Question 4 of 6: Silicon Photodiode — Responsivity, Bandwidth and Noise

Question 4: Silicon Photodiode — Responsivity, Bandwidth and Noise (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A silicon PIN photodiode whose measured responsivity curve is reproduced below, operated with the electrical parameters collected here.

Find. An explanation of the responsivity curve; the quantum efficiency at 850 nm with and without bias; the light intensity that matches the dark current; the detection bandwidth and response time; the quantum- and thermal-noise-limited SNR and the NEP at 10 µW; and the operating mode with its load line.

[Figure not reproduced: Measured responsivity of the silicon photodiode, redrawn from the printed figure. At 850 nm the curve reads about 0.56 A/W with 30 V of reverse bias and about 0.54 A/W unbiased. See the official exam paper.]

Approach. Read the responsivity off the curve and convert it to quantum efficiency through $R = \eta q\lambda/hc$; combine the $RC$ time constant with the transit time to get the response time and hence the bandwidth; then evaluate the shot- and thermal-noise-limited SNRs at that bandwidth.

Part (a) — reading the curve. Three features define it. At long wavelengths the response collapses beyond about 1100 nm, because a silicon photon must carry at least the indirect band gap to create a pair; the cut-off is $$\lambda_c = \frac{hc}{E_g} = \frac{1240\ \text{eV}\cdot\text{nm}}{1.11\ \text{eV}} = 1117\ \text{nm},$$ matching the printed roll-off exactly. At short wavelengths the response also falls, but for the opposite reason: the absorption coefficient of silicon is so large in the blue and ultraviolet that carriers are generated within a few tens of nanometres of the surface, inside the heavily doped contact layer, where they recombine at surface states before they can be collected. Between those two limits the responsivity rises almost linearly with wavelength, which is exactly what $R = \eta q \lambda / hc$ predicts at constant quantum efficiency — a longer-wavelength photon carries less energy, so a given optical power delivers more photons and hence more current. The peak near 900 nm is where the rising $\lambda$ factor meets the falling absorption. Finally, the 30 V curve lies above the unbiased one over the long-wavelength half of the range, because reverse bias widens the depletion layer and sweeps out carriers generated deeper in the silicon, precisely where the weakly absorbed near-infrared photons are stopped.

