode the inductor
current has three intervals per
cycle, rather than two. A more
complex model is required for
circuit analysis.
Electronic Applications 199 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 296. Buck converter (Fig 287 repeated).
t
i
c
t
C
t
O
= I
2
-

I
O
I
2
=

I
1
-

I
O
I
2
-
Fig. 297. Capacitor current in the buck converter.
C L O
i i i (310)
Estimating Peak-to-Peak Output Voltage Ripple:
Consider the following points:
G The steady-state output current is constant (because we
assumed constant output voltage).
G In steady-state, the average capacitor current must be zero.
This is because the average (dc) output voltage neither
increases nor decreases with time.
G From KCL we can write:
So we need only to shift the inductor current waveform down by I
O
to obtain the capacitor current:
Electronic Applications 200 Inductor-Based Converters
0
1
( ) ( ) (0)
t
O C O
v t i t dt v
C
+

(311)
= I
2
-

I
O
I
2
=

I
1
-

I
O
I
2
-
t
i
c
t
C
t
O
t
v
O
T
2
T
2
V
Fig. 298. The relationship between capacitor current and
output voltage ripple in a buck converter.
G The capacitor current waveform of Fig. 297 indicates that V
O
will not truly be constant, but will contain a ripple component
described by:
G Thus, when i
c
is greater than zero, v
O
rises slightly above its
average value of V
O
. Conversely, when i
c
is less than zero, v
O
falls slightly below its average value of V
O
.
G Because the output ripple is obtained by integrating linear
current segments, the ripple waveform is made up of parabolic
segments:
Electronic Applications 201 Inductor-Based Converters
= I
2
-

I
O
I
2
=

I
1
-

I
O
I
2
-
t
i
c
t
C
t
O
t
v
O
T
2
T
2
V
Fig. 299. The relationship between capacitor current and
output voltage ripple in a buck converter (Fig. 298
repeated).
1 1
and
2 2 2 2
C O
T I
b t t h

+
(312)
1
and
2 8 8
IT Q IT
Q bh V
C C


(313)
The change in output voltage V results from the charge that is
delivered to C during the interval when i
c
0.
(Equivalently we could have equated V with the charge taken from
C when i
C
0.)
The charge delivered is the area under a triangle with base and
height equal to:
Thus:
Electronic Applications 202 Inductor-Based Converters
= I
2
-

I
O
I
2
=

I
1
-

I
O
I
2
-
t
i
c
t
C
t
O
t
v
O
T
2
T
2
V
Fig. 300. The relationship between capacitor current and
output voltage ripple in a buck converter (Fig. 298
repeated).
O
O
V
I t
L

(314)
( )
1
S
DV
I D T
L

(315)
To complete our estimate we need to calculate I. From eq.(301):
But V
O
= DV
S
and t
O
= (1-D)T. Thus:
Electronic Applications 203 Inductor-Based Converters
= I
2
-

I
O
I
2
=

I
1
-

I
O
I
2
-
t
i
c
t
C
t
O
t
v
O
T
2
T
2
V
Fig. 301. The relationship between capacitor current and
output voltage ripple in a buck converter (Fig. 298
repeated).
( ) ( )
2
1 1
8 8
S S
DV D TT V D D
V
LC f LC


(316)
max
2
32
S
V
V
f LC

(317)
Finally, substituting eq.(315) into eq. (313) gives:
We can show that this reaches a maximum at D = 0.5, from which
we obtain an expression for the maximum ripple voltage:
In the following example we shall see that there is another
contribution to output ripple voltage - that due to the capacitor
current flowing through the capacitor ESR.
Electronic Applications 204 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ -
Fig. 302. Buck converter (Fig 287 repeated).
3V
0.25
12V
D (318)
( )
( )
6
2 1
6
3V 30 10 s
0.9A
100 10 H
O
O
V
I t I I
L

