The ideal diode model treats the Diode as a perfect switch: a short circuit (zero voltage drop, unlimited current) when forward biased, and an open circuit (zero current) when reverse biased. There is no threshold voltage at all — the diode turns on the instant goes positive.
The two states
The model has exactly two states, decided by the polarity of the would-be current or voltage:
- On (forward): the diode is a wire. , and is whatever the rest of the circuit forces through it.
- Off (reverse): the diode is a break in the circuit. , and is whatever reverse voltage the rest of the circuit applies (it stays negative).
There is no middle ground and no 0.7 V drop. Because the state is not known in advance, the standard procedure is to assume a state, solve the linear circuit, and then check the assumption: if you assumed “on”, the current must come out positive (forward); if you assumed “off”, the voltage across the diode must come out reverse-biased. If the check fails, you assumed wrong — flip the state and re-solve.
When to use it
The ideal model is valid whenever the circuit voltages are much larger than the real diode drop, so that neglecting 0.7 V introduces a small error. If you are rectifying a 170 V mains peak, the 0.7 V the diode actually drops is a 0.4 % effect and is fairly ignored. If you are working with a 1 V signal, dropping 0.7 V is most of the signal and the ideal model is useless — use the Constant-voltage-drop model instead.
It is the right tool for understanding the topology of rectifiers and switching circuits. The Half-wave rectifier, for instance, is cleanest to explain with an ideal diode first: positive half-cycles pass straight through, negative ones are blocked, and the transfer characteristic is exactly when on and when off — no diode drop cluttering the picture. The 0.7 V correction is then layered on with the Constant-voltage-drop model.