A Novel Integrated Power Module with Solid-State …
17
From Eqs. (26), (25a, 25b), and (18) we have the expression for threshold current
as:
I th | S D R =
θ
2
ω
2 C
2
d R s
3α (1 − cos θ)
(27a)
I th | D D R =
2θ
2
ω
2 C
2
d R s
3α (1 − cos θ)
(27b)
Again, as the DC bias current increases above the oscillation threshold, which
is the practical situation for oscillator operation, we continually adjust the load
conductance g L by using EH tuner placed between the oscillator and the load to
give maximum power output, this means that:
∂ P out
∂g
= 0
(28)
where P out is given by Eq. (24). Using Eq. (24) and (22) in Eq. (28) and resorting to
threshold approximation: (I dc – I th ) << I th we have:
For SDR IMPATT diode:
g L =
3
2
α
θ 2 (1 − cos θ)(I dc − I th )
V
2
r. f
=
4
9
θ
2
α 2
(I dc − I th )
I th
P out =
(1 − cos θ)
3α
(I dc − I th )
2
I th
(29)
For DDR IMPATT diode:
g L =
3
4
α
θ 2 (1 − cos θ)(I dc − I th )
V
2
r. f
=
16
9
θ
2
α 2
(I dc − I th )
I th
P out =
2(1 − cos θ)
3α
(I dc − I th )
2
I th
(30)
17
From Eqs. (26), (25a, 25b), and (18) we have the expression for threshold current
as:
I th | S D R =
θ
2
ω
2 C
2
d R s
3α (1 − cos θ)
(27a)
I th | D D R =
2θ
2
ω
2 C
2
d R s
3α (1 − cos θ)
(27b)
Again, as the DC bias current increases above the oscillation threshold, which
is the practical situation for oscillator operation, we continually adjust the load
conductance g L by using EH tuner placed between the oscillator and the load to
give maximum power output, this means that:
∂ P out
∂g
= 0
(28)
where P out is given by Eq. (24). Using Eq. (24) and (22) in Eq. (28) and resorting to
threshold approximation: (I dc – I th ) << I th we have:
For SDR IMPATT diode:
g L =
3
2
α
θ 2 (1 − cos θ)(I dc − I th )
V
2
r. f
=
4
9
θ
2
α 2
(I dc − I th )
I th
P out =
(1 − cos θ)
3α
(I dc − I th )
2
I th
(29)
For DDR IMPATT diode:
g L =
3
4
α
θ 2 (1 − cos θ)(I dc − I th )
V
2
r. f
=
16
9
θ
2
α 2
(I dc − I th )
I th
P out =
2(1 − cos θ)
3α
(I dc − I th )
2
I th
(30)
