More and more Indian homes now have blue-black panels on their roofs. A government scheme launched in 2024 aims to fit them to a crore of homes. The panels have no moving parts and no fuel, and no battery drives them. Yet on a sunny afternoon they run the fans and the fridge of the house below. The whole panel is made of thin slices of silicon.
One photon, one pair
In a semiconductor the valence band is full and the conduction band above it is nearly empty, with the band gap \(E_g\) between them. The chapter on photons showed that light of wavelength \(\lambda\) comes in photons of energy \(h\nu = hc/\lambda\). A photon can lift an electron across the gap only if its energy is at least \(E_g\). Then it makes an electron-hole pair, one free electron and one hole.
The reverse also happens. When a free electron falls back into a hole, the pair disappears. This is recombination, and it releases energy of about \(E_g\). In some materials that energy leaves as a photon. Absorbing and emitting are the same step, taken in opposite directions, so one number, \(E_g\), fixes the wavelength in both.
Silicon has a band gap of \(1.1\,\text{eV}\). If an electron falls across silicon's gap and the energy leaves as light, what colour is the light?
Play every chip of the figure above. Silicon's line appears at \(1127\,\text{nm}\), about \(1130\), and gallium arsenide's at \(867\,\text{nm}\); both are infrared. A gap of \(1.9\,\text{eV}\) gives \(653\,\text{nm}\), red, and \(2.76\,\text{eV}\) gives \(449\,\text{nm}\), blue. The visible band, \(380\) to \(750\,\text{nm}\), needs gaps between \(1.65\) and \(3.26\,\text{eV}\).
(a) A yellow light-emitting diode emits at \(590\,\text{nm}\). What is its band gap? (b) What is the longest wavelength that gallium arsenide, \(E_g = 1.43\,\text{eV}\), can absorb?
(a) \(E_g = 1240/590 = 2.10\,\text{eV}\).
(b) \(\lambda_{max} = 1240/1.43 = 867\,\text{nm}\), in the near infrared. Every shorter wavelength, all visible light included, carries enough energy to make pairs in it.
Values used: \(hc = 1240\,\text{eV nm}\), \(e = 1.6 \times 10^{-19}\,\text{C}\); band gaps Si \(1.1\,\text{eV}\), GaAs \(1.43\,\text{eV}\), red LED \(1.9\,\text{eV}\), blue LED \(2.76\,\text{eV}\).
A blue light-emitting diode emits at \(470\,\text{nm}\). What is the band gap of its material, in eV?
The photodiode: light in, current out
A photodiode is a p-n junction with a window, connected in reverse bias. In the dark only the tiny reverse current flows, the dark current. When light of energy \(h\nu \ge E_g\) falls on the junction, it makes electron-hole pairs in and near the depletion layer. The strong field there sweeps the electrons to the n side and the holes to the p side. That is an extra reverse current.
Each absorbed photon gives at most one pair, and each pair that reaches the terminals carries the charge \(e\) round the circuit. So the photocurrent is proportional to the number of photons arriving each second, that is, to the intensity. This is the same rule as in the photocell of the chapter on photons.
Why reverse bias and not forward? In forward bias the ordinary current is large, milliamperes, and a small photocurrent would be lost in it. In reverse bias the dark current is so small that even faint light makes a clear change. The reverse field also sweeps the pairs apart quickly, so the diode responds within a fraction of a microsecond.
A photodiode absorbs \(5 \times 10^{14}\) photons each second, and \(60\,\%\) of the pairs they make reach the terminals. Find the photocurrent in microamperes.
Pairs collected each second \(= 0.6 \times 5 \times 10^{14} = 3 \times 10^{14}\).
Each carries the charge \(e\), so \(I = 3 \times 10^{14} \times 1.6 \times 10^{-19} = 4.8 \times 10^{-5}\,\text{A} = 48\,\mu\text{A}\).
Fibre-optic broadband carries signals as infrared light of \(1550\,\text{nm}\). Can a silicon photodiode (\(E_g = 1.1\,\text{eV}\)) detect it?
The LED: current in, light out
A light-emitting diode, or LED, is a p-n junction connected in forward bias. The battery pushes electrons from the n side and holes from the p side into the junction, where they recombine. Each recombination gives a photon of energy about \(E_g\). So the material decides the colour: gallium arsenide gives the infrared of a television remote, and gallium nitride gives blue.
Silicon and germanium make poor LEDs. Their gaps are too small for visible light, and there is a deeper reason too. In these crystals an electron and a hole rarely recombine by emitting a photon. The energy usually goes into vibrations of the crystal, as heat. LEDs are made instead of compounds such as gallium arsenide, gallium arsenide phosphide and gallium nitride.
Why an LED needs a resistor
The forward voltage of an LED is roughly its gap in volts: about \(2\,\text{V}\) for red and \(3\,\text{V}\) for blue. Above that voltage the current rises very steeply, so an LED joined straight to a battery would draw a huge current and be destroyed. A resistor in series takes the extra voltage and fixes the current: \(R = (V_{supply} - V_{LED})/I\).
