649Microwave Transistors
E
m
, maximum allowable applied voltage V
m
, minimum value of distance
L (L
min
), collector–base capacitance C
o
and maximum current of the
deviceI
m
.
Solution
Given
υ
s
= 5 × 10
5
cm/s, f
T
= 5 GHz, X
C
= 1 Ω and L = 30 μm.
In view of Equation 14.6, V
m
f
τ
= (E
m
υ
s
)/2
π
= 2 × 10
11
V/s for silicon.
E
m
υ
s
/2
π
= 2 × 10
11
V/s or E
m
= 4
π
× 10
11
/5 × 10
5
= 2.51 × 10
6
V/cm
V
m
f
τ
= V
m
× 5 × 10
9
= 2 × 10
11
or V
m
= 2 × 10
11
/5 × 10
9
= 40 V
V
m
= E
m
L
min
or L
min
= V
m
/E
m
= 40/2.51 × 10
6
= 15.92 μm
In view of Equation 14.7, (I
m
X
c
) f
τ
= (E
m
υ
s
)/2
π
= 2 × 10
11
X
c
= 1/2
π
f
τ
C
o
Or
C
o
= 1/2
π
f
τ
X
c
= 1/(2
π
× 5 × 10
9
× 1) = 31.83 pF
(I
m
X
c
) f
τ
= 2 × 10
11
Or
I
m
= 2 × 10
11
/(X
c
f
τ
) = 2 × 10
11
/(5 × 10
9
) = 40 A
EXAMPLE 14.2
A silicon transistor ...