Bipolar JunctionBipolar Junction Transistor Circuit Analysis Transistor Circuit Analysis

EE314

Chapter 13: Bipolar Junction Transistors

1.Large signal DC analysis2.Small signal equivalent3.Amplifiers

BJT Transistor Circuit Analysis

Circuit with BJTs

Our approach: Operating point - dc operating pointAnalysis of the signals - the signals to be amplified

Circuit is divided into: model for large-signal dc analysis of BJT circuitbias circuits for BJT amplifiersmall-signal models used to analyze circuits for signals being amplified

Remember !

Large-Signal dc Analysis: Active-Region Model

Important: a current-controlled current source models the dependence of the collector current on the base current

The constrains for IB and VCE must be satisfy to keep BJT in the active-mode

VBE

forward bias

VCB

reverse bias ?

?

Large-Signal dc Analysis: Saturation-Region Model

VBE

forward bias

VCB

forward bias?

?

Large-Signal dc Analysis: Cutoff-Region Model

VCB

reverse bias

VBE

reverse bias

?

?

If small forward-bias voltage of up to about 0.5 V are applied, the currents are often negligible and we use the cutoff-region model.

Large-Signal dc Analysis: characteristics of an npn BJT

Large-Signal dc Analysis

Procedure: (1) select the operation mode of the BJT (2) use selected model for the device to solve the circuit

and determine IC, IB, VBE, and VCE (3) check to see if the solution satisfies the constrains for

the region, if so the analysis is done (4) if not, assume operation in a different region and repeat until a valid solution is found

This procedure is very important in the analysis and design of the bias circuit for BJT amplifier.

The objective of the bias circuit is to place the operating point inthe active region.

Bias point – it is important to select IC, IB, VBE, and VCE

independent of the and operation temperature.

Example 13.4, 13.5, 13.6

Large-Signal dc Analysis: Bias Circuit

From Example 13.6

Remember: that the Q point should be independent of the stability issue) VBB & VCC provide this stability, however this impractical solutionOther approach is necessary to solve this problem-resistor network

VBB acts as a short circuit for ac signals

Large-Signal dc Analysis: Four-Resistor Bias Circuit

1

2

3

4

Thevenin equivalent

21 RRRB 212 / RRRVV CCB

Equivalent circuit for active-region model

Solution of the bias problem:

Input Outpu

t

EEBEBBB IRVIRV

BE II 1 VVBE 7.0

EB

BEBB RR

VVI

1

EECCCCCE IRIRVV

Small-Signal Equivalent Circuit

Thevenin equivalent

Small signal equivalent circuit for BJT:

so

V

tvItiI

xx

V

tvI

V

tvvI

tiIi

T

beBQbBQ

T

beBQ

T

beBEQES

bBQB

)(1)(

,1)exp(

)(exp

)(exp1

)(

r

tv

V

tvIti be

T

beBQb

)()()(

BQ

T

I

Vr and

Common Emitter Amplifier

Find voltage gain:

First perform DC analysis to find small-signal equivalent parameters at the operating point.

Find input impedance:

Common Emitter Amplifier

Find power gain:

Find current gain

Find output impedance:

Problem 13.13: Suppose that a certain npn transistor has VBE = 0.7V for IE =10mA. Compute VBE for IE = 1mA.

Repeat for IE = 1µA. Assume that VT = 26mV.

VV

V

sidesbothdivide

and

BE

BE

64.010ln*026.07.0

7.010ln*0.026

0.026

V - 0.7exp =10

0.026

VexpI=1mA

0.026

0.7expI=10mA

V

VexpI 1-

V

Vexp I=I

BE

BEESES

T

BEES

T

BEESE

Problem 13.14: Consider the circuit shown in Figure P13.14. Transistors Q1 and Q2 are identical, both having IES = 10-14A and β = 100. Calculate VBE and IC2. Assume that VT = 26mV for both transistors.

Hint: Both transistors are operating in the active region. Because the transistors are identical and have identical values of VBE, their collector currents are equal.

VI

IVV

haveweV

VIIce

mAII

mAmA

ImAI

IImAIII

ES

ETBE

T

BEESE

CE

CC

BCCBB

658.010*99.0ln*026.0ln

expsin

99.01

1

98.002.1

111

2

&1

11

21

Problem 13.50: The transistors shown in Figure P13.50 operate in active region and have β = 100, VBE=0.7V. Determine IC and VCE for each transistor.

I1

IE2

VBE

VI

mAkV

VIkIIkV

mAIImAI

ImA

ImA

I

k

II

k

kI

mAIIAM

I

CCE

EECCE

ECE

E

EE

EC

E

C

6126.41*1015

213.7*99.115*115

8735.399.09126.3

10

1

101

1*43.0

1011

1010

3.14

110

7.01*15

11043.1

3.14

21

2222

222

2

22

21

2

111

I

I1IE

Problem 13.52: Analyze the circuit of Figure P13.52 to determine IC and VCE.

VmAkIIkVV

mAII

Akkk

kmAI

kkIkkI

VkIkIkIkI

VkIkII

IImAK

VI

IIIII

BBECE

BC

B

B

BB

B

BE

BCE

04.61137.0*477.0*47

8.1

0.97.991

413.53.14

7.94447

7.51*1047.03.14

7.4*20147*7.0157.447*

157.4**17.4*7.047*47*

157.4*7.047*

11047.0150

157.0

1

1

11

1

1

1

IB

IC

Problem 13.45: Analyze the circuits shown in Figure P13.45 to determine I and V. For all transistors, assume that β = 100 and |VBE| = 0.7V in both the active and saturation regions. Repeat for β = 300.

2.1878.23

42.4

43.42.2

8.98.9V Since

Incorrect73.152.2*15.7

300for

236.52.2*38.2

8.23390

3.9

3.97.0

100for(a)

maxmax

maxmax

A

mA

I

I

mAk

IV

VkIVmAI

VkIVmAII

Ak

I

VVV

B

C

CC

CBC

B

BBE

Problem 13.45: Contd.

5.124953.0

8.14

8.148.14V since and

Incorrect) ,8.85V give would(

8.85*286I ,300For

533.91*533.9*

33.959533.015

3.14

100For (d)

1

max2max

max2maxmax12

2

22B2

22

2111

A

mA

I

I

ImAIVII

V

mAIIA

VkIVmAIII

IAIIAM

I

B

C

CBC

BC

BC

BBCB

I1

Problem 13.67: Consider the emitter-follower amplifier of Figure P13.67 . Draw the dc circuit and find ICQ. Next, determine the value of rπ. Then, calculate midband values for Av, Avoc, Zin, Ai, G and Z0.

mAII

Ak

IkkI

onefirstthesubtractandbyequationndmultiply

VkIkIkIkI

VkIkIVkIIkI

BCQ

BB

BB

BBE

42.6*

2.64212

6.13156.2810202*

22

3.14101*10*1**17.010*15

1510*20*1510*10*

Analysis DC

11

111

npn

BJTs – Practical Aspects

R

VI

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