A full lesson: a cartoon movie to start, then how diffusion builds the depletion region, how drift balances it and leaves a barrier potential, what forward and reverse bias do to the barrier, which carriers really carry the current, a multimeter test you can do on a real diode, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Smell spreads from where it is crowded to where it is not: that is diffusion. Electrons and holes do the same across a p-n junction, until the fixed ions they leave behind build a hill (the barrier potential) that stops them. A battery can lower the hill (forward bias) or raise it (reverse bias).
Dadi lights an agarbatti in the puja room. Nobody blows the smoke, yet within a minute Priya smells it in the next room. Molecules simply wander from where they are crowded to where they are scarce. That is diffusion, and it needs no force at all.
Now picture two colonies next to each other inside one silicon crystal. The n-side is crowded with free electrons; the p-side is crowded with holes. The moment they meet, electrons diffuse into the p-side and holes into the n-side. But every carrier that leaves its home uncovers a fixed ion that cannot move. Those ions build an electric field, a hill, that finally stops the rush. This lesson is about that hill: how it forms, how tall it is, and how a battery can push it down or pile it up.
"Remember two words: diffusion and drift. Diffusion is the crowd spreading out. Drift is the push from the electric field. The p-n junction is the place where these two fight, and at equilibrium they exactly cancel."
"After electrons cross from n to p, which side of the junction is left positively charged?"
The n-side. Each electron that leaves uncovers a fixed positive donor ion. On the p-side, every hole that leaves (or every electron that arrives and fills a hole) leaves a fixed negative acceptor ion. So the n-side of the junction becomes positive, the p-side negative.
Put a drop of ink or a few crystals of potassium permanganate in one corner of a still glass of water. Do not stir. Watch the colour spread from the crowded corner to the clear water over 20 minutes. That slow, forceless spreading is exactly how carriers diffuse across a junction.
A p-n junction is made inside one crystal (for example by diffusing donors into part of a p-type wafer), not by pressing two pieces together. Press Join and watch electrons and holes cross and recombine. The fixed ions they leave behind form a layer with no mobile carriers: the depletion region. Change the doping on each side.
Each carrier diffuses from where it is the majority to where it is the minority.
Donor ions (+) are uncovered on the n-side, acceptor ions (−) on the p-side. They are fixed in the lattice.
NCERT: about one-tenth of a micrometre. Equal charges on both sides, so it reaches deeper into the lightly doped side: N_A x_p = N_D x_n.
"Depletion means emptied. The region is emptied of mobile carriers, but it is full of fixed charges. That is why it acts almost like an insulator sitting between two conductors."
The fixed ions create an electric field pointing from the n-side (+) to the p-side (−). This field pushes electrons back from p to n and holes from n to p: a drift current opposite to the diffusion current. Slide the time after the junction is made and watch the charge, the field, the potential and the two currents grow to equilibrium.
Large at first. It falls as the field grows, because fewer carriers can climb the growing hill.
Starts at zero and grows with the field. Electrons drift p → n, holes n → p.
The potential difference left across the junction is the barrier potential V₀, with the n-side positive. It depends on the dopant density, the material and the temperature.
"Zero net current does not mean nothing is moving. Carriers keep crossing both ways all the time; the two currents just cancel. And V₀ cannot be measured with a voltmeter across an open diode: the contact potentials at the metal ends cancel it."
"An electron on the n-side wants to cross into the p-side. Is it going uphill or downhill in potential energy?"
Uphill. The p-side is at lower potential, and for a negative charge lower potential means higher potential energy. It needs at least eV₀ (about 0.7 eV in Si) of energy to get over. Holes on the p-side face the same hill in the other direction.
Add metal contacts to the two ends and you have a diode. Connect a battery. P to Positive is forward bias: the applied voltage opposes V₀. Reverse the battery and it adds to V₀. Slide the applied voltage (positive = forward) and tap the battery (or the button) to flip it.
The depletion layer gets thinner. Once V approaches V₀ the barrier nearly vanishes and a large current (mA) flows.
The depletion layer widens. Only a tiny current (µA) of minority carriers drifts across.
V₀ = 0.7 V across W = 0.5 µm gives E ≈ 1.4 × 10⁶ V/m: an enormous field, which is why minority carriers are swept across instantly.
"Almost all the battery voltage appears across the depletion layer, because that is the region with no free carriers, the high-resistance part. So the hill height changes directly by V."
