The diode and its symbol
A semiconductor diode, as NCERT §14.6 states, is basically a p-n junction provided with metallic contacts at both ends so that an external voltage can be applied across it. It is a two-terminal device. The symbol is a triangle (the p-side, or anode) pointing into a bar (the n-side, or cathode). The direction of the arrow marks the conventional direction of current when the diode is forward biased — current flows easily in the direction the arrow points, and is choked off in the opposite direction.
At equilibrium, with no external voltage, the junction already has a built-in potential $V_0$ across its depletion region. Applying an external voltage $V$ alters this barrier. The whole study of the I–V characteristic is really the study of how the current responds as we change $V$ in each direction. Because the depletion region has very high resistance compared with the bulk p- and n-regions, almost all of the applied voltage drops across the junction itself.
Forward bias: barrier lowered
A diode is forward biased when the p-side is connected to the positive terminal of the battery and the n-side to the negative terminal. The applied voltage opposes the built-in potential, so the effective barrier height drops to $(V_0 - V)$ and the depletion layer narrows. When $V$ is small the barrier falls only slightly, so only a handful of carriers in the uppermost energy levels can cross — the current is negligible. Increase $V$ further and far more carriers acquire the energy to cross, so the current climbs sharply.
The carriers that cross become minority carriers on the far side — this is minority carrier injection. Electrons injected into the p-side and holes injected into the n-side set up concentration gradients and diffuse away from the junction. The total forward current is the sum of the hole-diffusion and electron-diffusion contributions, and its magnitude is typically in milliamperes. NIOS §28.4 notes that the forward-biased junction offers a low resistance, in the range of about 10 Ω to 30 Ω.
Reverse bias: barrier raised
A diode is reverse biased when the n-side is made positive and the p-side negative. Now the applied voltage acts in the same direction as the barrier potential, so the effective barrier rises to $(V_0 + V)$ and the depletion region widens. This suppresses the diffusion of majority carriers, so the diffusion current collapses compared with the forward case.
A small current still flows. The junction field sweeps any minority carriers that wander near it across to the other side — a drift current of the order of a few microamperes. Crucially, this current is limited by the supply of minority carriers, not by the applied voltage: even a small reverse voltage sweeps them all across. That is why the reverse current is almost independent of the applied voltage and stays nearly constant. NCERT confirms that the increase in depletion-region width is associated with reverse bias.
This voltage-independent reverse current persists only up to a critical reverse voltage called the breakdown voltage $V_{br}$. At $V = V_{br}$ the reverse current rises sharply, with a large change in current for a slight rise in bias. Breakdown does not by itself destroy the diode; the diode is destroyed only if the current is not limited by an external circuit and exceeds the rated value, causing overheating.
| Feature | Forward bias | Reverse bias |
|---|---|---|
| Connection | p-side to (+), n-side to (−) | n-side to (+), p-side to (−) |
| Effective barrier | $(V_0 - V)$ — lowered | $(V_0 + V)$ — raised |
| Depletion width | Decreases | Increases |
| Main current | Diffusion (minority injection) | Drift of minority carriers |
| Magnitude of current | mA (large) | µA / nA (very small) |
| Junction resistance | Low (~10–30 Ω) | Very high (megaohms) |
| Meter used | Milliammeter | Microammeter |
The I–V characteristic curve
To record the characteristic, the diode is connected to a battery through a potentiometer (or rheostat) so the applied voltage can be varied, and the current is read for each voltage. Because forward currents are large, a milliammeter is used in the forward branch, while a microammeter is used in the reverse branch. Plotting $I$ against $V$ — forward voltages positive, reverse voltages negative — gives the familiar single curve below.
In the forward region the current first increases very slowly, almost negligibly, until $V$ crosses the characteristic threshold; beyond it the current rises almost exponentially for a very small increase in voltage. In the reverse region the current is tiny and nearly constant (the reverse saturation current) until, at very high reverse bias, breakdown sets in and the reverse current shoots up.
The shape of this curve underlies the diode equation $I = I_0\left(e^{\,qV/kT} - 1\right)$, where $I_0$ is the reverse saturation current. For positive $V$ the exponential dominates and current grows; for negative $V$ the exponential term vanishes and $I \to -I_0$, the constant reverse value. The same equation explains why temperature changes the whole characteristic, since both $I_0$ and the exponent depend on $T$.
The forward/reverse behaviour here grows directly out of how the barrier forms. Revisit p-n Junction Formation to see where $V_0$ and the depletion region come from.
