The bias voltages of the circuit shown in Figure P7.67 are changed to V + = 3 V and V − = − 3 V . The input resistances are R i = 4 kΩ and R G = 200 kΩ . The transistor parameters are K P =0 .5mA/V 2 , V T P = − 0.5 V , λ = 0 , C g s = 0.8 pF , and C g d = 0.08 pF . (a) Design the circuit such that I D Q =0 .5mA and V S D Q = 2 V . (b) Find the midband voltage gain. (c) Determine the equivalent Miller capacitance. (d) Find the upper 3dB frequency.
The bias voltages of the circuit shown in Figure P7.67 are changed to V + = 3 V and V − = − 3 V . The input resistances are R i = 4 kΩ and R G = 200 kΩ . The transistor parameters are K P =0 .5mA/V 2 , V T P = − 0.5 V , λ = 0 , C g s = 0.8 pF , and C g d = 0.08 pF . (a) Design the circuit such that I D Q =0 .5mA and V S D Q = 2 V . (b) Find the midband voltage gain. (c) Determine the equivalent Miller capacitance. (d) Find the upper 3dB frequency.
Solution Summary: The author explains the design parameters of the circuit. Input resistances are given by, lR_i=4kOmega
The bias voltages of the circuit shown in Figure P7.67 are changed to
V
+
=
3
V
and
V
−
=
−
3
V
. The input resistances are
R
i
=
4
kΩ
and
R
G
=
200
kΩ
. The transistor parameters are
K
P
=0
.5mA/V
2
,
V
T
P
=
−
0.5
V
,
λ
=
0
,
C
g
s
=
0.8
pF
, and
C
g
d
=
0.08
pF
. (a) Design the circuit such that
I
D
Q
=0
.5mA
and
V
S
D
Q
=
2
V
. (b) Find the midband voltage gain. (c) Determine the equivalent Miller capacitance. (d) Find the upper 3dB frequency.
Use Gauss-Jordan Elimination method to solve the following system:
4x1+5x2 + x3 = 2
x1-2x2-3x3 = 7
3x1 x2 2x3 = 1.
-
3. As the audio frequency of Fig. 11-7 goes down, what components of Fig.
12-4 must be modified for normal operation?
OD
C₂ 100
HF
R₁ 300
Re 300
ww
100A
R
8
Voc
Rz
10k
reset
output 3
R7
8
Voc
3
reset
output
Z
discharge
VR₁
5k
2
trigger
2 trigger
7
discharge
R 3
1k
5
control
voltage
threshold 6
5 control
voltage
6
threshold
GND
Rs
2k
C.
C.
100
GND
Uz LM555 1
Ce
0.01
U, LM555
0.01
8.01.4
PRO
Fig. 11-7
Audio lutput
Pulse width modulator
R4 1k
ww
C7
Re 1k
ww
R7 100
VR
50k
10μ
Ra
R10
C₁.
R1
3.9k
3.9k
0.14 100k
TO
w
Rs 51
82
3
H
10
Carrier
U₁
Ca
Input
A741
2.2
Us
MC1496
PWM signal
input
R2
0.1100k
Uz
A741
41
Cs
1
Re
10k
VR2
50k
VR3
100k
14
12
C3.
3% +
Ce
0.1
10μ
5
1A
HH
C
+12V
0.1
O PWM
Output
C
0.02-
R
100k +12 V
Demodulated
output
6
Ca
0.33
w
R
10k
R12
100k
ww 31
о
+
4A741
-12 V
Fig. 12-4 PWM demodulator
C
1500p
DUC
1. In Fig. 12-4, what are the functions of the VR1 and VR2?
2. In Fig. 12-4, what is the function of the VR3?
VR₁
50k
C₁ R1
0.1 100k
Carrier
Input
U₁
A741
PWM signal
input
R41k
www
Re 1k
w
C7 ±
10μT
R7 100
ww
=L
H
C4
2.2
H
W82
Rs 51
3
10
U3
MC1496
C2
R2
U2
A741
22
0.1 100k
VR2
50k
VR3
100kr
14
C3
10μ
1k
0.1
4
5
6
12
m
Re
10k
R9 R102
3.9k 3.9k
HHI
C10
0.1
-0
+12V
C11
R
0.02 100k +12 V
Demodulated
output
C
R11
R12
A741
0.33 10k
100k
-12 V
Ca
1μ
C12
1500p
PRODUC
Fig. 12-4 PWM demodulator
PRODUCTS
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