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 postive 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.

 

[oswaler]

Hello everyone,

Thank you for sharing your findings on the analogue square root circuit. I find this topic very interesting and would like to provide some insight on the practical output of this circuit.

Firstly, it is important to note that in practical applications, it is common to use a diode or a transistor in the circuit to ensure that the output signal is always positive. This helps to avoid any non-real solutions that may occur when the input signal is positive, as mentioned in the previous figure.

Additionally, the output signal may also be affected by other factors such as noise, temperature, and component tolerances. These factors can lead to some deviation from the expected output, but they can be minimized through careful design and calibration of the circuit.

In terms of the expected practical output signal, it is important to consider the limitations and constraints of the circuit. The output signal may not be a perfect square root of the input signal, but it should still be a close approximation. The accuracy of the output will also depend on the precision of the components used in the circuit.

Overall, while the mathematical concept of the square root may suggest non-real solutions for positive input signals, in practical applications, the output can still be a useful approximation of the square root. Thank you for bringing up this interesting point and I hope my response has provided some insight on the practical output of the analogue square root circuit.

Best regards,