Microelectronic Circuits

Amplifier Basics

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Mixed signal System design

ADC- SubSystem Design

A to D Converter, D to A Converter

Flash ADC-100 Msps

Band width requirement of OPAMP

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DAC

Design issues

Diff amp, biasing circuit, CSA, C (if compensation is

reqd.)

Signals

Arbitrary in nature.

Obtained through sensors---variations

converted into current or voltage
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Sinusoid

Important signal in analysis , design, testing

For an LTI system, if input is sinusoid, output is
also sinusoid with modified amplitude and
phase. Hence analysis is easy

Every natural signal can be represented as sum

of sine waves of different frequencies and
amplitude.

Lab testing is possible

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Why amplifier first?

Reasons
Fundamental signal processing function
Employed in every electronic system
Easy to understand
Design techniques can be easily extended to
design of complex analog circuits.

Similar to NOT gate in Digital Electronics

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Amplifier circuit symbol

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Amplifiers

Need- Weak signals- energy too small for

reliable processing

Requirement---Information contained in the

signal should not get changed/ Output must

be exact replica of the input.

Relation ship of amplifiers

vo(t) = A vi(t)
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Characterizing parameters

Gain
Voltage swing
Linearity
Power efficiency
Frequency response
Power supply and dc bias

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Analog Design tradeoffs

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Gain/ transfer characteristics

Voltage Gain

Current gain

Power gain
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VTC

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Information from VTC

Highest and Lowest signal amplitude

Gainsteepness of transition
Inverting/ non inverting nature
Single/ dual power supply
Offset

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Measuring Unit---Use of Decibel

unit

The decibel, or dB, is a means of expressing

either the gain of an active device (such as
an amplifier) or the loss in a passive device
(such as an attenuator or length of cable).

The decibel was developed by the telephone

company to conveniently express the gain or
loss in telephone transmission systems.
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Input >---- Amp #1 ---- Amp #2 ------> Output

A1 = 275, A2 = 55
The total gain factor At = 275 x 55 = 15,125.
Use logarithmslog (A x B) = log A + log B
log (A/B) = log A - log B

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Invented a unit of gain measurement called a

"Bel," named after Alexander Graham Bell.

They defined the Bel as

Gain in Bels = log A = log (Po / Pi )
where A = Power amplification factor

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log 275 = 2.4393326 and

log 55 = 1.7403626,
15,125

so the total gain in our cascade is

2.4393326 + 1.7403626 = 4.179,695,289
Bels
Rounding problem---

4.179 Bels
15124.99----4.2 Bels
15,849

5% error

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it was decided to express power gain in

units which were equal to one-tenth of a
Bel, or in deci-Bels
1 Bel=10 decibels
Gain in decibels (dB) = 10 log A

2.4393326 + 1.7403626 = 4.179,695,289 Bels

24.39 + 17.40 = 41.79 decibels

41.79 dB is a power gain of 15,101,

while 41.8 dB is a power gain of 15,136,
so the error is only 0.23%.

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Voltage gain--

Gain in dB---10 log[ (Vo2/RL) / (Vi2/Ri)]

If RL= Ri

Gain

in dB= 20 log (Vo/ Vi)

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Linearity

Amplifier follows a relationship.--- linear

amplifier

vo(t) = A vi(t)

Any deviation (higher powers of vi) --nonlinear distortion

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THD

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Amplifier Power Supplies

Important part of the circuit

Power balance equation--Pdc + Pin = P load + P dissipated
Maximum power must be delivered to the
load
Figure of merit---Amplifier Power efficiency

PL
=
100
Pdc
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Classes of amplifiers

Class A----max = 25 %
Class B (~70%)
Class AB (~70%)
Class C (~80%)
Class D (~100%)

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Need to set DC Bias

Practical VTC is non
linear-- Saturation,
varying slope
output will be distorted
--Operate at a point where
VTC is close to linearmiddle
--Keep input small
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Frequency responsebandwidth

Ideal frequency response---gain does not

change with frequency

Practical frequency response

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Poles, Zeros and Bode Plots

Characterization:

K ( s + z1 )( s + z 2 ) .. . ( s + z m )
G (s) =
s ( s + p1 )( s + p 2 ) .. . ( s + pn )
s
s
s
( + 1)( + 1) ... ( + 1)
K ( z1 z 2 z m )
z1
z2
zm
G (s) =
( p1 p2 pn ) s ( s + 1)( s + 1) ... ( s + 1)
p1
p2
pn
K ( z1 z 2 z m )
KB =
( p1 p2 pn )
( z1s + 1)( z 2 s + 1) .. . ( zm s + 1)
G (s) = K B
s ( p1 s + 1)( p 2 s + 1) .. . ( pn s + 1)

(Time Constant Form.)

