Light-emitting diodes (LED) operation — Hardware notes

WHAT is an LED?

Key facts:

Emits light only when forward biased (electrons and holes must be pushed into the junction to recombine).
The color is fixed by the material's bandgap energy $E_g$ , not by the current.
Current controls brightness, not color.

WHY does an LED emit light?

WHY (the physics): When an electron in the conduction band drops down to fill a hole in the valence band, it must lose energy equal to the bandgap $E_g$ . That energy has to go somewhere:

Indirect bandgap (Si, Ge): the electron's momentum ≠ hole's momentum, so a lattice vibration (phonon) must also be involved → energy leaks as heat. Bad for light.
Direct bandgap (GaAs, GaN): momenta match, so the whole energy comes out as a single photon → good for light.

WHY forward bias? Only forward bias lowers the depletion barrier so that majority carriers flood across the junction and meet carriers of the opposite type. In reverse bias, carriers are pulled away — nothing recombines, no light.

HOW to get the photon energy & color

Derivation from first principles

Step 1 — Energy released. A recombining electron loses exactly the bandgap: $E_{\text{photon}} = E_g$ Why this step? The band edges are the only allowed states; the drop is from conduction-band bottom to valence-band top = $E_g$ .

Step 2 — Photon energy in terms of frequency. Planck: $E_{\text{photon}} = h\,f$ Why? Quantum mechanics: light comes in packets of energy $hf$ .

Step 3 — Frequency to wavelength. Light travels at $c$ , and $c = f\lambda$ , so $f = c/\lambda$ : $E_g = \frac{hc}{\lambda}$ Why? Substituting $f=c/\lambda$ links energy directly to the color we see.

Step 4 — Solve for the emitted wavelength: $\boxed{\lambda = \dfrac{hc}{E_g}}$

Figure — Light-emitting diodes (LED) operation

The circuit side: current-limiting resistor

An LED is a diode → its current rises exponentially with voltage. Connect it straight to a battery and it self-destructs. So we always add a series resistor.

Derive the resistor value. Kirchhoff's Voltage Law around the loop: $V_S = I_F R + V_F$ Why? Supply voltage = drop across resistor + forward drop across LED.

Solve for $R$ : $\boxed{R = \dfrac{V_S - V_F}{I_F}}$

$V_S$ = supply voltage
$V_F$ = LED forward voltage (roughly its color: red ≈ 1.8 V, green ≈ 2.2 V, blue ≈ 3.2 V)
$I_F$ = desired forward current (often 10–20 mA)

Worked Examples

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine a tiny electron sitting on a high shelf. When it jumps down to a lower shelf, it has to get rid of its extra "height energy." In an ordinary chip that energy turns into warmth. But in an LED the shelves are lined up perfectly, so instead of warmth the electron shoots out a little flash of light — a photon! Push more electrons in (more current) and you get more flashes = brighter. Make the shelves further apart (bigger bandgap) and each flash is more powerful = bluer light. Small gap = red, big gap = blue.

Active Recall Flashcards

What bias must an LED be in to emit light?

Forward bias.

Why does silicon make a poor LED?

It is an indirect bandgap material, so recombination releases heat (phonons) instead of photons.

What physical property sets the color of an LED?

The semiconductor's bandgap energy

E_g

Formula linking bandgap to emitted wavelength?

\lambda = hc/E_g

; shortcut

\lambda[\text{nm}] = 1240/E_g[\text{eV}]

A GaN LED has

E_g \approx 3.4

eV. What color region?

\lambda = 1240/3.4 \approx 365

nm → ultraviolet/blue.

What does increasing LED current change — color or brightness?

Brightness (color is fixed by the material).

Why is a series resistor required with an LED?

LED current rises exponentially with voltage; the resistor limits current to prevent thermal runaway/destruction.

Formula for the series resistor?

R = (V_S - V_F)/I_F

Compute R for a 2 V LED at 10 mA from a 6 V supply.

R = (6-2)/0.01 = 400\ \Omega

Name the light-producing mechanism in an LED.

Electroluminescence (radiative recombination).

Roughly, red vs blue LED forward voltage?

Red ≈ 1.8 V, blue ≈ 3.2 V (bigger gap ⇒ bigger

V_F

Connections

PN Junction Diode — the LED is a special p–n junction.
Forward and Reverse Bias — LED needs forward bias to work.
Semiconductor Bandgap — $E_g$ decides color.
Direct vs Indirect Bandgap — why GaAs glows and Si doesn't.
Photodiodes and Solar Cells — the reverse process (light → current).
Ohms Law and KVL — used to size the series resistor.
Planck Relation E=hf — quantum basis of $\lambda = hc/E_g$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek normal diode aur ek LED dono me electron aur hole recombine karte hain — matlab electron neeche gir ke hole ko bhar deta hai. Fark sirf ye hai ki us gire hue electron ki energy kahan jaati hai. Silicon jaise indirect bandgap material me ye energy heat ban jaati hai (isliye Si se light nahi milti). Lekin GaAs, GaN jaise direct bandgap materials me wahi energy ek photon — yaani light — ban ke bahar aati hai. Isi ko electroluminescence kehte hain.

LED ka color kaun decide karta hai? Sirf uska bandgap $E_g$ . Formula simple hai: $\lambda = 1240/E_g$ (jab $E_g$ eV me ho to $\lambda$ nm me milega). Bada gap → zyada energy wala photon → chhoti wavelength → blue. Chhota gap → red ya infrared. Yaad rakho: current sirf brightness badalta hai, color nahi. Aur light sirf forward bias me aati hai, kyun ki tabhi carriers junction ke andar ghuste hain aur recombine karte hain.

Circuit me ek cheez kabhi mat bhoolna — series resistor. LED ka current voltage ke saath exponentially badhta hai, to seedha battery se joda to LED jal jaayega. KVL lagao: $V_S = I_F R + V_F$ , isse $R = (V_S - V_F)/I_F$ . Jaise 5 V supply, red LED ( $V_F \approx 1.8$ V), 15 mA chahiye → $R = (5-1.8)/0.015 \approx 213\,\Omega$ , nearest standard 220 Ω le lo. Bas itna samajh liya to LED chapter ka 80% ho gaya.

Test yourself — Diodes & Applications

Connections