Chapter 5: BJT AC Analysis Two-Port Systems Approach

This approach: • Reduces a circuit to a two-port system • Provides a “Thévenin look” at the output terminals • Makes it easier to determine the effects of a changing load

With Vi set to 0 V:

ZTh  Zo  Ro

The voltage across the open terminals is:

ETh  AvNLVi

where AvNL is the no-load voltage gain.

Electronic Devices and Circuit Theory, 10/e Copyright ©2009 by Pearson Education, Inc. Robert L. Boylestad and Louis Nashelsky Upper Saddle River, New Jersey 07458 • All rights reserved. Effect of Load Impedance on Gain

This model can be applied to any current- or voltage- controlled amplifier.

Adding a load reduces the gain of the amplifier:

Vo RL A v   A vNL Vi RL  Ro Zi Ai  A v R L

The fraction of applied signal that reaches the input of the amplifier is: R i Vs Vi  R i  Rs

The internal resistance of the signal source reduces the overall gain:

Vo Ri A vs   A vNL Vs Ri  Rs

Effects of RL: Vo RLA vNL A v   Vi RL  Ro Ri Ai  A v RL

Effects of RL and RS:

Vo Ri RL A vs   A vNL Vs Ri  Rs RL  Ro Rs  Ri Ais  A vs RL

• The output of one amplifier is the input to the next amplifier • The overall voltage gain is determined by the product of gains of the individual stages • The DC bias circuits are isolated from each other by the coupling capacitors • The DC calculations are independent of the cascading • The AC calculations for gain and impedance are interdependent

This image cannot currently be displayed. Input impedance, first stage:

Zi  R1 || R 2 || re

Output impedance, second stage: Zo  RC

Voltage gain:

RC || R1 || R 2 || re A v1  re RC AV2  re A v A 1v A v2

Draw the ac equivalent circuit and calculate the voltage gain, input resistance and output resistance for the cascade BJT amplifier in above Figure. Let the parameters are:

VCC  20V ,  Q1   Q2  200,VBE(ON )  7.0 V ,r0  

R1  R3  15k, R2  R4  7.4 k, RC1  RC 2  2.2 k, RE1  RE 2  1k

Solution

Dc Analysis

I  19.89A I BQ1  19.89A BQ2

I CQ1  .3 979mA I CQ2  .3 979mA

1re1  2re2  .1 307 k

Ac Analysis: The voltage gain at Q1 is:

AV 1   .0 153 2.2( k // 957.36)  102.06

The voltage gain at Q2 is

AV 2   .0 153 2.2( k)  336 6.

The overall gain is then,

AV  AV1 AV 2  (102.06)(336 )6.  34,353

The input resistance is:

Ri  R1 // R2 // Br e 957.36

The o/p resistance is: Ro  RC 2  2.2 k

This example is a CE–CB combination. This arrangement provides high input impedance but a low voltage gain.

The low voltage gain of the input stage reduces the Miller input capacitance, making this combination suitable for high- frequency applications.

The Darlington circuit provides a very high current gain—the product of the individual current gains:

bD = b1b2 The practical significance is that the circuit provides a very high input impedance.

This image cannot currently be displayed. Base current: VCC  VBE IB  R B   DR E

Emitter current: IE  (D  I)1 B  DIB

Emitter voltage: VE  IER E

Base voltage: VB  VE  VBE

This is a two-transistor circuit that operates like a Darlington pair, but it is not a Darlington pair.

It has similar characteristics: • High current gain • Voltage gain near unity • Low output impedance • High input impedance

The difference is that a Darlington uses a pair of like transistors, whereas the feedback-pair configuration uses complementary transistors.

Current mirror circuits provide constant current in integrated circuits.

Find I?

Constant-current sources can be built using FETs, BJTs, and combinations of these devices.

VZ  VBE IE  IC I  I E  R E more…

VGS = 0V ID = IDSS = 10 mA