(319)
2 1
2 2A
O
I I I + (320)
Buck Converter Example
We will analyze a buck converter with the following:
V
S
= 12 V L = 100 H f = 25 kHz C = assumed large
The required output is 3 V at 1 A.
G First, we know that T = 40 s, and the required duty ratio is:
G From D, we can determine that t
C
= 10 s and t
O
= 30 s.
G Note that an output of 3 V at 1 A corresponds to R
L min
= 3 .
G From eq. (314):
G And from eq. (302):
G Adding the two previous equations gives 2I
2
= 2.9 A, from
which we obtain I
2
= 1.45 A and I
1
= 0.55 A.
Electronic Applications 205 Inductor-Based Converters
( )
( )( )( )
( ) ( )( )
2
2
3 6 3
1
8
12V 0.25 0.75
4.50mV
8 25 10 Hz 100 10 H 10 F
S
V D D
V
f LC

(321)
( )( ) 0.9A 0.075 67.5mV
(322)
Now lets examine the output ripple voltage . . . suppose we have
a capacitance of 1000 F. From eq. (316):
The calculated output ripple, 4.50 mV, is the value for our idealized
circuit, but it is misleading . . . to see this we need to look at a
typical filter capacitor.
G A Panasonic HFS, 1000 F, 50 V capacitor:
Equivalent Series Resistance (ESR) of 0.075 .
Maximum allowable rms ripple current of 2.7 A.
G Our peak-to-peak ripple current, I
2
- I
1
, flows through the
capacitor ESR, producing a peak-to-peak ripple voltage of:
In this case, the ripple due to the drop across the capacitor
ESR is much larger than that produced by charging the
capacitor.
In every converter configuration we will examine, the output ripple
voltage component due to capacitor ESR will completely dominate
converter operation. The component due to charging the capacitor
will be negligible.
(Note also that we are well within the capacitor current rating, so we
need not be concerned with overheating the capacitor.)
Electronic Applications 206 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ - i
C
i
L
i
D
Fig. 303. Boost converter.
t
i
L
I
2
I
1
t
C
t
O
I = I
2
-

I
1
i
L ave
=

I
S
Fig. 304. Inductor current in boost converter.
S L
C
V di I
dt L t

(323)
S O L
O
V V di I
dt L t

(324)
Boost Converter
G Inductive energy stored when S is closed is transferred to the
capacitor when S is open.
G Diode prevents C from discharging through S.
Inductor Current:
G i
L
increases when S closed, decreases when S is open:
S closed: S open:
Electronic Applications 207 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ - i
C
i
L
i
D
Fig. 305. Boost converter (Fig. 303 repeated).
( ) ( )( )
1
O S O O S
S C S
V V t V V D T
V t V DT
I
L L L L


(325)
( )( ) 1
S O S O O S S
V D V V D V V D V V D +
(326)
1
S
O
V
V
D

(327)
( )
( )
( )
1
1
i.e., 1
S S
S S
O O S S O S
O S
O S
V I D
V I
V I V I I I D
V V
I I D

(328)
Output Voltage:
We find the output voltage in exactly the same fashion as we did for
the buck converter, i.e., we solve eqs. (323) and (324) for I, and
set them equal:
from which:
Solving yields:
Input Current:
We know that the average input current, I
S
, equals the average
inductor current, I
L ave
, and from our assumption of a lossless circuit:
Electronic Applications 208 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ - i
C
i
L
i
D
Fig. 306. Boost converter (Fig. 303 repeated).
t
i
L
I
2
I
1
t
C
t
O
I
I
S
Fig. 307. Inductor current in boost
converter.
t
i
D
I
2
I
1
I
S
t
C
t
O
Fig. 308. Diode current in boost converter.
t
i
C
I
2
-

I
O
-I
O
t
C
t
O
I
1
-

I
O
Fig. 309. Capacitor current in boost
converter.
S S S
C
V V DV
I t DT
L L fL

(329)
Current Waveforms:
I is the peak-to-peak inductor
ripple current. From eq. (323):
The diode is reverse-biased when S
is closed.
When S is open i
L
= i
D
.
From i
C
= i
D
- I
O
, and because weve
assumed I
O
is constant, we obtain
the waveform at left.
We use this waveform to find the
output ripple voltage.
Electronic Applications 209 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ - i
C
i
L
i
D
Fig. 310. Boost converter (Fig. 303 repeated).
t
i
C
I
2
-

