Photodiodes and solar cells — Hardware notes

WHY does a PN junction respond to light at all?

WHY the bandgap threshold? An electron sitting in the valence band is "bonded." To free it you must supply at least the energy gap $E_g$ . A photon is a packet of exactly $h\nu$ energy; if that packet is too small, it cannot bridge the gap and just passes through (the material is transparent to it).

$E_{ph}=h\nu=\frac{hc}{\lambda}\ge E_g \quad\Rightarrow\quad \lambda \le \lambda_{max}=\frac{hc}{E_g}$

HOW the junction turns EHPs into current

So the total diode current becomes the ordinary diode equation minus a light-generated term:

$I = \underbrace{I_0\left(e^{V/nV_T}-1\right)}_{\text{normal diode}} \;-\; \underbrace{I_L}_{\text{light}}$

Why the minus sign? The photocurrent is a reverse (n→p external) current, opposite in sign to the forward diode current. More light ⇒ larger $I_L$ ⇒ the whole I–V curve shifts downward.

Mode 1 — The PHOTODIODE (reverse-biased detector)

WHY reverse bias?

Widens the depletion region ⇒ larger collection volume ⇒ more EHPs caught.
Widens ⇒ lower junction capacitance ⇒ faster response (high-speed optical comms).
Linear relation $I_L \propto$ optical power ⇒ a clean measurement.

Mode 2 — The SOLAR CELL (power generator, zero external bias)

Two key operating points, both derived from $I=I_0(e^{V/nV_T}-1)-I_L$ :

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

What condition must a photon satisfy to create an electron–hole pair?

E_{ph}=h\nu\ge E_g

(photon energy ≥ bandgap).

Formula for cut-off wavelength and the engineer's shortcut?

\lambda_{max}=hc/E_g\approx 1.24/E_g(\text{eV})\ \mu m

Why does adding reverse bias help a photodiode?

Widens depletion region (more collection, faster) and lowers junction capacitance (faster response).

Full I–V equation covering both devices?

I=I_0(e^{V/nV_T}-1)-I_L

Short-circuit current of a solar cell?

I_{sc}=I_L

(set

V=0

, diode term vanishes).

Open-circuit voltage of a solar cell?

V_{oc}=nV_T\ln(I_L/I_0+1)

; grows only logarithmically with light.

Definition of responsivity?

\mathcal{R}=I_L/P_{opt}=\eta q\lambda/(hc)

; current per watt of light.

Why is

V_{oc}

of one Si cell only ~0.5 V?

Because it's

nV_T\ln(I_L/I_0)

— a log of a large ratio still gives a small number.

What is the fill factor?

FF=V_{mp}I_{mp}/(V_{oc}I_{sc})

; how square/ideal the I–V curve is.

Which quadrant does a solar cell operate in?

The fourth quadrant (V>0, I<0): the power-delivering region.

Recall Feynman: explain it to a 12-year-old

Imagine a tiny gate on a hill. Sunlight is like little energy-bullets. If a bullet is strong enough, it knocks a marble loose. The hill (built-in field) always rolls loose marbles down one side, so they pile up and can push a toy car (electricity!). If you just want to count the bullets, you tilt the hill hard (reverse bias) and watch how fast marbles roll — that's a photodiode. If you want the marbles to power your toy car, you let them pile up and turn a wheel — that's a solar cell. Same hill, two jobs. And a weak, wimpy bullet (too-red light) can't knock any marble loose no matter how many you throw.

Connections

PN Junction Diode — same device physics, built-in field & depletion region.
Diode I-V Characteristic Equation — parent equation we subtracted $I_L$ from.
Reverse Bias and Depletion Width — why reverse bias speeds up detection.
Bandgap and Semiconductor Materials — sets $\lambda_{max}$ ; why Si vs GaAs vs Ge differ.
LEDs — the inverse device (current → light instead of light → current).
Photoelectric Effect — the quantum root of the $E_{ph}\ge E_g$ threshold.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, photodiode aur solar cell — dono basically ek hi PN junction hain, bas use karne ka tarika alag hai. Jab light ka photon junction pe girta hai aur uski energy $h\nu$ bandgap $E_g$ se zyada hoti hai, tabhi ek electron free hota hai aur ek hole banta hai (EHP). Agar photon ki energy kam hai (matlab wavelength bahut lambi), to chahe kitni bhi tez light maar do, koi electron free nahi hoga — yahi reason hai Silicon sirf 1.1 μm tak ki light dekh pata hai.

Ab depletion region me ek built-in field hota hai jo electron ko n-side aur hole ko p-side push karta hai. Ye directed flow hi photocurrent $I_L$ hai, aur ye reverse direction me flow karta hai — isliye equation me minus aata hai: $I = I_0(e^{V/nV_T}-1) - I_L$ . Photodiode me hum reverse bias lagate hain taaki depletion region chauda ho (zyada light catch, fast response) aur current se light ki intensity maapte hain. Solar cell me koi battery nahi lagate — light khud current banati hai aur load ke across voltage develop karke power deliver karti hai.

Do important points yaad rakhna: short circuit ( $V=0$ ) pe $I_{sc}=I_L$ , aur open circuit ( $I=0$ ) pe $V_{oc}=nV_T\ln(I_L/I_0)$ . Yahan trick ye hai — current to light ke saath linear badhta hai, par voltage sirf logarithmically badhta hai. Isliye ek Si cell ka voltage bas ~0.5 V hota hai, aur panels me bahut saari cells series me lagani padti hain. Exam me ye sign convention aur log-vs-linear wala point hi sabse zyada marks aur samajh deta hai.

Test yourself — Diodes & Applications

Connections