Special purpose diodes

A family, not a single device. LED, photodiode and solar cell are all p-n junctions. Nothing about their basic construction is exotic — what distinguishes them is (a) how the junction is biased, and (b) whether the device is meant to be looked at (emitting light), looked into (absorbing light as a signal), or exposed to sunlight (absorbing light as power).

LED — forward bias, current in, light out. The junction is made from a compound semiconductor such as gallium arsenide (GaAs) or gallium arsenide phosphide (GaAsP), chosen so the energy gap $E_g$ matches a visible-range photon. Under forward bias, electrons are injected from the n-side and holes from the p-side; they recombine inside and near the depletion region. In these direct-gap materials, recombination releases its energy mainly as a photon rather than as heat — unlike ordinary silicon, where recombination is largely non-radiative (heat only, no useful light). The photon energy is set by the gap:

where: $E$ is the photon energy, $E_g$ the semiconductor's band gap, $h$ Planck's constant, $\nu$ the photon frequency, $c$ the speed of light, and $\lambda$ the emitted wavelength. A larger $E_g$ pushes the emitted photon toward blue (higher energy, shorter $\lambda$); a smaller $E_g$ pushes it toward red. This single relation is why different LED materials give different showroom colours, and it is the only quantitative tool this topic asks you to use.

Photodiode — reverse bias, light in, current out. Here the junction sits behind a transparent window and is deliberately reverse biased. In the dark, reverse bias allows only a minuscule reverse saturation current from thermally generated minority carriers. When light with photon energy $h\nu > E_g$ strikes the junction, it creates extra electron-hole pairs right where the field is strongest; that field sweeps them across immediately, adding to the reverse current. Because each absorbed photon contributes at most one electron-hole pair, the resulting current tracks the rate of photon arrival, i.e. light intensity, almost independent of how large the reverse voltage is (once it is large enough to sweep carriers out efficiently). This makes the photodiode a fast, linear light sensor.

Solar cell — no external bias, light in, power out. Structurally similar to a photodiode but built with a large-area, unencapsulated junction so as much light as possible reaches it. With no battery connected, incident photons ($h\nu > E_g$) generate electron-hole pairs near the junction, and the junction's own built-in field — the same field that exists at any unbiased p-n junction — separates them before they can recombine: electrons drift to the n-side, holes to the p-side. This charge separation creates a potential difference across the terminals; connect an external circuit and current flows, driven entirely by light. Solar panels are simply many such cells wired in series and parallel to raise the usable voltage and current.

Question: An LED is fabricated from a semiconductor with band gap $E_g = 1.9\ \text{eV}$. Find the wavelength of light it emits, and check which part of the visible spectrum it lies in. Then, for a second material with a smaller gap $E_g = 1.5\ \text{eV}$, find its emitted wavelength and compare. (Use $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$, $c = 3\times10^8\ \text{m/s}$, $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.)

Step 1. Convert the first band gap to joules: $E_g = 1.9 \times 1.6\times10^{-19} = 3.04\times10^{-19}\ \text{J}$.

Step 2. Apply $\lambda = hc/E_g$: $\lambda = \dfrac{(6.63\times10^{-34})(3\times10^8)}{3.04\times10^{-19}} \approx 6.54\times10^{-7}\ \text{m} = 654\ \text{nm}$. This falls in the red band of the visible spectrum (roughly $620$–$750\ \text{nm}$).

Step 3. Repeat for the smaller gap: $E_g = 1.5 \times 1.6\times10^{-19} = 2.40\times10^{-19}\ \text{J}$, so $\lambda = \dfrac{(6.63\times10^{-34})(3\times10^8)}{2.40\times10^{-19}} \approx 8.29\times10^{-7}\ \text{m} = 829\ \text{nm}$.

Step 4. Compare and conclude. The $829\ \text{nm}$ result lies just beyond the red end of the visible range, in the near-infrared — this material would make an infrared LED (used in remote controls), not a visible one. The $1.9\ \text{eV}$ device emits red light at about $654\ \text{nm}$; the $1.5\ \text{eV}$ device emits invisible infrared light at about $829\ \text{nm}$ — confirming that a smaller band gap always shifts emission toward longer, redder (or infrared) wavelengths.

Competitive stretch

A photodiode with the $1.5\ \text{eV}$-gap material above is reverse biased through a $5\ \text{k}\Omega$ load on a $9\ \text{V}$ supply. In darkness the current is negligible; under illumination it becomes $40\ \mu\text{A}$. The voltage drop across the load is $V = IR = (40\times10^{-6})(5\times10^3) = 0.2\ \text{V}$, essentially unchanged whether the reverse supply is $6\ \text{V}$ or $9\ \text{V}$, because the field is already strong enough to sweep out all photogenerated carriers — the current is set by light intensity, not by bias voltage.

• Mixing up the bias: LED is forward biased, photodiode is reverse biased, solar cell needs no external bias — it makes its own EMF.
• Calling a photodiode a power source. It is not: it needs a battery and only modulates a small reverse current with light. Only the solar cell generates net electrical power from light.
• Forgetting the eV-to-joule conversion when using $\lambda = hc/E_g$ — this single missed factor of $1.6\times10^{-19}$ throws the wavelength off by eighteen orders of magnitude.
• Assuming any LED material can be forced to emit any colour by changing the current or voltage — emitted wavelength is fixed by $E_g$, a property of the material, not of the operating point.

Ask the tutor: Want to work through how solar cell efficiency depends on choosing the right band gap, or try a few more LED wavelength-to-colour problems?