Biasing, the Diode Equation and I-V Characteristics

The syllabus marks the diode equation (Derivation). It follows from one idea carried over from the last topic: an applied voltage changes the height of the barrier, and the diffusion current depends exponentially on that height.

Forward bias

Connect the p side to positive, n side to negative.

The applied voltage opposes the built-in field, so the barrier falls from $eV_{bi}$ to $e(V_{bi} - V)$. Three consequences:

• The depletion region narrows — the external supply provides carriers that neutralise some exposed ions.
• Far more majority carriers now have enough energy to climb the reduced barrier. Because the population in the tail is exponential in energy, a modest drop in barrier height produces an enormous rise in current.
• Current grows exponentially with applied voltage.

Reverse bias

Connect the p side to negative.

The applied voltage reinforces the built-in field, so the barrier grows to $e(V_{bi} + V)$.

• The depletion region widens.
• Diffusion current is choked off almost completely.
• What remains is the drift current of minority carriers — thermally generated, swept across as soon as they appear.

This reverse saturation current $I_0$ is tiny, typically nanoamps or less in silicon. Crucially it is independent of the applied voltage: it is limited by how fast minority carriers are generated, not by how hard they are pulled. Doubling the reverse voltage does not double it.

It does depend strongly on temperature, since generation is thermal. A useful rule: $I_0$ roughly doubles for every 10 °C rise.

Deriving the diode equation

At equilibrium the two currents balance. Write the diffusion current at zero bias as $I_{diff}(0)$; it is set by the fraction of carriers able to climb $eV_{bi}$, which by Boltzmann statistics is proportional to

The drift current $I_0$ flows down the barrier and does not care about its height. At equilibrium:

Apply a bias $V$. The barrier becomes $e(V_{bi} - V)$, so the diffusion current is multiplied by the ratio of the new Boltzmann factor to the old:

The drift current is unchanged at $I_0$ and flows the other way. The net current is the difference:

the Shockley diode equation. Often written with the thermal voltage $V_T = k_BT/e = 25.9\ \text{mV}$ at 300 K:

Real diodes carry an ideality factor $\eta$ between 1 and 2 in the exponent, $I = I_0(e^{V/\eta V_T} - 1)$, accounting for recombination inside the depletion region. Take $\eta = 1$ unless told otherwise.

Where the $-1$ comes from and why it matters. It is the drift current, present at every bias. In forward bias beyond about 0.1 V the exponential dwarfs it and it can be dropped. In reverse bias the exponential collapses to nearly zero and the $-1$ is the entire answer, giving $I \to -I_0$. So the term you may ignore in one direction is the only term that survives in the other — which is precisely how one equation describes both halves of the characteristic.

Reading the I-V characteristic

Forward. Almost nothing until a few tenths of a volt, then a sharp knee and a near-vertical rise. The cut-in or knee voltage is about 0.7 V for silicon and 0.3 V for germanium — set by the built-in potential, hence by the band gap.

There is no true threshold. The curve is exponential everywhere; it merely crosses from microamps to milliamps over a narrow range and looks like a threshold on a linear axis. On a logarithmic current axis it is a straight line with no knee at all.

Reverse. A flat line just below zero at $-I_0$, extending until breakdown, where current rises abruptly. Breakdown is not in itself destructive — the Zener diode in Module 4 is designed to operate there. What destroys an ordinary diode is the power dissipated if the current is not limited.