Small resistor value for a differentiator

[renzoms]

I’m having a hard time understanding “pick a resistor value that is small enough so dVout/dt << dVin/dt” for a differentiator. I’m graphing 5 V DC supplied to this circuit, and even at 1M ohm and 1pF, the voltage across the capacitor reaches Vin in 10us. I don’t experiment with AC & ”high speed” yet

I get how a small resistor (R2) connected to another resistor (R1) in series, would cause a large portion of the supplied voltage to be across (R1). I’m trying to connect that to this technique and I need help. 🙏 Thank you.

[rstofer]

When they  calculate I, they assume dVin/dt >> dVout/dt and summarily drop dVout/dt from the remainder of the calculation.  What I think they aren't pointing out is that dVin/dt is the result of a driving circuit with a very low impendance compared to R such that the source can charge the capacitor much faster than R can discharge it.

https://www.electronics-tutorials.ws/rc/rc-differentiator.html

This article also doesn't get into the driving impedance but I imagine they are talking about a perfect signal source which would have zero impedance or, more practically, 50 Ohms.  So, if R is 10K, 10K is >> 50 so it should work.  But that assumes that Vout is driving into a circuit with very high impedance.

All of which points to using an op amp on the output to buffer the voltage and provide a low impedance to the next stage.

https://www.electronics-notes.com/articles/analogue_circuits/operational-amplifier-op-amp/analogue-differentiator-circuit.php

In the realm of analog computing, we go out of our way to avoid differentiators because they are high pass filters and allow noise to propagate.  An integrator, by definition, accumulates the noise and averages it out like a low pass filter.

[renzoms]

When they  calculate I, they assume dVin/dt >> dVout/dt and summarily drop dVout/dt from the remainder of the calculation.  What I think they aren't pointing out is that dVin/dt is the result of a driving circuit with a very low impendance compared to R such that the source can charge the capacitor much faster than R can discharge it.

Wow dude. Thanks I really understand more of this information now.