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What would be the oscillation frequency?

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Hello maninnet,
our phase plots are the same, its just that my simulator rounded off any phase after 180 deg to start from 0 deg. I have also realised the opamp using multiple VCVS and R C components. I have attached the schematic.
 

i brought this problem to my instructor, he pointed out the trick:

close loop: H(s)/(1+H(s))
oscilliation at: 1+H(s)=0
as you can see from your open loop plot, at 180deg, magnitude of H(s) is not 0dB, so the frequency at 180 deg does not satsify the equation.

the root of this equation is:

f1,2=+/-12498269-722403j, f3=101544964j

f1,2 roughly =+/- 12.5M, which matches your simulation
 

    A

    aryajur

    Points: 2
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Thanks Maninnet,
Problem resolved. In my theoretical evaluation I did find the roots of 1+H(s) and still I could not identify the solution which was staring at my face :D
 

to aryajur:

1. when you solve the roots, did you get the same result? ( i use TI calculator to get it and not confirm with other math program, and I skip all the digit after decimation point)?

2. why f1 and f2 are not complex conjugate?
 

I solved my roots using MATLAB. The results are on Page 1 of 3pole system-2 PDF document I attached earlier. My roots are in rad/sec they are:

P1 = -6.3803*10^8 rad/sec
P2 = (0.0454 + 0.7853 i)*10^8 rad/sec
P3 = (0.0454 - 0.7853 i)*10^8 rad/sec

So I got complex conjugate roots and the imaginary part corresponds to 12.5MHz the oscillation frequency.
 

oh, it it just a math trick, we are both right, i assum f to be real, so i include i in the calculation, and you assume f to be complex and do not include the i, and sorry to mislead dardmand to think the oscillition f is 100MHz
 

The point which the phase changes from -180 to 180
you should run open loop analysis at first & measure the phase margin it should be not less than 60 to have a stable feedback amplifier circuits
best regards,
Rania
 

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