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Three devices, one p-n junction, three different graphs. Students who learnt the ordinary diode curve often draw all three the same way and lose easy marks. The extra difficulty is that the current NCERT chapter no longer covers them. This post sets out where each curve sits and why, and works three problems: an LED's colour, a photodiode's cut-off and a solar cell's efficiency.
Which syllabus still has these devices?
The JEE (Main) 2026 syllabus lists, under Electronic Devices, "I-V characteristics of LED, the photodiode, solar cell, Zener diode". The NEET (UG) 2026 syllabus has the same line.
The current NCERT Semiconductor Electronics chapter (the PDF on the NCERT site is marked reprint 2025-26) ends with the junction diode as a rectifier. Section 14.8.2, Optoelectronic junction devices, is in the 2022-23 edition (archived copy), and the device facts below come from it. So this is a topic dropped from NCERT but still in both entrance syllabi. Check your year's syllabus.
Which quadrant does each curve sit in?
Photodiode: reverse bias, third quadrant. Light with photon energy hν greater than the band gap Eg creates electron-hole pairs in or near the depletion region. The junction field separates them before they recombine, and the photocurrent is proportional to light intensity. Brighter light pushes the curve further from the axis. Why reverse bias, when forward current is much larger? Illumination changes the small minority-carrier current by a large fraction, so the change is easier to measure.
LED: forward bias, first quadrant. It is a heavily doped junction. Electrons and holes recombine near the junction and give out photons with energy equal to or slightly less than Eg. The curve looks like a silicon diode's, but the threshold voltage is higher and differs by colour. Reverse breakdown is low, typically around 5 V.
Solar cell: no bias, fourth quadrant. It works on the same photovoltaic effect as the photodiode, with a much larger junction area. The old NCERT text explains that the curve is drawn in the fourth quadrant because the cell supplies current rather than drawing it. Materials with a band gap close to 1.5 eV are ideal; Si (1.1 eV), GaAs (1.43 eV) and CdTe (1.45 eV) are used.
Worked example 1: what colour is this LED?
The photon wavelength is λ = hc/E. Using h and c from NIST, hc = 1,239.8 eV nm.
GaAs0.6P0.4, which old NCERT gives as Eg ≈ 1.9 eV: λ = 1,239.8/1.9 = 653 nm. Red, as the book says.
GaAs, Eg ≈ 1.4 eV: λ = 886 nm. Beyond 700 nm, so infrared, as used in remote controls.
Visible light runs from about 0.4 µm to 0.7 µm, so a visible LED needs at least about 1.8 eV.
Now a circuit with sample values: a 5.0 V supply drives an LED that drops 2.0 V, at 20 mA. The series resistor is R = (5.0 − 2.0)/0.020 = 150 Ω. It dissipates 3.0 × 0.020 = 60 mW; the LED takes 40 mW.
Worked example 2: can a silicon photodiode see this light?
A photon is detected only if hν > Eg. For silicon, Eg = 1.1 eV, so the cut-off wavelength is 1,239.8/1.1 = 1,127 nm.
850 nm light: E = 1.46 eV. Above 1.1 eV, so it is detected.
1,550 nm light: E = 0.80 eV. Below the gap, so silicon cannot detect it.
Worked example 3: reading a solar cell curve (sample numbers)
A cell's curve crosses the voltage axis at Voc = 0.60 V and the current axis at Isc = 3.0 A. The point of maximum power on the curve is at 0.50 V and 2.8 A.
Maximum power: 0.50 × 2.8 = 1.40 W.
Fill factor, how "square" the curve is: 1.40/(0.60 × 3.0) = 0.78.
If the cell is 12.5 cm × 12.5 cm in light of 1,000 W/m², the incident power is 0.125² × 1,000 = 15.6 W, so the efficiency is 1.40/15.6 = 9.0%.
There is a ceiling. Shockley and Queisser (1961) calculated a detailed-balance limit for an ideal p-n junction cell and found a maximum efficiency of 30% for an energy gap of 1.1 eV.
Which mistakes cost marks?
Drawing the photodiode in forward bias.
Drawing the solar cell curve in the first quadrant.
Giving an LED the 0.7 V silicon threshold. It is higher and depends on colour.
Assuming any light works. Photons below Eg create no pairs.
Applying a large reverse voltage to an LED. Its breakdown is low.
For curves that are easier to understand moving, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Sketch all three curves on one set of axes, labelling quadrant and bias.
Compute λ for Eg = 2.4 eV and 3.0 eV and name the colours (about 517 nm green, 413 nm violet).
Make up a solar cell curve and find its fill factor and efficiency.
Revise the ordinary diode curve with our guide to p-n junction diode characteristics, and photon energy with our photoelectric effect guide.
Tonight, draw the three curves from memory and check each quadrant.
