The Semiconductor Diode: a one-way valve made of silicon
Two slabs of doped silicon touching. That's it — and it changed the century.
Start here: a one-way valve
You know what a one-way valve is. Water goes through it in one direction and is blocked in the other. A bicycle pump has one; so does your heart.
A diode is a one-way valve for electricity. Current passes one way and is stopped the other. That is the entire job, and everything else in this chapter is about how a lump of silicon manages to do it.
How silicon can possibly do that
Pure silicon is a poor conductor — its electrons are all locked into place. But you can change that by mixing in tiny amounts of other elements, and there are two useful ways to do it.
🧪 Two kinds of doctored silicon
n-typeAdd an element with spare electrons. Now there are loose negative charges that can move. Think of a crowded room with a few people standing free of the seats.
p-typeAdd an element that leaves empty spaces where electrons should be. These gaps are called holes, and they behave like loose positive charges. Think of the same room with a few empty seats.
A hole is worth pausing on, because the word sounds strange. When an electron shuffles sideways into an empty seat, the empty seat has effectively moved the other way. Nothing positive is really travelling — but the gap moves, and it moves exactly as a positive charge would. So we track the gap.
Why joining them makes a valve
Press a piece of n-type against a piece of p-type. Right at the join, spare electrons fall into nearby empty seats and both disappear. That leaves a thin strip in the middle with no loose charges at all — an insulating barrier that formed itself.
🔋 Which way you connect it decides everything
Battery + to the p sideCharges are pushed toward the join. The barrier is squeezed thin and collapses. Current flows — this is forward bias.
Battery + to the n sideCharges are pulled away from the join. The barrier grows wider. Almost nothing flows — this is reverse bias.
💡 That is the valve. Not a moving part, not a switch — just a barrier that gets thinner or thicker depending on which way you push. Reverse the battery and the same lump of silicon changes from conductor to insulator.
The 0.7 V you will use in every calculation
Even forward-biased, the barrier does not vanish for free. It takes about 0.7 V to push it out of the way in silicon, and until you supply that, essentially nothing flows.
A silicon diode in series with 1 kΩ, on 10 V:
Ignoring the diode: I = 10 / 1000 = 10.0 mA
Allowing 0.7 V: I = 9.3 / 1000 = 9.3 mA
The diode swallows 0.7 V before the resistor gets any.
💡 That is a 7% error from ignoring it — small enough that the "ideal diode" model is useful for rough work, and large enough that exam questions punish it. If a question says silicon, subtract 0.7 V. If it says germanium, subtract 0.3 V. If it says ideal, subtract nothing.
Take pure silicon and add a trace of phosphorus to one side (giving spare electrons — n-type) and a trace of boron to the other (giving spare holes — p-type). At the boundary something remarkable happens on its own, with no external help: a region forms that allows current one way and blocks it the other. Understanding why makes the V–I curve obvious rather than something to memorise.
The depletion region forms itself
I = I_S(e^(V/ηV_T) − 1) With V = 0.65 V, eta = 1.5, Is = 1 nA.
Step the voltage from 0.60 to 0.66 V — a 10% change — and the current goes from about 5 mA to about 25 mA, five times higher. That steepness is what makes the constant-drop model useful, and what makes a diode fed from a voltage source without a series resistor destroy itself.
Electrons from the n-side diffuse across to fill holes on the p-side. But each electron that leaves exposes a fixed positive donor ion behind it, and each hole filled exposes a fixed negative acceptor ion. Those exposed ions build an electric field that opposes further diffusion — and diffusion stops when the field is strong enough to balance it. What's left is a thin zone stripped of mobile carriers: the depletion region, with a built-in potential across it.
Barrier (built-in) potential at 25 °C:
Silicon ≈ 0.7 V
Germanium ≈ 0.3 V
Temperature dependence:
dV/dT ≈ −2 mV/°C (barrier falls as it warms)
🔀 Three regions of operation
Forward biasP to +, N to −. The applied voltage opposes the built-in field, narrowing the depletion region. Below ~0.7 V almost nothing flows; above it, current rises steeply. Diode ≈ closed switch with a 0.7 V drop.
Reverse biasP to −, N to +. Depletion region widens. Only a tiny saturation current I_S flows (nanoamps in silicon), carried by thermally generated minority carriers. Diode ≈ open switch.
BreakdownPush reverse voltage past V_BR and current suddenly floods through — by avalanche (high V) or Zener tunnelling (low V). Destructive in an ordinary diode; the operating point in a Zener regulator.
