RC circuit and opamp used as a comparator.

[DBoulanger]

Hi every one,

I've seen quite a few methods here and there allowing us to estimate the value of a given capacitor with a somehow basic RC low-pass or hi-pass filter circuit.

In one example I found, there is a suggestion to vary the frequency such that the voltage accross the capacitor is half of the source.

Under these circumtances, it would be legitimate to conclude that the voltage accross the resistor and capacitor are the same.  This is where the comparator comes in the picture.  How interface the opamp with the RC circuit in such a way that when the equilibrium is reached, the output of the opamp is "flat".

Obviously, due to the nature of the opamps, a perfect match of Vc and Vr is pratically impossible, but if that could be done within few millivolts, I think that would be acceptable.

So ideally, a led would lit up if the frequency is too high and another one would be lit if the frequency needs to be increased.  Having both lit or off would be acceptable, indicating that we're within the difference of few millivolts we've preset.

This is potentially not something too complicated, however I'm somehow uncertain where to start.

BTW, when the equilibrium is achieved, the formula sqrt(3)/(2*pi*f*r) would provide us with an estimated value of C.

This little circuit/project is simply to get more familiar with opamps and RC circuits and at the same time, give me the opportunity to create something that could be usefull, who knows.

Your help would be greatly appreciated.

[Paul Price]

BTW, when the cutoff freq. of a r-c filter (-3dB attenuation)  is reached, the voltage across the capacitor is .707 of the input voltage, not .500 as you state, when the applied signal is a sinewave.


If a dual op-amp follower is sniffing the voltage across the capacitor to be measured, the output of this op-amp could be compared to the second op-amp fed by divide by 2 voltage of the input voltage and then by rectifying both outputs  a comparator can then show the above/below/match of the two rectified voltages with LED indicators.

[DBoulanger]

Thanks Paul for the information.

I understand clearly that at the cutoff frequency, the amplitude attenuation will be of 3db, therefore 1/sqrt(2) or 0.707 for short.

The formula I'm using is not using the cutoff frequency as such, but the frequency at which an attenuation of 6db is achieved, or if you prefer, when Vc is half of Vs.

Based on the circuit description you provide, I will attempt to create a prototype.  Meantime, if you or someone else could actually draw the circuit you describe, that would be great.

Thanks for your help.

P.S.: Keep in mind that I'm fairly new to that gymnastic, so somethig that may seem obvious for some, might not be so natural for others.