Junction capacitance (depletion + diffusion)

A PN junction stores charge in two different ways, so it has two capacitances that add together depending on bias: $C_j = C_{dep} + C_{diff}$ Reverse bias → depletion capacitance dominates. Forward bias → diffusion capacitance dominates.

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

2. Depletion (Junction / Transition) Capacitance

WHAT is stored

On each side sit uncompensated dopant ions: $-qN_A$ (acceptors, p-side) and $+qN_D$ (donors, n-side). The total charge per side has magnitude $Q_{dep} = qN_A x_p A = qN_D x_n A$ (charge neutrality: $N_A x_p = N_D x_n$ ).

HOW we derive it (from first principles)

Step 1 — Depletion width vs voltage. Solving Poisson's equation across an abrupt junction (see Depletion Region Width) gives $W = \sqrt{\frac{2\varepsilon_s}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)(V_{bi}-V)}$ Why this step? The built-in potential $V_{bi}$ must be dropped across $W$ ; more reverse bias ( $V<0$ ) widens $W$ , more forward bias narrows it.

Step 2 — Charge per unit area. Using the effective doping $N_{eff}=\left(\frac{1}{N_A}+\frac{1}{N_D}\right)^{-1}$ , $Q_{dep} = A\sqrt{2\varepsilon_s q\,N_{eff}\,(V_{bi}-V)}$ Why this step? Substituting $W$ into the depletion-charge relation ties the exposed ionic charge directly to voltage.

Step 3 — Differentiate. Because $C=dQ/dV$ (and $V$ appears under a square root): $\boxed{C_{dep}=\left|\frac{dQ_{dep}}{dV}\right| = \frac{A\,\varepsilon_s}{W} = A\sqrt{\frac{q\,\varepsilon_s N_{eff}}{2(V_{bi}-V)}}}$

3. Diffusion (Charge-Storage) Capacitance

WHY it appears only in forward bias

Under forward bias, huge numbers of minority carriers are injected and pile up in the neutral regions (an exponential decay away from the junction). This stored mobile charge $Q_{diff}$ grows exponentially with $V$ . In reverse bias there's essentially no injected minority charge, so $C_{diff}\approx 0$ .

HOW we derive it

Step 1 — Stored charge. The injected minority charge equals current × how long carriers survive. If $\tau$ is the minority-carrier transit/lifetime and $I$ the diode current: $Q_{diff}=\tau I$ Why this step? In steady state, charge in = (recombination rate)×(charge), so $I = Q/\tau$ , i.e. $Q=\tau I$ . Charge injected per second ( $I/q$ carriers) lives for time $\tau$ .

Step 2 — Diode current. From the ideal diode equation (Diode Equation): $I = I_S\left(e^{V/V_T}-1\right),\qquad V_T=\frac{kT}{q}$

Step 3 — Differentiate. $C_{diff}=\dfrac{dQ_{diff}}{dV}=\tau\dfrac{dI}{dV}$ . In forward bias $I\gg I_S$ so $\dfrac{dI}{dV}=\dfrac{I}{V_T}=g_d$ (the diode conductance): $\boxed{C_{diff}=\frac{\tau I}{V_T}=\tau\,g_d}$

4. Putting It Together

Figure — Junction capacitance (depletion + diffusion)

Regime	Dominant $C$	Scaling	Physical charge
Reverse bias	$C_{dep}$	$\propto (V_{bi}-V)^{-1/2}$	fixed dopant ions
Small forward	comparable	—	both
Strong forward	$C_{diff}$	$\propto e^{V/V_T}$	injected minority carriers

5. Worked Examples

Worked example Example 1 — Depletion capacitance of an abrupt junction

Si junction, $N_A=10^{16}$ , $N_D=10^{16}\text{ cm}^{-3}$ , $A=10^{-4}\text{ cm}^2$ , $V_{bi}=0.7$ V, $\varepsilon_s=1.05\times10^{-12}$ F/cm, at $V=-3.3$ V (reverse). Take $q=1.6\times10^{-19}$ C.

Step 1 $N_{eff}=(1/10^{16}+1/10^{16})^{-1}=5\times10^{15}\text{ cm}^{-3}$ . Why? Series-combine the two dopings. Step 2 $V_{bi}-V = 0.7-(-3.3)=4.0$ V. Why? Reverse bias subtracts a negative → adds. Step 3 $C_{dep}=A\sqrt{\dfrac{q\varepsilon_s N_{eff}}{2(V_{bi}-V)}}$ $=10^{-4}\sqrt{\dfrac{1.6\times10^{-19}\cdot1.05\times10^{-12}\cdot5\times10^{15}}{2\cdot4}}$ $=10^{-4}\sqrt{1.05\times10^{-16}}\approx10^{-4}\cdot1.02\times10^{-8}\approx1.0\times10^{-12}$ F $=1.0$ pF.

