A p-n junction diode is forward biased when the p-side is joined to the positive terminal of the battery and the n-side to the negative terminal. This external voltage pushes against the built-in barrier, so the barrier potential drops, the depletion region gets thinner, and once the applied voltage crosses the knee (threshold) voltage the diode conducts a large current with very low resistance. Memory hook: "P to Plus, current flows" - p-side to the plus terminal means forward bias and easy conduction.
In forward bias the p-side connects to the battery positive and the n-side to negative. The barrier and depletion region shrink, so a large forward current flows through the diode.
Your doubts, answered
In forward bias, is the p-side connected to the positive or the negative terminal?
The p-side goes to the positive terminal and the n-side to the negative terminal. This is the key rule NEET tests. Because the p-side is at higher potential, the applied voltage opposes the built-in barrier field and helps the majority carriers cross the junction. If you accidentally connect p to negative, that is reverse bias, not forward bias.
Why does the depletion region become thinner in forward bias?
The built-in barrier field points from n to p and keeps carriers away from the junction. In forward bias the battery field points the opposite way (p at higher potential), so it partly cancels the built-in field. With a weaker net field, majority electrons and holes move closer to the junction and neutralise some of the fixed ions. Fewer uncovered ions means a thinner depletion layer and a lower barrier potential.
Why does current flow easily in forward bias but hardly at all in reverse bias?
In forward bias the barrier is lowered, so a large number of majority carriers (electrons from n-side, holes from p-side) get enough energy to cross the junction. This gives a large diffusion current in milliamperes. In reverse bias the barrier is raised, so majority carriers cannot cross; only a tiny current from minority carriers flows (microamperes). So forward resistance is low and reverse resistance is very high.
What is the knee or threshold voltage and why does the diode barely conduct below it?
The knee (threshold) voltage is the forward voltage at which current starts to rise sharply: about 0.7 V for silicon and about 0.3 V for germanium. Below this value the applied voltage has not yet cancelled enough of the barrier, so very few carriers cross and current is almost zero. Above it, the barrier is nearly gone and current increases very fast for a small rise in voltage.
Which charge carriers actually cross the junction in forward bias?
The majority carriers cross the junction: electrons diffuse from the n-side into the p-side and holes diffuse from the p-side into the n-side. This flow is the forward (diffusion) current. This is opposite to reverse bias, where the small current is carried by minority carriers.
Which way does the conventional current flow in a forward-biased diode?
Conventional current flows from the p-side to the n-side inside the diode, which is the direction the arrow in the diode symbol points. So current enters at the p-side (anode, arrow tail) and leaves at the n-side (cathode, the bar). If current would have to flow against the arrow, the diode is reverse biased and blocks it.
⚠️ The NEET trap ✗ Forward bias increases the width of the depletion region and raises the barrier potential. ✓ Forward bias decreases the depletion width and lowers the barrier potential; it is reverse bias that widens the depletion region and raises the barrier. 🧠 Forward = Thinner. Remember F-T: Forward bias makes the depletion layer thinner and the barrier smaller. NEET 2020 asked the opposite (reverse bias widens it) to catch students who mix these up.
Real NEET questions
2016
Consider the junction diode as ideal. The value of current flowing through a forward-biased ideal diode with V_A = +4 V on the p-side, V_B = -6 V on the n-side, and a 1000 ohm resistor in the loop is:
A · A. 0 A
B · B. 10 raised to -2 A ✓
C · C. 10 raised to -1 A
D · D. 10 raised to -3 A
Solution: Step 1: Check bias. The p-side (A) is at +4 V and the n-side (B) at -6 V, so p is at higher potential. The diode is forward biased and, being ideal, acts as a short (zero drop). Step 2: Voltage across the resistor = V_A - V_B = 4 - (-6) = 10 V. Step 3: Current i = V / R = 10 / 1000 = 0.01 A = 10 raised to -2 A. Answer: B.
2017
Which one of the following represents a forward biased diode?
A · A. p-side at higher potential than n-side ✓
B · B. n-side at higher potential than p-side
C · C. both sides at equal potential
D · D. p-side grounded, n-side at higher potential
Solution: Rule: A diode is forward biased when the p-type region is at a higher potential than the n-type region, so the applied field opposes the barrier and the diode conducts. Only the option where the p-side is at higher potential (Option A in the original figure) satisfies this. Answer: A.
2026
Statement A: When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly. Statement B: This current is called reverse saturation current. Choose the correct answer:
A · A. Both A and B are true
B · B. Statement A is true, but Statement B is false ✓
C · C. Both A and B are false
D · D. Statement A is false, but Statement B is true
Solution: Statement A is true: above the threshold (knee) voltage the barrier is nearly cancelled, so forward current rises sharply. Statement B is false: this large forward current is the forward conduction (diffusion) current, not the reverse saturation current. Reverse saturation current is the tiny minority-carrier current in reverse bias. So A true, B false. Answer: B.
What is forward bias of a p-n junction diode in one line?
It is connecting the p-side to the positive terminal and the n-side to the negative terminal, which lowers the barrier, thins the depletion region, and lets a large current flow.
What is the knee voltage for silicon and germanium diodes?
About 0.7 V for a silicon diode and about 0.3 V for a germanium diode. The diode conducts strongly only after the forward voltage crosses this value.
Does forward bias increase or decrease the depletion width?
It decreases the depletion width and lowers the barrier potential. Reverse bias does the opposite.
Is the resistance of a diode high or low in forward bias?
Low. Forward resistance is small (a few ohms to tens of ohms), so a large current flows for a small voltage. Reverse resistance is very high.
Which current dominates in forward bias, diffusion or drift?
Diffusion current dominates. The lowered barrier lets many majority carriers diffuse across the junction, giving a large forward current in milliamperes.
