How is an emf generated by a solar cell due to the three basic processes involved ? Explain.

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Q. How is an emf generated by a solar cell due to the three basic processes involved ? Explain.

CBSECBSE Class XII Board 2022Subjective· 2mImportance★★★★★est

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Concept understanding — Optoelectronic Devices

Optoelectronic Devices: When Light Meets a p-n Junction

You already know that a p-n junction diode lets current flow one way. But there is a deeper story: the junction can also talk to light. It can either emit light when current passes through it, or generate current when light falls on it. Devices that do this are called optoelectronic devices — and they are simply p-n junctions designed to make that light–electricity conversion efficient.

The Intuition: Two Sides of the Same Coin

Think of a p-n junction as a tiny energy staircase. On the p-side, holes are at a high energy level (the valence band). On the n-side, electrons are at a higher energy level (the conduction band). When an electron falls down that staircase — from the n-side to the p-side — it loses energy. That energy has to go somewhere.

In an ordinary silicon diode, the lost energy turns into heat (vibrations of the crystal lattice). But in a direct bandgap semiconductor like gallium arsenide, that energy can instead be released as a photon — a particle of light. That is an LED (light-emitting diode).

Now reverse the process. A photon carrying just the right amount of energy can hit the junction and kick an electron up the staircase — from the valence band to the conduction band. That creates an electron–hole pair. If you connect the diode to a circuit, those carriers get swept away by the built-in electric field, producing a current. That is a photodiode (or, if you do it on a large area with no external bias, a solar cell).

Note

The key requirement for efficient light emission or absorption is a direct bandgap. Silicon has an indirect bandgap — it barely emits light and absorbs it weakly. That is why LEDs and laser diodes are made from compounds like GaAs, GaN, or InP, not from silicon.

The Precise Statement

An optoelectronic device is a p-n junction (or a related semiconductor structure) that converts electrical energy into light (LED, laser diode) or light into electrical energy (photodiode, solar cell). The conversion happens because electrons and photons exchange energy across the bandgap of the semiconductor.

Device	Input	Output	Bias condition	Key use
LED	Electric current	Light (spontaneous emission)	Forward bias	Indicators, displays, lighting
Laser diode	Electric current	Coherent light (stimulated emission)	Forward bias (high current)	Optical communication, barcode readers
Photodiode	Light	Electric current	Reverse bias (or zero bias)	Light detection, optical receivers
Solar cell	Sunlight	Electric power	Zero bias (photovoltaic mode)	Power generation

How Each One Works (Briefly)

LED: Under forward bias, electrons from the n-side and holes from the p-side meet in the depletion region. They recombine. In a direct-gap semiconductor, that recombination emits a photon whose energy equals the bandgap. The colour of the LED depends on the bandgap — wider gap gives shorter wavelength (blue), narrower gap gives longer wavelength (red).

Photodiode: Under reverse bias, the depletion region widens and the electric field is strong. When a photon with energy $h\nu \ge E_g$ is absorbed, it creates an electron–hole pair. The field immediately sweeps them apart — electron to n-side, hole to p-side — producing a photocurrent proportional to the light intensity.

Solar cell: Same physics as a photodiode, but no external battery. The built-in field separates the photo-generated carriers, creating a voltage across the junction. Connect a load, and current flows. The cell is essentially a large-area photodiode optimised for sunlight.

Important

All three devices rely on the same fundamental process: electron–hole recombination (light out) or photon absorption creating electron–hole pairs (light in). The difference is only in the bias and the design optimisation.

The One Formula That Connects Them All

The energy of the emitted or absorbed photon is tied to the bandgap $E_g$ :

E_g = h\nu = \frac{hc}{\lambda}

where $h$ is Planck’s constant, $\nu$ is the frequency of light, $c$ is the speed of light, and $\lambda$ is the wavelength. This means:

For an LED, the colour is determined by $E_g$ .
For a photodiode, only photons with $\lambda \le hc/E_g$ can be detected.
For a solar cell, the bandgap must be matched to the solar spectrum for maximum efficiency.

