Reverse saturation current — Hardware notes

WHAT is it?

Why "saturation"? Because once you reverse-bias even slightly, every minority carrier that reaches the depletion edge is swept across. You can't sweep more than the supply of minority carriers reaching the junction — so increasing $|V|$ doesn't increase the current. It saturates at $I_S$ .

Why "minority carriers"? The depletion-region field points from n → p. This field pushes electrons toward n and holes toward p — the wrong way for majority carriers, but exactly the direction that carries minority carriers across. So $I_S$ is a minority-carrier show.

WHY does any current flow at all?

At any $T>0$ , thermal energy breaks bonds and creates electron–hole pairs everywhere. In the neutral p-region a few electrons (minorities) are constantly created; some diffuse to the depletion edge, feel the field, and are swept into the n-side. Same for holes on the n-side. This trickle is $I_S$ .

Recall Feynman: explain to a 12-year-old

Imagine a hill (the junction) with a slide going down one side. Most kids (majority carriers) are on the wrong side to use the slide, so almost nobody goes across. But every now and then heat "kicks" a rare kid (a minority carrier) to the top of the slide, and whoosh they slide down instantly. It doesn't matter how steep you make the slide — the number of kids sliding is limited by how many get kicked to the top, not by the steepness. That trickle of kids is the reverse saturation current.

HOW to derive $I_S$ from scratch

We build it from minority-carrier diffusion. Consider the n-side (holes are minorities there).

Step 1 — Equilibrium minority concentration. Mass-action law: $np = n_i^2$ . On the n-side, majority electrons $\approx N_D$ , so equilibrium minority holes: $p_{n0} = \frac{n_i^2}{N_D}$ Why this step? It sets the "supply" of the rare carriers that will feed the current.

Step 2 — Diffusion length. A minority hole injected into the n-region diffuses an average distance before it recombines. Solving the steady-state diffusion equation $D_p\frac{d^2 p}{dx^2} = \frac{p - p_{n0}}{\tau_p}$ gives an exponential decay with characteristic length $L_p = \sqrt{D_p\,\tau_p}$ Why this step? The gradient over $L_p$ sets how fast carriers diffuse to the junction.

Step 3 — Diffusion current density at the edge. Under reverse bias the minority concentration at the depletion edge is driven to ~0, so the excess is $\Delta p \approx -p_{n0}$ over length $L_p$ . Fick's law $J_p = -qD_p\,dp/dx$ gives the magnitude: $J_p = \frac{qD_p\,p_{n0}}{L_p} = \frac{qD_p\,n_i^2}{L_p N_D}$

Step 4 — Add the electron contribution (from the p-side) symmetrically: $J_n = \frac{qD_n\,n_{p0}}{L_n} = \frac{qD_n\,n_i^2}{L_n N_A}$

Step 5 — Total. Multiply the total current density by junction area $A$ :

Why $n_i^2$ matters most. With $n_i \propto T^{3/2}e^{-E_g/2kT}$ , we get $n_i^2 \propto T^3 e^{-E_g/kT}$ . So $I_S$ is extremely temperature-sensitive — it roughly doubles every ~10 °C for silicon. This is the true reason diodes/transistors drift with temperature.

Worked examples

Worked example 1 — Compute

$I_S$ (one-sided junction) Given (Si, $n_i=1.5\times10^{10}\,\text{cm}^{-3}$ ): $A=10^{-4}\,\text{cm}^2$ , $D_p=12\,\text{cm}^2/\text{s}$ , $L_p=10^{-3}\,\text{cm}$ , $N_D=10^{16}\,\text{cm}^{-3}$ . p-side term negligible ( $N_A$ huge). $I_S \approx qA\frac{D_p n_i^2}{L_p N_D}$ Why drop $J_n$ ? Heavily doped p-side → tiny minority electrons → negligible. $= (1.6\times10^{-19})(10^{-4})\frac{(12)(1.5\times10^{10})^2}{(10^{-3})(10^{16})}$ Numerator inside: $12\times2.25\times10^{20}=2.7\times10^{21}$ ; denom $=10^{13}$ → ratio $=2.7\times10^{8}$ . $I_S = 1.6\times10^{-19}\times10^{-4}\times2.7\times10^{8} \approx 4.3\times10^{-15}\,\text{A}$ (a few femtoamps). Tiny — as expected.

