0
\$\begingroup\$

I am building a proportional-Differentiator circuit based non-inverting OP-amp. But I am confused with the stability condition of OP-amp.

enter image description here

The system will be unstable if the loop gain |T(s)|=|a(s)B(s)|>1 at the -180 phase shift frequency (Is this thought wrong?). Here is the example from TI's cookbook "Differentiator Circuit (Rev. C)" (sboa276)

enter image description here

The frequency range is set from 100/3.5 to 2.5k*3.5.(28.6 to 8.75k). The loop gain |T(s)|(dB) can be obtained from the difference between |a(s)|(dB) and |1/B(s)|(dB). And It can be seen from the bode plot that the phase shifts -180 over the range from 286 to 875, and the corresponding loop gain is greater than 1. enter image description here

The cookbook only considered the GBP should be sufficiently large to assure that the crossover frequency of a(s) and B(s) exceeds the upper bound of differentiator.

enter image description here

How to understand the stability of differentiator correctly?

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

No - as far as I can see, the loop gain phase never reaches the 180deg limit. However, it is very close to it.

This can also be observed in the magnitude response because the positive slope of the 1/B curve is reduced only a little before crossing the gain magnitude a(s). This is the point where the loop gain T is 0 dB. The red line (PD behaviour) claerly shows the stability limit.

As a consequence, the capacitor should be increased to shift the corresponding zero of the 1/B response to a smaller frequency.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.