I
O
-I
O
t
C
t
O
I
1
-

I
O
t
v
o
V
linear
parabolic
Fig. 311. Relationship between capacitor
current and output voltage ripple in a boost
converter.
dv
i C
dt

(330)
O
C
V C V
I C
t DT

(331)
O O O
I DT V DT V D
V
C RC fRC

(332)
O
V D
V fRC

(333)
Estimating Peak-to-Peak Output Voltage Ripple:
We first concentrate on the ripple
component resulting from charging
the capacitor, and ignore that from the
voltage drop across the capacitor
ESR.
We will use the linear region shown in
the figure to determine V. From:
we have:
from which:
This peak-to-peak output ripple component is usually written in
normalized fashion:
Electronic Applications 210 Inductor-Based Converters
Fig. 312. Normalized output voltage vs. duty ratio for a boost converter.
( )
1
S
O
V
V
D

(334)
Output Voltage Sensitivity:
From:
we can see that the output voltage becomes very sensitive to small
changes in duty ratio as D becomes large. This is illustrated in the
figure below.
As a result of this sensitivity, operation of a boost converter at large
duty ratios is generally avoided.
Electronic Applications 211 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+ - i
C
i
L
i
D
Fig. 313. Boost converter (Fig. 303 repeated).
( ) ( )
1A
2.4A
1 1 0.583
O
S
I
I
D


(335)
( )
( )
max
2 2 3 A 2.4 A 1.2 A
L S
I i i
(337)
max
2
L S
I
i I

+
(336)
Boost Converter Example
We will analyze a boost converter with the following:
V
S
= 5 V and f
switch
250 kHz
We require:
V
O
= 12 V I
O max
= 1A i
L max
= 3.0 A minimum ripple
G From , we obtain D = 0.583.
( )
1
S
O
V
V
D

G Assuming a lossless circuit, the maximum source current is:

G Now, from:
we obtain:
Electronic Applications 212 Inductor-Based Converters
S
DV
I
fL

(338)
( )( )
( )
( )
3
0.583 5V
9.72 H
250 10 1.2A
S
DV
L
f I

(339)
( )( )
( )
( )
( )
3 6
0.583 12V
23.3mV
250 10 Hz 12 100 10 F
O
DV
V
fRC

(340)
G And, from:
we obtain:
Note that this inductance value is quite small . . .
G Now, the peak-to-peak output ripple voltage due to i
C
charging
and discharging C is a different waveform than the ripple due
to i
C
flowing through the capacitor ESR.
Also, the peaks of these two waveforms do not occur at the
same time.
As a result the total ripple voltage will be less than the direct
sum of these two terms.
G We will assume that the direct sum is a worst case value, and
use the Sprague Type 550D line of solid tantalum electrolytic
capacitors (www.vishay.com).
At 20 WVDC (Working Volts DC) the largest capacitance in
this line is 100 F, with an ESR of 0.075 .
G The peak-to-peak output ripple voltage due to i
C
charging and
discharging C is:
Electronic Applications 213 Inductor-Based Converters
( )( ) 3A 0.075 225mV
(341)
G The peak-to-peak capacitor current (see, for example, Fig.
311) is just I
2
, which is the same as i
L max
, i.e., 3 A.
The peak-to-peak output ripple voltage due to this current is:
G Clearly, the contribution due to capacitor ESR dominates. We
can improve this somewhat by choosing a larger inductor,
which will reduce the peak-to-peak capacitor current.
For example, if we increase the inductance to 100 H
(approximately a factor of 10), the peak-to-peak capacitor
current is 2.46 A, and the peak-to-peak ripple due to the
capacitor current is reduced to 184 mV.
However, because the average inductor current is 2.4 A, a
further increase in L does little to further reduce the ripple.
Before we leave this example, let us note again that the output
ripple is dominated by the drop across the capacitor ESR. The
other ripple component is, at best, small, and perhaps even
negligible.
Electronic Applications 214 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+
-
i
L
i
D
i
C
I
S
Fig. 314. The buck-boost converter, also called a flyback
converter, or an inverting converter.
t
i
L
I
2
I
1
t
C
t
O
I = I
2
-

