Reverse saturation current in pn junction

Reverse saturation current ($I_0$ or $I_s$) is the small, nearly constant current that flows through a reverse-biased pn junction. It is caused by minority carriers that are thermally generated near the junction and swept across by the junction field.

How it arises

• Under reverse bias the barrier rises from $eV_0$ to $e(V_0+V_r)$; majority carriers cannot cross it, so diffusion current becomes almost zero.
• Minority carriers — holes in the n region and electrons in the p region — that reach the depletion edge by diffusion are pulled across by the field (drift).
• Their number depends only on how fast they are thermally generated, not on the applied voltage. So once $V_r$ exceeds a few $kT/e$ (about 0.1 V), the current saturates.

Expression

From the ideal diode (Shockley) equation $I = I_0(e^{eV/kT}-1)$, for large reverse $V$: $I \to -I_0$, where

$$I_0 = Ae n_i^2\left(\frac{D_h}{L_hN_d} + \frac{D_e}{L_eN_a}\right)$$

$A$ = junction area, $D$ = diffusion coefficients, $L$ = diffusion lengths. There is also a thermal generation current in the depletion layer, proportional to $n_iW/\tau$, which is important in Si.

     I
     |        /  forward
     |       /
 ----+------/------- V
 ____|_____ -I0  (reverse, flat)
     |

Main features

• Very small: nA for Si, µA for Ge (Ge has a smaller $E_g$, hence larger $n_i$).
• Strongly temperature dependent because $I_0 \propto n_i^2 \propto T^3e^{-E_g/kT}$; roughly doubles every 10°C.
• Nearly independent of reverse voltage until breakdown.
• Increases with junction area and decreases with heavier doping.
• Light shining on the junction generates more minority carriers and increases it (photodiode principle).