Oscillation Explanation using S-parameters

[mdocampo1]

Im designing a microwave oscillator using an S2P file. Because of this linear model, I cannot do any nonlinear analysis, so I'm stucked with the S-parameter output.

How can I know if my oscillator is oscillating?

My adviser told me that if S11>1, my oscillator is now oscillating.

What is the reason behind this? Thanks

 

[RCinFLA]

It only means it has the ability to oscillate. Actual oscillation will depend on tank circuit losses.

 

[BigBoss]

This is Negative Resistance Principle Oscillation.When you look through input port of your oscillator, the real part must be negative because of positive feedback.
Tank circuit will always have an input impedance with positive real part because of losses and these two parts will cancel each other and reactive parts will be added algebratically and the circuit will have
RHS poles to oscillate.
But you should carefully read some books and papers about this principle to find the right oscillator topology.

 

[tony_lth]

In AWR or ADS, there are many examples on oscillators.

 

[darcyrandall2004]

Using the linear model, to determine if I have oscillations, I plot the closed loop input reflection coefficient S11 using a special terminator. I think it is called OSCPORT in the ADS liobraries. Plot S11 on a polar plot. If magnitude of S11 is greater than one and the phase = 0 then we have oscillations.

 

[biff44]

S11 magnitude >1 does not necessarily guarantee oscillation. For example, you can have an IMPATT diode that has S11>1 and it can make a very nice amplifier--no oscillations.

To oscillate, you have to have a resonator circuit attached the active device. The resonator has a loss (due to Q, and due to coupling to get some energy out of the output connector). If the loss is higher than the amount of reflection coefficient feedback, then the oscillations can not build up.

A good way to think of it is to convert your active device's S11 into an impedance. The impedance will have a real and imaginary part. The real part will be negative if S11 > 1. So if you a real part of say -20 ohms, and your load is <20 ohms real part, you have ONE condition of oscillation.

A 2nd condition of oscillation is that the feedback has to be positive. In terms of impedances once again, the imaginary part of your active device's impedance X, should be equal to but negative in sign to the resonators imaginary part of impedance -X. This is what determines exactly WHAT frequency the oscillator will oscillate at.

 

[jayce3390]

That's a very good explanation !
Thank you

 

[mdocampo1]

@biff44

yes sir i designed my oscillator using these conditions that you said.

The impedance from s11 of my active device has a negative real part. The real part of my resonator impedance is negative 1/3 of the input impedance of my active device, so it is positive. [1/3 rule of thumb]

The imaginary part of the input impedance of my active device is equal but opposite sign of the imaginary part of the resonator impedance.

I just wanted to know how can I know if there are oscillations at the output port by only looking at the s11 of the whole oscillator?

 

[Scrts]

The basic theory is like this (I will use T as greek capital gama as reflection coefficient), the signal, which has amplitude U_signal goes to its destination and reflects with amplitude U_signal*T_dest, then goes back to the source and reflects with the amplitude U_signal*T_dest*T_source and does the same as before. If the T_dest and T_source is >1, which means that amplitude every period gets U_signal*(T_dest*T_source)^n, this causes oscillation. So if You understood me correctly, U_dest=|S22|>1 and U_source=|S11|>1 in partial cases, when the unstable part of Smith chart includes the required termination center.