Experiment 8Bench 8MZU-PE / LED-8Ready

Semiconductor Band Gaps from LEDs

LED Characterisation Bench

Ramp each LED to its turn-on knee, read the threshold voltage, and extract Planck's constant from your own points.

Duration: 3 hours
Kittel: Chapter 8
Topics: Photon emissionPlanck's constantDirect band gap

Optical bench

Red LED · AlGaInP · 660.0 nm

1.00 mA

Standby

Spectrometer

Emission spectrum of the device under test

Peak wavelength

660.00nm

Photon energy hc/λ

1.8785eV

Band gap E_g

1.8785eV

Planck plot

V₀ against 1/λ, from your own recorded readings

Standby

Measure V₀ for each LED and record it. Every reading you log is plotted here.

Your readingsLeast-squares fit0 points

At least two readings are needed before a gradient can be taken.

Bench meters

Regulated supply · voltmeter across the LED · series milliammeter

Supply voltage

Voltage across LED

Current

mASET

Junction temperature

1/λ

nm⁻¹

V₀ at 1 mA

Output

Aluminium gallium indium phosphide. Raising the aluminium fraction widens the gap and shortens the wavelength.

LED bench control

MZU-PE / LED-8

Ready

Turret position

Supply

Supply voltage1.200 V

Hold Shift while dragging for millivolt steps — you will need them to sit exactly on 1.000 mA.

Circuit

Series resistor220 Ω

Junction temp300 K

EmittingThreshold

Key Equations

An LED emits when an electron in the conduction band recombines with a hole, giving up its energy as a photon:

eV_0 \approx E_g = \frac{hc}{\lambda}

so plotting $V_0$ against $1/\lambda$ gives a straight line of gradient $hc/e$ :

h = \frac{e}{c}\times\text{gradient}

Why the intercept is not zero

Turn-on is not a sharp event: the diode current grows exponentially, so “V₀” only means anything once you fix the current at which you read it. The junction also starts conducting slightly below E_g/e. Both push the line down by a constant, which changes the intercept but leaves the gradient — and so your value of h — intact.

Procedure

0/4

ABand gap and colourTurn the turret through all six devices at 300 K. For each, note the peak wavelength on the spectrometer and the photon energy, and confirm that hc/λ matches the band gap readout.
BMeasure the turn-on voltageFor each LED, raise the supply slowly until the milliammeter reads exactly 1.000 mA, then record the voltage across the LED. That is V₀. Record all six.
CDetermine Planck's constantThe Planck plot builds itself from your recorded readings. Once you have all six, read the gradient and the value of h, and compare it with 6.626 × 10⁻³⁴ J·s.
DTemperature dependenceFit the blue LED and take the junction from 200 K to 420 K. Record the peak wavelength at each step and describe which way the band gap moves.

Table 8 — LED thresholds

0 entries

Set the current to exactly 1.000 mA before recording — the Planck plot uses these numbers directly.

#	LED	Material	λ peak/ nm	1/λ/ nm⁻¹	V₀ at 1 mA/ V	I/ mA	E_g/ eV	hc/λ/ eV	T/ K
No readings yet — set the controls, then press “Record reading”.

Analysis & Reflection

Analysis questions

Quote the value of h you obtained, with its uncertainty from the fit, and the percentage difference from the accepted value. Which single measurement contributed most of the error?
Your fitted line has a non-zero intercept. Explain why $eV_0$ is not exactly $E_g$ , and why the gradient is still the right way to get $h$ .
Silicon has a band gap of 1.12 eV, which would put it in the infrared. Why does silicon nevertheless make a very poor LED?
From Task D, describe how the peak wavelength changed as you heated the junction. What does this tell you about the temperature dependence of the band gap?
The green LED has the widest spectral line of the six. Suggest a reason connected to the InGaN alloy from which it is made.

Physics problems

Show that a photon of wavelength 660 nm has an energy of 1.88 eV, using $E = hc/\lambda$ and $hc = 1240\ \text{eV nm}$ .
Explain the difference between a direct and an indirect band gap, and why only direct-gap materials are used for LEDs.
An InGaN LED emits at 525 nm. Estimate the band gap, and comment on how the indium fraction would have to change to make it emit blue instead.
Using your gradient, work out $hc/e$ in V·nm and compare with the accepted 1239.8 V·nm.