A Common Mistake to Avoid

Watch out

Do not think that a forward-biased p-n junction always emits light. In silicon, recombination is non-radiative — the energy goes to heat, not photons. Only direct-bandgap semiconductors give useful light emission. Similarly, a reverse-biased silicon diode can detect light, but its efficiency is poor compared to a dedicated photodiode made from a direct-gap material.

The Big Picture

Optoelectronic devices are the reason you have LED bulbs in your home, the laser in your optical mouse, the camera sensor in your phone, and the solar panels on rooftops. They are all the same idea — a p-n junction that has learned to speak the language of light.

Optoelectronic devices — LEDs, photodiodes, and solar cells — form a scoring CBSE Class 12 Physics NCERT topic, often searched as optoelectronic devices class 12 physics important questions or difference between LED and photodiode. Because these p-n junction applications connect directly to semiconductor bandgap physics tested in JEE Main and NEET, this topic rewards understanding the underlying photon-energy relation, not just memorising the device table.

A solar cell generates emf by absorbing photons to create electron-hole pairs, separating these charges via the built-in electric field of a p-n junction, and collecting them at the electrodes — producing a potential difference that drives current in an external circuit.

The key to understanding a solar cell is that it’s not a battery — it doesn’t store energy. Instead, it converts light energy directly into electrical energy. The emf (electromotive force) arises because the device creates and maintains a separation of positive and negative charges, which then want to recombine through an external path. That “wanting” is the voltage.

Three fundamental processes work together to make this happen. Let’s walk through them in the order they occur.

Photon absorption and generation of electron-hole pairs

When sunlight (photons) strikes the solar cell, a photon with energy equal to or greater than the semiconductor’s band gap ( $E_g$ ) is absorbed. This energy kicks an electron from the valence band into the conduction band, leaving behind a vacancy called a hole. The result is an electron-hole pair — a mobile negative charge and a mobile positive charge.

Note

If the photon energy is less than $E_g$ , it passes through without being absorbed. If it’s greater, the extra energy is lost as heat — this is why solar cells have a theoretical efficiency limit (the Shockley-Queisser limit).
Charge separation by the built-in electric field

A solar cell is built around a p-n junction. In the depletion region (near the junction), there is a built-in electric field pointing from the n-type side to the p-type side. This field is crucial: it pushes electrons toward the n-side and holes toward the p-side. Without this field, the electron-hole pairs would simply recombine, releasing heat and producing no current.

Watch out

A common mistake is to think the light itself creates the electric field. It doesn’t — the field is already there from the junction. Light only creates the mobile charges that the field then separates.
Charge collection and emf generation

Once separated, electrons accumulate on the n-side (negative terminal) and holes on the p-side (positive terminal). This charge buildup creates a potential difference — the emf — across the cell. When an external circuit is connected, electrons flow from the n-side through the load to the p-side, where they recombine with holes. This flow is the electric current. The emf is the voltage you measure across the open terminals; under load, the voltage drops slightly due to internal resistance.

Tip

Think of the solar cell as a “light-driven charge pump.” The photon provides the energy to lift an electron to a higher energy state, and the junction’s field provides the directional “push” that prevents it from falling back immediately.

The emf is not a single number — it depends on the material’s band gap. For a silicon solar cell, the open-circuit voltage is typically around 0.6–0.7 V. The current depends on the light intensity and the cell area.

✓Final answer

The emf in a solar cell is generated by the sequential processes of photon absorption creating electron-hole pairs, their separation by the built-in electric field of the p-n junction, and the resulting charge accumulation at the terminals — producing a voltage that drives current through an external load.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL1 mark
$Q. True/False : LED works under reverse bias condition.$
›Reveal solutionSolution

False — an LED emits light only under forward bias, not reverse bias.

A light-emitting diode is a heavily doped p-n junction operated in forward bias. Forward bias drives electrons and holes into the junction where they recombine, releasing the energy as photons (light). Under reverse bias no such recombination current flows, so no light is emitted (and enough reverse voltage would only damage it). This is standard NCERT/CBSE Class 12 semiconductor-device behaviour.

