I have a simple circuit that drive a LED:


simulate this circuit – Schematic created using CircuitLab

The input PWM is created as follow:

  • base frequency: 200 kHz
  • PWM duty-cycle: 0..100% (to change light intensity)
  • low freq modulation: 10 Hz..10 kHz

I mean, the low frequency modulation turn on and off the PWM signal. I'm looking for a way to filter out the 200 kHz base frequency, leaving the low frequency modulation there.

In other works, the PWM duty-cycle should be "levelled" ( = filtered) to obtain a constant DC voltage to drive the LED when it's active. I tried to put a capacitor in parallel to the led but I don't know if it's the best place to put it.

Increasing the capacity will filter a bit the 200 kHz but also "levels-out" the low-frequency modulation.

Is that position the best one? Is the x20 PWM base frequency enough to easily filter that and only that?

  • \$\begingroup\$ Start over with specs. What is your LED power and voltage? What is your best matching supply spec? What control input-output limits or specs? duty cycle vs power, Show the specs 1st not the fuzzy questions and arbitrary R values. What efficiency do you expect? Did you know that choke can store energy and now you have a SMPS current regulator? \$\endgroup\$ Commented Nov 19, 2017 at 15:18
  • \$\begingroup\$ Efficiency is not important. The LED maximum current is 100 mA DC, its Vf is about 2.5V. The R values are not arbitrary, are the actual ones I see on the board. +V is 3.3V. \$\endgroup\$
    – Mark
    Commented Nov 19, 2017 at 15:21
  • \$\begingroup\$ What LED spec? what board? \$\endgroup\$ Commented Nov 19, 2017 at 15:22
  • \$\begingroup\$ Right now is a white one, but it might change (and the R will change according to that). Would you mind to explain me why the color of the LED is important to filter out the 200 kHz PWM frequency? \$\endgroup\$
    – Mark
    Commented Nov 19, 2017 at 15:22
  • 1
    \$\begingroup\$ Trev. you need a cap with that L to make a current filter \$\endgroup\$ Commented Nov 19, 2017 at 19:18

1 Answer 1


Establish a low-pass=filter at 10% of the chopping frequency. Thus 20KHz.

A filter at 20KHz has TAU of 1/(2*PI*20KHx) or 50,000 microseconds/2*PI = 9 uSeconds.

You already have the R at 15 ohms. Simply place a 0.47uF cap from FET output to ground. Or any larger cap. This forms a parallel RC low pass filter.


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