Bipolar Junction
Transistor Circuit Analysis
EE314
BJT Transistor Circuit Analysis
1.Large signal DC analysis
2.Small signal equivalent
3.Amplifiers
Chapter 13: Bipolar
Junction Transistors
Circuit with BJTs
Our approach: Operating point - dc operating point
Analysis of the signals - the signals to be amplified
Circuit is divided into: model for large-signal dc analysis of BJT circuit
bias circuits for BJT amplifier
small-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
VCB
reverse bias
VBE
forward bias
?
?
The constrains for IB and VCE must be satisfy to keep BJT in the
active-mode
Large-Signal dc Analysis: Saturation-Region Model
VCB
forward bias
VBE
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 in
the active region.
Bias point – it is important to select IC, IB, VBE, and VCE
independent of the b and operation temperature.
Example 13.4, 13.5, 13.6
Large-Signal dc Analysis: Bias Circuit
From Example 13.6
VBB acts as a short
circuit for ac signals
Remember: that the Q point should be independent of the b
(stability issue)
VBB & VCC provide this stability, however this impractical solution
Other approach is necessary to solve this problem-resistor network
Large-Signal dc Analysis: Four-Resistor Bias Circuit
Solution of the bias problem:
1
VB  RB I B  VBE  RE I E
I E  (b  1I B
3
VBE  0.7V
VB  VBE
IB 
RB  (b  1RE
Thevenin
equivalent
4
2
Equivalent
circuit for
active-region
model
Input
Output
RB  R1 R2 VB  VCC (R2 / (R1  R2 
VCE  VCC  RC I C  RE I E
Small-Signal Equivalent Circuit
iB  I BQ  ib (t ) 
Small signal equivalent circuit
for BJT:
Thevenin
equivalent
  vBEQ  vbe (t ) 

 (1   I ES exp 
VT

 
  vbe (t ) 

 I BQ exp 
  VT 
exp( x)  1  x,
 vbe (t ) 

I BQ  ib (t )  I BQ 1 
VT 

so
ib (t )  I BQ
vbe (t ) vbe (t )

VT
r
and
VT
r 
I BQ
Common Emitter Amplifier
First perform DC analysis to find
small-signal equivalent
parameters at the operating point.
Find voltage gain:
Find input impedance:
Common Emitter Amplifier
Find current gain
Find power 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.

 VBE  
V 
 - 1  I ESexp  BE 
I E = I ES  exp 

 VT  
 VT 

 0.7 
 V 
10mA = I ESexp 
 and 1mA = I ESexp  BE 
 0.026 
 0.026 
 0.7 - VBE 
divide both sides  10 = exp 

 0.026 
 0.026 * ln 10  0.7  VBE
VBE  0.7  0.026 * ln 10  0.64V 
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.
I B1  I B 2  I C  1mA & I C  b I B
2 
1mA
 I C   1  1mA  I C 
 0.98mA
1.02
b

 1
I E  1   I C  0.99mA
 b
 VBE 
 we have
sin ce I E  I ES exp 
 VT 
I
VBE  VT ln E  0.026 * ln 0.99 *1011  0.658V
I ES
(

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.
14.3
 10 A
I C1  bI1  1mA
1.43M
(15  (I E 2 *1k  0.7   I  I E 2
C1
10k
b 1
14.3 I E 2
IE2

 1mA 
10k 10
101
1
 1
0.43mA  I E 2 * 
 
 101 10 
I E 2  3.9126mA  I C 2  0.99 I E 2  3.8735mA
I1 
I1
IE2
VBE
VCE 2  15  1k * (I C 2  I E 2   15  1.99k * I E 2  7.213V

I 
VCE1  15  10k * 1mA  C 2   4.6126V
b 

Problem 13.52:
Analyze the circuit of Figure P13.52 to determine IC and VCE.
I
I1 
IC
I1
I C  bI B
I  I1  I E
(0.7  15V
 0.1047mA
I E  (b  1I B
150 K
(I1  I B * 47k  0.7  I * 4.7k  15V
I1 * 47k  I B * 47k  0.7  I1 * 4.7k  (b  1* I B * 4.7k  15V
IB
IE
I1 * (47k  4.7k   15  0.7  I B * (47k  201* 4.7k 
14.3  0.1047mA * 51.7k 14.3  5.413
 9.0A

991.7k
47k  944.7 k
I C  bI B  1.8mA
IB 
VCE  VBE  47k * (I1  I B   0.7  47k * 0.1137mA  6.04V
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.
(a) for b  100
VBE  0.7  VB  9.3V
9.3
 23.8A
390k
I C  bI B  2.38mA  V  I C * 2.2k  5.236 V
IB 
for b  300
I C  7.15mA  V  I C * 2.2k  15.73V
Since Vmax  9.8V
 b max 
 I C max 
(Incorrect 
9.8
 4.43mA
2.2k
I max 4.42mA

 187.2
IB
23.8A
Problem 13.45: Contd.
(d) For b  100
14.3
I B1 
 0.9533A
15M
I  I C 2  I B 2 * b  9.533mA
I C1  bI B1  95.33A  I B 2

V  I *1k  9.533 V
For b  300, I B2  286 A

I C 2  I B 2 * b  85.8mA
( would give V  85.8V , Incorrect)
I C 2  b 2 I B1
b max 
and since Vmax  14.8V
I C 2 max
14.8mA

 124.5
I B1
0.953A
I max  14.8mA  I C 2 max
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.
I1
DC Analysis
I1 *10k  (I1  I BE *10k  15 V

15  I1 *10k  0.7  (1  b * I B *1k 
I1 * 20k  I B *10k  15 V
I1 *10k  I B *101k  14.3 V
multiply 2nd equation by 2 and subtract the first one
I B * (202k  10k   28.6  15
I CQ  I B * b  6.42mA
IB 
13.6
 64.2A
212k
BJTs – Practical Aspects
npn
V
I
R
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