Special Diodes: LED, Photodiode & Solar Cell

You are surrounded by these three devices. Every screen you own is a grid of red, green and blue LEDs or a blue-LED backlight, and the general-lighting revolution that replaced the incandescent bulb is Holonyak's LED grown up — now lighting streets and homes at a fraction of the old energy cost. Photodiodes are the sensing layer of the digital world: the receiver at the far end of every fibre-optic internet link, the light meter in every camera, the pulse-oximeter on a fingertip, and the detector that catches your TV remote's infrared GaAs beam. Solar cells, first used to power satellites (no fuel needed in orbit), are now the fastest-growing source of new electricity on Earth. The frontier pushes all three: deep-ultraviolet LEDs for water sterilisation, single-photon avalanche photodiodes for quantum communication, and tandem solar cells that stack different band gaps to beat the single-junction ceiling. The next article turns to a fourth special diode — the Zener — run in reverse breakdown to hold a voltage steady.

Everything in this article rests on one sentence from the band-theory chapter: an electron in a semiconductor can sit in the low-energy valence band or the high-energy conduction band, and to cross between them it must gain or lose exactly the band-gap energy $E_g$. In silicon that is $1.1\ \text{eV}$; in the compound semiconductors used for LEDs it ranges from about $1.4$ to $3.4\ \text{eV}$. The whole trade is this: one electron crossing the gap is worth one photon of energy $E_g$. Which direction the trade runs — electron-to-photon or photon-to-electron — is what separates the three devices. Start with the light-emitting diode. Forward-bias an ordinary junction and the barrier drops: electrons flood in from the n-side, holes from the p-side, and they meet in the junction region. When a conduction-band electron falls into a valence-band hole — recombination — the energy it loses, about $E_g$, has to go somewhere. In silicon it mostly leaks away as heat (lattice vibrations), which is why silicon diodes just get warm. But in direct-gap compound semiconductors like gallium arsenide phosphide, that energy comes out as a photon instead. Pump a steady forward current through such a junction and you get a steady stream of photons: light. The junction has become a lamp.

Now run the trade the other way. Reverse-bias the junction and, in the dark, almost no current flows — only the tiny saturation leakage you already know. But shine light of the right colour onto the depletion region and each photon that carries more than $E_g$ can be absorbed, lifting an electron across the gap and leaving a hole behind. The strong reverse field immediately sweeps this new electron–hole pair apart, adding to the reverse current. Brighter light means more pairs per second means more current — so the reverse current becomes a faithful measure of the light intensity. That is a photodiode: a light meter that speaks in microamps. Finally, take away the battery altogether. A junction sitting in sunlight still absorbs photons and still makes electron–hole pairs, and the built-in field of the depletion region — the barrier potential that was there all along — sweeps electrons to the n-side and holes to the p-side. Charge piles up, the n-side goes negative and the p-side positive, and a voltage appears across the terminals all by itself. Connect a load and current flows: the cell is now a battery charged by light. This is the photovoltaic effect, and the device is a solar cell. No bias, no fuel, no moving parts — just a junction and the Sun.

An LED is a forward-biased junction engineered so that recombination gives up its energy as light rather than heat. The energy balance is the simplest possible statement of conservation: an electron falling across the gap releases $E_g$, and in a good LED that appears as a single photon. Two or three sentences of physics, then the equation. A conduction electron at the bottom of the conduction band drops into a hole at the top of the valence band; the energy difference is the gap; and a photon carries that difference away. So the emitted photon energy is

where: $h = 6.626\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ is Planck's constant, $\nu$ the light's frequency, and $E_g$ the band gap. The physical reading: the LED is monochromatic-ish — it emits a narrow band of colour centred on the gap energy, not a broad glow like a hot filament. Its colour is a material property, chosen by the chemist who set $E_g$, not by how hard you drive it. Because we usually want a wavelength (a colour) rather than a frequency, combine $\nu = c/\lambda$ with equation (1):

where: $c = 3.0\times10^8\ \text{m/s}$ is the speed of light and $\lambda$ the emitted wavelength. In exam-friendly units this collapses to $\lambda(\text{nm}) = 1240 / E_g(\text{eV})$ — memorise the $1240$. Feed in the gaps and the colours fall out: a $1.9\ \text{eV}$ gap gives $653\ \text{nm}$ (red), $2.3\ \text{eV}$ gives $539\ \text{nm}$ (green), $2.7\ \text{eV}$ gives $459\ \text{nm}$ (blue). And the punchline for silicon: $E_g = 1.1\ \text{eV}$ gives $\lambda = 1127\ \text{nm}$, deep in the infrared — invisible. Silicon cannot make a visible LED. That is why LEDs are built from compound semiconductors — GaAsP and GaP for red/green, indium gallium nitride (InGaN, a GaN alloy) for blue — whose gaps land in the visible window of roughly $1.8$ to $3.1\ \text{eV}$.

