# Analyzing the voltage output of clampers

Let us say that we have the following negative clamper circuit with a 50 volt peak-to-peak square wave as an input:

The output from the circuit is a -50 volt square wave with the peak positive point referenced to 0 volts.

I want an accurate interpretation of the output voltage. As far as I understand at $T_o$, the -25 v appears across the diode and resistor. When the voltage drop across $C$ reaches -25 volts, the value of the output is 0 volts. At $T_1$, the -25 volts input signal and -25 volts signal on the $C$ are in series which explains why we see the -50 volts appearing across the resistor and diode. This is how I've made sense of the output voltage so far.

First, for simplicity, assume the diode is ideal. For an ideal diode, the voltage across cannot be positive (the voltage at the anode is either equal to or less than the voltage at the cathode).

Since the cathode of the diode is connected to the 0V reference, and since the anode is connected to the output node, it follows that the output voltage cannot be positive.

Second, we assume that the time constant $\tau = R_LC$ is much greater than the period of the square wave. If this isn't so, the output voltage will not resemble a square wave.

Then, when the source voltage is +25V, the capacitor very quickly charges through the diode to 25V (note that the left most terminal is 25V and the right most terminal is 0V due to the diode).

When the source voltage changes to -25V, the output voltage is, by KVL, -50V since the capacitor is now in series with the source and load (the diode is 'off').

Because of our 2nd assumption, the capacitor discharges through the load resistor only slightly before the source voltage changes back to 25V.

Thus, the output is approximately the source voltage shifted by -25V. By our 2nd assumption, the capacitor has an approximately constant voltage across of 25V.