8 min readConcept sheet1-mark MCQ and assertion-reason, 2-mark effect of forward bias on the barrier, 3-mark circuit, I-V curve and dynamic resistance

Junction diode in forward bias

Forward bias joins p to + and n to −, lowering the barrier to V₀ − V. Current rises steeply after the threshold voltage (about 0.7 V for Si).

Forward biasp-side to the positive terminal, n-side to the negative terminal

Effect on the junctionBarrier falls from V₀ to (V₀ − V); the depletion region becomes narrower

Threshold (cut-in) voltageAbout 0.7 V for silicon, 0.2 V for germanium

Dynamic resistancer = ΔV/ΔI, read from the slope of the I-V curve; small in forward bias

What forward bias does

In forward bias, the applied voltage V opposes the barrier potential. The effective barrier becomes V₀ − V, and the depletion region narrows. Majority carriers can now cross: holes go from p to n and electrons from n to p. On the other side they are minority carriers, so this is called minority carrier injection. The total current, usually in milliamperes (mA), is the sum of the hole and electron diffusion currents.

The diode and its symbol

A semiconductor diode is a p-n junction with metal contacts on its two ends.
Its symbol is a triangle pointing at a bar. The triangle side is p, the bar side is n.
The arrow shows the direction of conventional current in forward bias.

How to draw the forward-bias I-V curve

Draw the x-axis for voltage V (volts) and the y-axis for current I (mA).
Keep the curve almost flat near zero until about 0.7 V for silicon (0.2 V for germanium).
After this, draw it rising very steeply; a small increase in V gives a large increase in I.
Mark the point where it turns up as the threshold (cut-in) voltage.
Dynamic resistance at any point is ΔV/ΔI on the steep part of the curve.

Worked example: dynamic resistance

Problem

For a silicon diode, the forward current is 10 mA at 0.70 V and 30 mA at 0.80 V. Find its dynamic resistance in this range.

Working

ΔV = 0.80 V − 0.70 V = 0.10 V.
ΔI = 30 mA − 10 mA = 20 mA = 0.020 A.
Formula: r = ΔV/ΔI = 0.10 V ÷ 0.020 A = 5 Ω.

Answer

r = 5 Ω

Worked example: current in a simple diode circuit

Problem

A silicon diode is forward biased by a 3.7 V cell through a 300 Ω resistor in series. Taking the voltage drop across the conducting diode as 0.7 V, find the current.

Working

Voltage across the resistor = 3.7 V − 0.7 V = 3.0 V.
Formula: I = V/R.
I = 3.0 V ÷ 300 Ω = 0.01 A = 10 mA.

Answer

I = 10 mA

In real life

A red LED indicator on a TV or a charger is a diode in forward bias. It lights only when current flows from its p-side to its n-side, which is why an LED put in the wrong way round stays dark.

Exam angle

2 marks: “What happens to the width of the depletion region and to the barrier height when a p-n junction is forward biased?” The applied voltage opposes the barrier, so the barrier falls to V₀ − V and the depletion region becomes narrower.

3 marks: “Draw the circuit used to study the forward-bias characteristic of a p-n junction diode, and draw the curve you expect. Mark the threshold voltage.” Battery, rheostat as potential divider, diode with p to +, milliammeter in series, voltmeter across the diode. Curve flat to about 0.7 V, then rising steeply.

2 marks: “From a forward I-V curve, the current changes from 10 mA to 30 mA as the voltage changes from 0.7 V to 0.8 V. Find the dynamic resistance.” r = ΔV/ΔI = 0.1 V ÷ 0.02 A = 5 Ω.

Remember it

P to Positive, then Pass: forward bias lets current pass once V crosses about 0.7 V (Si).