Electronic Circuits I

Lecture 12: BJT & BJT Amp

Junwon Jeong
 Review of Last Lecture
- Early effect
- Early effect in large/small-signal modeling
- Bipolar TR in saturation mode
- PNP transistor: Operation principle, operating
modes, large/small-signal modeling

2
 This Lecture
Chapter 4
- PNP transistor: Operation principle, operating
modes, large/small-signal modeling

Chapter 5
- What are input/output impedances?

3
Course Objectives

Lumped Abstraction Amp & Analog

• Diode transistor • R+C+Diode → Rectifier
• Bipolar transistor • R+Diode+Battery →
• MOS transistor Limiter
• Amplifiers: Input/output
impedances
• Operational amplifier
• Current mirror
• Large/small signal analysis

4
 This Lecture
Chapter 4
- PNP transistor: Operation principle, operating
modes, large/small-signal modeling

Chapter 5
- What are input/output impedances?

5
Why PNP?

• What if input DC bias (0.8V) decreases? (ex. 0.3V)

6
PNP Bipolar Transistor Structure and Operation

Fig 4.38 (a) Structure of pnp transistor, (b) current flow in pnp transistor, (c) proper biasing, (d)
more intuitive view of (c).
With the polarities of emitter, collector, and base reversed, a PNP transistor is formed.
All the principles that applied to NPN's also apply to PNP’s, with the exception that emitter is at a higher
potential than base and base at a higher potential than collector (for active mode operation).

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Large-Signal Model of PNP Transistor

n
p
n

p
n
p

Fig 4.39 (a) Voltage and current polarities in npn and pnp transistors, (b) illustration of active
and saturation regions.

Current flows from collector to emitter in npn devices and from emitter to collector in PNP.
The distinction between active and saturation regions is based on the B-C junction bias.

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Large-Signal Model of PNP Transistor

PNP Equations
For the pnp device, the collector must not be higher than the base.

VEB
I C = I S exp (4.100)
VT

IS VEB
IB = exp (4.101)
 VT

 +1 V
IE = I S exp EB (4.102)
 VT
Fig 4.40 Large-signal model of pnp transistor.

 V  V 
I C =  I S exp EB  1 + EC 
 VT   VA 

The Early effect is included.

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Small-Signal Model of PNP Transistor
The small signal model for PNP transistor is exactly IDENTICAL to that of NPN.
This is not a mistake because the current direction is taken care of by the
polarity of VBE.

Fig 4.43 (a) Small-signal model of pnp transistor, (b) more intuitive view of (a).

How to make a BJT operate as a diode? → Example 4.19 !!

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Small-Signal Model (NPN & PNP)
< NPN BJT >

Small signal model Large signal model

< PNP BJT >

Small signal model Large signal model

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Example 4.20
Draw the small-signal equivalent circuits for the topologies shown in
Figs. 4.45(a)-(c) and compare the results.

Fig 4.45 (c) pnp stage

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CH. 4 Summary
Voltage-dependent current source NPN Current Equation
V  IC
I C = I S exp  BE 
 VT 
1 V 
IB
IB = I S exp  BE 
  VT 
Amplifier Vout = − KRLVin  +1 V 
IE = I S exp  BE 
  VT  IE

Large signal vs. Small signal

dI C IC
gm = =
dVBE VT
VBE 
r = =
I B gm

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CH. 4 Summary
• Without Early Effect

Large signal model I/V Characteristic Small signal model

• With Early Effect

Non ideal case !

Large signal model I/V Characteristic Small signal model

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CH. 4 Summary

PNP BJT
Saturation Region

VBC < 0

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CH. 4 Summary
PNP Equation
Large Signal model Small Signal model
VEB
I C = I S exp
VT
IS VEB
IB = exp
 VT
 +1 V
IE = I S exp EB
 VT

 V  V 
I C =  I S exp EB  1 + EC 
 VT   VA 

ac ground !
Early effect !

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 This Lecture
Chapter 4
- PNP transistor: Operation principle, operating
modes, large/small-signal modeling

Chapter 5
- What are input/output impedances?

17
 Chapter 5.
Bipolar Amplifier
5.1 General Considerations
5.2 Operating Point Analysis and Design
5.3 Bipolar Amplifier Topologies

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Chapter 5: Overview

• a voltage-controlled current source along with a load resistor can form an amplifier.
• an amplifier produces an output (V or I) that is a magnified version of the input (V or I).
• Considerations to design amplifiers
• power dissipation
• speed
• noise
• The input and output (I/O) impedances

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Backgrounds: Equivalent Circuits

Thevenin: A (linear) one-port Norton: A (linear) one-port

network can be represented by network can be represented by
one voltage source and one one current source and one
series impedance parallel impedance

• Circuit with voltage output → Thevenin equivalent circuit

• Circuit with current output → Norton equivalent circuit
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Background: Lumped Abstraction-Resistor
I
R=10Ω
Object in nature
V

I
R=0
V

I
R=∞
V

• Resistance: a measure of its opposition to the low of electric current

• Every object in nature can be modeled by lumped resistor
Ideal vs. Nonideal Voltage Source <1>
<Ideal> <Nonideal>
I
I 1/R O UT
+ I + Iide al I
Iid ea l
V V ROUT
- -
V V

