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