One practical consequence worth carrying: because a photon of energy $E_g$ costs the electron a voltage of about $E_g/e$ to cross, an LED's forward turn-on voltage tracks its colour. A red LED ($1.9\ \text{eV}$) lights at about $1.9\ \text{V}$, but a blue LED ($2.7\ \text{eV}$) needs about $2.7\ \text{V}$ — noticeably more than the $0.7\ \text{V}$ of a plain silicon diode. This is why a blue or white LED will not light from a single $1.5\ \text{V}$ cell while a red one might flicker to life.

Reverse the arrangement and you get a detector. A photodiode is a junction run in reverse bias, where in darkness only the tiny saturation current $I_0$ flows. Photons entering the depletion region are absorbed one at a time, each lifting an electron across the gap and creating an electron–hole pair, which the reverse field pulls apart into extra current. But a photon can only do this if it carries at least the gap energy — a photon with $h\nu < E_g$ cannot bridge the gap and simply passes through, unabsorbed. Setting $h\nu = E_g$ gives the longest wavelength the diode can detect, the cut-off wavelength:

where: $\lambda_c$ is the cut-off (maximum) wavelength; light with $\lambda > \lambda_c$ is invisible to the device. For a silicon photodiode, $\lambda_c = 1240/1.1 = 1127\ \text{nm}$: silicon happily detects all visible light and near-infrared out to about $1127\ \text{nm}$, but is blind to the $1550\ \text{nm}$ light used in optical fibres — which is why long-haul fibre links use germanium ($\lambda_c = 1771\ \text{nm}$) or indium-gallium-arsenide detectors instead. The total reverse current is

where: $I_0$ is the small dark (saturation) current and $I_{\text{ph}}$ the photocurrent, which is directly proportional to the light intensity — double the illumination, double the photocurrent. That linearity is exactly what makes a photodiode a good light meter: read the current, and you have read the brightness. Photodiodes are the eyes of cameras, barcode scanners, pulse oximeters, TV-remote receivers, and the fibre-optic internet. A close cousin worth naming here is the Zener diode, the subject of the next article: it too is run in reverse, but instead of responding to light it is deliberately operated in reverse breakdown at a fixed voltage to act as a voltage regulator. Same junction, yet another job.

A solar cell is a photodiode with the battery thrown away — a large-area junction that generates its own power from light. The mechanism up to the point of pair creation is identical to the photodiode: sunlight absorbed in and near the junction frees electron–hole pairs. The difference is that there is no external bias at all; the built-in field of the depletion region does the sweeping, pushing electrons to the n-side and holes to the p-side. That separation of charge is a voltage — the cell becomes a source, not a sink. This is the photovoltaic effect, and it is what distinguishes a solar cell from a photodiode: a photodiode measures light, a solar cell is powered by it. Two numbers characterise a single silicon cell. Its open-circuit voltage — the voltage with no load — is only about $0.5$ to $0.6\ \text{V}$, always a good deal less than the $1.1\ \text{V}$ band gap, because some energy is unavoidably lost to recombination and heat before the carriers reach the terminals. Its current scales with area and brightness. Because $0.6\ \text{V}$ is far too little to be useful, cells are wired in series to add their voltages: about $36$ silicon cells in series give roughly $36 \times 0.6 \approx 21\ \text{V}$ open-circuit, which comfortably charges a nominal $12\ \text{V}$ battery. A rooftop module is dozens of these cells; a solar farm is millions. Silicon's $1.1\ \text{eV}$ gap is not an accident — it is close to the sweet spot for the solar spectrum, which is why silicon dominates photovoltaics. But that same gap sets two hard limits on efficiency. First, any photon with $\lambda > 1127\ \text{nm}$ (energy below $1.1\ \text{eV}$) is not absorbed at all — that near-infrared sunlight passes straight through, wasted. Second, any photon with energy above $1.1\ \text{eV}$ delivers only $E_g$ of useful energy; the excess is dumped instantly as heat (thermalization). A green $550\ \text{nm}$ photon carries $2.25\ \text{eV}$ but hands the circuit only $1.1\ \text{eV}$, wasting more than half. These two losses together cap a single silicon junction near the Shockley–Queisser limit of about $33\%$; real commercial panels reach roughly $20$–$22\%$. Escaping that ceiling — with stacked "tandem" cells of different gaps — is the frontier of solar research today.

