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.