Knee & saturation numbers
The forward voltage at which conduction begins in earnest is the threshold voltage, also called the cut-in or knee voltage. NCERT gives it as about 0.2 V for a germanium diode and about 0.7 V for a silicon diode; NIOS quotes the corresponding knee values as about 0.3 V for Ge and about 0.7 V for Si. Below this voltage the characteristic effectively hugs the axis, which is why the forward curve appears to meet the voltage axis at the knee rather than passing through the origin.
| Quantity | Silicon | Germanium |
|---|---|---|
| Threshold / knee voltage | ≈ 0.7 V | ≈ 0.2–0.3 V |
| Forward current scale | mA (30–80 mA near 1 V) | mA |
| Reverse saturation current | µA to nA, nearly constant | µA to nA, nearly constant |
The reverse saturation current is the small, almost voltage-independent current carried by minority drift across the junction. In most commercial diodes it is of the order of microamperes (and can be nanoamperes in silicon). General-purpose diodes are operated only within this reverse saturation region — they are not used beyond it, since pushing into breakdown without current limiting risks destruction.
Four facts that examiners flip on you
The diode I–V topic is dense with near-miss distractors. Keep these straight:
- Bias polarity: forward = p-side to (+), n-side to (−). The p-side (arrowhead) must be at the higher potential to conduct.
- Knee voltage: ~0.7 V for Si, ~0.2–0.3 V for Ge — not 0.1–0.3 V for Si. NEET 2021 marked "barrier 0.1–0.3 V" as the wrong statement.
- Reverse current: a few µA/nA and nearly constant with voltage — carried by minority carriers, not majority carriers.
- Dynamic resistance: $r_d = \dfrac{\Delta V}{\Delta I}$ (slope-based), not the static $V/I$.
Forward = low resistance, large mA current after the knee. Reverse = high resistance, tiny constant µA current until breakdown.
Dynamic resistance
Because the I–V curve is not a straight line, a diode does not have a single fixed resistance. NCERT defines the dynamic resistance as the ratio of a small change in voltage to the small change in current it produces:
$$r_d = \frac{\Delta V}{\Delta I}$$This is the inverse slope of the characteristic at the chosen operating point. Where the forward curve is steep (just past the knee) a small $\Delta V$ produces a large $\Delta I$, so $r_d$ is small — a few ohms to tens of ohms. In the reverse region the curve is almost flat, so a large $\Delta V$ gives a minuscule $\Delta I$, and $r_d$ runs into megaohms. The forward dynamic resistance is therefore far smaller than the reverse dynamic resistance, which is the quantitative form of "a diode conducts one way."
Worked example (NCERT 14.4)
The V–I characteristic of a silicon diode is given. Calculate the resistance of the diode at (a) $I_D = 15$ mA and (b) $V_D = -10$ V.
(a) Forward dynamic resistance. Treating the characteristic as a straight line between $I = 10$ mA and $I = 20$ mA: at $I = 20$ mA, $V = 0.8$ V; at $I = 10$ mA, $V = 0.7$ V. So
$$r_{fb} = \frac{\Delta V}{\Delta I} = \frac{0.8 - 0.7}{(20 - 10)\times 10^{-3}} = \frac{0.1\,\text{V}}{10\,\text{mA}} = 10\ \Omega$$(b) Reverse resistance. From the curve, at $V = -10$ V the reverse current is about $I = -1\ \mu\text{A}$. So
$$r_{rb} = \frac{10\,\text{V}}{1\,\mu\text{A}} = 1.0 \times 10^{7}\ \Omega$$The reverse resistance is roughly a million times the forward resistance — the numerical signature of one-way conduction.
A one-way device
Pulling the threads together: a p-n junction diode primarily allows current in one direction only. Its forward resistance is low, its reverse resistance is very high, and the transition between the two is governed by the knee voltage. NIOS uses a vivid image — a diode is a one-way turnstile for electrons, letting them through from the n-region to the p-end in forward bias but blocking them otherwise.
This single property is exactly what makes the diode useful: feed it an alternating voltage and current flows only during the half-cycles in which it is forward biased. That is the basis of rectification, taken up in the next subtopic.
See how the I–V curve turns ac into dc in Junction Diode as a Rectifier.
Semiconductor diode I–V in one screen
- A diode is a p-n junction; the symbol's arrow points the way of forward current (p-side → n-side).
- Forward bias (p to +, n to −): barrier $(V_0 - V)$ lowered, depletion narrows, large mA diffusion current after the knee.
- Reverse bias (n to +, p to −): barrier $(V_0 + V)$ raised, depletion widens, tiny µA/nA reverse saturation current, nearly constant.
- Knee/threshold voltage ≈ 0.7 V (Si), ≈ 0.2–0.3 V (Ge); the forward curve meets the V-axis at the knee.
- At breakdown voltage $V_{br}$ the reverse current rises sharply; the diode survives only if current is externally limited.
- Dynamic resistance $r_d = \Delta V/\Delta I$: low in forward (ohms), very high in reverse (megaohms).
- Diode equation $I = I_0(e^{qV/kT}-1)$ — temperature changes the whole V–I curve.