6. Common Mistakes (Steel-manned)

7. Feynman

Recall Explain to a 12-year-old

Imagine two crowds (electrons and holes) separated by an empty no-man's-land in the middle of the diode. That empty gap acts like the space between two capacitor plates — that's the depletion capacitance. If you push the crowds apart (reverse bias), the gap gets wider and the "capacitor" gets weaker. Now if you let the crowds rush across into each other's territory (forward bias), lots of people pile up near the border and hang around for a while before disappearing (recombining). That stored crowd is extra charge that changes when you change the push — that's the diffusion capacitance, and it gets huge fast, which is why the diode is "sluggish" to turn off.

8. Active Recall

Definition of small-signal capacitance

C=\dfrac{dQ}{dV}

(slope of charge–voltage curve), not

Q/V

Two capacitances of a PN junction

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

Physical charge behind

C_{dep}

Fixed ionized dopant atoms exposed in the depletion region.

Physical charge behind

C_{diff}

Mobile minority carriers injected across the junction in forward bias.

Depletion capacitance formula (parallel-plate form)

C_{dep}=\dfrac{\varepsilon_s A}{W}

Depletion capacitance vs voltage (abrupt)

C_{dep}=A\sqrt{\dfrac{q\varepsilon_s N_{eff}}{2(V_{bi}-V)}}\propto (V_{bi}-V)^{-1/2}

Diffusion capacitance formula

C_{diff}=\dfrac{\tau I}{V_T}=\tau g_d

Effective doping

N_{eff}

\left(\dfrac{1}{N_A}+\dfrac{1}{N_D}\right)^{-1}

(series combination).

Why

C_{dep}

falls with reverse bias

Reverse bias widens

W

, and

C=\varepsilon_s A/W

, so larger

W

→ smaller

C

Why

C_{diff}\approx 0

in reverse bias

I\approx-I_S\approx0

, so almost no injected minority charge is stored.

Grading coefficient meaning

C_{dep}\propto(V_{bi}-V)^{-m}

;

m=1/2

abrupt,

m=1/3

linearly graded.

Application of voltage-dependent

C_{dep}

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

Why forward-biased diodes switch slowly

Large stored diffusion charge

Q=\tau I

must be removed before the diode turns off.

9. Connections

Depletion Region Width — supplies $W(V)$ used in $C_{dep}$ .
Diode Equation — supplies $I(V)$ used in $C_{diff}$ .
Built-in Potential — the $V_{bi}$ term.
Poisson's Equation in Semiconductors — origin of the $W\propto\sqrt{V_{bi}-V}$ law.
Varactor Diode — engineering use of $C_{dep}$ .
Diode Switching / Reverse Recovery — consequence of $C_{diff}$ .
Small-Signal Diode Model — $g_d$ , $C_{dep}$ , $C_{diff}$ combined.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek PN junction do tarike se charge store karta hai, isliye uski do capacitance hoti hain. Yaad rakhna capacitance ka matlab yahan $C = dQ/dV$ hai — yaani charge aur voltage curve ka slope, seedha $Q/V$ nahi, kyunki relation non-linear hai.

Pehli hai depletion capacitance. Junction ke beech ek khaali depletion region hoti hai jismein fixed dopant ions expose hote hain. Yeh bilkul parallel-plate capacitor jaisa hai: $C_{dep}=\varepsilon_s A/W$ , jahan $W$ depletion width hai. Reverse bias lagao to $W$ badh jaata hai, gap wide ho jaata hai, aur capacitance kam ho jaati hai — isi property se varactor diode banta hai (voltage se tunable capacitor). Scaling yaad rakho: $C_{dep}\propto (V_{bi}-V)^{-1/2}$ abrupt junction ke liye.

Doosri hai diffusion capacitance, jo sirf forward bias mein important hai. Forward bias mein bahut saare minority carriers junction ke paar inject hote hain aur neutral region mein pile up ho jaate hain. Yeh stored charge $Q=\tau I$ hota hai, jahan $\tau$ carrier lifetime hai. Differentiate karo to $C_{diff}=\tau I/V_T=\tau g_d$ milta hai — aur yeh voltage ke saath exponentially badhta hai. Isi wajah se diode forward mein slow ho jaata hai: itna sara stored charge nikaalna padta hai jab diode band karte ho (reverse recovery).

Bottom line: reverse bias mein depletion capacitance chalti hai (chhoti, pF range), forward bias mein diffusion capacitance haavi ho jaati hai (badi, nF range). Sign ka dhyaan rakho — reverse bias mein $V$ negative daalna, taaki $V_{bi}-V$ badh jaaye.

Test yourself — Doping & PN Junctions

Connections