The Shockley equation
I = I_S [ e^(V/(η V_T)) − 1 ]
I_S = reverse saturation current
η = ideality factor (1 for Ge, ≈2 for Si)
V_T = thermal voltage = kT/q ≈ 26 mV at 300 K
Key consequences:
· Forward: the exponential dominates → I ≈ I_S·e^(V/ηV_T)
· Reverse: e^(negative) → 0, so I ≈ −I_S
· I_S roughly DOUBLES for every 10 °C rise
V_T = 26 mV at room temperature is worth memorising — it appears in the diode equation, in transistor small-signal models (r_e = 26 mV/I_E), and in dozens of numericals. It comes from kT/q, so it is a genuine physical constant at a given temperature, not an arbitrary figure.
Worked numerical 1 — using the Shockley equation
A silicon diode has I_S = 10 nA and η = 2 at 300 K. Find the current at V = 0.6 V forward, and at V = −5 V.
Forward, V = 0.6 V:
V/(ηV_T) = 0.6/(2 × 0.026) = 0.6/0.052 = 11.54
e^11.54 = 1.03 × 10⁵
I = 10 × 10⁻⁹ × (1.03 × 10⁵ − 1)
= 10 × 10⁻⁹ × 1.03 × 10⁵
= 1.03 × 10⁻³ A = 1.03 mA
Reverse, V = −5 V:
e^(−5/0.052) = e^(−96) ≈ 0
I = 10⁻⁸ × (0 − 1) = −10 nA
Just the saturation current, regardless of how negative V
goes (until breakdown).
The asymmetry in numbers: 1.03 mA forward vs 10 nA reverse
— a ratio of 100 000:1. That is what "one-way valve" means
quantitatively.
Worked numerical 2 — a diode in a circuit
A silicon diode in series with 1 kΩ is connected to 10 V. Find the current using (a) the ideal model, (b) the 0.7 V constant-drop model.
(a) Ideal model (diode = perfect short when forward):
I = 10/1000 = 10 mA
(b) Constant-voltage-drop model (0.7 V):
Applying KVL: 10 = 0.7 + I × 1000
I = (10 − 0.7)/1000 = 9.3/1000 = 9.3 mA
Error from using the ideal model: 7.5%
Power dissipated in the diode: P = 0.7 × 9.3 mA = 6.5 mW
Note: with a 2 V supply instead of 10 V, the difference
matters far more:
ideal: 2 mA real: (2−0.7)/1000 = 1.3 mA (35% error)
The 0.7 V is a bigger fraction of a small supply.
A bridge rectifier is fed from a transformer whose secondary is 24 V RMS, 50 Hz, feeding a 100 Ω load. Find V_m, V_dc, I_dc, ripple factor, output frequency and PIV.
Peak of the secondary:
V_m = √2 × 24 = 33.94 V
Allowing for two conducting diodes in a bridge (2 × 0.7 V):
V_m(load) = 33.94 − 1.4 = 32.54 V
(ideal-diode answer would use 33.94 V)
DC output:
V_dc = 2V_m/π = 2 × 32.54/3.1416 = 20.72 V
I_dc = 20.72/100 = 0.207 A = 207 mA
RMS output:
V_rms = V_m/√2 = 32.54/1.414 = 23.0 V
Ripple factor:
γ = √((23.0/20.72)² − 1) = √(1.232 − 1) = √0.232
= 0.482 ✔ (matches the standard full-wave value)
Output frequency = 2 × 50 = 100 Hz
PIV per diode (bridge) = V_m = 33.94 V
Power in the load: P = I²R = (0.207)² × 100 = 4.28 W
The full-wave output ripples at twice the input frequency, and that is its real advantage. A 100 Hz ripple is far easier to smooth with a capacitor than a 50 Hz one, because the capacitor has only half as long to discharge between peaks. Combined with double the DC output and double the efficiency, it's why almost no practical supply uses half-wave.
💡 Exam angle: two reliable question types. (1) Explain the depletion region and V–I characteristic with a diagram — draw all three regions and label the 0.7 V knee and breakdown. (2) A rectifier numerical: memorise the table (0.318/0.637 for V_dc, 1.21/0.482 for ripple, 40.6%/81.2% for efficiency) and always state whether you included the diode drops. The PIV of a centre-tap full-wave rectifier being 2V_m — unlike a bridge — is a favourite trap.
Syllabus points
V-I characteristics
Diode as rectifier
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