The circuit parameters for the circuit in Figure 6.3 are V C C = 3.3 V , V B B = 0.850 V , R B = 180 kΩ , and R C = 15 kΩ . The transistor parameters are β = 120 and V B E (on) = 0.7 V . (a) Determine the Q −point values I C Q and V C E Q . (b) Find the small−signal hybrid− π parameters g m and r π . (c) Calculate the small−signal voltage gain. (Ans. (a) I C Q =0 .1mA , V C E Q = 1.8 V ; (b) g m = 3.846 mA/V , r π = 31.2 kΩ ; (c) A υ = − 8.52 ).
The circuit parameters for the circuit in Figure 6.3 are V C C = 3.3 V , V B B = 0.850 V , R B = 180 kΩ , and R C = 15 kΩ . The transistor parameters are β = 120 and V B E (on) = 0.7 V . (a) Determine the Q −point values I C Q and V C E Q . (b) Find the small−signal hybrid− π parameters g m and r π . (c) Calculate the small−signal voltage gain. (Ans. (a) I C Q =0 .1mA , V C E Q = 1.8 V ; (b) g m = 3.846 mA/V , r π = 31.2 kΩ ; (c) A υ = − 8.52 ).
The circuit parameters for the circuit in Figure 6.3 are
V
C
C
=
3.3
V
,
V
B
B
=
0.850
V
,
R
B
=
180
kΩ
, and
R
C
=
15
kΩ
. The transistor parameters are
β
=
120
and
V
B
E
(on)
=
0.7
V
. (a) Determine the Q−point values
I
C
Q
and
V
C
E
Q
. (b) Find the small−signal hybrid−
π
parameters
g
m
and
r
π
. (c) Calculate the small−signal voltage gain. (Ans. (a)
I
C
Q
=0
.1mA
,
V
C
E
Q
=
1.8
V
; (b)
g
m
=
3.846
mA/V
,
r
π
=
31.2
kΩ
; (c)
A
υ
=
−
8.52
).
(a)
Expert Solution
To determine
The quiescent collector current ICQ and the Q -point value VCEQ for the given transistor.
Answer to Problem 6.1EP
The quiescent collector current ICQ is 0.1 mA and Q -point VCEQ is 1.8 V .
Explanation of Solution
Given:
The circuit for common emitter is shown in Figure 1.
The circuit parameters for the transistor circuit shown in Figure 1 are as follows:
VCC=3.3 VRC=15 kΩRB=180 kΩVBE(on)=0.7 VVBB=0.85 V
The value of current gain β is 120 .
Concept used:
The expression for quiescent collector current is written below:
ICQ=βIBQ ...... (1)
The expression for quiescent value VCEQ is written below.
VCEQ=VCC−ICQRC ...... (2)
Calculation:
From DC analysis the ac voltage source is reduced to zero and the equation can be written as,
IBQ=VBB−VB(on)RB ...... (3)
Substitute 0.85 for VBB , 0.7 for VB(on) and 180×103 for RB in equation (3).
IBQ=0.85−0.7180×103=11200 mA
Substitute 120 for β and 11200 for IBQ in equation (1).
ICQ=120⋅11200 mA=110 mA=0.1 mA
Therefore, the quiescent collector current ICQ is 0.1 mA .
Substitute 3.3 for VCC , 0.1×10−3 for ICQ and 15×103 for RC in equation (2).
VCEQ=3.3−(0.1×10−3)(15×103)=3.3−1.5=1.8 V
Therefore, the Q -point VCEQ is 1.8 V .
Conclusion:
Thus, the quiescent collector current ICQ is 0.1 mA and Q -point VCEQ is 1.8 V .
(b)
Expert Solution
To determine
The transconductance gm and the diffusion resistance rπ for small signal analysis.
Answer to Problem 6.1EP
The transconductance gm is 3.846mA/V and the diffusion resistance rπ is 31.2 kΩ .
Explanation of Solution
Concept used:
The expression for transconductance gm is written below.
gm=ICQVT ...... (4)
Here, VT is thermal voltage and its value is 26 mV .
The expression for diffusion resistance is written below.
rπ=βVTICQ ...... (5)
Calculation:
Substitute 0.1×10−3 for ICQ and 26×10−3 for VT in equation (4).
gm=0.1×10−326×10−3=3.846mA/V
Therefore, the transconductance gm is 3.846mA/V .
Substitute 120 for β , 26×10−3 for VT and 0.1×10−3 for ICQ in equation (5).
rπ=120(26×10−3)0.1×10−3=31.2×103Ω
Therefore, the diffusion resistance rπ is 31.2 kΩ .
Conclusion:
Thus, the transconductance gm is 3.846mA/V and the diffusion resistance rπ is 31.2 kΩ .
(c)
Expert Solution
To determine
The value of small signal voltage gain Av .
Answer to Problem 6.1EP
The value of small signal voltage gain Av is −8.52 .
Explanation of Solution
Concept used:
The expression for small signal voltage gain Av is written below.
Av=−(gmRC)⋅(rπrπ+RB) ...... (6)
Calculation:
Substitute 3.846 for gm , 15 for RC , 31.2 for rπ and 180 for RB in equation (6).
Av=−[(3.846)(15)]⋅(31.231.2+180)≈−8.52
Therefore, the voltage gain Av is −8.52 .
Conclusion:
Thus, the value of small signal voltage gain Av is −8.52 .
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6:39
Draw the input waveform and output waveform for the circuit given below with proper values marked in the figure. Assume D1 as
silicon and D2 as germanium diodes. Input Vpp=20V, V1=5 V and V2=9 V.
www
Vin
Vout
Maximum voltage of output waveform
Minimum voltage of output waveform
B: Find the value of RB for the Si transistor circuit as shown in the figure, where the
operating point is exactly at the center of the load line, Vcc = 14V, Rc
RE = 2KN and B=75.
Vcc
Rc
RB
RE
Design a CE voltage-divider configuration with the indicated ac parameters. That is, determine the required values of R1, R2, Rc, and RE. Also, determine the expected output voltage if the input voltage is equal to 0.001 Vrms
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