Shockley diode equation — Hardware notes

What is it?

Deriving it from first principles (Derivation-from-scratch)

WHAT we build: current = charge that diffuses across the depletion edge per second.

Step 1 — The barrier and Boltzmann statistics. In equilibrium the junction has a built-in potential $V_{bi}$ . The number of majority carriers with enough energy to climb the barrier follows the Boltzmann factor $e^{-qV_{bi}/kT}$ . Why this step? Carriers must thermally hop the electrostatic hill; the fraction that can is exponential in barrier height, a universal statistical-mechanics result.

Step 2 — Apply a bias. Forward bias $V$ lowers the barrier to $V_{bi}-V$ . So the population that can cross scales as: $e^{-q(V_{bi}-V)/kT} = e^{-qV_{bi}/kT}\cdot e^{qV/kT}$ Why this step? Applied voltage shifts the electrostatic hill directly; a lower hill lets exponentially more carriers over.

Step 3 — Injected minority-carrier concentration. The equilibrium minority concentration $n_{p0}$ already carries the $e^{-qV_{bi}/kT}$ factor. So under bias the injected excess at the depletion edge is: $n_p(0) = n_{p0}\,e^{qV/kT}$ Excess above equilibrium: $\Delta n = n_{p0}\left(e^{qV/kT}-1\right)$ Why this step? Current is driven by carriers beyond equilibrium — we subtract the equilibrium $1$ .

Step 4 — Diffusion current from the gradient. These injected carriers diffuse and recombine over a diffusion length $L_n$ . The diffusion current density (Fick's law) is: $J_n = q\,D_n\frac{\Delta n}{L_n} = \frac{qD_n n_{p0}}{L_n}\left(e^{qV/kT}-1\right)$ Adding the symmetric hole term: $J = \underbrace{\left(\frac{qD_n n_{p0}}{L_n}+\frac{qD_p p_{n0}}{L_p}\right)}_{J_S}\left(e^{qV/kT}-1\right)$ Why this step? A concentration gradient always drives diffusion; the injected excess decays over $L$ , giving slope $\Delta n / L$ .

Step 5 — Bundle constants. Multiply by area $A$ and insert $n$ and $V_T=kT/q$ : $\boxed{I = I_S\left(e^{V/(nV_T)}-1\right)}$

Reading the equation (the three regimes)

Worked examples

Common mistakes (Steel-man then fix)

Recall Feynman: explain to a 12-year-old

Imagine a hill with kids on one side wanting to slide down to the other. The hill is high, so only a few make it over — those few are the "leakage." Now you push the whole slide down a bit (that's the voltage). Each little push doesn't just help a few more kids — it lets ten times more kids over for every small push! That's why a diode goes from "almost no current" to "lots of current" over a tiny voltage range. And if you tilt the hill the wrong way, still only that same tiny handful trickles over — no matter how hard you tilt it.

Active-recall flashcards

Write the Shockley diode equation.

I = I_S\left(e^{V/(nV_T)}-1\right)

What is

V_T

and its value at 300 K?

Thermal voltage

V_T=kT/q \approx 26\,\text{mV}

at 300 K.

Why is there a "

-1

" term?

It forces

I=0

V=0

and gives the constant

-I_S

leakage in reverse bias.

What does

I_S

physically come from?

Diffusion of minority carriers:

I_S\propto \frac{qD_n n_{p0}}{L_n}+\frac{qD_p p_{n0}}{L_p}

(times area).

Approximate current in strong forward bias?

I\approx I_S e^{V/nV_T}

(drop the

-1

Approximate current in strong reverse bias?

I\approx -I_S

(saturates at leakage).

How much extra voltage multiplies current by 10 (

n=1

V_T\ln 10 \approx 60\,\text{mV}

Invert the equation to get V from I.

V = nV_T\ln\!\left(\frac{I}{I_S}+1\right)

What is the ideality factor

n

and its range?

Correction for recombination;

n=1

ideal,

1

–

2

real diodes.

Statistical factor that governs barrier crossing?

Boltzmann factor

e^{-qV_{bi}/kT}

; bias lowers barrier by

V

Connections

PN Junction — the device this equation models
Depletion Region — where $I_S$ and injection happen
Built-in Potential — the barrier $V_{bi}$ the Boltzmann factor climbs
Diffusion Current — Fick's law used in Step 4
Minority Carriers — set the magnitude of $I_S$
Thermal Voltage — $V_T=kT/q$
Diode I-V Characteristic — the plotted curve
Reverse Breakdown — where the ideal model fails

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, diode ek one-way valve hai — current ek hi direction me easily jaane deta hai. Shockley equation batati hai ki kitna current bahega given voltage pe: $I = I_S(e^{V/nV_T}-1)$ . Yahan asli magic exponential me hai. Junction pe ek "hill" (barrier) hota hai, aur carriers ko us hill ko cross karna padta hai. Jab aap forward voltage lagate ho, wo hill choti ho jaati hai, aur Boltzmann statistics ke hisaab se exponentially zyada carriers cross kar paate hain. Isiliye thoda sa voltage badhao aur current chhalaang maar deta hai.

$V_T = kT/q$ ko thermal voltage kehte hain, room temperature (300K) pe iska value ~26 mV hota hai — ye number yaad rakhna, exam me kaam aata hai. Ek killer rule: har extra 60 mV voltage current ko 10 guna kar deta hai (kyunki $V_T \ln 10 = 60$ mV). Yahi reason hai ki practical diode almost 0.7 V pe "clamp" ho jaata hai — usse aage voltage badhana mushkil, current bahut tezi se badhta hai.

Ab wo " $-1$ " ko mat bhoolna. Strong forward bias me to $e^{V/nV_T}$ itna bada ho jaata hai ki $-1$ ignore kar sakte ho. Lekin $V=0$ pe $e^0-1=0$ — matlab bina voltage ke koi current nahi, bilkul sahi. Aur reverse bias me exponential zero ho jaata hai, to sirf $-I_S$ (tiny leakage) bachta hai — current saturate ho jaata hai, resistor ki tarah badhta nahi. Common galti: log samajhte hain reverse me zyada voltage se zyada current — nahi bhai, wo constant leakage rehta hai.

Practical tip: agar current se voltage nikaalna ho to inverse use karo — $V = nV_T \ln(I/I_S + 1)$ . Voltage current ka logarithm hai, isiliye current 1000 guna badhe to voltage sirf thoda sa (~ $3V_T\ln10 \approx$ 180 mV) badhta hai. Ye samajh gaye to diode ki poori feel aa jaati hai.

Test yourself — Doping & PN Junctions

Connections