Photodiode (A-level only) — AQA A-Level Physics

The photodiode is a semiconductor p-n junction device that converts light into electrical current. It is a key component in the AQA A-Level Physics specification, linking the photoelectric effect to practical applications such as light sensors, remote controls, and fibre-optic receivers. Understanding the photodiode requires knowledge of energy bands, the depletion layer, and the internal electric field.

In the wider subject, the photodiode exemplifies the particle nature of light and the quantisation of energy. It provides a concrete context for calculating photon energies, quantum efficiency, and responsivity, and it reinforces the concept that light intensity determines the rate of photon arrival, while wavelength determines whether the photoelectric effect can occur.

• →A photodiode is a p-n junction diode operated in reverse bias, which creates a wide depletion layer with a strong internal electric field.
• →Photons with energy greater than the band gap energy (E = hc/λ > E_gap) are absorbed in the depletion layer, creating electron-hole pairs.
• →The internal electric field separates the electron-hole pairs, sweeping electrons to the n-side and holes to the p-side, producing a photocurrent.
• →The photocurrent is directly proportional to the intensity of incident light (for a given wavelength) because intensity determines the number of photons per second.
• →Quantum efficiency is the ratio of the number of electrons emitted to the number of photons incident, and responsivity is the ratio of photocurrent to incident optical power.

• States that a photodiode is operated in reverse bias, so the p-n junction has a wide depletion region and a strong internal field.
• Describes the characteristic curve as current against voltage for a set of constant irradiance values, with the reverse-bias region nearly flat.
• Explains that in reverse bias the photocurrent is limited by the rate at which photons generate electron-hole pairs, not by the applied voltage.
• States that increasing irradiance increases the number of electron-hole pairs per second and so increases the reverse photocurrent.
• Describes the spectral response curve as responsivity or photocurrent against wavelength at fixed irradiance and fixed reverse bias.
• Explains the long-wavelength cut-off in terms of photon energy falling below the energy needed to excite an electron across the band gap.
• Explains the short-wavelength fall in terms of strong absorption close to the surface, where carriers recombine before reaching the junction.
• Identifies the peak of the spectral response as the wavelength giving the best compromise between photon energy and penetration depth.
• States that photo-conductive mode uses reverse bias, giving a wide depletion region and a strong electric field across the junction.
• Explains that absorbed photons create electron-hole pairs, which are separated by the field and produce a photocurrent.
• Describes the photocurrent as approximately proportional to irradiance over the intended operating range.
• Explains that a load resistor or amplifier converts the photocurrent into a voltage that can be processed by the rest of the system.
• Relates the choice of detector material to the spectral response matching the wavelength of the optical source.
• Discusses response time in terms of junction capacitance and load resistance, noting the trade-off between speed and signal size.
• Applies the detector to a named optical system, such as a fibre-optic receiver, barcode reader or beam-break alarm.
• A scintillator absorbs energy from an incident atomic particle and re-emits visible or near-visible photons.
• A photodiode is a p-n junction operated in reverse bias, creating a wide depletion region and strong internal electric field.
• Photons entering the depletion region generate electron-hole pairs, producing a photocurrent proportional to light intensity.
• The photodiode signal is processed by external circuitry such as a transimpedance amplifier to produce a measurable voltage pulse.
• The scintillator and photodiode must be optically coupled and spectrally matched so emitted photons are efficiently detected.
• The system detects particles indirectly via scintillation light, not by direct charge collection from the particle.

