(a) Three photo diodes $D_1$, $D_2$ and $D_3$ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm? (b) Why photodiodes are required to operate in reverse bias? Explain.

(a) Three photo diodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV resp

Question

D_1

(b) Why photodiodes are required to operate in reverse bias? Explain.

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★est

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

A photodiode detects light only if the photon energy is at least equal to the semiconductor’s band gap. For 600 nm light, the photon energy is about 2.07 eV. $D_1$ (2.5 eV) and $D_3$ (3 eV) have band gaps larger than this, so they cannot detect it. Only $D_2$ (2 eV) can. Photodiodes are reverse-biased to create a wide depletion region and reduce dark current, making detection possible.

The core idea: photon energy vs band gap

A photodiode works because a photon absorbed in the semiconductor can excite an electron from the valence band to the conduction band — but only if the photon carries at least as much energy as the band gap $E_g$ . If the photon energy is less than $E_g$ , it passes right through or is wasted as heat; no electron-hole pair is created, and no signal is generated.

So the first step is always: compute the photon energy for the given wavelength, then compare it with each band gap.

Step-by-step solution

1. Find the energy of a 600 nm photon

The energy of a photon is given by

E = \frac{hc}{\lambda}

where $h = 4.1357 \times 10^{-15} \text{ eV·s}$ (Planck’s constant in eV·s), $c = 3 \times 10^8 \text{ m/s}$ , and $\lambda$ is the wavelength in metres.

Convert 600 nm to metres:

\lambda = 600 \times 10^{-9} \text{ m} = 6 \times 10^{-7} \text{ m}

Now compute:

E = \frac{(4.1357 \times 10^{-15})(3 \times 10^8)}{6 \times 10^{-7}}

First, the numerator:

4.1357 \times 10^{-15} \times 3 \times 10^8 = 1.24071 \times 10^{-6} \text{ eV·m}

Divide by $6 \times 10^{-7}$ :

E = \frac{1.24071 \times 10^{-6}}{6 \times 10^{-7}} = 2.06785 \text{ eV}

So the photon energy is approximately 2.07 eV.

Tip

A handy shortcut: $hc \approx 1240 \text{ eV·nm}$ . Then $E = \frac{1240}{\lambda (\text{in nm})}$ gives the energy directly in eV. For 600 nm: $E = 1240/600 \approx 2.07 \text{ eV}$ . This is a time-saver in exams.

2. Compare with each band gap

Photodiode	Band gap $E_g$	Photon energy (2.07 eV) ≥ $E_g$ ?	Can detect?
$D_1$	2.5 eV	No (2.07 < 2.5)	No
$D_2$	2.0 eV	Yes (2.07 ≥ 2.0)	Yes
$D_3$	3.0 eV	No (2.07 < 3.0)	No

So $D_1$ and $D_3$ will not detect 600 nm light. Only $D_2$ will.

Watch out

A common mistake is to think that a larger band gap is better for detection. Actually, the band gap must be smaller than or equal to the photon energy. A photodiode with too large a band gap simply cannot respond to low-energy (long-wavelength) light.

3. Answer for part (a)

The photodiodes that will not detect 600 nm light are $D_1$ (2.5 eV) and $D_3$ (3 eV).

Part (b): Why reverse bias?

A photodiode is always operated in reverse bias (the p-side connected to the negative terminal and the n-side to the positive terminal). Here’s why.

1. Reverse bias widens the depletion region …

Start your 14-day free trial to unlock the full solution →

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 volta …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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

$\lambda_{\text{Si}} = \frac{1240 \, \text{eV·nm}}{1.1 \, \text{eV}} \approx 1130 \, \text{nm} \quad (\text{near-IR}),$
$\lambda_{\text{Ge}} = \frac{1240}{0.66} \approx 1880 \, \text{nm} \quad (\text{mid-IR}).$ …
Start your 14-day free trial to see the full solution →
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 reve …

Start your 14-day free trial to see the full solution →
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,

…

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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 mate …

Start your 14-day free trial to see the full solution →

Q.(a) Three photo diodes D1D_1D1​, D2D_2D2​ and D3D_3D3​ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?

The core idea: photon energy vs band gap

Step-by-step solution

Part (b): Why reverse bias?

🎓Unlock everything free for 14 days

Q.(a) Three photo diodes $D_1$ , $D_2$ and $D_3$ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?

Unlock everything free for 14 days