Why is there a voltage drop across the load resistor in an ideal diode clamper circuit?

Consider a clamper circuit with an ideal diode, a capacitor, a load resistor and an AC input supply. When diode is in forward bias configuration, it must conduct and behave like a short. We must then get zero voltage across load resistor. Then why do we get a voltage drop across the load resistor?

• Circuit diagram? – Dave Tweed Sep 12 '13 at 17:31
• The clamp is used to dissipate energy from an inductive load. If it was pure resistive then no clamping is needed. Current flows through the diode and through the load. The load will have a voltage across it based on its impedance and the current flowing. SO the ideal inductor has voltage across it and the ideal diode has none. Clearly by idealising both load and inductor the situation has been oversimplified. – Russell McMahon Sep 12 '13 at 17:54
• Perhaps reading this would help. en.wikipedia.org/wiki/Clamper_(electronics) – JIm Dearden Sep 12 '13 at 18:16

From your description, I believe this is what you want to say (and be disciplined enough to tell the details properly).

This is on basic diode circuit analysis. For an ideal diode in this configuration, the load resistor has a zero voltage across it when the AC input is in the positive half cycle. Why is this the case? Your load is in parallel with a short.

What happens? You can look at it in two ways. First, electrical current takes the path of least resistance. It will only run through the short. No current means no voltage for an Ohmic resistor (remember Ohm's Law). And second, your load resistor is in a parallel configuration with a short. A voltage across a short is always zero volts. And in a parallel configuration, the voltage must be the same for each branch.

What is the consequence then for the negative half cycle?

Your diode is in a reverse bias configuration. It means your diode behaves like an open. There is a voltage drop across the load resistor because the current passes through the load resistor and not through the diode. simulate this circuit – Schematic created using CircuitLab