Junction capacitance (depletion + diffusion)

2.2.11 · Hardware › Doping & PN Junctions

Ek PN junction charge ko do alag-alag tarike se store karti hai, isliye uske paas do capacitances hote hain jo bias ke hisaab se add ho jaate hain: $C_{j} = C_{d e p} + C_{d i f f}$ Reverse bias → depletion capacitance dominate karti hai. Forward bias → diffusion capacitance dominate karti hai.

1. The Big Picture (WHY does a diode have capacitance at all?)

2. Depletion (Junction / Transition) Capacitance

WHAT is stored

Har side par uncompensated dopant ions baithe hote hain: $- q N_{A}$ (acceptors, p-side) aur $+ q N_{D}$ (donors, n-side). Har side ka total charge per side ka magnitude hota hai $Q_{d e p} = q N_{A} x_{p} A = q N_{D} x_{n} A$ (charge neutrality: $N_{A} x_{p} = N_{D} x_{n}$ ).

HOW we derive it (from first principles)

Step 1 — Depletion width vs voltage. Ek abrupt junction mein Poisson's equation solve karne par (dekho Depletion Region Width) milta hai $W = \frac{2 ε _{s}}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}}) (V_{bi} - V)$ Yeh step kyun? Built-in potential $V_{bi}$ ko $W$ ke across drop hona padta hai; zyaada reverse bias ( $V < 0$ ) se $W$ wide hoti hai, zyaada forward bias se narrow hoti hai.

Step 2 — Charge per unit area. Effective doping $N_{e f f} = (\frac{1}{N _{A}} + \frac{1}{N _{D}})^{- 1}$ use karte hue, $Q_{d e p} = A 2 ε_{s} q N_{e f f} (V_{bi} - V)$ Yeh step kyun? $W$ ko depletion-charge relation mein substitute karne se exposed ionic charge directly voltage se tied ho jaata hai.

Step 3 — Differentiate karo. Kyunki $C = d Q / d V$ hai (aur $V$ square root ke andar appear karta hai): $C_{d e p} = \frac{d Q _{d e p}}{d V} = \frac{A ε _{s}}{W} = A \frac{q ε _{s} N _{e f f}}{2 ( V _{bi} - V )}$

3. Diffusion (Charge-Storage) Capacitance

WHY it appears only in forward bias

Forward bias mein, bahut saare minority carriers inject hote hain aur neutral regions mein pile up karte hain (junction se door exponential decay ke saath). Yeh stored mobile charge $Q_{d i f f}$ $V$ ke saath exponentially badhta hai. Reverse bias mein practically koi injected minority charge nahi hota, isliye $C_{d i f f} \approx 0$ .

HOW we derive it

Step 1 — Stored charge. Injected minority charge barabar hoti hai current × kitni der tak carriers survive karte hain. Agar $τ$ minority-carrier transit/lifetime hai aur $I$ diode current hai: $Q_{d i f f} = τ I$ Yeh step kyun? Steady state mein, charge in = (recombination rate)×(charge), isliye $I = Q / τ$ , yani $Q = τ I$ . Har second mein inject hue carriers ( $I / q$ carriers) $τ$ time tak jeete hain.

Step 2 — Diode current. Ideal diode equation se (Diode Equation): $I = I_{S} (e^{V / V_{T}} - 1), V_{T} = \frac{k T}{q}$

Step 3 — Differentiate karo. $C_{d i f f} = \frac{d Q _{d i f f}}{d V} = τ \frac{d I}{d V}$ . Forward bias mein $I ≫ I_{S}$ hota hai isliye $\frac{d I}{d V} = \frac{I}{V _{T}} = g_{d}$ (diode conductance): $C_{d i f f} = \frac{τ I}{V _{T}} = τ g_{d}$

4. Putting It Together

Figure — Junction capacitance (depletion + diffusion)

Regime	Dominant $C$	Scaling	Physical charge
Reverse bias	$C_{d e p}$	$\propto (V_{bi} - V)^{- 1/2}$	fixed dopant ions
Small forward	comparable	—	dono
Strong forward	$C_{d i f f}$	$\propto e^{V / V_{T}}$	injected minority carriers

5. Worked Examples

6. Common Mistakes (Steel-manned)

7. Feynman

Recall Ek 12-saal ke bachche ko samjhao

Socho diode ke beech mein do bheed (electrons aur holes) hain jo ek khaali no-man's-land se alag hain. Woh khaali gap do capacitor plates ke beech ki jagah ki tarah kaam karta hai — yahi depletion capacitance hai. Agar tum bheedo ko alag dhakelo (reverse bias), gap aur wide ho jaata hai aur "capacitor" kamzor ho jaata hai. Ab agar tum bheedo ko ek doosre ki territory mein rush karne do (forward bias), toh bahut saare log border ke paas pile up ho jaate hain aur thodi der wahan ruk jaate hain aur phir gaayab ho jaate hain (recombine karte hain). Woh stored bheed extra charge hai jo push change hone par change hoti hai — yahi diffusion capacitance hai, aur yeh bahut tezi se badi ho jaati hai, isliye diode "sluggish" hota hai turn off hone mein.

8. Active Recall

Small-signal capacitance ki definition

C = \frac{d Q}{d V}

(charge–voltage curve ka slope),

Q / V

nahi.

PN junction ki do capacitances

Depletion (junction) capacitance + diffusion (charge-storage) capacitance.

C_{d e p}

ke peechhe physical charge

Depletion region mein expose hue fixed ionized dopant atoms.

C_{d i f f}

ke peechhe physical charge

Forward bias mein junction ke across inject hue mobile minority carriers.

Depletion capacitance formula (parallel-plate form)

C_{d e p} = \frac{ε _{s} A}{W}

Depletion capacitance vs voltage (abrupt)

C_{d e p} = A \frac{q ε _{s} N _{e f f}}{2 ( V _{bi} - V )} \propto (V_{bi} - V)^{- 1/2}

Diffusion capacitance formula

C_{d i f f} = \frac{τ I}{V _{T}} = τ g_{d}

Effective doping

N_{e f f}

(\frac{1}{N _{A}} + \frac{1}{N _{D}})^{- 1}

(series combination).

C_{d e p}

reverse bias ke saath kyun girati hai

Reverse bias

W

ko wide karta hai, aur

C = ε_{s} A / W

hai, isliye bada

W

→ chhota

C

Reverse bias mein

C_{d i f f} \approx 0

kyun

I \approx - I_{S} \approx 0

, isliye almost koi injected minority charge store nahi hota.

Grading coefficient ka matlab

C_{d e p} \propto (V_{bi} - V)^{- m}

;

m = 1/2

abrupt,

m = 1/3

linearly graded.

Voltage-dependent

C_{d e p}

ka application

Varactor (varicap) diode — tuning/VCOs ke liye voltage-tunable capacitor.

Forward-biased diodes switch karne mein slow kyun hote hain

Stored diffusion charge

Q = τ I

ko remove karna padta hai pehle diode off ho sake.

9. Connections

Depletion Region Width — $C_{d e p}$ mein use hone wala $W (V)$ provide karta hai.
Diode Equation — $C_{d i f f}$ mein use hone wala $I (V)$ provide karta hai.
Built-in Potential — $V_{bi}$ term.
Poisson's Equation in Semiconductors — $W \propto V_{bi} - V$ law ki origin.
Varactor Diode — $C_{d e p}$ ka engineering use.
Diode Switching / Reverse Recovery — $C_{d i f f}$ ka consequence.
Small-Signal Diode Model — $g_{d}$ , $C_{d e p}$ , $C_{d i f f}$ combined.