5. Intro to BJT Small Signal

Sem I 0809/rosdiyanaSem I 0809/rosdiyana

Introduction to BJT Introduction to BJT Small Signal AnalysisSmall Signal Analysis

CHAPTER 5CHAPTER 5

22

IntroductionIntroduction•To begin analyze of small-signal AC response of BJT amplifier the knowledge of modeling the transistor is important.•The input signal will determine whether it’s a small signal (AC) or large signal (DC) analysis.•The goal when modeling small-signal behavior is to make of a transistor that work for small-signal enough to “keep things linear” (i.e.: not distort too much) [3]•There are two models commonly used in the small signal analysis:

a) re modelb) hybrid equivalent model

33

DisadvantageDisadvantage RRee model model

• Fails to account the output impedance Fails to account the output impedance level of device and feedback effect from level of device and feedback effect from output to inputoutput to input

Hybrid equivalent modelHybrid equivalent model• Limited to specified operating condition Limited to specified operating condition

in order to obtain accurate resultin order to obtain accurate result

IntroductionIntroduction

44

The transistor can be employed as an The transistor can be employed as an amplifying device. That is, the output amplifying device. That is, the output sinusoidal signal is greater than the sinusoidal signal is greater than the input signal or the ac input power is input signal or the ac input power is greater than ac input power.greater than ac input power.

How the ac power output can be How the ac power output can be greater than the input ac power?greater than the input ac power?

Amplification in the AC domainAmplification in the AC domain

55

Amplification in the AC domainAmplification in the AC domain

Conservation; output Conservation; output power of a system power of a system cannot be large than its cannot be large than its input and the efficiency input and the efficiency cannot be greater than 1cannot be greater than 1

The input dc plays the The input dc plays the important role for the important role for the amplification to amplification to contribute its level to the contribute its level to the ac domain where the ac domain where the conversion will become conversion will become as as ηη=P=Po(ac)o(ac)/P/Pi(dc)i(dc)

66

The superposition theorem is applicable for the The superposition theorem is applicable for the analysis and design of the dc & ac components of analysis and design of the dc & ac components of a BJT network, permitting the separation of the a BJT network, permitting the separation of the analysis of the dc & ac responses of the system.analysis of the dc & ac responses of the system.

In other words, one can make a complete dc In other words, one can make a complete dc analysis of a system before considering the ac analysis of a system before considering the ac response.response.

Once the dc analysis is complete, the ac response Once the dc analysis is complete, the ac response can be determined using a completely ac can be determined using a completely ac analysis.analysis.

Amplification in the AC domainAmplification in the AC domain

77

BJT Transistor ModelBJT Transistor Model Use equivalent circuitUse equivalent circuit Schematic symbol for the device can be replaced Schematic symbol for the device can be replaced

by this equivalent circuits.by this equivalent circuits. Basic methods of circuit analysis is applied.Basic methods of circuit analysis is applied. DC levels were important to determine the Q-pointDC levels were important to determine the Q-point Once determined, the DC level can be ignored in Once determined, the DC level can be ignored in

the AC analysis of the network.the AC analysis of the network. Coupling capacitors & bypass capacitor were Coupling capacitors & bypass capacitor were

chosen to have a very small reactance at the chosen to have a very small reactance at the frequency of applications.frequency of applications.

88

The AC equivalent of a network isThe AC equivalent of a network isobtained by:obtained by:1.1. Setting all DC sources to zero & replacing them Setting all DC sources to zero & replacing them

by a short-circuit equivalent.by a short-circuit equivalent.2.2. Replacing all capacitors by a short-circuit Replacing all capacitors by a short-circuit

equivalent.equivalent.3.3. Removing all elements bypassed by short-Removing all elements bypassed by short-

circuit equivalent.circuit equivalent.4.4. Redrawing the network. Redrawing the network.

BJT Transistor ModelBJT Transistor Model

99

1010

1111

ExampleExample

1212

ExampleExample

1313

ExampleExample

1414

The re transistor modelThe re transistor model

Common Base PNP ConfigurationCommon Base PNP Configuration

1515

Common Base PNP ConfigurationCommon Base PNP Configuration

Transistor is replaced by Transistor is replaced by a single diode between E a single diode between E & B, and control current & B, and control current source between B & Csource between B & C

Collector current Ic is Collector current Ic is controlled by the level of controlled by the level of emitter current Ie.emitter current Ie.

For the ac response the For the ac response the diode can be replaced by diode can be replaced by its equivalent ac its equivalent ac resistance.resistance.

1616

Common Base PNP ConfigurationCommon Base PNP Configuration

The ac resistance The ac resistance of a diode can be of a diode can be determined by the determined by the equation;equation;

Where IWhere IDD is the dc is the dc current through current through the diode at the Q-the diode at the Q-point. point.