• 💡Sketch both axes with quantity and unit, and label each characteristic curve with its irradiance value rather than leaving unlabelled lines.
• 💡When explaining a cut-off, name the band gap and compare photon energy with it, using E = hf or E = hc/λ as appropriate.
• 💡Link each feature of a curve to a physical process: generation rate for irradiance, photon energy and absorption depth for wavelength.
• 💡Name the mode explicitly and state the bias condition before describing how the photocurrent is produced.
• 💡Include a simple circuit diagram with the reverse bias supply, photodiode and load resistor, and mark the output voltage across the load.
• 💡When discussing speed, refer to the RC time constant of the junction capacitance and load resistance rather than giving a vague claim about fast response.
• 💡Describe the sequence clearly: particle energy → scintillator photons → photodiode electron-hole pairs → photocurrent → amplified voltage pulse.
• 💡State that the photodiode is reverse-biased and explain why this improves detection.
• 💡When comparing detectors, mention that the scintillator plus photodiode is indirect and may have lower energy resolution than direct semiconductor detectors.
• 💡Always state that the photodiode is reverse-biased and refer to the depletion layer and internal electric field when explaining its operation.
• 💡In calculations, show all steps: convert wavelength to metres, calculate photon energy, find number of photons per second, then use quantum efficiency to find photocurrent. Include units at every stage.
• 💡When explaining why a photocurrent is not produced, compare the photon energy to the band gap energy explicitly, and state that no electron-hole pairs are created.

• Treating the photodiode as forward biased: correct this by stating that the detector is reverse biased, so the dark current is small and the photocurrent is the dominant reverse current.
• Reading the flat reverse-bias region as a constant current for all irradiances: correct this by labelling each curve with its own irradiance and noting that higher irradiance gives a larger reverse current.
• Assuming the spectral response rises without limit as wavelength decreases: correct this by explaining that very short wavelengths are absorbed near the surface, so fewer carriers reach the junction.
• Confusing photo-conductive mode with photovoltaic mode: correct this by stating that photo-conductive mode uses reverse bias and an external supply, whereas photovoltaic mode uses no external bias.
• Assuming the output voltage is independent of the load resistor: correct this by using V = IR to show that the same photocurrent gives a larger voltage across a larger load resistance.
• Ignoring the wavelength match between source and detector: correct this by checking the spectral response curve and choosing a material whose peak response covers the source wavelength.
• Thinking the photodiode directly absorbs the atomic particle and produces current; correct understanding is that the scintillator converts particle energy to light first.
• Assuming the photodiode operates in forward bias; it must be reverse-biased to widen the depletion region and increase sensitivity.
• Ignoring spectral matching between scintillator emission and photodiode responsivity; poor matching reduces signal and may prevent detection.
• Students often think the photodiode is forward-biased. Correction: It must be reverse-biased to widen the depletion layer and create a strong internal electric field for efficient charge separation.
• Students confuse the photodiode with an LED or solar cell. Correction: An LED emits light when forward-biased; a solar cell generates power without an external bias; a photodiode requires reverse bias and produces a current proportional to light intensity.
• Students believe any wavelength of light will produce a photocurrent. Correction: The photon energy must exceed the band gap energy; otherwise, no electron-hole pairs are created, regardless of intensity.

1. 1Day 1-2: Review the structure of a p-n junction and the effect of reverse bias. Learn the definitions of depletion layer, internal electric field, and band gap energy.
2. 2Day 3-4: Study the operation of the photodiode. Draw a diagram showing the p-n junction, depletion layer, and charge separation. Write a step-by-step explanation of how a photocurrent is generated.
3. 3Day 5-6: Practice calculations involving photon energy, number of photons per second, quantum efficiency, and photocurrent. Use past paper questions from AQA A-Level Physics.
4. 4Day 7-8: Learn the differences between photodiodes, LEDs, and solar cells. Create a comparison table to avoid confusion.
5. 5Day 9-10: Complete a timed exam-style question on photodiodes, including a 6-mark explanation and a calculation. Review mark schemes to understand what examiners expect.

• 📋Explanation of operation (6 marks)
• 📋Calculation of photocurrent or quantum efficiency (4-5 marks)
• 📋Comparison with LED or solar cell (3-4 marks)
• 📋Graph analysis (e.g., I-V characteristics under illumination) (3-4 marks)