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Microwave Engineering
book

Microwave Engineering

by Ahmad Shahid Khan
May 2014
Intermediate to advanced content levelIntermediate to advanced
800 pages
24h 50m
English
CRC Press
Content preview from Microwave Engineering
654 Microwave Engineering
Solution
Given N
a
= 2 × 10
23
m
3
,
ε
r
= 13.1,
ε
ir
= 4, d = 0.02 μm and T = 300 K.
a. The surface potential (
ψ
i
) for strong inversion is given by the
relation
ψ
i
= V
T
ln|N
a
/n
i
|, where V
T
(= 26 mV at 300 K)
= 26 × 10
3
ln|(2 × 10
23
/1.5 × 10
10
| = 0.786 V
b. The insulator capacitance per unit area is given by the relation
C
i
=
ε
ir
/d = 4 × 8.854 × 10 12/2 × 10 8 = 1.77 mF/m
2
c. The threshold voltage is given by Equation 14.52b
V
th
= 2
ψ
i
+ (2/C
i
) (
ε
s
q N
a
ψ
i
)
1/2
= 2 × 0.786 + {2/(1.77 × 10
3
)} (8.854 × 10
12
` × 13.1 × 1.6 × 10
19
× 2 × 1023 × 0.786)
1/2
= 3.5 V
EXAMPLE 14.10
Calculate the (a) insulator capacitance, (b) saturation drain current, (c)
transconductance in saturation region and (d) maxim
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Publisher Resources

ISBN: 9781466591417