✓Final answer
$False — an LED works under forward bias, not reverse bias.$
CBSE 2022Set ANNUAL1 mark
$Q. What is an LED?$
›Reveal solutionSolution

An LED is a heavily-doped p-n junction that converts electrical energy directly into light when forward biased, via radiative recombination of injected charge carriers.

Working

An LED is a p-n junction diode made from a direct-band-gap compound semiconductor (e.g. GaAs, GaAsP, GaP), and is very heavily doped compared to an ordinary rectifying diode. When it is forward biased, electrons are injected from the n-region into the p-region and holes from the p-region into the n-region. Near the junction, these excess minority carriers recombine with the majority carriers. In a direct-band-gap material, this recombination releases the electron's excess energy as a photon of light (rather than as heat/lattice vibrations, as in ordinary silicon diodes), so light is emitted from the junction region.

The wavelength (colour) of the emitted light depends on the band gap $E_g$ : $\lambda=hc/E_g$ , so different compound semiconductors give red, yellow, green, or blue LEDs. LEDs are used as compact, low-power, fast-switching indicator lamps and displays.

✓Final answer
$An LED is a heavily-doped, forward-biased p-n junction that emits light when injected electrons and holes recombine.$
CBSE 2020Set 55/1/11 mark
$Q. Why cannot we use Si and Ge in fabrication of visible LEDs ?$
›Reveal solutionSolution
Silicon and germanium are indirect bandgap semiconductors, meaning radiative recombination is extremely inefficient; photons emitted also fall in the infrared range, not visible light. Only direct bandgap materials like GaAs or GaN produce visible LEDs efficiently.

Why bandgap type matters for light emission

When an electron and hole recombine in a semiconductor, the energy released can take two paths: it can emerge as a photon (light) or dissipate as heat through lattice vibrations (phonons). Whether recombination produces useful light depends critically on the band structure of the material.

In a direct bandgap semiconductor, the conduction band minimum and valence band maximum occur at the same crystal momentum $k$ . An electron can drop straight down in energy, emitting a photon that carries away the bandgap energy $E_g$ while conserving momentum trivially. This radiative process is fast and efficient.

In an indirect bandgap semiconductor like Si or Ge, the conduction minimum and valence maximum sit at different $k$ -values. For an electron to recombine, it must simultaneously change both energy and momentum. A photon alone cannot provide the momentum change (photons carry negligible momentum compared to crystal lattice scales), so the transition requires a phonon to conserve momentum. This three-particle process—electron, hole, and phonon—is far less probable. Most recombinations become non-radiative, releasing heat instead of light.

Watch out

Even when an indirect-gap material does emit a photon, the quantum efficiency is typically $< 10^{-4}$ , meaning fewer than 1 in 10,000 recombinations produce light. Direct-gap LEDs achieve efficiencies above 50%.

The two problems with Si and Ge

Radiative efficiency is abysmal

Silicon has $E_g \approx 1.1 \, \text{eV}$ and germanium $E_g \approx 0.66 \, \text{eV}$ , both indirect. The probability of radiative recombination is so low that injected carriers overwhelmingly recombine non-radiatively. You pump in electrical power and get heat, not light.

Photon energy lies in the infrared

Even if Si or Ge did emit efficiently, the photon energy $E = h\nu = E_g$ corresponds to wavelengths

$\lambda_{\text{Si}} = \frac{1240 \, \text{eV·nm}}{1.1 \, \text{eV}} \approx 1130 \, \text{nm} \quad (\text{near-IR}),$ $\lambda_{\text{Ge}} = \frac{1240}{0.66} \approx 1880 \, \text{nm} \quad (\text{mid-IR}).$
The visible spectrum runs from roughly 400 nm (violet, $\sim 3.1 \, \text{eV}$ ) to 700 nm (red, $\sim 1.8 \, \text{eV}$ ). Both Si and Ge fall well outside this range.

