I have an analog signal in the range 0.5 ~ 4.5 V. I'm less interested in the signal's absolute value than I am in its rate of change. Is there a relatively simple passive circuit that I can put between the signal and an ADC that will do this? Is it as simple as throwing in a capacitor? The input range of the ADC is 0 ~ 3 V.

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    \$\begingroup\$ You can use a passive differentiator (R/C) or a better active differentiator using an op amp. (Or you could compute the difference digitally if you have A/D with a microcontroller.) For more details see en.wikipedia.org/wiki/Differentiator \$\endgroup\$ Commented Mar 30, 2017 at 19:06

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Using an ideal op amp (the part number was just the default), the difference between the output and ref will be inversely proportional to the rate of change multiplied by R and C. Multiplying voltage/time times capacitance yields current, and multiplying current by resistance yields voltage.

Depending upon the amplifier, it may be necessary to add some capacitance in parallel with R1, and depending upon the characteristics of what's driving the input it may be necessary to add some resistance in series with C1. Such changes would shift the behavior of the circuit away from reporting a pure derivative, but if the series resistance is small relative to r1 and the parallel capacitance is small relative to C1, they may help make things more stable.


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