Lecture13 BJT Transistor Circuit Analysis (2)
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Transcript of Lecture13 BJT Transistor Circuit Analysis (2)
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: 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
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:
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