The circuit is driven by an LDR, but the output signal is too noisy. Is there a way hysteresis can clean up the noise in such circuit?

  • 2
    \$\begingroup\$ Circuit diagram would help. Are you looking for an on/off output? \$\endgroup\$
    – Icy
    Commented Jan 6, 2016 at 15:00
  • \$\begingroup\$ Google is your friend. Putting the following string into Google brings up more than 100,000 hits: "non-inverting schmitt trigger" \$\endgroup\$ Commented Jan 6, 2016 at 15:06
  • \$\begingroup\$ I have a programmed micro-controller that will read the change in Vout when there's a dip in the LDR. So, I need a cleaner output signal for the micro-controller. Not necessarily an on/off \$\endgroup\$
    – Mena Labib
    Commented Jan 6, 2016 at 15:07
  • \$\begingroup\$ Then you are looking for a low pass filter not a schmitt trigger \$\endgroup\$
    – Icy
    Commented Jan 6, 2016 at 15:09
  • \$\begingroup\$ How does the word "differentiating" have an impact on this question? Why do you mention "op-amp" when your comments refer to using an MCU? \$\endgroup\$
    – Andy aka
    Commented Jan 6, 2016 at 15:44

1 Answer 1


It is possible to give a differential amplifier hysteresis by adding positive feed-back, and this will create a Schmitt trigger. However: A Schmitt trigger produces a digital on/off signal as its output from an input analogue input - the trigger has hysteresis - that is the output turns on when the input level is at a higher than the trigger level for it to turn off again. As such it is not suitable for your application.

What you are looking for is a low-pass filter. Probably a simple first order filter consisting of a single resistor and capacitor on the output of the amplifier will suffice.


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