Common mistakes

Flashcards

What carriers are responsible for reverse saturation current?

Thermally generated minority carriers swept across by the junction field.

Why is

I_S

called a "saturation" current?

Because it is limited by the supply of minority carriers, so it stays ~constant regardless of how large the reverse voltage is.

Write the full expression for

I_S

I_S = qA\,n_i^2\left(\dfrac{D_p}{L_p N_D} + \dfrac{D_n}{L_n N_A}\right)

How does

I_S

scale with intrinsic concentration?

I_S \propto n_i^2

(not

n_i

), because minority concentration

= n_i^2/N

How does

I_S

depend on temperature and by roughly how much?

\propto T^3 e^{-E_g/kT}

; roughly doubles every ~10 °C in silicon.

In the diode equation, what does

I

approach under strong reverse bias?

I \to -I_S

(the exponential term → 0).

Why do minority not majority carriers cross under reverse bias?

The depletion field direction pushes majorities away but sweeps minorities across.

What is the equilibrium minority hole concentration in the n-region?

p_{n0} = n_i^2/N_D

(from mass-action

np=n_i^2

What is the diffusion length

L_p

and why does it appear?

L_p=\sqrt{D_p\tau_p}

; it sets the concentration gradient feeding diffusion current.

Why does a more heavily doped side contribute less to

I_S

Higher doping → fewer minority carriers (

n_i^2/N

smaller) → smaller diffusion current.

Connections

PN Junction Diode Equation — $I_S$ is its pre-factor.
Depletion Region & Built-in Field — the field that sweeps minorities.
Minority & Majority Carriers — the supply source of $I_S$ .
Mass-Action Law (np = ni²) — gives $p_{n0}, n_{p0}$ .
Intrinsic Carrier Concentration ni(T) — drives the $T$ dependence.
Diffusion Current & Fick's Law — the transport mechanism.
Reverse Breakdown (Zener/Avalanche) — where the flat curve finally breaks.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab hum PN junction ko reverse bias karte hain, toh sabko lagta hai current bilkul zero ho jaata hai. Lekin actually ek bahut chhota current flow hota rehta hai — usko hi reverse saturation current $I_S$ bolte hain. Ye current minority carriers ke wajah se hota hai: heat ke kaaran har jagah electron-hole pairs bante rehte hain, aur p-side ke thode electrons aur n-side ke thode holes junction tak diffuse karke depletion field se "sweep" ho jaate hain. Isiliye thoda sa current cross karta hai.

"Saturation" naam isliye pada kyunki chahe aap reverse voltage kitna bhi badha do, current nahi badhta — kyunki current supply-limited hai, field-limited nahi. Matlab jitne minority carriers ban rahe hain sirf utna hi current ho sakta hai. Extra voltage unko fast sweep karega par naye carriers toh paida nahi karega, isliye current flat rehta hai $-I_S$ par.

Formula derive karne ka core idea: minority concentration $=n_i^2/N$ (mass-action law se), diffusion length $L=\sqrt{D\tau}$ , aur Fick's law se diffusion current. Sab jodke milta hai $I_S = qA\,n_i^2(\frac{D_p}{L_pN_D}+\frac{D_n}{L_nN_A})$ . Yaad rakhna sabse important baat: $I_S \propto n_i^2$ , aur $n_i^2$ temperature ke saath tezi se badhta hai — har ~10°C par $I_S$ almost double ho jaata hai. Isiliye diode aur transistor garmi mein "drift" karte hain. Exam mein $n_i^2$ (na ki $n_i$ ) aur temperature doubling ka point pakka poochha jaata hai.

Test yourself — Doping & PN Junctions

Connections