Photodiode Characteristics and Operation Revision Notes | A Level AQA Physics

1. Why Photodiodes Matter in Optical Systems

Every time a barcode scanner reads a label, a fibre-optic link carries data, or a remote control sends a signal, a photodiode is converting light into an electrical current. Understanding how that conversion works — and how to read the curves that describe it — is exactly what AQA expects you to be able to do at A Level.

The two core ideas you need to master are the characteristic curves (how current responds to voltage and illumination) and the spectral response curve (how sensitivity varies with wavelength). Both feed directly into understanding why photodiodes are used in photo-conductive mode as detectors in optical systems.

2. The p-n Junction and How Light Generates Current

Picture a p-n junction diode. At the boundary between the p-type and n-type semiconductor, mobile charge carriers have diffused across and recombined, leaving behind a thin region stripped of free carriers — the depletion region. This region has a built-in electric field pointing from the n-side to the p-side.

When a photon with energy $E = hf$ enters the depletion region and its energy exceeds the semiconductor's band gap energy $E_g$, it can excite an electron from the valence band to the conduction band. This creates an electron-hole pair. The built-in electric field immediately sweeps the electron toward the n-side and the hole toward the p-side — producing a small but measurable current without any external voltage being needed.

The key point for exam answers: photocurrent flows because photons supply the energy to create charge carriers, not because an external voltage drives them. The external circuit provides a path for those carriers to flow around.

3. Characteristic Curves of a Photodiode

The characteristic curve of a photodiode plots current $I$ against voltage $V$ across the device — the same style of $I$–$V$ graph you have seen for ordinary diodes, but now with illumination as an extra variable. Each curve on the family of characteristics corresponds to a different level of illumination (irradiance), often labelled $E_0$, $E_1$, $E_2$ in increasing order of brightness.

In the dark, the curve sits close to the voltage axis in the reverse-bias region — only a tiny dark current flows, caused by thermally generated carriers. As illumination increases, each successive curve shifts downward (in the conventional current direction) into the third quadrant of the $I$–$V$ graph, because photocurrent adds to the reverse current flowing through the device.

Notice two critical features of these curves. First, in the reverse-bias region (negative voltage), the photocurrent is almost completely independent of the applied voltage — the curves are nearly horizontal. Second, the photocurrent is directly proportional to illumination: doubling the irradiance approximately doubles the photocurrent. These two properties make the reverse-biased photodiode an excellent linear detector.

In the forward-bias region (positive voltage), the diode begins to conduct heavily in the conventional sense, and the photocurrent becomes a small fraction of the total current — making it much harder to extract a clean signal proportional to light intensity. This is why forward bias is not used for detection.

4. Spectral Response Curves

A photodiode does not respond equally to all wavelengths of light. The spectral response curve shows how the device's sensitivity — typically expressed as responsivity in amperes per watt (A W$^{-1}$) — varies with the wavelength of the incident light. This curve is one of the most important graphs you need to be able to read and interpret.

The curve has a characteristic shape: sensitivity rises from zero at short wavelengths, reaches a peak at a wavelength determined by the semiconductor material, then falls back to zero at longer wavelengths. For a silicon photodiode, the peak responsivity typically occurs around 800–900 nm, in the near-infrared region.

To understand the shape, consider what happens at each extreme. At short wavelengths (high photon energy), photons are absorbed very close to the surface of the semiconductor — often before reaching the depletion region. Carriers generated there recombine before being swept apart, so they do not contribute to photocurrent. This is why sensitivity is low at short wavelengths despite the high photon energy.

At long wavelengths (low photon energy), the photon energy $E = hf = \frac{hc}{\lambda}$ eventually falls below the band gap energy $E_g$ of the semiconductor. When $hf < E_g$, photons pass straight through the material without being absorbed — no electron-hole pairs are generated, and sensitivity drops to zero. The long-wavelength cut-off is therefore set by the band gap of the material.

The peak of the spectral response curve sits between these two extremes, at the wavelength where photons are absorbed efficiently within or close to the depletion region. Choosing a photodiode for an optical system therefore requires matching the spectral response peak to the wavelength of the light source being detected.

Worked example: Calculating the Long-Wavelength Cut-Off from Band Gap Energy

A photodiode is made from a semiconductor with a band gap energy of $E_g = 1.12 \text{ eV}$. Calculate the long-wavelength cut-off $\lambda_{max}$ for this device. (Use $h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^{8}$ m s$^{-1}$, $1 \text{ eV} = 1.60 \times 10^{-19}$ J)

