Author Topic: Differentiator output  (Read 1165 times)

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Offline renzomsTopic starter

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Differentiator output
« on: October 23, 2019, 10:02:47 pm »
So I’ve wrapped my head around Differentiator. I understand how a small R will produce allow Vc to be ~Vin, for DC. And adding a load will mean that the current across R (not labeled but there) will decrease, and increase Vc. So there is: an output proportional to the rate of change of input.

The output proportional to the rate of change of input is dVc/dt in I = CdVc/dt. Here the proportion is times C. Attaching a load, it would be a fraction of I (aCdVc/dt) and then the voltage would be Rload * aCdVc/dt. Correct? And now, I imagine there are mA to uA and a resistor of 100k to get some voltage drop. Thinking quickly to get help, I’d love to know the uses of an output proportional to the rate of change of input and if this is how that output is used (i.e. Rload * a*CdVc/dt).

Kind of a tangent: a while ago I considered using an integrator to produce an LED that glows brighter.
« Last Edit: October 26, 2019, 03:15:49 am by renzoms »
 

Offline rstofer

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Re: Differentiator output
« Reply #1 on: October 24, 2019, 01:52:52 am »
Differentiators are avoided, but sometimes used, for analog computing.

I used to see clock pulses created by differentiating a square wave and keeping just the positive going edge.  This was done a long time ago when the square wave was used to gate a signal and the edge was used to latch it.

Otherwise, these differentiators are nothing more than high pass filters.

It remains to be seen if the derivative of a sine wave is equal to a cosine wave.  i think there will be some fiddling around to find the proper values.

 
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Offline pwlps

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Re: Differentiator output
« Reply #2 on: October 24, 2019, 01:00:34 pm »
Attaching a load, it would be a fraction of I (aCdVc/dt) and then the voltage across it would be Rload * aCdVc/dt. Correct?

Just replace R by R||Rload in the calculations. So actually your "a" will be a=R/(R+Rload).
 
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Offline rstofer

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Re: Differentiator output
« Reply #3 on: October 24, 2019, 04:40:51 pm »
In the case of differentiating a sine wave, we expect a cosine.  So, I grabbed some values out of thin air and stuffed them into LTspice.  Output attached.

Because there is no initial condition, the very beginning of the waveform is not correct.  By 90 degrees, everything is pretty close to correct.  The values aren't exactly right but close enough to show the effect.

ETA:  As a consequence of requiring dVin/dt to be much larger than dVout/dt, we can assume that the output voltage will be considerably less than the input voltage because the dt in both cases is the same.  Therefore, dVin must also be much greater than dVout.

ETA(again):  A value of 10k is probably a better match.  In order to get 90 degrees of phase shift, the input frequency must be MUCH less than the 3dB cuttoff frequency of the high pass filter.

https://www.electronics-tutorials.ws/filter/filter_3.html
« Last Edit: October 24, 2019, 05:46:44 pm by rstofer »
 
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