mine the the circuit of Fig. (1) in the ON state. (20 Marks] For the circuit shown in Fig. (2), determine the low cut-off frequencies FLS, FLC, and FLE. Assume that the beta value of the BJT transistor (B) is 120. [20 Marks] For the circuit shown in Fig. (3), determine VG, VGSQ, IDQ, VD, Vs, Vos, and VDG. [20 Marks] 20 V 18 V- $4kQ די $2.5kQ $80kQ 2.1MQ loss-8 mA Vp -4 V 5000 1μF VV2-15 V 1.5kQ Izm = 60 mA 10µF ΣΤΟΚΩ Vi 20ΚΩ 260kQ 2KQ 50μF $2k0 + Fig. (1) Fig. (2) Fig. (3) 04 For the circuit shown in Fig. (4), identify the type of the feedback and determine the gain without feedback (A), the feedback value (B), the gain with feedback (Af), and the voltage gain (Avf). Assume that the beta value of the BJT transistor (B) is 100. [20 Marks] 03 For the circuit shown in Fig. (5), determine the exact value of the total voltage gain (Avt). Assume that the beta value of the BJT transistor (B) is 100. [20 Marks] 0-6 For the circuit shown in Fig. (6), if Vs(t) = 375 x 10-3 +0.1 sin 12560t+ 0.3 cos 18840t, determine the voltage expression across the 15 kQ resistor. 20 V 12 V Ne Ve 250kQ 8=100 → w 10kQ ww 25KQ Vs Vo w 80kQ 9V loss = 8 mA Vp = -6 V 20kQ 20 [20 Marks] + w 50kQ W 15kQ 12V VE Vi 3KQ 3KQ 5600 =9V -20V I 12V 4 60kQ 40ΚΩ | 20ΚΩ www 20ΚΩ ww Flip (4 Fig. (5) Fig. (6) Lecturer Mohammed H. Ibrahi Signature: اللجنة الامتحانية Lecturer Ahmed M. Sana Signature: Head of Dept. Yaseen kh. Yaseen Signature:

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Solve Q/4 

mine the
the circuit of Fig. (1) in the ON state.
(20 Marks]
For the circuit shown in Fig. (2), determine the low cut-off frequencies FLS, FLC, and FLE. Assume
that the beta value of the BJT transistor (B) is 120.
[20 Marks]
For the circuit shown in Fig. (3), determine VG, VGSQ, IDQ, VD, Vs, Vos, and VDG.
[20 Marks]
20 V
18 V-
$4kQ
די
$2.5kQ
$80kQ
2.1MQ
loss-8 mA
Vp -4 V
5000
1μF
VV2-15 V
1.5kQ
Izm = 60 mA
10µF
ΣΤΟΚΩ
Vi
20ΚΩ
260kQ
2KQ 50μF
$2k0
+
Fig. (1)
Fig. (2)
Fig. (3)
04 For the circuit shown in Fig. (4), identify the type of the feedback and determine the gain without
feedback (A), the feedback value (B), the gain with feedback (Af), and the voltage gain (Avf).
Assume that the beta value of the BJT transistor (B) is 100.
[20 Marks]
03 For the circuit shown in Fig. (5), determine the exact value of the total voltage gain (Avt). Assume
that the beta value of the BJT transistor (B) is 100.
[20 Marks]
0-6 For the circuit shown in Fig. (6), if Vs(t) = 375 x 10-3 +0.1 sin 12560t+ 0.3 cos 18840t, determine
the voltage expression across the 15 kQ resistor.
20 V
12 V
Ne
Ve
250kQ
8=100
→
w
10kQ
ww
25KQ
Vs
Vo
w
80kQ
9V
loss = 8 mA
Vp = -6 V
20kQ
20
[20 Marks]
+
w
50kQ
W
15kQ
12V
VE
Vi
3KQ
3KQ 5600
=9V
-20V
I
12V
4
60kQ
40ΚΩ | 20ΚΩ
www
20ΚΩ
ww
Flip (4
Fig. (5)
Fig. (6)
Lecturer
Mohammed H. Ibrahi
Signature:
اللجنة الامتحانية
Lecturer
Ahmed M. Sana
Signature:
Head of Dept.
Yaseen kh. Yaseen
Signature:
Transcribed Image Text:mine the the circuit of Fig. (1) in the ON state. (20 Marks] For the circuit shown in Fig. (2), determine the low cut-off frequencies FLS, FLC, and FLE. Assume that the beta value of the BJT transistor (B) is 120. [20 Marks] For the circuit shown in Fig. (3), determine VG, VGSQ, IDQ, VD, Vs, Vos, and VDG. [20 Marks] 20 V 18 V- $4kQ די $2.5kQ $80kQ 2.1MQ loss-8 mA Vp -4 V 5000 1μF VV2-15 V 1.5kQ Izm = 60 mA 10µF ΣΤΟΚΩ Vi 20ΚΩ 260kQ 2KQ 50μF $2k0 + Fig. (1) Fig. (2) Fig. (3) 04 For the circuit shown in Fig. (4), identify the type of the feedback and determine the gain without feedback (A), the feedback value (B), the gain with feedback (Af), and the voltage gain (Avf). Assume that the beta value of the BJT transistor (B) is 100. [20 Marks] 03 For the circuit shown in Fig. (5), determine the exact value of the total voltage gain (Avt). Assume that the beta value of the BJT transistor (B) is 100. [20 Marks] 0-6 For the circuit shown in Fig. (6), if Vs(t) = 375 x 10-3 +0.1 sin 12560t+ 0.3 cos 18840t, determine the voltage expression across the 15 kQ resistor. 20 V 12 V Ne Ve 250kQ 8=100 → w 10kQ ww 25KQ Vs Vo w 80kQ 9V loss = 8 mA Vp = -6 V 20kQ 20 [20 Marks] + w 50kQ W 15kQ 12V VE Vi 3KQ 3KQ 5600 =9V -20V I 12V 4 60kQ 40ΚΩ | 20ΚΩ www 20ΚΩ ww Flip (4 Fig. (5) Fig. (6) Lecturer Mohammed H. Ibrahi Signature: اللجنة الامتحانية Lecturer Ahmed M. Sana Signature: Head of Dept. Yaseen kh. Yaseen Signature:
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