1. $E_g = 1.12 \times 1.60 \times 10^{-19} = 1.792 \times 10^{-19}$ J — The band gap is given in electron-volts. Multiply by the charge of one electron to convert to joules, since $1 \text{ eV} = 1.60 \times 10^{-19}$ J.
2. $hf = E_g$ at the cut-off, and $f = \frac{c}{\lambda}$, so $\frac{hc}{\lambda_{max}} = E_g$ — At the long-wavelength cut-off, the photon energy exactly equals the band gap energy. Photons with longer wavelength (lower frequency) cannot generate electron-hole pairs.
3. $\lambda_{max} = \frac{hc}{E_g}$ — Rearranging gives the cut-off wavelength directly in terms of known quantities.
4. $\lambda_{max} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{1.792 \times 10^{-19}}$ — Insert $h$, $c$, and $E_g$ in SI units. Keep all values in scientific notation to avoid arithmetic errors.
5. $6.63 \times 10^{-34} \times 3.00 \times 10^{8} = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$ J m — $6.63 \times 3.00 = 19.89$; exponents: $10^{-34} \times 10^{8} = 10^{-26}$. Rewrite as $1.989 \times 10^{-25}$.
6. $\lambda_{max} = \frac{1.989 \times 10^{-25}}{1.792 \times 10^{-19}} = 1.11 \times 10^{-6}$ m $= 1110$ nm — $1.989 \div 1.792 \approx 1.110$; exponents: $10^{-25} \div 10^{-19} = 10^{-6}$. The result is in metres; converting gives 1110 nm, which is in the near-infrared — consistent with a silicon photodiode.

Answer: $\lambda_{max} \approx 1.11 \times 10^{-6}$ m (1110 nm)

5. Photo-Conductive Mode: Operation as a Detector

Now that you understand the characteristic curves, you can see exactly why photo-conductive mode is the preferred operating mode for detection. In photo-conductive mode, the photodiode is connected with a reverse bias applied — the p-side is connected to the negative terminal of the supply and the n-side to the positive terminal.

Applying reverse bias widens the depletion region. A wider depletion region means a larger volume in which photon absorption can generate electron-hole pairs that are immediately swept apart by the electric field. The result is a larger photocurrent for the same illumination level, and a faster response time — because carriers spend less time drifting before being collected.

The reverse bias also reduces the junction capacitance. Lower capacitance means the device can respond to rapidly changing light signals — important in high-speed optical communication links where the light intensity may be switching at millions of times per second.

In an optical detection circuit, the photodiode in photo-conductive mode is typically connected in series with a load resistor $R_L$. The photocurrent $I_{ph}$ flowing through $R_L$ produces an output voltage $V_{out} = I_{ph} R_L$. Because $I_{ph}$ is proportional to illumination, $V_{out}$ is also proportional to illumination — giving a linear electrical output that faithfully represents the optical input signal.

6. Comparing Characteristic Curves: Dark vs Illuminated

Returning to the characteristic curves with fresh eyes, you can now interpret them at a deeper level. In the dark, the reverse current is the dark current — a small leakage current caused by thermally generated electron-hole pairs in the depletion region. This sets a noise floor for the detector: any photocurrent must be larger than the dark current to produce a detectable signal.

Under illumination, the total reverse current is the sum of the dark current and the photocurrent: $I_{total} = I_{dark} + I_{ph}$. Since $I_{dark}$ is essentially constant for a given temperature and reverse voltage, changes in $I_{total}$ directly reflect changes in $I_{ph}$, which in turn reflect changes in illumination. This linearity is the key property that makes the photodiode useful as a detector.

When comparing curves for different illumination levels, the spacing between curves tells you about linearity. If the curves are equally spaced for equal increments of illumination, the device is responding linearly — a desirable property for accurate measurement. Non-uniform spacing would indicate a non-linear response, which complicates signal processing.

7. Exam Technique and Mark-Winning Strategies

AQA questions on photodiodes tend to fall into a small number of patterns. Recognising the pattern early lets you structure your answer to hit every marking point efficiently.

For describe questions about characteristic curves, work through the graph systematically: state what the axes represent, describe the shape of the curve in the reverse-bias region (nearly horizontal), state how the curve shifts with illumination, and note the dark current. Do not use the word 'because' — describe questions want observations, not explanations.

For explain questions about photo-conductive mode, build a chain: reverse bias applied → depletion region widens → photons absorbed in depletion region → electron-hole pairs generated → electric field sweeps carriers apart → photocurrent flows in external circuit → photocurrent proportional to illumination. Each arrow in that chain is a potential marking point.

For spectral response questions, always link the long-wavelength cut-off explicitly to the band gap energy using $hf = E_g$, and link the low-sensitivity at short wavelengths to surface absorption and recombination before carriers reach the depletion region. Examiners reward precise physical reasoning over vague statements like 'the photons do not have enough energy' — specify that it is the band gap energy that sets the threshold.

Across both specification requirements — characteristic curves and spectral response curves, and use in photo-conductive mode — the underlying theme is the same: the photodiode's behaviour is governed by the physics of the p-n junction and the interaction of photons with the semiconductor band structure. Master those two ideas and the exam questions become predictable.