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BIPOLAR JUNCTION TRANSISTOR AMPLIFIER
(AC ANALYSIS)
1. Refer to Figure 1, given RE=0.6kΩ, RC=5.6kΩ, β=120, VBE(on)=0.7V, R1=250kΩ,
R2=75kΩ.
a)
For VA=∞, determine small-signal voltage gain, Av
b)
Determine the input resistance looking into the base of the transistor.
(Ans: Av= -8.27 and Rib=80.1kΩ)
Figure 1
Draw Thevenin equivalent circuit. Refer to Amplifier dc equation slide in BJT Commonemitter lecture note. Label all the current and voltages for the circuit.
We know that,
I CQ   I BQ
I EQ  (1   ) I BQ
The given circuit has RE connected at emitter terminal. Apply KVL to base-emitter loop,
we will get:
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VTH  I BQ RTH  V BE ( on)  I EQ R E  0
VTH  I BQ RTH  V BE ( on)  (1   ) I BQ R E  0
I BQ RTH  (1   ) R E   VTH  V BE ( on)
I BQ 
VTH  V BE ( on)
RTH  (1   ) R E
If we need to find VCEQ,
Apply KVL around collector-emitter loop for common-emitter circuit yield to,
VCC  I CQ RC  VCEQ  I EQ RE  0
VCEQ  VCC  I CQ RC  I EQ R E
 VCC   I BQ RC  (1   ) I BQ R E
 VCC  I BQ  RC  (1   ) R E 
Thevenin resistance, RTH and thevenin voltage, VTH are given as below:
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2. Let β=100, VBE(on)=0.7V and VA=100V.
Determine:
a) Small-signal voltage gain
b) Input resistance seen by the signal source, Rin
c) Output resistance looking back into the output terminal, Ro
(Ans: ICQ = 0.439mA, Av= -148, Rin=6.09kΩ, Ro = 9.58kΩ)
Figure 2
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3. Assume that β=100, VA=∞, R1=10kΩ and R2=50kΩ for the circuit in Figure 3.
a. Plot the Q-point on dc load line
b. Determine the small-signal voltage gain
(Ans: ICQ=1.19mA, VECQ = 8.42V and Av= -1.94)
Figure 3
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For small-signal equivalent circuit shown above, make sure label all the current, voltages
and Ri and Rib.
100(0.026)
 2.18k
1.19
VO  g mV RC
Write KVL from input around B-E loop,
V

V S  V     g mV  R E
 r

r 
given g m r   ,
V 
 VS
1  
 R E
1  
 r 
Voltage gain, Av
 RC
Av 
 1.94
r  (1   ) RE
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4. Refer to Figure 4, given β=180, VA=∞.
a. Find ICQ and VCEQ
b. Plot dc and ac load lines
c. Calculate small-signal voltage gain.
(Ans: ICQ= 15.6mA, VCEQ= 10.1V, rπ=0.30kΩ, Rib=34.2kΩ and Av= 0.806)
Figure 4
RTH  R1 R2  5k
 R2  
 10 
(V  V  )  V   
VTH  
18  9  0
 10  10 
 R 1  R2 
V  V BE  V 
0  0.7  9
I BQ  TH

 0.0869mA
RTH  (1   ) R E 5  181(0.5)
I CQ  15.6mA , I EQ  (1   ) R E  15.7 mA
VCEQ  18  (15.7)( 0.5)  10.1V
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