Why a diode lets current through one way only: the p–n junction

How to use this simulation

1. Start with silicon at a bias of 0.8 V. The diode is forward biased: the depletion layer is narrow, red holes and blue electrons stream across the junction, and the dot on the I–V graph sits high on the steep part of the curve.
2. Drag the bias down to 0 V. The depletion layer widens to 0.40 µm, the barrier rises to 0.7 V, the carriers stop streaming and the current falls to almost nothing.
3. Keep going into negative values. The layer grows wider still (1.14 µm at −5 V), yet the current stays almost flat at a tiny leakage value. This is reverse bias, the closed direction.
4. Push the slider to its negative end. Beyond −4.5 V, inside the red shaded zone, the current suddenly rushes the other way. That is breakdown.
5. Switch the semiconductor to germanium. The curve slides left: the knee moves from about 0.70 V to about 0.30 V, and the dashed curve shows the silicon one for comparison.
6. Open the half-wave rectifier view. A sine wave goes in, only the positive humps come out, each one a little lower than the input because the diode keeps back its own knee voltage. Change the peak voltage and read the average (DC) output.

p-type, n-type and the junction in the middle

Pure silicon or germanium is a poor conductor. Add a trace of a few atoms from a different group and it changes completely. This process is called doping. If the added atoms have one electron too few, each one leaves a vacancy called a hole, and the crystal is p-type, because the moving carriers are positive holes. If the added atoms have one electron too many, spare electrons wander about, and the crystal is n-type.

Neither piece is charged overall. The p-type block has as many negative ions as free holes. The n-type block has as many positive ions as free electrons. The ions are locked into the crystal, but the holes and electrons are free to move.

Now join the two blocks into a single crystal. The surface where they meet is the p–n junction, and a diode is little more than that. Everything interesting happens in a layer only a fraction of a micrometre wide.

How the depletion layer forms

At the moment of joining, electrons are crowded on the n side and almost absent on the p side, so they diffuse across the junction. In the same way, holes diffuse from p to n. When an electron meets a hole they recombine and both disappear as free carriers.

That leaves a strip on each side of the junction with no free carriers at all, only fixed ions: negative ions on the p side, positive ions on the n side. This strip is the depletion layer. The charges of the ions create an electric field pointing from n to p, which pushes electrons back to the n side and holes back to the p side. Diffusion stops when this field exactly balances it.

The potential difference across the layer at that point is the built-in potential, or barrier potential. For silicon it is about 0.7 V and for germanium about 0.3 V. You cannot measure it with a voltmeter across the diode, because the metal contacts have their own potentials that cancel it exactly. But it decides how much voltage you must apply before current flows.

Forward bias: opening the gate

Connect the positive terminal of a battery to the p side and the negative terminal to the n side. This is forward bias. The battery pushes holes towards the junction from one side and electrons from the other, so they squeeze the depletion layer thinner and the barrier shrinks to V_bi − V.

At first almost nothing happens, because the barrier is still high. Then, as the bias approaches the knee voltage, a large number of carriers have enough energy to climb over the barrier, and the current climbs very steeply. In the simulation the red and blue carriers start to cross the junction, and the more current, the more carriers you see.

With the width formula, a bias of 0.6 V on silicon gives W = 0.40 × √((0.7 − 0.6) ÷ 0.7) = 0.40 × 0.38 = 0.15 µm. The layer has shrunk to about a third of its zero-bias width.

Reverse bias: closing the gate

Swap the battery. Now the positive terminal is on the n side. The battery pulls holes and electrons away from the junction, so the depletion layer grows wider and the barrier rises to V_bi + |V|. Majority carriers cannot cross anything this high.

With silicon at −5 V the width is 0.40 × √((0.7 + 5) ÷ 0.7) = 0.40 × 2.85 = 1.14 µm, about 2.9 times the zero-bias value. Germanium, with its lower built-in potential, widens even more, to 1.68 µm.

A tiny current still flows. Heat keeps creating a few electron–hole pairs near the junction, and the field sweeps these minority carriers across. This is the reverse saturation current I_s. It barely depends on the voltage, but it depends strongly on temperature, and it is much larger in germanium than in silicon: 10 nA against 2 fA, a ratio of about 5,000,000.

Breakdown: when the gate bursts open

If the reverse voltage keeps growing, the field in the thin layer becomes enormous. Two things can happen. In avalanche breakdown, a fast carrier knocks more carriers loose, which knock out still more, and the current multiplies. In Zener breakdown (in heavily doped diodes with a very thin layer) the field itself pulls electrons straight out of the bonds.

In the animation the breakdown voltage is 4.5 V. Go past it with the slider and the carriers flow backwards at full speed; at −5 V the current is -16.7 mA. The curve becomes almost vertical, so the voltage across the diode stays nearly constant while the current changes a lot. A Zener diode is built to work exactly there and hold a steady voltage. An ordinary rectifier diode must never be pushed that far.

The Shockley equation and the I–V curve

Physicists summarise the whole forward and reverse behaviour in one short equation, named after William Shockley. I is the diode current, V the voltage across the junction, I_s the reverse saturation current, V_T the thermal voltage and n the ideality factor, between 1 and 2 (the animation uses 1).

The thermal voltage is V_T = kT ÷ q. At room temperature (300 K) it is 0.0259 V, or 25.9 mV. It is the voltage scale on which the exponential changes: each extra 25.9 mV multiplies the current by e ≈ 2.718, so each extra 0.1 V multiplies it by about 47.5.