A blue LED drops \(3.0\,\text{V}\) and should carry \(20\,\text{mA}\). It is run from a \(5\,\text{V}\) USB supply. Find the series resistance, and the power in the resistor and in the LED.
The resistor takes \(5 - 3.0 = 2.0\,\text{V}\) at \(20\,\text{mA}\), so \(R = 2.0/0.020 = 100\,\Omega\).
Power in the resistor: \(2.0 \times 0.020 = 0.040\,\text{W} = 40\,\text{mW}\). Power in the LED: \(3.0 \times 0.020 = 60\,\text{mW}\).
Without the resistor the LED would have \(5\,\text{V}\) across it, far above its forward voltage, and the current would destroy it at once.
A red LED drops \(2\,\text{V}\) and should carry \(10\,\text{mA}\). What series resistance, in ohms, lets it run from a \(9\,\text{V}\) battery?
Red and green LEDs were glowing in calculators and watches by the 1970s. Blue resisted for decades, because it needs a wide-gap material, gallium nitride, that nobody could grow as good crystals. Around 1990 Akasaki and Amano in Nagoya, and Nakamura at a small company in Tokushima, finally made bright blue LEDs from it. With blue, white light became possible. They shared the Nobel Prize in Physics for 2014.
An LED bulb holds a few blue LEDs coated with a yellow phosphor. The phosphor absorbs part of the blue light and emits yellow; the blue and the yellow together look white. Since 2015 the UJALA scheme has put crores of LED bulbs into Indian homes. A \(9\,\text{W}\) LED bulb gives about the light of a \(60\,\text{W}\) filament bulb. Burning four hours a day, it saves about \(74\,\text{kWh}\) of electricity a year.
The solar cell: light in, power out
A solar cell is a p-n junction with a large area and no battery at all. Light makes pairs near the junction, and the built-in field of the depletion layer separates them. Electrons are swept to the n side and holes to the p side. The p side becomes positive and the n side negative. A voltage appears across the cell, and joined to a lamp it drives a current.
The LED is the third member of the family. It is forward-biased, absorbs electrical energy and emits light, the reverse of the other two. In each device, one photon goes with one electron-hole pair.
Read the curve at each end. With the terminals shorted, \(V = 0\) and the current is the short-circuit current, \(I_{sc}\). With nothing connected, \(I = 0\) and the voltage is the open-circuit voltage, \(V_{oc}\). In the sim, cutting the light tenfold cuts \(I_{sc}\) tenfold, from \(3.5\) to \(0.35\,\text{A}\). But \(V_{oc}\) falls only from \(0.60\) to \(0.53\,\text{V}\).
Neither end gives power, because \(P = VI\) is zero when \(V\) or \(I\) is zero. The power is largest at the knee of the curve, where the rectangle under it is largest. In full sunlight this cell gives \(1.70\,\text{W}\) from the \(10\,\text{W}\) of light falling on it, an efficiency of \(17\,\%\).
A cloud halves the light falling on a solar cell. What happens to its short-circuit current and its open-circuit voltage?
A rooftop panel of area \(1.6\,\text{m}^2\) receives \(1000\,\text{W/m}^2\) of sunlight and converts \(20\,\%\) of it into electricity. It holds \(60\) silicon cells in series, each giving about \(0.6\,\text{V}\). Find the power and the open-circuit voltage of the panel.
Light falling on the panel: \(1000 \times 1.6 = 1600\,\text{W}\). Electrical power: \(0.20 \times 1600 = 320\,\text{W}\).
Cells in series add their voltages: \(60 \times 0.6 = 36\,\text{V}\).
One cell gives too little voltage to be useful alone. Joining many in series gives a useful voltage, and a larger area gives more current.
A house has a \(3\,\text{kW}\) rooftop system. A sunny Indian day gives the equivalent of about five hours of full sunlight. Estimate the energy it produces in a day, in kWh.
Return to the panels on the roof. Each silicon cell in them is a p-n junction with no battery. Sunlight makes pairs, the junction's own field separates them, and the cell becomes a small source of about \(0.6\,\text{V}\). Sixty cells in series give about \(36\,\text{V}\). At \(20\,\%\) a \(1.6\,\text{m}^2\) panel gives about \(320\,\text{W}\), and ten such panels run a house.
Silicon is the most common semiconductor, yet it is not used to make LEDs. Give two reasons.
Its band gap, \(1.1\,\text{eV}\), gives photons of about \(1130\,\text{nm}\), in the infrared, so no visible light could come out. Also, in silicon an electron and a hole usually recombine without emitting a photon; the energy goes into vibrations of the crystal as heat.
- Biasing all three light diodes the same way.LED forward, photodiode reverse, solar cell no bias at all.
- Thinking brighter light can make a photon of too little energy count.Only \(h\nu \ge E_g\) makes a pair. Brightness brings more photons, not bigger ones.
- Thinking the solar cell's voltage grows in proportion to the light.The current follows the light; the voltage stays near \(0.6\,\text{V}\) per silicon cell.
- Leaving out the LED's series resistor.Above its forward voltage an LED's current rises steeply. The resistor sets the current.