Electronic – Diode in parallel with resistor

diodesresistors

SE,

I've used the SE:EE forum search and several search engines internet and haven't found a single question similar to mine. But I don't think that it's that complicated, I'm just having a hard time grasping what's going on.

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The output from pin 1 is a square wave signal with a peak of 14V. The Scope picture shows the reading at pin 5.

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Why is there a diode parallel to the resistor? What does it do? If I'm putting a square wave through this circuit, why is it coming back as saw tooth?

I'm having a hard time visualizing when the diode opens, what happens to the rest of the circuit and why I'm getting the output at pin 5.

Best Answer

The diode is there to discharge the capacitor quickly.

If the output of U3A is high the diode is reversed polarized, so you can ignore it then. Then C7 gets charged via R22 with a time constant of 4 ms. That means it takes about 20 ms to charge completely.

If the output of U3A is low the diode conducts and discharges C7 very quickly, which you can see as the sharp falling edge on the scope.