Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
Look at which side is at higher potential. A diode is forward biased (ON) when its p-side (the arrow / anode, the flat bar side points away) is at a higher potential than its n-side (cathode, the bar). Trick for circuits: temporarily assume the diode is removed, find the potential the current source would push, and check if that pushes current in the arrow direction. If the battery/source drives conventional current in the direction the diode arrow points, it is forward biased. If it opposes the arrow, it is reverse biased and stays OFF.
Zero. An ideal diode has zero forward voltage (no 0.7 V knee) and zero forward resistance. So when it is ON, both its ends are at the same potential, exactly like a piece of connecting wire. NEET problems say 'ideal diode' or 'assume the diode as ideal' to tell you to use 0 V drop. Only when a problem gives a knee voltage (like 0.7 V for Si or 0.3 V for Ge) do you subtract that.
No. For an ideal diode the reverse resistance is infinite, so the reverse (leakage) current is exactly zero. Treat that branch as a broken wire (open switch). Even if there is a resistor in series with an OFF diode, no current flows through that whole branch, so that resistor is effectively out of the circuit.
Test each diode's branch separately by seeing which way the source pushes current in that branch. The branch whose diode is forward biased conducts; the branch whose diode is reverse biased is open. In the classic NEET two-diode problem, one diode ends up reverse biased (open), so current is forced through the other path only, and you apply Ohm's law to just that path.
It depends on bias, and that is the whole point. Forward biased ideal diode = short circuit (a wire, zero resistance). Reverse biased ideal diode = open circuit (a gap, infinite resistance). It is a one-way switch: it lets current pass in one direction only and blocks it completely in the other.
Consider the junction diode as ideal. The value of current flowing through AB is (a 4 V and a -6 V source feed a 1000 ohm resistor in series with a forward-biased ideal diode from A to B):
The given circuit has two ideal diodes connected as shown. Diode D1 is in one branch and D2 in another, with two 2 ohm resistors, fed by a 10 V source. The current flowing through the resistance R1 will be:
The current I in the circuit shown is (all diodes are ideal and identical), with a 15 V source and diode branches:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It means the diode has zero resistance and zero voltage drop when forward biased (ON), and infinite resistance with zero current when reverse biased (OFF). It behaves as a perfect one-way switch, so you never subtract a knee voltage.
Step 1: Decide the bias of each diode (forward or reverse). Step 2: Replace every forward-biased diode with a plain wire and every reverse-biased diode with an open gap. Step 3: Apply Ohm's law (and series/parallel rules) to the simplified circuit to find current or voltage.
No. Since no current flows through a reverse-biased ideal diode, that entire branch is dead. The series resistor carries zero current and does not affect the rest of the circuit.
A real silicon diode needs about 0.7 V (0.3 V for germanium) before it conducts and has a small forward resistance and a tiny reverse leakage current. An ideal diode conducts with 0 V drop, has zero forward resistance, and zero reverse current.
Because an ideal forward-biased diode has zero forward resistance. By V = IR, with R = 0 the voltage across it is 0 no matter how much current flows, so both terminals sit at the same potential like a wire.
