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