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Capacitors act as a short for high frequency signals. If I place a capacitor on a digital 5V PWM line or perhaps a motor encoder line does the cap still act as a short to ground for the square wave signal? My PWM signal is about 50kHz.

Here is the circuit diagramenter image description here

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    \$\begingroup\$ EE.SE has a built-in schematic editor -- just edit your post, position the cursor, and hit control-M to bring it up. The results will be a lot easier for all of us to discuss. \$\endgroup\$ – Dave Tweed Jul 16 '14 at 17:05
  • \$\begingroup\$ ^^^this helps us immensly \$\endgroup\$ – Funkyguy Jul 16 '14 at 20:27
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With a lot of hand-waving you can say that capacitor acts as a short for the AC component of your PWM signal, thus reducing the signal to a constant voltage.

This is however only a crude analogy to what really happens. A capacitor is not an ideal 'short', so the AC component is only attenuated. By how much depends can be calculated from the R and C values, and the frequency.

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It depends on the capacitor value, the output impedance of what is driving it... and in your schematic the value of the series resistor. (I assume that is a resistor.)
I also assume the R to ground is large compared to the other impedances involved. The series R and C to ground make a low pass filter, you just need to calculate it's corner frequency.

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Assumptions:

1) Micro controller input is high impedance 2) R to GND is very large compared to R in series

What this circuit effectively does is to filter the high frequency components of the rising edge of the PWM pulse. To the micro this is in effect a delay at the sample point. That is if you look at the PWM LO-to-Hi Threshold, the micro sees that threshold with some delay.

You have to be careful if the micro input does NOT have hysteresis. It might see multiple transitions at the switching point.

Why are you doing this? Are you reducing the edge rate of the PWM signal because you have signal integrity issues?

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