Optical detectors: photodiodes and avalanche photodiodes

Why it matters

The photodetector turns light back into current at the end of every optical link and inside every optical instrument — power meters, spectrophotometers, encoders, smoke detectors, lidar and fibre sensors. Its responsivity, speed and noise decide the weakest signal you can measure and the bit rate you can receive. Choosing between a PIN photodiode and an avalanche photodiode (APD) is a classic sensitivity-versus-cost-and-complexity decision.

Key ideas

Absorption and photocurrent. A photon with energy above the bandgap (hν ≥ E_g) can create an electron–hole pair. In a reverse-biased junction the electric field in the depletion region sweeps the pair apart, producing a current proportional to the optical power. Photons with λ longer than the cut-off λ_c = hc/E_g (λ_c in µm ≈ 1.24/E_g in eV) are not absorbed: silicon (E_g = 1.12 eV) works up to about 1.1 µm, so 1.3 µm and 1.55 µm links use InGaAs or germanium.

Quantum efficiency and responsivity. Quantum efficiency η is electrons collected per incident photon (0 to 1). Responsivity is current per watt: R = I_p/P = ηqλ/(hc), i.e. R ≈ η·λ(µm)/1.24 A/W. For a fixed η, R rises linearly with wavelength up to λ_c and then falls sharply.

PN and PIN photodiodes. In a simple PN diode much light is absorbed outside the thin depletion region, and carriers there move slowly by diffusion. A PIN diode inserts a wide, lightly doped intrinsic layer so nearly all absorption happens in the high-field region: higher η, faster drift-limited response and lower junction capacitance. Speed is limited by the transit time across the depletion layer and by the RC time constant of junction capacitance with the load: f_RC = 1/(2πR_L·C_j). A thicker intrinsic layer improves η but increases transit time — the basic trade-off.

Operating modes. Photoconductive mode (reverse bias) gives a wide depletion region, low capacitance and fast linear response, at the cost of dark current. Photovoltaic mode (zero bias) gives no dark current and the lowest noise for low-frequency, precision light measurement (power meters, solar cells), but is slower. A transimpedance amplifier holds the diode near zero volts and converts current to voltage.

Avalanche photodiodes. An APD adds a high-field multiplication region, biased just below breakdown (tens to hundreds of volts). Photo-generated carriers gain enough energy to create further pairs by impact ionisation, giving an internal gain M (typically 10–100): I = M·R·P. The multiplication is random, adding an excess noise factor F(M) ≈ M^x, where x is about 0.3 for Si, 0.7 for InGaAs and 1.0 for Ge. APDs help when the receiver is limited by amplifier thermal noise; there is an optimum M beyond which multiplied shot noise dominates. The gain is strongly temperature dependent: breakdown voltage rises with temperature, so the bias must be temperature-compensated.

Noise. Shot noise of the photocurrent and dark current, i_s² = 2q(I_p + I_d)B (times M²F for an APD), and thermal noise of the load, i_t² = 4kTB/R_L. Their sum against the signal power I_p² gives the receiver SNR, which sets the bit-error rate in a digital link. Other figures of merit: noise-equivalent power (NEP, the optical power giving SNR = 1 in 1 Hz) and detectivity D* for infrared detectors.

Other detectors. Phototransistors (gain but slow), photomultiplier tubes (very high gain, single-photon, used in spectroscopy), photoconductive cells (LDRs), CCD and CMOS arrays, and thermal detectors (bolometers, thermopiles, pyroelectric) whose response does not depend on photon energy.

Formulas

λ_c = hc/E_g, λ_c(µm) ≈ 1.24/E_g(eV) λ_c cut-off wavelength, E_g bandgap.

η = (I_p/q)/(P/hν), R = I_p/P = ηqλ/(hc) ≈ η·λ(µm)/1.24 η quantum efficiency, I_p photocurrent (A), P incident optical power (W), R responsivity (A/W), q = 1.602 × 10⁻¹⁹ C.

I_p = q·η·Φ Φ incident photon flux (photons/s).

I_APD = M·R·P, F(M) ≈ M^x M avalanche gain, F excess noise factor, x material exponent.

i_s² = 2q(I_p + I_d)B·M²F (M = F = 1 for PIN), i_t² = 4kTB/R_L I_d dark current (A), B bandwidth (Hz), k = 1.381 × 10⁻²³ J/K, T temperature (K), R_L load (Ω); noise currents in A².