Characterization:
Considering the transfer function in the time constant form.
we have 4 different types of terms in the time constant form,
these are:

1
1
KB, ,
, (s / z + 1)
s (s / p + 1)
Expressing the transfer function dB:
j
KB (

G ( jw) =
(j

+ 1)

j
)(
+ 1)
0 p

20 log | G ( j ) |
= 20 log K B + 20 log | (

j
+ 1) | 20 log | j
| 20 log |
+1|
o
z
p

Mechanics:

We have 4 distinct terms to consider:

20logKB

----- ( constant gain in time constant format)

- 20log|j /0 |

----- (Pole at origin if w=1)

- 20log|(j /p + 1)| ------ (Pole at 0 = p )

20log|(j/z +1)|

----- (zero at 0 = z )

wlg

This is a sheet of 5 cycle, semi-log paper.

This is the type of paper usually used for
preparing Bode plots.

dB Mag
Phase
(deg)

wlg

(rad/sec)

Frequency response plots

Different types of transfer functions--K

w
j
wo

1
j

w
wo

jw
1+
wo

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1
jw
1+
wo

K constant
Gain in dB

Log w

= 0
Log w

jw/wo
Gain in dB

20 dB/ dec

wo

Log w

= 90
Log w

-j w/wo = 1/ [jw/wo]-- pole at the origin,

jw/wo

Gain in dB

20 dB/ dec

wo

Log w
rad./ sec

For a pole at the origin draw a line with a slope of -20

dB/decade that goes through 0 dB at 1 rad/sec

Log w
= -90

1+ j (w/wo)
Gain in dB
Corner plot
20 dB/ dec

wo

Corner frequency

Log w

90

= tan-1 (w/wo)

45

0.1 wo

Log w
~10 wo

Magnitude and phase

jw
1+
wo

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1 / [1+ j (w/wo)]
Gain in dB

wo
Log w
20 dB/ dec

= -tan-1 (w/wo)

~0.1 wo

Log w

-45

-90

~10 wo

Using Matlab For Frequency Response

Instruction: We can use Matlab to run the frequency response for the
previous example. We place the transfer function in the
form:
5000 ( s + 10)
[ 5000s + 50000 ]
=
( s + 1) ( s + 500)
[ s 2 + 501s + 500]

The Matlab Program

num = [5000 50000];
den = [1 501 500];
Bode (num,den)

Using Matlab For Freq. Response

Instruction: We can use Matlab to run the frequency response for the
previous example. We place the transfer function in the
form:
5000 ( s + 10)
[ 5000s + 50000 ]
=
( s + 1) ( s + 500)
[ s 2 + 501s + 500]

The Matlab Program

num = [5000 50000];
den = [1 501 500];
Bode (num,den)

Bode Diagrams
From: U(1)
40
30

10
0
-10
1

10

100

500

0
-20
To: Y(1)

Phase (deg); Magnitude (dB)

20

-40
-60
-80
-100
10-1

Bode for:
100

G ( jw) =
101

100(1 + jw / 10)
(1 + jw)(1 + jw / 500)
102

Frequency (rad/sec)

G( j) = tan1 ( / 10) tan1 ( / 1) tan1 ( / 500)

Initial angle=00 and final angle -900

103

104

Evaluating the frequency

response

Single time constant circuits

Vo (s) = 1/ [1+sCR] vi (s)

Vo (s) = sCR/ [1+sCR] vi (s)

[vo./ vi]= K / [1+{s/wo}]

[vo./ vi]= K s / [1+{s/wo}]

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Amplifier circuit models-analysis tools

Amplifiers must be characterized for its

terminal behavior first to be used as block in
system design
For analysis purpose, complex circuits are
replaced by their (models)--- simple circuits
Voltage amplifier model (v,v)
Current amplifier (i,i)
Trans-conductance (v,i)
Trans-resistance (i,v)
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Use 2 port network theory

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Z parameter model

8/9/2012

Anu Gupta BITS PILANI

CLASSIFICATION OF AMPLIFIERS
Voltage amplifier
Ro
+

vi

Ri

Avvi

+
vo
-

Using the voltage divider rule open circuit voltage gain is

Av = vo/vi|io=0
Unit (V/V)
Ideal conditions: R0 = 0

Ri = condition for no loss

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Current amplifier
io

ii

Ri

Aivi
-

Short-Circuit current gain

Ais = io/ii |vo=0
Unit

(A/A)

Ideal conditions

Ri = 0 ; R0 =
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+
Ro vo
-

Transconductance amplifier
io
+

+
vi

Ri

Gmvi

Short-Circuit Transconductance
Gm = io/vi |vo=0
Unit

Ro vo

(A/V)

Ideal conditions Ri = ; R0 =
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Transresistance amplifier
Ro

ii

+
Ri

Rmii

vo
-

Open-Circuit Transresistence
Rm = vo/ii |io=0
Unit

(V/A)

Ideal conditions Ri = 0 ; R0 = 0
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Relations between parameters

Av0 = Ais(Ro /Ri)

Av0 = GmRo
Av0 = Rm/Ri

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END

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