12 min readConcept sheetSection A MCQ (1 mark) and objective (2 marks, two parts); Section B very short answer (2 marks) on breakdown; Section B long answer (5 marks) on the V-I characteristics with circuit diagrams, often with a choice; also a NEET favourite

Forward and reverse bias: V-I characteristics of a diode

In forward bias a diode conducts in mA once the knee voltage is crossed. In reverse bias only a tiny leakage current flows until breakdown.

Forward biasp-side to + and n-side to − of the battery; depletion region narrows; resistance about 10 to 30 Ω

Knee voltageForward voltage at which the current starts rising sharply: about 0.7 V for Si, 0.3 V for Ge

Reverse biasp-side to − and n-side to +; depletion region widens; tiny reverse saturation current (μA in Ge, nA in Si)

Static and dynamic resistanceStatic R = V/I at a point; dynamic r = ΔV/ΔI on the straight part of the curve

Forward bias and reverse bias

Forward biasp to +, n to −. The battery pushes holes and electrons towards the junction, so the depletion region narrows. Once the bias beats the barrier, majority carriers cross and a forward current in mA flows. The junction offers low resistance, about 10 to 30 Ω.

Reverse biasp to −, n to +. Holes and electrons are pulled away from the junction, so the depletion region widens. Only a few heat-made minority carriers cross. They give a tiny reverse saturation (leakage) current: μA for Ge, nA for Si. It hardly changes with voltage, so the resistance is very high.

So a diode lets current pass easily in one direction only. This one-way action is used in rectifiers.

Breakdown in reverse bias

If the reverse voltage is made very large (a few hundred volts for an ordinary diode), the reverse current suddenly shoots up. This is breakdown, and the voltage is the breakdown voltage. It happens in two ways. In the avalanche effect, the strong field speeds up minority carriers. They hit atoms and knock out more electrons, which knock out still more. In the Zener effect, the field is strong enough to pull electrons straight out of their bonds. The large current and heat at breakdown damage an ordinary diode.

Activity: forward characteristic, static and dynamic resistance

Aim: plot the forward V-I curve of a p-n junction diode and find its static and dynamic resistance. You need a germanium diode (such as OA79), a 2 V cell, a rheostat of about 25 Ω, a 0 to 1.5 V voltmeter, a 0 to 30 mA milliammeter, a key and wires.

1
Note the zero error and least count of both meters.
2
Join the rheostat across the cell as a potential divider. Connect the diode (anode to the + side) and the milliammeter in series from its slider. Put the voltmeter across the diode.
3
Start with the slider at zero. Raise the voltage in steps of about 0.1 V. Record V and I each time, until I is about 25 to 30 mA.
4
Observe: the current stays near zero until the knee voltage (about 0.3 V for Ge), then rises sharply.
5
Plot I (mA) against V (V). At a point P, find the static resistance R = V/I. Repeat at two more points: R changes from point to point.
6★
Take two close points on the straight part, equally spaced either side of P. Find the dynamic resistance r = ΔV/ΔI. It is nearly constant there and much smaller than R. Record the knee voltage, three values of R and three of r.

★ key step

How to draw the circuits and the V-I curve

Forward circuit: battery B, rheostat Rh, protective resistor R, diode D (p-side towards +) and a milliammeter in series; a voltmeter across D.
Reverse circuit: the same, but turn the diode round and use a microammeter in place of the milliammeter, with a higher-voltage battery.
Axes: forward V to the right and forward I (mA) upward; reverse V to the left and reverse I (μA) downward. Write the two current scales.
First quadrant: flat near zero up to the knee voltage (0.7 V Si, 0.3 V Ge), then rising steeply and almost straight.
Third quadrant: a tiny, almost flat reverse current, then a sharp drop at the breakdown voltage. Label it.

Worked example: static and dynamic resistance

Problem

A germanium diode gives these forward readings: 0.4 V, 5 mA; 0.5 V, 12 mA; 0.6 V, 20 mA; 0.7 V, 28 mA. Find (a) the static resistance at 0.4 V and at 0.6 V, and (b) the dynamic resistance between 0.5 V and 0.7 V.

Working

(a) Static R = V/I.
At 0.4 V: R = 0.4 V ÷ 0.005 A = 80 Ω.
At 0.6 V: R = 0.6 V ÷ 0.020 A = 30 Ω. So the static resistance changes along the curve.
(b) ΔV = 0.7 − 0.5 = 0.2 V; ΔI = 28 − 12 = 16 mA = 0.016 A.
Dynamic r = ΔV/ΔI = 0.2 V ÷ 0.016 A = 12.5 Ω.

Answer

(a) 80 Ω and 30 Ω; (b) 12.5 Ω, smaller than the static values

In real life

A diode works like the valve in a bicycle tube: air goes in but not out. Phone chargers use this one-way action to turn AC mains into DC.

Exam angle

5 marks: “A p-n junction diode is to be tested first in forward bias and then in reverse bias. Sketch the two circuits, say which meter changes between them and why, and sketch and explain the V-I curve you expect.” Forward: battery, rheostat, protective resistor, diode, mA meter, voltmeter across the diode. Reverse: diode turned round, and a μA meter replaces the mA meter because the reverse current is only microamperes. Forward curve: near zero up to the knee voltage (0.7 V Si), then a steep rise in mA. Reverse: tiny, nearly constant μA current, then a sharp rise at breakdown. So conduction is one way.

2 marks: “Distinguish between avalanche breakdown and Zener breakdown.” Avalanche: the strong field speeds up minority carriers, which knock out more electrons by collision. Zener: a very strong field in a thin depletion layer pulls electrons directly out of their bonds.

3 marks: “A diode passes 20 mA at 0.6 V and 28 mA at 0.7 V. Find its static resistance at 0.7 V and its dynamic resistance in this range.” Static R = V/I = 0.7 V ÷ 0.028 A = 25 Ω. Dynamic r = ΔV/ΔI = 0.1 V ÷ 0.008 A = 12.5 Ω.

Remember it

Forward: thin and flowing; reverse: wide and waiting (for breakdown). Knee voltage: 0.7 Si, 0.3 Ge.