Choose the bias and watch the carriers. In forward bias electrons cross into the p-side and holes into the n-side, where each is a minority carrier: this is minority carrier injection. In reverse bias the hill stops the majority carriers, and only the few minority carriers near the junction are swept across by the field.
| Bias | Barrier | Depletion layer | Main current | Size |
|---|---|---|---|---|
| Forward | V₀ − V (lower) | thinner | diffusion of majority carriers → injected minority carriers | mA |
| Reverse | V₀ + V (higher) | wider | drift of minority carriers | µA |
"Forward current: majority carriers go over the hill and become minority carriers on the other side, so it is huge. Reverse current: only the minority carriers roll down the hill, and there are very few of them, so it is tiny. That is the whole secret of one-way conduction."
"If I heat a reverse-biased diode, does the reverse current go up or down?"
Up. Heat breaks more bonds and makes more minority carriers, and the reverse current is just those minority carriers drifting across. It roughly doubles for every 10 °C rise in silicon.
A digital multimeter set to the diode symbol pushes a small current through the part and shows the voltage across it. The red probe is positive. Pick a part and tap the diode (or use the button) to swap the probes.
The silver band on a diode marks the n-side. The arrow in the symbol points from p (anode) to n (cathode): the direction of easy conventional current.
A bridge of four p-n junctions turns ac from the socket into one-way current. That is Lesson 7.
Light makes pairs in the depletion region of a solar cell and the field separates them. In an LED, forward current makes electrons and holes recombine and give light.
Or: the arrow of the diode symbol points the same way as the conventional current from the cell's + terminal.
Depletion width shrinks in forward bias and grows in reverse bias.
From + donor ions to − acceptor ions. The n-side is at the higher potential.
0.5 V across 0.5 µm: 10⁶ V/m. Always convert µm to m.
An electron crossing a 0.7 V barrier loses 0.7 eV of kinetic energy: KE_after = KE_before − eV₀.
Forward current is due to majority carriers (injected as minority); reverse current is drift of minority carriers.
"Four one-liners collect most marks here: the depletion region has only fixed ions; the field points from n to p; forward bias lowers the barrier to V zero minus V; and reverse current is the drift of minority carriers."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
A silicon junction has a barrier potential of 0.7 V and a depletion width of 0.5 µm. Find the average electric field in the depletion region. Which way does it point?
E ≈ V₀ / W with W in metres.
The barrier potential of a p-n junction is 0.4 V. An electron from the n-side approaches the junction with speed 5 × 10⁵ m/s. With what speed does it enter the p-side? (mₑ = 9.1 × 10⁻³¹ kg)
½mv₂² = ½mv₁² − eV₀.
A Si diode has V₀ = 0.7 V. Find the effective barrier when (a) 0.5 V forward bias and (b) 5 V reverse bias is applied. In which case is the depletion layer wider?
Forward subtracts, reverse adds.
For an abrupt junction the depletion width varies as W ∝ √(barrier). A junction with V₀ = 0.7 V has W = 0.2 µm with no bias. Find W at a reverse bias of 6.3 V.
New barrier = 0.7 + 6.3 = 7 V, ten times the old one.
A junction has N_A = 10²² m⁻³ on the p-side and N_D = 4 × 10²² m⁻³ on the n-side. Its depletion width is 0.5 µm. How far does it extend into each side?
Equal and opposite charge: N_A x_p = N_D x_n and x_p + x_n = W.
Take e = 1.6 × 10⁻¹⁹ C and mₑ = 9.1 × 10⁻³¹ kg. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "depletion region contains" | fixed (immobile) ions only | Saying it contains electrons and holes |
| "direction of field at junction" | from n-side to p-side | p to n, because "p is positive" |
| "at equilibrium, unbiased" | I_diff = I_drift, I_net = 0 | Saying both currents are zero |
| "forward biased" / "p connected to +" | barrier V₀ − V, layer thinner | Thinking forward bias widens the layer |
| "reverse current is due to" | drift of minority carriers | Choosing diffusion of majority carriers |
| "barrier V₀ and width d, find field" | E = V₀ / d (d in m) | Leaving d in µm |
| "electron with speed v crosses the junction" | ½mv² − eV₀ | Adding eV₀ instead of subtracting |
| "potential barrier depends on" | doping, material, temperature | Saying it depends on the applied current |
Lesson 6 plots the full V-I characteristic of this diode: the cut-in (knee) voltage where the barrier is overcome in forward bias, the tiny reverse saturation current, breakdown, and dynamic resistance.
Values used: e = 1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10⁻³¹ kg, barrier potential V₀ ≈ 0.7 V (Si) and 0.3 V (Ge), depletion width 0.1–1 µm. Animations show carrier numbers and widths schematically, not to scale.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.