The ideal diode model treats the Diode as a perfect switch: a short circuit (zero voltage drop, unlimited current) when forward biased, an open circuit (zero current) when reverse biased. 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).
No middle ground, no 0.7 V drop. Since the state isn’t known in advance, you assume a state, solve the linear circuit, 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, so 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.