What works instead

Visible LEDs use direct bandgap III-V or II-VI compounds whose $E_g$ can be tuned across the visible range:

Material $E_g$ (eV) $\lambda$ (nm) Color
GaN 3.4 ~365 UV/Blue
InGaN 2.4–3.4 365–520 Blue–Green
GaP 2.26 ~550 Green
AlGaInP 1.9–2.2 565–650 Yellow–Red

These materials combine high radiative efficiency (direct gap) with the right photon energy (visible range). Alloying lets manufacturers dial in the exact color needed.

Tip

Infrared LEDs and detectors (remote controls, fiber optics) do use materials near Si's bandgap—like GaAs ( $E_g = 1.43 \, \text{eV}$ , direct)—but still not Si itself, because the direct gap is essential even for IR applications.

✓Final answer
$Silicon and germanium are indirect bandgap semiconductors with extremely low radiative recombination efficiency, and their bandgaps correspond to infrared wavelengths, not visible light—making them unsuitable for visible LED fabrication.$
CBSE 2019Set ANNUAL1 mark
$Q. LED emits light when connected in ______. (Fill in the blank using 'forward bias' or 'reversed bias')$
›Reveal solutionSolution

An LED emits light only when forward biased.

An LED is a specially doped p-n junction diode. Under forward bias the junction conducts: electrons and holes are injected across the junction and recombine, releasing energy as photons (light). Under reverse bias the junction is non-conducting, so there is no recombination and no light.

✓Final answer
$Forward bias.$
CBSE 2019Set ANNUAL1 markMCQ
$Q. In amplitude modulated wave, the amplitude of each side band frequency is$
(a) A_c
(b) μA_c
(c) μA_c/2
(d) 2μA_c
›Reveal solutionSolution

Each side band amplitude in an AM wave is μA_c/2: option (c).

An amplitude-modulated wave can be written as a sum of three sinusoids: the carrier (angular frequency ω_c, amplitude A_c) and two side bands at frequencies (ω_c + ω_m) and (ω_c − ω_m):

c_m(t) = A_c sinω_c t + (μA_c/2) cos(ω_c − ω_m)t − (μA_c/2) cos(ω_c + ω_m)t,

where μ is the modulation index. From this expansion the amplitude of each side band frequency is clearly (μA_c)/2.

✓Final answer
$(c) μA_c/2 — the amplitude of each side band in an AM wave.$
CBSE 2018Set ANNUAL1 mark
$Q. Write the name of any one dopped semiconductor used for making light emitting diode(LED).$
›Reveal solutionSolution
LEDs use compound semiconductors such as Gallium Arsenide (GaAs).

An LED is a heavily doped p-n junction which emits light when forward biased. Because plain silicon and germanium are indirect-band-gap materials that do not efficiently emit light, LEDs use direct-band-gap compound semiconductors whose band gap corresponds to visible/infrared photon energies.

Common materials:

Gallium Arsenide (GaAs) — infrared/red

Gallium Arsenide Phosphide (GaAsP) — red to yellow

Gallium Phosphide (GaP) — green

✓Final answer
$Gallium Arsenide (GaAs) is one doped semiconductor used for making LEDs.$
CBSE 2016Set ANNUAL1 mark
$Q. What is the full form of LED?$
›Reveal solutionSolution

LED = Light Emitting Diode.

An LED (Light Emitting Diode) is a heavily-doped p-n junction diode which, when forward biased, emits spontaneous light (electroluminescence) as electrons and holes recombine across the junction, releasing energy roughly equal to the semiconductor's band gap as a photon. The colour (wavelength) of light emitted depends on the band gap of the semiconducting material used (e.g. GaAsP for red, GaP for green, GaN for blue). LEDs are used as indicator lamps, in displays, and in optical communication.

✓Final answer
$Light Emitting Diode.$

Material	$E_g$ (eV)	$\lambda$ (nm)	Color
GaN	3.4	~365	UV/Blue
InGaN	2.4–3.4	365–520	Blue–Green
GaP	2.26	~550	Green
AlGaInP	1.9–2.2	565–650	Yellow–Red

Q.How is an emf generated by a solar cell due to the three basic processes involved ? Explain.

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