Analogue Square Root: Practical Output Explained

[desmal]

Hi everybody...


http://www3.0zz0.com/2008/07/02/08/264668819.jpg
As shown in the previose figure, the output from the op amp is sequre root the input signal.
Since the feedback of our circuit is sequare the feedback signal ( the output), we shall expect the output to be sequare root for the input signal (since the output function should be the inverse of the feedback function).

What is strange to me is:-
If the input signal is positive we will have non-real solution, this is from the respect of the mathmatical point of view. But what do you expect about the practical output signal? Also, Why do you expect that?

Regards...

 

[rbj]

assuming that the ideal multiplier is only that, when ther input voltage goes positive by a small amount, you will have a situation of positive feedback, not negative feed back. the virtual ground (what you call "0V") won't be 0V. since V₁ and V_o² are both positive, the input to the op-amp will be negative (since it is wired with the "+" terminal to ground) and the op-amp will saturate negatively. it will also get stuck pinned to the negative rail (a little hysteresis) until the input again gets negative enough so that the virtual ground gets back to 0V, and then nice, stable negative feedback will kick in again.

the reason you do not always have negative feedback (which is meant to adjust your output to whatever it has to be to maintain 0V at the op-amp input terminals) is because your squaring circuit is not always an increasing operator. sometimes it's a decreasing operator and when it's a decreasing operator, your feedback turns from negative feedback into positive feedback and the op-amp will race toward the negative rail (since the "-" terminal will be more positive than the "+" terminal) and get stuck there, like a Schmidt trigger.