Board level — the colour of an LED from its band gap (CBSE and ICSE pattern)

An LED is made from a semiconductor whose band gap is $E_g = 1.9\ \text{eV}$. Taking $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$, (a) find the wavelength of the light it emits, and (b) state its colour.

Step 1. Energy conservation gives one photon of energy equal to the gap, $h\nu = E_g$, so the wavelength is $\lambda = hc/E_g$.

Step 2. Substitute: $\lambda = 1240 / 1.9 = \boxed{653\ \text{nm}}$.

Step 3. The visible spectrum runs from about $400\ \text{nm}$ (violet) to $700\ \text{nm}$ (red); $653\ \text{nm}$ sits near the red end, so the LED glows red. A quick sanity check on the whole method: a larger gap would give a shorter wavelength (bluer), a smaller gap a longer (redder) — consistent with $\lambda \propto 1/E_g$.

Board level — why silicon cannot make a visible LED (CBSE and JEE pattern)

Silicon has a band gap of $1.1\ \text{eV}$. (a) Find the wavelength a silicon LED would emit and explain why you could not see it. (b) What is the smallest band gap a semiconductor needs for its LED to be visible? Take $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$ and the visible range as $400$–$700\ \text{nm}$.

Step 1. For silicon, $\lambda = 1240/1.1 = 1127\ \text{nm}$. This is well beyond $700\ \text{nm}$, in the near-infrared, so the human eye cannot detect it — a "silicon LED" would emit invisibly.

Step 2. To be visible the wavelength must be at most $700\ \text{nm}$. Since $\lambda = 1240/E_g$, requiring $\lambda \le 700\ \text{nm}$ means $E_g \ge 1240/700 = \boxed{1.77\ \text{eV}}$.

Step 3. For the full visible band down to $400\ \text{nm}$ (violet) the gap must reach $1240/400 = 3.1\ \text{eV}$. So visible LEDs need $E_g$ between about $1.8$ and $3.1\ \text{eV}$ — the range of the compound semiconductors GaAsP, GaP and GaN, and precisely why silicon and germanium (gaps of $1.1$ and $0.7\ \text{eV}$) are ruled out.

Competitive level — the cut-off wavelength of a photodiode (JEE and NEET pattern)

A silicon photodiode has a band gap of $1.1\ \text{eV}$. (a) Find the longest wavelength of light it can detect. (b) Can it be used to detect the $1550\ \text{nm}$ light of a long-haul optical fibre? If not, suggest a material that can. Take $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$.

Step 1. A photon is only absorbed if it carries at least the gap energy, $h\nu \ge E_g$. The longest detectable wavelength (smallest energy) is set by $h\nu = E_g$, giving the cut-off $\lambda_c = hc/E_g = 1240/1.1 = \boxed{1127\ \text{nm}}$.

Step 2. For the fibre light, find its photon energy: $E = 1240/1550 = 0.80\ \text{eV}$. Since $0.80\ \text{eV} < 1.1\ \text{eV}$, the photon cannot bridge silicon's gap — the diode is blind to $1550\ \text{nm}$.

Step 3. A smaller-gap material is needed so that $\lambda_c$ extends past $1550\ \text{nm}$. Germanium ($E_g = 0.7\ \text{eV}$) has $\lambda_c = 1240/0.7 = 1771\ \text{nm}$, which comfortably covers $1550\ \text{nm}$ — germanium and indium-gallium-arsenide are the standard fibre-optic detectors. Timing note: about a minute. The whole item is the single relation $\lambda_c = 1240/E_g$ used forwards (part a) and the same $1240/\lambda$ used to get a photon energy (part b). Carry the $1240$ constant and both parts are one division each; do not detour through joules — staying in eV and nm is far faster under exam pressure.

Competitive level — energy lost as heat in a silicon solar cell (JEE Advanced pattern)

A green photon of wavelength $550\ \text{nm}$ from sunlight is absorbed by a silicon solar cell of band gap $1.1\ \text{eV}$. (a) Find the photon's energy. (b) How much of that energy is delivered to the circuit, and how much is wasted as heat? (c) What fraction is wasted? Take $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$.

Step 1. Photon energy: $E = hc/\lambda = 1240/550 = 2.25\ \text{eV}$.

Step 2. Absorbing the photon lifts one electron across the gap, so at most the gap energy $E_g = 1.1\ \text{eV}$ can be delivered to the external circuit. The rest is dumped instantly as lattice heat (thermalization): $E - E_g = 2.25 - 1.1 = \boxed{1.15\ \text{eV}\ \text{wasted}}$.

