Electronic – Analyzing RC-circuit with diode

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I'm trying to make an explanation for the behavior of the following RC-circuit with a diode pulled up to 12V. The input signal is a square wave alternating from -7V to 12V. My intuition is that the diode pulls up the output to 12V – Vff, and then reflects the input signal at the output offset by that voltage, so that the output is a square wave from 11.4V to 30.4V.

However, I have a hard time solving this mathematically, so help would be appreciated.

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

Well the lowest voltage on the output cannot be less than 11.3 volts (one diode drop from 12V) and given that your 19 volt p-p square wave attaches itself to the output when the lowest point hits 11.3 volts, the final output has to range from 11.3 volts to 19 volts higher at 30.3 volts.

Virtually same answer as you - just a difference in what I assumed the forward volt drop of the diode to be. Given that the square wave is 50 MHz the 100k pull down resistor might reduce the peak output by around a milli volt or so.

Also no problem with the capacitor at 50MHz - it'll sail straight through. Given that the diode is really fast you won't get much of an issue with the diode's reverse recovery time response.