1. Part (b): quantum efficiency at 850 nm. Responsivity and quantum efficiency are related by $R = \eta\,q\lambda/(hc)$, so $$\eta = \frac{R\,hc}{q\lambda} = \frac{1240\ \text{eV}\cdot\text{nm}} {850\ \text{nm}}\cdot\frac{R}{1\ \text{V}} = 1.459\,R .$$ Reading the curve at 850 nm gives $$\eta(30\ \text{V}) = 1.459(0.56) = \boxed{\;0.82\ \ (82\ \%)\;}, \qquad \eta(0\ \text{V}) = 1.459(0.54) = \boxed{\;0.79\ \ (79\ \%)\;}$$ The three-point gain from reverse bias is the deeper depletion region collecting carriers that would otherwise recombine.
2. Part (c): the optical power that matches the dark current. Setting $I_p = I_d = 10$ nA and inverting the responsivity, $$P = \frac{I_d}{R} = \frac{10\times10^{-9}}{0.56} = 1.786\times10^{-8}\ \text{W} = 17.9\ \text{nW}.$$ Spread over the 5 mm$^2$ window this is an irradiance of $$E = \frac{P}{A} = \frac{1.786\times10^{-8}}{5\times10^{-6}\ \text{m}^{2}} = \boxed{\;3.57\ \text{mW/m}^{2} = 0.357\ \mu\text{W/cm}^{2}\;}$$ This is the illumination at which the detector is exactly as noisy as it is useful — a convenient practical definition of the point where dark current starts to matter.
3. Part (d): the two time constants. The junction and amplifier capacitances appear in parallel across the load, so $$C_T = C_j + C_a = 3 + 7 = 10\ \text{pF}, \qquad \tau_{RC} = R_L C_T = (50)(10\times10^{-12}) = 0.50\ \text{ns}.$$ This happens to equal the quoted transit time, so neither mechanism can be neglected.
4. Combine them and convert to bandwidth. Independent broadening mechanisms add in quadrature: $$\tau = \sqrt{\tau_{tr}^{2}+\tau_{RC}^{2}} = \sqrt{0.5^{2}+0.5^{2}} = \boxed{\;0.71\ \text{ns}\;}$$ and the corresponding $-3$ dB electrical bandwidth of a single-pole response is $$B = \frac{1}{2\pi\tau} = \frac{1}{2\pi(0.707\times10^{-9})} = \boxed{\;225\ \text{MHz}\;}$$ Quoted as a 10–90 % rise time this is $0.35/B = 1.56$ ns. Reducing the load to, say, 10 Ω would make the detector transit-time-limited at about 440 MHz — but at the cost of a five-fold larger thermal noise current, which is the trade the next part quantifies.
5. Part (e): photocurrent at 10 µW. $$I_p = R\,P = (0.56)(10\times10^{-6}) = 5.6\ \mu\text{A}.$$
6. Quantum (shot) noise limit. If the detector were perfect and only the discreteness of the photoelectrons limited it, $$\mathrm{SNR}_q = \frac{I_p^{2}}{2q I_p B} = \frac{I_p}{2qB} = \frac{5.6\times10^{-6}}{2(1.602\times10^{-19})(2.251\times10^{8})} = \boxed{\;7.76\times10^{4} = 48.9\ \text{dB}\;}$$
7. Thermal noise limit. With the real 50 Ω load, $$\langle i_T^{2}\rangle = \frac{4kTB}{R_L} = \frac{4(1.381\times10^{-23})(300)(2.251\times10^{8})}{50} = 7.46\times10^{-14}\ \text{A}^{2},$$ $$\mathrm{SNR}_T = \frac{I_p^{2}}{\langle i_T^{2}\rangle} = \frac{(5.6\times10^{-6})^{2}}{7.46\times10^{-14}} = \boxed{\;420 = 26.2\ \text{dB}\;}$$ The gap of nearly 23 dB between the two limits is the price of the resistive front end: this detector is nowhere near the quantum limit, and a transimpedance or high-impedance amplifier (Question 5) is the standard remedy.
8. Noise-equivalent power. The NEP is the optical power that would produce a signal equal to the noise in a 1 Hz band. With no signal present the noise density is thermal plus dark-current shot noise, $$i_n = \sqrt{\frac{4kT}{R_L}+2qI_d} = \sqrt{3.314\times10^{-22}+3.20\times10^{-27}} = 1.82\times10^{-11}\ \text{A}/\sqrt{\text{Hz}},$$ so $$\mathrm{NEP} = \frac{i_n}{R} = \frac{1.82\times10^{-11}}{0.56} = \boxed{\;3.25\times10^{-11}\ \text{W}/\sqrt{\text{Hz}} \ \ (32.5\ \text{pW}/\sqrt{\text{Hz}})\;}$$ Over the full 225 MHz band that corresponds to $\mathrm{NEP}\sqrt{B} = 0.49\ \mu\text{W}$ — consistent with the 26.2 dB found above for a 10 µW signal, since $(10/0.49)^{2} = 417$.

Part (f) — operating mode and load line. The diode is held at 30 V of reverse bias, so it is operating in the photoconductive mode, not the photovoltaic mode. In the photovoltaic mode the diode is unbiased or open-circuited and works in the fourth quadrant near $I = 0$, delivering power like a solar cell: the response is logarithmic in illumination, the depletion layer is narrow and the junction capacitance large, so the device is slow but has no dark current and hence very low noise. Reverse biasing moves the operating point deep into the fourth quadrant along the near-vertical load line drawn below, where the reverse current is essentially independent of voltage and therefore linear in optical power — the property this whole question relies on. The wider depletion layer also reduces $C_j$ and raises the field, giving the 0.5 ns transit time and the 225 MHz bandwidth of part (d), at the cost of the 10 nA dark current. Because $R_L$ is only 50 Ω, the load line is very nearly vertical: the largest photocurrent considered here, 5.6 µA, drops just 0.28 mV across the load, so the diode voltage stays at $-30$ V for every practical illumination.