<Ideal> <Nonideal>
V id ea l
I
+ I I
V ? 1/R O UT

-
V V

• Nonideal current source: Changing I according to V → ROUT in

parallel with the current source (Norton eq. circuit)
• Ideal voltage source: Constant V regardless of I
• Nonideal voltage source: Changing V according to I → How to
represent the nonideal voltage source?
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Ideal vs. Nonideal Voltage Source <2>
<Ideal> <Nonideal>
I
I 1/R O UT
+ I + Iide al I
Iid ea l
V V ROUT
- -
V V

<Ideal> <Nonideal>
ROUT
V id ea l
I
+ I + I + I
V Vide al V 1/R O UT

- - -
V V

• Nonideal voltage source: Changing V according to I → ROUT in

series with ideal voltage source (Thevenin eq. circuit)
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Nonideal Voltage Source Example: Func. Gen. & V. Amp
<Ideal> <Nonideal>
ROUT
V id ea l
I
+ I + I + I
V Vide al V 1/R O UT

- - -
V V

<Nonideal>
ROUT
V id ea l
+ I + I
Vide al V 1/R O UT

- -
V

Gain (Av)

• Function generator, Voltage amplifier: voltage output → modeled

by nonideal voltage source (=Thevenin eq. circuit)
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Background: Voltage/Current Dividers
<Voltage Divider> <Current Divider>
R1

+ + IIN IOUT
VIN R2 VOUT R1 R2
- -

R1 R2
VOUT = VIN IOUT = IIN
R1 +R2 R1 +R2

• Voltage divider
– What is desired?: Large VOUT/Small VOUT
– How to increase VOUT?
• Current divider: How to increase IOUT?
– What is desired?: Large IOUT/Small IOUT
– How to increase IOUT?
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Function Generator Example

ROUT

+ +
VIN RL VOUT
= - -
+
Function Generator
RL VOUT
- RL
VOUT = V
ROUT +RL IN

• Function generator wants to apply voltage (VIN) to resistor (RL)

• What happens if RL is too small or ROUT is too large?
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Voltage Amplifier Example <1>

ROUT
Av
Volt. + +
+ Avvin RL vout
Amp +
vin
RL vo ut = - -
- -
Voltage Amplifier

RL
vout = vin
ROUT +RL

• Voltage amplifier wants to apply the amplified voltage (Avvin) to

resistor (RL)
• What happens if RL is too small or ROUT is too large?
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Voltage Amplifier Example <2>

Av Av
1st 2nd 1st
+ Amp + Amp + Amp +
vin vout vin vout
- - - -

• Voltage amplifier wants to apply the amplified voltage (Avvin) to

2nd stage amplifier
• How to model the 2nd stage amplifier?
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Voltage Amplifier Example <3>

Av Av
1st 1st
+ Amp + = + Amp +
vin vout vin
RL=rπ vout
- - - -

• 2nd stage amplifier can be modeled by load resistor RL.

• What is RL value in this case?

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Why We Learn Input/Output Impedance?

<Voltage Divider> <Current Divider>

R1

+ + IIN IOUT
VIN R2 VOUT R1 R2
- -

R1 R2
VOUT = VIN IOUT = IIN
R1 +R2 R1 +R2

• Electronics: Manage information with the movement of electron

(Voltage, current, etc.) → To deliver the information as much as
possible, delivered voltage, current should be the maximum. →
Input/Output impedance should be figured out
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Input & Output Impedances of Amplifier

Gain (Av)

Gain * Signal input

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What are Ideal Input and Output Impedances of Voltage Amp?

• Ideal voltage amplifier

• Input impedance : To sense a voltage w/o loading the preceding stage
• Ideal input impedance = ∞
• Output impedance : To deliver a constant signal level to any load.
• Ideal output impedance = 0

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Example 5.1 <1>
• An amplifier w/ a voltage gain of 10

Fig. 5.1 (a) Simple audio system

(a) Determine the signal level sensed by the amplifier if the circuit has an input
impedance of 2 k or 500 .
(b) Determine the signal level delivered to the speaker if the circuit has an output
impedance of 10 or 2 .

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Example 5.1 <2>

• Solution (a),
Rin
Sensed voltage: v1 = vm
Rin + Rm

For Rin=2kΩ, v1 = 091vm

For Rin=500Ω, v1 = 071vm

Fig. 5.1 (b) signal loss due to amplifier
input impedance.

• Solution (b),
The interface between the amp and the speaker as
in Fig.5.1(c) RL
vout = vamp
RL + Ramp

For Ramp= 10Ω, vout = 044vamp

Fig. 5.1 (c) signal loss due to amplifier For Ramp= 2Ω, vout = 08vamp 
output impedance

• Input impedance: the ?, the better

• Output impedance: the ?, the better
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Appendix: Microphone System

Power
Supply

Processor

Microphone Amplifier Filter Analog-to Digital

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Appendix: Amplifier Design Example
Power
Supply
Microphone

Processor

Amplifier Filter Analog-to Digital

DC Bias
Generator

<Q1 large-signal model>

Signal 15mA
from mic
3mA
1mA
DC Bias

• Determine DC bioas for voltage gain > 10, (VT=0.026V)

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