For the network of figure below, if the mid- band voltage gain is (-2.47), answer the following: a. Draw the ac equivalant circuit. b. Calculate Ci and Co. d. Determine fH and fHo e. determine the cutoff frequency. f. Sketch the frequency response for the high-frequency region using a Bode plot. 12 V Cw Cw. 8 pF = 4 pF Cgd 12 pF = 6 pF Ces Cás 3 pF %3D 2.7 k2 10 uF Rig Ipss = 10 mA Vp = -6 V 0.6 k2 10μF RL 4.7 k2 1 M2 0.51 k2 20 µF
For the network of figure below, if the mid- band voltage gain is (-2.47), answer the following: a. Draw the ac equivalant circuit. b. Calculate Ci and Co. d. Determine fH and fHo e. determine the cutoff frequency. f. Sketch the frequency response for the high-frequency region using a Bode plot. 12 V Cw Cw. 8 pF = 4 pF Cgd 12 pF = 6 pF Ces Cás 3 pF %3D 2.7 k2 10 uF Rig Ipss = 10 mA Vp = -6 V 0.6 k2 10μF RL 4.7 k2 1 M2 0.51 k2 20 µF
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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KVL and KCL
KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.
Sign Convention
Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.
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Transcribed Image Text:For the network of figure below, if the mid-
band voltage gain is (-2.47), answer the
following:
a. Draw the ac equivalant circuit.
b. Calculate Ci and Co.
d. Determine fHi and fHo-
e. determine the cutoff frequency.
f. Sketch the frequency response for the
high-frequency region using a Bode plot.
12 V
Cw;
Cw, = 6 pF
Cyd = 8 pF
4 pF
%3D
%3D
Ces
12 pF
%3D
%3D
Cás = 3 pF
2.7 k2
10 uF
Rsis
Ipss
= 10 mA
0.6 k2
Vp = -6 V
10µF
Zo RL24.7 kN
1 M2
0.51 k2
20 μF
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