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Two-Port Negative-Resistance Oscillators


Stanley Chien

Jan 1, 1970
Hi! Everyone,

Currently I am stuck on solving one of oscillator-design example
used from the textbook. The name of the book is Microwave Transistor
Amplifiers Analysis and Design, 2nd edition from Gonzalez.

The following is the example that I try to solve and understand.

Example 5.3.1 (on page 399 for these people who might use the same
Design an 8-GHz GaAs FET oscillator using the reverse-channel
configuration shown in Fig. 5.3.2a (this figure shows that gate port
connects to terminating network, drain connects to load network, and
drain connects to ground). The S parameters of the transistor, in the
reverse-channel configuration, at 8 GHz are

S_11 = 0.98 /_ 163 degree
S_21 = 0.675 /_-161 degree
S_12 = 0.39 /_-54 degree
S_22 = 0.465 /_ 120 degree

Then, the solution shown in the book is

The transistor is potentially unstable at 8 GHz (i.e. K=0.529), and
the stability circle at the gate-to-drain port is shown in Fig. 5.3.2b
(this figure shows the stability circle on the smith chart.).

As shown in Fig. 5.3.2b, any gamma_T (terminating reflection
coefficient) in the shaded region produces |gamma_IN| > 1 (i.e., a
negative resistance at the input port). Selecting gamma_T at point at
1 /_-163 degree, associated impedance is Z_T = -j7.5 ohm. This
reactance can be implemented by an open-circuited 50 ohm line of
length 0.226 lambda. With Z_T connected, the input reflection
coefficient is found to be gamma_IN = 12.8 /_ -16.6 degree (this is
where I have the question. How do they obtain this value?), and the
associated impedance is Z_IN = -58-j2.6 ohm.

From the book, it says we can use the following formula to find input
reflection coefficient if we know the terminating reflection
coefficient (can be calculated from terminating impedance).

gamma_L = 1/gamma_IN = (1 - S_22*gamma_T) / (S_11 - delta*gamma_T)
where delta = S_11*S_22 - S_12*S_21


gamma_IN = S_11 + [(S_12*S_21*gamma_T) / (1-S_22*gamma_T)

However, the gamma_IN that I found by using above two equations is
0.678 /_ 176.76 instead of 12.8 /_ -16.6. Where would I do wrong?
Thank you very much for the help in advance!!

Best regards,
Stanley Chien