High Order Harmonic Oscillators in Microwave and Millimeter-wave ...

2 Oscillators
Recall from circuit analysis that if the ith node of the circuit is grounded, so that V i = 0,
the matrix of (2.3) will be modified by eliminating the ith row and column, reducing the
modified by adding the corresponding rows and columns.
2.2.1 Oscillators Using a FET
Next consider an oscillator using an FET in a common gate configuration. In this case
V 1 = 0, and again V 3 = V 4 provides the feedback path. For an FET the input admittance
can be neglected, so we set G i = 0. Then the matrix of (2.3) reduces to
[
(Y1 + Y 2 + g m + G 0 ) −(Y 2 + G 0 )
−(G 0 + g m + Y 2 ) (Y 2 + Y 3 + G 0 )
] [
V2
V
]
= 0, (2.4)
where V = V 3 = V 4 . Again we assume the feedback network is composed of loss less
reactive elements, so that Y 1 ,Y 2 and Y 3 can be replaced with their susceptances. Setting
the determinant of (2.4) to zero then gives
∣
(g m + G 0 ) + j(B 1 + B 2 ) −G 0 − jB 2
−(G 0 + g m ) − jB 2 G 0 + j(B 2 + B 3 )
= 0. (2.5)
∣
Equating the real and imaginary parts to zero gives two equations:
1
+ 1 + 1 B 1 B 2 B 3
= 0, (2.6)
G 0
+ g m
+ G 0
B 3 B 1 B 1
= 0. (2.7)
As before, let X 1 ,X 2 , and X 3 be the reciprocals of the corresponding susceptances. Then
(2.6) can be rewritten as
X 1 + X 2 + X 3 = 0. (2.8)
Using (2.6) to eliminate B 3 from (2.7) reduces that equation to
X 2
X 1
= g m
G 0
. (2.9)
Since g m and G 0 are positive, (2.9) shows that X 1 and X 2 must have the same sign,
while (2.8) indicates that X 3 must have the opposite sign. If X 1 and X 2 are chosen to be
negative, then these elements will be capacitive and X 3 will be inductive. This conclusion
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