What is the impedance of the 0.01-μF coupling and bypass capacitors as shown at a frequency of 5 MHz? +VDDRpL100 kΩ10 μΗ100 pF오+VDp0.01 μFM1R3M10.01 μFRGvoRsCs100 k2RGRs100 k20.01 μF500 2(b)(a)Figure 17.62 (a) Tuned amplifier using a depletion-mode MOSFET. (b) dc equivalent circuit for the tuned amplifier in (a).

Answered: +VDD Rp L 100 kΩ 10 μΗ 100 pF 오+VDp…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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What is the impedance of the 0.01-μF coupling and bypass capacitors as shown at a frequency of 5 MHz?

+VDD
Rp
L
100 kΩ
10 μΗ
100 pF
오+VDp
0.01 μF
M1
R3
M1
0.01 μF
RG
vo
Rs
Cs
100 k2
RG
Rs
100 k2
0.01 μF
500 2
(b)
(a)
Figure 17.62 (a) Tuned amplifier using a depletion-mode MOSFET. (b) dc equivalent circuit for the tuned amplifier in (a).

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Step 1

Concept:

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Step 2

Given:

To calculate the coupling and bypass capacitors when the operating frequency and circuit diagram of the amplifier arrangement is given as follows,

Step 3

Formula used:

$X_{C 1} = \frac{1}{2 π f C_{s}} Z_{C 1} = - j X_{C 1}$

Step 4

$T h e r e a c \tan c e v a l u e o f t h e b y p a s s c a p a c i t o r i s g i v e n b y t h e f o r m u l a X_{C 1} = \frac{1}{2 π f C_{s}} S u b s t i t u t i n g t h e c o r r e s p o n d i n g v a l u e, X_{C 1} = \frac{1}{2 π (5 M H z) (0.01 μ F)} X_{C 1} = \frac{1}{0.3175} X_{C 1} = 3.18 o h m$

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