I
1
Fig. 315. Inductor current in a buck-boost converter.
S
C
V I
L t

(342)
O
O
V I
L t

(343)
Buck-Boost Converter
G Inductive energy is stored when S is closed, and transferred
to C when S is open.
G Note that C never sees the source V
S
directly, i.e., steady-
state V
O
is zero for either switch position.
Thus whether the circuit operates in buck or boost mode is
determined solely by the energy stored in L (i.e., dependent
upon D).
G Note also that V
O
is negative.
Inductor Current:
S closed:
S open:
Electronic Applications 215 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+
-
i
L
i
D
i
C
I
S
Fig. 316. Buck-boost converter (Fig. 314
repeated)
( )
1
O
S C O O S
V D T
V t V t V DT
I
L L L L

(344)
( )
1
1
S O O S
D
V D V D V V
D
j \

, (

( ,
(345)
Output Voltage:
Solving eqs. (342) and (343) for I and equating allows us to
determine the output voltage:
from which:
Notice that the dividing line between buck operation and boost
operation occurs for D = 0.5.
Current waveforms are shown on the following page.
Electronic Applications 216 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+
-
i
L
i
D
i
C
I
S
Fig. 317. Buck-boost converter (Fig. 314
repeated)
t
i
L
I
2
I
1
t
C
t
O
I
Fig. 318. Inductor current.
t
i
S
I
2
I
1
t
C
t
O
I
Fig. 319. Input current.
t
i
D
I
2
I
1
t
C
t
O
I
Fig. 320. Diode current.
S C S
V t V D
I
L fL

(346)
1
O O S O S S
D
V I V I V I
D
j \

, (

( ,
(347)
1
O S
D
I I
D

j \

, (
( ,
(348)
Current Waveforms:
Peak-to-peak inductor ripple current:
Dc input and output current are
related by equating P
O
and P
S
:
from which:
This equation could also be derived
by noting that I
O
= i
D ave
and I
S
= i
S ave
.
Electronic Applications 217 Inductor-Based Converters
L
C
I
O
V
S
+
-
V
O
S
v
L
+
-
i
L
i
D
i
C
I
S
Fig. 321. Buck-boost converter (Fig. 314
repeated)
t
i
C
I
2
-

I
O
-I
O
t
C
t
O
I
1
-

I
O
t
v
o
V
linear
parabolic
Fig. 322. Relationship between capacitor
current and output voltage ripple in a buck-
boost converter.
dv
i C
dt

(349)
O
C
V V
I C C
t DT

(350)
O O O
I DT V DT V D
V
C RC fRC

(351)
O
V D
V fRC

(352)
Estimating Peak-to-Peak Output Voltage Ripple:
This development is identical to that
for the boost converter. Again we
concentrate on the ripple component
resulting from charging the capacitor,
and ignore that from the voltage drop
across the capacitor ESR.
From:
we have, for the linear segment of the
voltage waveform:
from which:
The peak-to-peak output ripple, written in normalized fashion, is:
Electronic Applications 218 Inductor-Based Converters
0
2
4
6
8
10
0 0.2 0.4 0.6 0.8 1
Duty Ratio, D
buck
boost
buck-boost
Voltage Ratio, V
O
/V
S
Fig. 323. Normalized output voltage vs. duty ratio
for the buck, boost, and buck-boost converters.
As in the other converters we have examined, the output ripple due
to capacitor ESR has been neglected in the previous calculation.
However, for all but the lowest-power converters, the output ripple
due to capacitor ESR will dominate.
Output Voltage Sensitivity:
The normalized output voltage is plotted below. For comparison,
outputs of the buck converter and the boost converter are plotted
also.
As with the boost converter, the buck-boost converter suffers from
high sensitivity to duty ratio at high values of D. Thus, operation at
high duty ratios generally should be avoided.