80/20 core: Reverse bias widens the depletion region, raises the potential barrier, and lets only a tiny temperature-dependent saturation current IS flow — until the field gets so strong the junction breaks down.
Start from the Shockley diode equation (derived from carrier injection + diffusion):
I=IS(eV/(nVT)−1)
where VT=qkT is the thermal voltage (≈25.85 mV at 300 K) and n is the ideality factor.
WHY doesn't more voltage give more current? The reverse current comes from minority carriers that are thermally generated near the junction and then swept across by the field. The rate at which they are generated depends on temperature, not on the field strength. Once the field is strong enough to sweep every generated carrier across, adding more voltage cannot increase the current — the supply of carriers is the bottleneck.
The depletion region stores charge; treat it as a one-sided problem via Poisson's equation. With total voltage across the junction Vbi−V (for reverse bias V<0, so this increases):
W=q2ε(NA1+ND1)(Vbi−V)
Because W increases, the junction acts like a parallel-plate capacitor with a growing gap, so the junction capacitance decreases:
Cj=WεA=1−V/VbiCj0
This is exploited in varactor diodes (voltage-controlled capacitor).
What happens to the depletion width under reverse bias?
It widens (W∝Vbi−V).
What is the reverse current magnitude far from breakdown?
The saturation current IS (small, roughly constant).
Why is reverse current independent of voltage?
It's limited by thermal generation of minority carriers, not by the field.
How does IS change with temperature?
Roughly doubles every 10 °C (IS∝ni2∝e−Eg/kT).
Reverse-bias diode current from Shockley eq for ∣V∣≫VT?
I≈−IS.
What is thermal voltage VT at 300 K?
kT/q≈25.85 mV.
Why does junction capacitance drop with reverse bias?
Wider depletion region = bigger capacitor gap, Cj=εA/W.
Zener breakdown mechanism?
Direct field tunneling of electrons out of bonds (thin, low-V junctions).
Avalanche breakdown mechanism?
Impact ionization chain reaction of carriers (wide, high-V junctions).
Temp coefficient sign: Zener vs avalanche VBR?
Zener decreases with T; avalanche increases with T.
Recall Feynman: explain to a 12-year-old
Imagine a river (current) with a small gate. Forward bias props the gate open — lots of water flows. Reverse bias slams the gate shut and piles up sandbags (the wider depletion region), so almost no water passes. The only trickle that gets through is from rain falling right at the gate (heat making a few charges) — and hotter days mean more rain, so more trickle. Push way too hard on the gate and it bursts open all at once — that's breakdown.
Reverse bias ka matlab hai battery ka plus terminal N-side pe aur minus terminal P-side pe lagana. Aisa karne se majority carriers (P mein holes, N mein electrons) junction se door bhaag jaate hain. Isse depletion region chaudi ho jaati hai aur andar ka barrier voltage badh jaata hai — isliye normal current lagbhag rukh jaata hai.
Lekin current bilkul zero nahi hota. Thoda sa saturation current IS hamesha behta hai, jo garmi se bane minority carriers ke wajah se aata hai. Yaad rakho — chahe tum voltage kitna bhi negative kar do, yeh current constant rehta hai (saturate ho jaata hai), kyun ki bottleneck carriers ki supply hai, field ki strength nahi. Aur ek zaroori baat: yeh IS har 10 °C pe double ho jaata hai, isliye garmi mein diodes zyada leak karte hain.
Formula side dekho: I=IS(eV/nVT−1). Reverse mein V negative, toh enegative bada→0, isliye I≈−IS. Depletion width W∝Vbi−V — matlab jitna zyada reverse voltage, utni chaudi region, aur junction capacitance Cj=εA/Wkam ho jaati hai (yahi varactor diode ka funda hai).
Aakhir mein, agar reverse voltage bahut zyada ho jaye toh breakdown hota hai — current achanak upar chala jaata hai. Do type: Zener (patli junction, tunneling, kam voltage) aur Avalanche (chaudi junction, chain reaction, zyada voltage). Zener diodes isi breakdown ko jaan-boojh ke use karte hain voltage regulate karne ke liye. Yeh sab exam aur real circuit dono mein bahut kaam aata hai!