• For V large and positive, the exponential wins and I ≈ I_s e^(V/(nV_T)): a very steep rise.
• For V = 0 the two terms cancel and I = 0, as it must be with no battery.
• For V negative by a few V_T or more, the exponential vanishes and I ≈ −I_s, the small flat reverse current.

Silicon against germanium

Silicon wins almost everywhere today. It has a far smaller leakage current, works at higher temperatures, and makes a much better insulating oxide, which helps manufacturing. Germanium turns on at a lower voltage, which can matter for weak signals such as an old AM radio detector.

The half-wave rectifier

Mains electricity is alternating: the voltage swings positive and negative. Most electronics need direct current. A diode in series with a load fixes this. During the positive half of each cycle the diode is forward biased and conducts, so the load sees the voltage. During the negative half the diode is reverse biased and blocks, so the load sees almost zero.

The result is a pulsing DC that never goes negative. It is not smooth yet, but it only flows one way. Adding a capacitor across the load will smooth it later; this is the first step of every phone charger.

Because the diode keeps back its knee voltage, the output peak is a little lower than the input peak. With a 5 V peak and a 100 Ω load, a silicon diode gives an output peak of 3.83 V, while germanium gives 4.19 V. Both lose the knee voltage plus a little more across the diode’s own resistance.

Worked example 1: how touchy the exponential is

A silicon diode with I_s = 2 fA is at 300 K, so V_T = 0.0259 V. Find the current at 0.6 V and 0.7 V.

At 0.6 V: I = I_s e^(0.6 ÷ 0.0259) = 23.0 µA. At 0.7 V: I = 1.09 mA. A change of just 0.1 V multiplied the current by 47.5.

Go to 0.8 V and the pure equation gives 51.9 mA. In a real diode the series resistance of the silicon and the contacts limits that, which is why the graph tops out sensibly. It also shows why the voltage across a conducting silicon diode seems to sit at about 0.7 V whatever the current: to change the current ten times you only need 59.6 mV more.

Worked example 2: a diode in series with a resistor

A 5 V battery drives a silicon diode in series with a 1 kΩ resistor, forward biased. Find the current.

Use the knee as the diode’s drop: 0.7 V. The resistor gets the rest, 5 − 0.7 = 4.3 V. By Ohm’s law, I = 4.3 V ÷ 1 kΩ = 4.30 mA.

The power in the resistor is I²R = 4.30² × 1 = 18.5 mW, and the diode itself dissipates 0.7 × 4.30 = 3.01 mW.

Replace the silicon diode by germanium and the drop falls to 0.3 V, so the current rises to 4.70 mA. Reverse the battery and the current is only the leakage value, -2.00 fA. For practical purposes it is zero.

At a current of 10 mA, the dynamic resistance of an ideal junction is r = V_T ÷ I = 0.0259 ÷ 0.01 = 2.59 Ω. The steeper the curve, the smaller r.

Worked example 3: reading the rectifier

Feed a 5 V peak sine wave through the silicon diode into a 100 Ω load, then check the readings in the rectifier view.

Output peak: 3.83 V, about 1.17 V below the input peak. Average (DC) output: 1.08 V, whereas an ideal diode would give 5 ÷ π = 1.59 V. The diode conducts for 45% of each cycle, a bit less than half, because it needs the knee voltage first.

Now drop the peak to 1 V. The output peak becomes 0.25 V and the average 0.04 V, and the diode is on for only 20% of the cycle. A small signal loses a large share to the knee. With germanium at 1 V, the output peak is 0.60 V, much better. This is why germanium diodes were used as radio detectors.

Rules of thumb for exams

Keep these in your pocket for multiple-choice questions and short numericals.

• Forward bias: p to the positive terminal, n to the negative. The depletion layer narrows, the barrier falls, current is large.
• Reverse bias: the opposite. The layer widens, the barrier rises, only tiny leakage current flows.
• Knee voltage: 0.70 V for silicon, 0.30 V for germanium (the usual textbook values are about 0.7 V and 0.3 V).
• The depletion width follows √(V_bi − V). It shrinks forward and grows in reverse, but never as fast as the current changes.
• In a half-wave rectifier the output frequency equals the input frequency, and the average output is V_peak ÷ π.
• The diode symbol arrow points from p to n, the direction of conventional current.

Common mistakes

These cost the most marks.

• Thinking no current flows in reverse bias. A small leakage current always flows, and breakdown makes it large.
• Saying the depletion layer is empty of everything. It is empty of free carriers, but full of fixed ions and a strong electric field.
• Forgetting that the diode needs about 0.7 V (silicon) before it conducts well. A 0.5 V source cannot drive a silicon diode.
• Mixing up the polarity: forward bias needs the p side connected to the positive terminal.
• Using the output peak as the DC value. The average of a half-wave signal is far below the peak.

Revision list

Say these out loud before the exam.

• Doping makes p-type (holes) and n-type (electrons). Both are electrically neutral.
• At the junction, diffusion leaves a depletion layer of fixed ions and a built-in field.
• Forward bias narrows the layer and lets current flow above the knee voltage.
• Reverse bias widens the layer and passes only the tiny saturation current, until breakdown.
• I = I_s (e^(V/(nV_T)) − 1), with V_T ≈ 0.0259 V at 300 K.
• A half-wave rectifier passes one half of every cycle and its DC average is V_peak ÷ π.

Frequently asked questions