Step 3. Fraction wasted: $(E - E_g)/E = 1.15/2.25 = 0.51$, so about 51 % of this photon's energy is lost as heat — only $1.1/2.25 = 49\%$ is harvested. And this is the good case: photons redder than $1127\ \text{nm}$ are not absorbed at all. Together, thermalization of the blue-green photons and transmission of the infrared ones cap a single silicon junction near the Shockley–Queisser limit of about $33\%$. Timing note: under two minutes. Get the photon energy with $1240/\lambda$, then the loss is a single subtraction against the $1.1\ \text{eV}$ gap. The headline lesson — "useful energy per photon is capped at $E_g$, the rest is heat" — is worth remembering as the reason a single-junction cell cannot beat roughly a third efficiency, a favourite JEE conceptual hook.

Thinking an LED glows because it gets hot. It does not — an LED emits by band-gap recombination and stays cool; an incandescent bulb glows by heating a filament. Writing "current heats the diode until it emits light" loses marks on CBSE and NEET. The efficiency of LEDs comes precisely from skipping the heat stage.

Saying the drive current or voltage sets the colour. Colour is fixed by the band gap alone, $\lambda = hc/E_g$; changing the current only changes brightness. To change colour you change the semiconductor. Examiners love the distractor "increase the voltage to make it blue" — it is wrong.

Getting the bias backwards. An LED is forward-biased (it must conduct to inject carriers that recombine); a photodiode is reverse-biased (so the dark current is tiny and the photocurrent stands out); a solar cell has no bias (it generates its own). Mixing these up is the single most common error on this topic.

Forgetting the $h\nu \ge E_g$ absorption threshold. A photodiode or solar cell responds only to photons whose energy meets or beats the gap — equivalently $\lambda \le hc/E_g$. Longer-wavelength (lower-energy) light passes straight through. Questions about "why can't a silicon detector see infrared fibre light" are testing exactly this.

Confusing a photodiode with a solar cell. Both absorb light and both make electron–hole pairs, but a photodiode is reverse-biased and measures light (its current reports intensity), while a solar cell is unbiased and generates power (the photovoltaic effect). They are not interchangeable in an answer.

Board scope. LEDs, photodiodes and solar cells as special junction diodes are core on CBSE (Ch. 14), ICSE/ISC, JEE Main, JEE Advanced and NEET, and the qualitative story — what each device does and how it is biased — is on IGCSE 0625 and A-Level 9702. IGCSE does not require the $\lambda = hc/E_g$ colour calculation or the solar-cell efficiency numbers; give those in words for an IGCSE answer. AP Physics 2 does not carry semiconductor electronics at all, and current IB Physics has moved most electronics out of its core — so do not expect this as a compulsory AP or IB topic. The Shockley–Queisser efficiency limit is enrichment, not a standard board requirement anywhere.

Texts. NCERT Physics Class 12, Chapter 14 ("Semiconductor Electronics"), the special-purpose p–n junction diodes section (LED, photodiode, solar cell) · H. C. Verma, Concepts of Physics Vol. 2, "Semiconductors and Semiconductor Devices" · Sze, Physics of Semiconductor Devices, for the university treatment of photodetectors and photovoltaics (enrichment only, well beyond this article).

Provenance. Standard band gaps and the constant $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$. Recompute: $hc = 6.626\times10^{-34}\times3.0\times10^{8}/1.602\times10^{-19}\times10^{9} = 1241 \approx 1240\ \text{eV}\,\cdot\,\text{nm}$; LED/cut-off wavelengths $1240/1.1 = 1127$, $1240/1.42 = 873$, $1240/1.9 = 653$, $1240/2.3 = 539$, $1240/2.7 = 459$, $1240/0.7 = 1771\ \text{nm}$; visible-band gaps $1240/700 = 1.77\ \text{eV}$, $1240/400 = 3.10\ \text{eV}$; fibre photon $1240/1550 = 0.80\ \text{eV}$; solar thermalization $1240/550 = 2.25\ \text{eV}$, waste $2.25 - 1.1 = 1.15\ \text{eV}$, wasted fraction $1.15/2.25 = 0.51$, useful $1.1/2.25 = 0.49$; series module $36\times0.6 = 21.6\ \text{V}$; forward voltages $\approx E_g/e$ ($1.9\ \text{V}$ red, $2.7\ \text{V}$ blue); Shockley–Queisser $\approx 33\%$, commercial $\approx 20$–$22\%$.

Prerequisites. The Junction Diode and its I–V Curve — the forward/reverse-bias behaviour and the depletion region that all three special diodes exploit; without the ordinary diode there is no LED, photodiode or solar cell. Leads to. The Zener Diode as a Voltage Regulator, the next special diode, run in reverse breakdown. For the university seam, the go-deeper articles on semiconductor diode lasers and photodetectors and photodiodes treat stimulated emission and detector noise properly (enrichment by name only, never an exam dependency here).