Diode equation

The diode equation (also called the Shockley equation) is the exponential current–voltage relation of a forward-biased PN junction:

$i_D=I_S\left(e^{v_D/V_T}-1\right)\approx I_S{e}^{v_D/V_T}\text{for }{v}_D\gg V_T$

$i_D$ is the current through the Diode in the forward (anode→cathode) direction, $v_D$ is the voltage across it (anode minus cathode), $I_S$ is the Reverse saturation current, and $V_T$ is the Thermal voltage. This is the $n=1$ form: the full Shockley equation carries an ideality factor $n$ as $i_D=I_S(e^{v_D/n{V}_T}-1)$, with $n\approx 1$ for an ideal junction and rising toward $\approx 2$ when recombination dominates. This course takes $n=1$ throughout.

Reading the equation

The $-1$ term is the contribution of the reverse saturation. In reverse bias $v_D$ is negative, $e^{v_D/V_T}\to 0$, and $i_D\to -I_S$: a tiny constant reverse leakage. In forward bias, once $v_D$ climbs a few $V_T$ above zero, the exponential is enormous compared with 1, so the $-1$ is negligible and we use the simplified form $i_D\approx I_S{e}^{v_D/V_T}$. That simplified form is what we work with for all forward-bias analysis.

$I_S$ is a device parameter set by the cross-sectional area, doping levels, and minority-carrier diffusion lengths. For a typical small-signal silicon diode it lies somewhere between $10^{-15}$ A and $10^{-9}$ A — extraordinarily small. $V_T=kT/q\approx 25$ mV at room temperature, where $k$ is Boltzmann’s constant, $T$ the absolute temperature, and $q$ the electron charge.

Inverting the equation to solve for the voltage given a current:

$v_D=V_T\ln \left(\frac{i_D}{I_S}\right)$

This inverse form is exactly what Iterative diode analysis uses to refine a voltage estimate, and it makes the steepness of the curve concrete.

The diode equation in forward bias; inverting gives $v_D=V_T\ln (i_D/I_S)$.

Why the curve is so steep — and why this gives a 0.7 V “constant” drop

The exponential is brutally steep. Increase $v_D$ by one $V_T$ (25 mV) and the current is multiplied by $e\approx 2.7$. Increase $v_D$ by 60 mV and the current is multiplied by about 10. Check that second claim: a factor-of-10 change needs

$e^{\Delta v_D/V_T}=10\Rightarrow \Delta v_D=V_T\ln 10\approx 25\text{mV}\times 2.303\approx 60\text{mV}$

So every decade of current costs only ~60 mV of extra voltage. Run that the other way: changing the current by six orders of magnitude (say 1 µA to 1 A) moves $v_D$ by only about $6\times 60=360$ mV. That is why a forward-biased silicon diode sits between roughly 0.6 V and 0.8 V across a gigantic range of currents — the voltage is almost pinned. This near-constancy is the entire justification for the Constant-voltage-drop model and is the reason forward diodes make decent voltage references.

The same equation underlies the Exponential diode model for precise hand analysis, and the inverse form drives Iterative diode analysis.