LED, photodiode and solar cell

Light in, light out

A lamp that glows without a filament, a remote control that sends invisible light, a roof that makes electricity from the sun: these are all p-n junctions. In all of them, the idea is the one of the first course. A photon and an electron-hole pair have energies of the same order, the band gap. Three devices use this idea in different ways.

The light emitting diode (LED)

An LED is a p-n junction made from a compound such as gallium arsenide phosphide or gallium nitride, and it is forward biased.

• The forward current brings electrons into the p side and holes into the n side.
• Near the junction, an electron meets a hole and they recombine. The electron falls from the conduction band to the valence band.
• The energy of the fall, about $E_g$, leaves as a photon. The wavelength is $\lambda = \dfrac{hc}{E_g}$.

So the colour depends on the gap. A gap of $1.9$ eV gives red light, $2.3$ eV gives green and $2.8$ eV gives blue. Silicon is not suitable: in silicon the recombination mostly gives heat and not light, because it has an indirect gap. LEDs are made from compounds with a direct gap. In practice:

• An LED conducts only above a turn-on voltage, about $1.5$ V to $2.5$ V for red and green LEDs, and about $3$ V for blue.
• It needs a series resistor to limit the current: $R = \dfrac{V_s - V_f}{I}$, where $V_f$ is the forward drop and $I$ is the working current, often $10$ mA to $20$ mA.
• It is damaged by a reverse voltage above about $5$ V.
• It has a low power, a fast response and a long life.

A worked example. A red LED has $V_f = 1.8$ V and works at $20$ mA. It is joined to a $9.0$ V battery. Find the series resistor and the power in it.

1. The voltage across the resistor. $9.0\ \text{V} - 1.8\ \text{V} = 7.2$ V.
2. The resistance. $R = \dfrac{7.2\ \text{V}}{0.020\ \text{A}} = 360\ \Omega$.
3. The power in the resistor. $P = I^2 R = (0.020\ \text{A})^2 \times 360\ \Omega = 0.144$ W. A resistor rated for $0.25$ W is enough.

The photodiode

A photodiode is a p-n junction in a case with a window, and it is reverse biased.

• A photon with energy of at least $E_g$ makes an electron-hole pair in the depletion layer.
• The field there sweeps the electron to the n side and the hole to the p side. The reverse current rises.
• The extra current is in proportion to the intensity of the light.

In the dark there is only a small reverse current, the dark current. When light falls, the current rises by a large factor, which is easy to measure. This is why the diode is reverse biased: the minority-carrier current is small, so the relative change is large. Photodiodes are used to detect light signals, in remote controls, optical communication and light meters.

The current for each watt of light is the responsivity. The number of photons each second is the power $P$ divided by the energy $E$ of one photon. If a share $\eta$ of the photons each give one electron to the current, then

Here $P$ is in W, $E$ is in eV and $I$ is in A. The responsivity is in A/W, with $\lambda$ in nm.

The solar cell

A solar cell is a p-n junction used with no bias. It has a very thin p layer on top, so that light reaches the depletion layer.

• Photons with $E \ge E_g$ make electron-hole pairs near the junction.
• The built-in field separates them: electrons go to the n side and holes to the p side.
• The n side becomes the negative terminal and the p side the positive terminal. A voltage appears, and a load connected across the cell gets a current.

The cell works as a source of power. The current is nearly in proportion to the light intensity. Its two main numbers are:

• the short-circuit current $I_{sc}$ (the terminals joined by a wire, $V = 0$);
• the open-circuit voltage $V_{oc}$ (nothing connected, $I = 0$). For silicon it is about $0.6$ V.

In between, the power $P = VI$ is greatest at one point. Its ratio to $V_{oc} I_{sc}$ is the fill factor. The efficiency is $\eta = \dfrac{P_{max}}{G \times A}$, with the intensity $G$ and the area $A$. For silicon cells for roofs it is of the order of $15\ \%$ to $22\ \%$, measured with $G = 1000\ \text{W/m}^2$. The materials must be chosen with a gap near $1.1$ to $1.5$ eV, so that the cell uses most of the photons of sunlight.

One cell gives only about $0.5$ V. So many cells are joined in series to add the voltage, and in parallel to add the current.

A worked example. A cell of area $100\ \text{cm}^2$ has $I_{sc} = 3.4$ A and $V_{oc} = 0.60$ V in full sunlight of $1000\ \text{W/m}^2$. Its fill factor is $0.75$. Find the maximum power and the efficiency.

1. The maximum power. $P_{max} = 0.75 \times 0.60\ \text{V} \times 3.4\ \text{A} = 1.53$ W.
2. The light power on the cell. $G A = 1000\ \text{W/m}^2 \times 0.0100\ \text{m}^2 = 10.0$ W.
3. The efficiency. $\eta = \dfrac{1.53\ \text{W}}{10.0\ \text{W}} = 15.3\ \%$.

Other courses. A-Level and IB courses meet the diode mostly through rectification and through the light-dependent devices, and the solar cell through energy sources. The ideas above are the same. Photon energy and the band gap link the three devices.

Set the voltage to $0.1$ V and note the power. Then move to $0.5$ V, and then to $0.6$ V. Where does the cell give its greatest power? What happens when the sunlight is halved?

Practise with new numbers

Try it in Python

The program draws the current-voltage curve of a solar cell of $100\ \text{cm}^2$ in full sun, finds the point of greatest power, and gives the fill factor and the efficiency.

Common mistakes

• Forgetting the series resistor of an LED. Without it, the current is limited only by the supply, and the LED burns out.
• Reverse biasing an LED to light it. It lights in forward bias, and a reverse voltage above about $5$ V can damage it.
• Biasing a photodiode forward. It is used in reverse bias, where the current follows the light.
• Thinking a solar cell needs a battery. It makes its own voltage from light.
• Using the efficiency of the cell for the efficiency of the whole panel. Gaps, wires and dust lower it.
• Using light of too long a wavelength. A photon with $E < E_g$ makes no pair in any of the three devices.

Remember these

• LED: forward biased, light of wavelength $\lambda = hc/E_g$, a series resistor $R = (V_s - V_f)/I$.
• Photodiode: reverse biased, the current grows with the intensity of the light.
• Solar cell: no bias, it gives $V_{oc}$ and $I_{sc}$, with $P_{max} = FF \times V_{oc} I_{sc}$ and $\eta = P_{max}/(G A)$.
• All three work through electron-hole pairs and the photon energy $E \approx E_g$.

References

• NCERT Physics Class 12, Semiconductor Electronics, the section on special purpose p-n junction diodes (LED, photodiode, solar cell).
• OpenStax University Physics Volume 3, the chapter Condensed Matter Physics, on semiconductor devices.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An LED gives out light of wavelength $620$ nm. What is the gap of its material, to two figures?

2. A green LED, $V_f = 2.2$ V, works at $10$ mA from a $5.0$ V supply. Find the series resistor.

3. The light on a photodiode, with a fixed reverse voltage, becomes three times as strong. What happens to the photocurrent?

4. A solar cell of area $0.040\ \text{m}^2$ is in sunlight of $900\ \text{W/m}^2$ and has an efficiency of $16\ \%$. What power does it give?