SNR = (M·I_p)²/[2q(I_p + I_d)B·M²F + 4kTB/R_L]

f_RC = 1/(2πR_L·C_j) C_j junction capacitance (F).

Worked examples

Example 1 (standard). An InGaAs PIN photodiode has η = 0.8 at 1.3 µm. Find its responsivity and the photocurrent for 10 µW incident. Can a silicon diode detect this light?

1. R = η·λ/1.24 = 0.8 × 1.3/1.24 = 0.839 A/W.
2. I_p = R·P = 0.839 × 10 × 10⁻⁶ = 8.39 µA.
3. Silicon cut-off = 1.24/1.12 = 1.11 µm < 1.3 µm, so silicon cannot absorb it. Answer: R ≈ 0.84 A/W, I_p ≈ 8.4 µA; silicon is unusable at 1.3 µm.

Example 2 (GATE level). A receiver with responsivity 0.9 A/W receives 1 µW in a bandwidth of 100 MHz with a 1 kΩ load at 300 K (ignore dark current). Find the SNR for (a) a PIN diode and (b) an APD with M = 10 and x = 0.7 at the same responsivity.

1. I_p = 0.9 × 10⁻⁶ = 0.9 µA.
2. Shot noise: 2qI_pB = 2 × 1.602 × 10⁻¹⁹ × 0.9 × 10⁻⁶ × 10⁸ = 2.88 × 10⁻¹⁷ A².
3. Thermal noise: 4kTB/R_L = 4 × 1.381 × 10⁻²³ × 300 × 10⁸/10³ = 1.66 × 10⁻¹⁵ A².
4. (a) SNR = (0.9 × 10⁻⁶)²/(2.88 × 10⁻¹⁷ + 1.66 × 10⁻¹⁵) = 8.1 × 10⁻¹³/1.686 × 10⁻¹⁵ = 480 → 26.8 dB.
5. (b) F = 10^0.7 = 5.01; signal (M·I_p)² = 8.1 × 10⁻¹¹ A²; noise = 2.88 × 10⁻¹⁷ × 100 × 5.01 + 1.66 × 10⁻¹⁵ = 1.44 × 10⁻¹⁴ + 0.17 × 10⁻¹⁴ = 1.61 × 10⁻¹⁴ A².
6. SNR = 8.1 × 10⁻¹¹/1.61 × 10⁻¹⁴ = 5030 → 37.0 dB. Answer: about 26.8 dB with the PIN and 37.0 dB with the APD — a 10 dB gain because the PIN receiver was thermal-noise limited.

Example 3 (photon counting). A detector with η = 0.8 receives 5 × 10¹² photons/s. Find the photocurrent.

1. I_p = q·η·Φ = 1.602 × 10⁻¹⁹ × 0.8 × 5 × 10¹². Answer: I_p ≈ 6.4 × 10⁻⁷ A = 0.64 µA.

Common mistakes

• Using λ in nm in R = ηλ/1.24 (it needs µm).
• Saying responsivity is constant with wavelength; it rises with λ up to cut-off for fixed η.
• Assuming more APD gain always helps; excess noise makes an optimum gain.
• Forgetting that APD noise multiplies by M²F, not M².
• Adding noise amplitudes instead of mean-square values.
• Saying an unbiased photodiode gives much less current; at low frequencies the photocurrent is the same, but the response is slower and less linear at high power.

For GATE IN

• Responsivity, quantum efficiency and cut-off wavelength numericals.
• Photocurrent from optical power or photon flux; APD output current with gain.
• Shot and thermal noise, receiver SNR, and when an APD improves SNR.
• Bandwidth from RC time constant of the junction capacitance and load.

Quick check

1. A detector has η = 0.6 at 0.85 µm. What is R?
2. What is the cut-off wavelength of germanium (E_g = 0.66 eV)?
3. A PIN gives 2 µA; an APD with M = 25 sees the same light. What current does the APD give?
4. Which noise does an APD help to overcome? Answers: 1. 0.41 A/W. 2. About 1.88 µm. 3. 50 µA. 4. Thermal (amplifier) noise.