EI

mVre

26

1717

Input impedance is Input impedance is relatively small and relatively small and output impedance output impedance quite high.quite high.

range from a few range from a few ΩΩ to max 50 to max 50 ΩΩ

Typical values are in Typical values are in the M the M ΩΩ

CBi reZ

CBZo

Common Base PNP ConfigurationCommon Base PNP Configuration

1818

The common-base The common-base characteristicscharacteristics

1919

Voltage GainVoltage Gain

re

RA

re

R

rI

RI

V

VA

rI

ZI

ZIV

RI

RI

RIV

LV

L

ee

Le

i

OV

ee

ie

iii

Le

LC

Loo

:gain voltage

: ageinput volt

)(

: tageoutput vol

2020

Current GainCurrent Gain

1

i

e

e

e

C

i

oi

A

I

I

I

I

I

IA

The fact that the polarity of the VThe fact that the polarity of the Voo as determined as determined by the current Iby the current ICC is the same as defined by figure is the same as defined by figure below.below.

It reveals that VIt reveals that Voo and V and Vii are in phase for the are in phase for the common-base configuration.common-base configuration.

2121

Common Base PNP ConfigurationCommon Base PNP Configuration

Approximate model for a common-base npn transistor configuration

2222

Example 1: For a common-base configuration in figurebelow with IE=4mA, =0.98 and AC signal of 2mV isapplied between the base and emitter terminal:a) Determine the Zi b) Calculate Av if RL=0.56kc) Find Zo and Ai

e

b b

c

ec I αI

IcIe

common-base re equivalent cct

re

2323

Solution:

5.6m4

m26

I

26mr Za)

Eei

43.845.6

)k56.0(98.0

r

RA b)

e

Lv

98.0I

IA

Ω Zc)

i

oi

o

2424

Example 2: For a common-base configuration in previous example with Ie=0.5mA, =0.98 and AC signal of 10mV is applied, determine:a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib

20m5.0

m10

I

V Za)

:Solution

e

ii

88mV5(1.2k)0.98(0.5m)

RIRIV b) LeLco

8.58m10

m588

V

VA c)

i

ov

98.0A d) i

A10

)98.01(m5.0

)1(m5.0

I-I

I-II e)

ee

ceb

Common Emitter NPN ConfigurationCommon Emitter NPN Configuration

Base and emitter Base and emitter are input are input terminalterminal

Collector and Collector and emitter are emitter are output terminalsoutput terminals

2525

Common Emitter NPN ConfigurationCommon Emitter NPN Configuration

Substitute rSubstitute re e

equivalent circuitequivalent circuit

Current through Current through diodediode

2626

bc II

bbe

bbbce

III

IIIII

)1(

2727

Input impedanceInput impedance

ei

ei

b

ebi

eb

eei

b

be

i

ii

rZ

rZ

I

rIZ

rI

rIV

I

V

I

VZ

; 1an greater thusually

)1(

)1( that so

)1(

:ageinput volt

:impedanceinput

2828

The output graphThe output graph

2929

bI

c

e

bIi=Ib

re model for the C-E transistor configuration

rero

e

0AbI

c

e

bIi=Ib

rero

e

Vs=0V

= 0A

oZ

impedance)high cct,(open ΩZ

the thusignored is r if

rZ

o

o

oo

Output impedance Zo

3030e

LV

eb

Lb

i

oV

eb

iii

Lb

Lco

Loo

r

RA

rI

RI

V

VA

rI

ZIV

RI

RIV

RIV

that so

:ageinput volt

: tageoutput vol

i

b

b

b

C

i

oi

A

I

I

I

I

I

IA

Voltage GainVoltage Gain Current GainCurrent Gain

rree model for common-emitter model for common-emitter

3131

3232

Example 3: Given =120 and IE(dc)=3.2mA for a common-emitter configuration with ro= , determine:

a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load

975)125.8(120rZ

125.8m2.3

m26

I

26mr a)

ei

Ee

:Solution

15.246125.8

k2

r

Rb)A

e

Lv

120I

IA c)

i

oi

3333

Example 4: Using the npn common-emitter configuration, determine the following if =80, IE(dc)=2 mA and ro=40 k

a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k

k04.1)13(80rZ

13m2

m26

I

26mr a)

ei

Ee

:Solution

bI

cbIi=Ib

re model for the C-E transistor configuration

rero

e

RL

Io

3434

67.77

)80(k2.1k40

k40

Rr

r

IRr

)I(r

A

Rr

)I(rI

I

I

I

IiAb)

(cont)Solution

Lo

o

b

Lo

bo

i

Lo

boL

b

L

i

o

6.8913

k40k2.1

r

rRvAc)

e

oL

3535

Common Collector ConfigurationCommon Collector Configuration

For the CC configuration, the model For the CC configuration, the model defined for the common-emitter defined for the common-emitter configuration is normally applied configuration is normally applied rather than defining a model for the rather than defining a model for the common-collector configuration.common-collector configuration.