Question 2 of 7: The pin photodiode — structure, fields, photoconductive operation, responsivity and response time

Question 2: The pin photodiode — structure, fields, photoconductive operation, responsivity and response time (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 operated under reverse bias into a load resistance $R_{L}$, with the page-1 constants $h$, $c$, $q$ and the silicon band gap $E_{g} = 1.11\ \text{eV}$ (and $\varepsilon_{r} = 11.8$ for the illustrative optimum-width calculation in part (d)).

Find. The device cross-section with its charge and field profiles, the photocurrent equation and the name of the operating mode, the shape of the silicon responsivity curve and its band-edge wavelength, and the physical origin of the optimum intrinsic-layer width.

(a) Structure, charge and field, and how the device works

The device is a sandwich: a thin, heavily doped $\text{p}^{+}$ entrance layer carrying an annular (ring) electrode and an anti-reflection window, a wide lightly doped intrinsic layer of width $w$, and a heavily doped $\text{n}^{+}$ layer with a full ohmic back contact. The ring geometry matters — the metal must contact the $\text{p}^{+}$ sheet without shadowing the optical window it defines.

Under reverse bias, essentially the whole applied voltage is dropped across the intrinsic layer, which is fully depleted even at modest bias because it contains almost no dopant. The space charge is therefore confined to two thin sheets: an uncovered acceptor sheet $-qN_{A}$ at the $\text{p}^{+}/\text{i}$ boundary and a donor sheet $+qN_{D}$ at the $\text{i}/\text{n}^{+}$ boundary, with $\rho \approx 0$ in between. Integrating Poisson’s equation $d\mathcal{E}/dx = \rho/\varepsilon$ across a region of zero charge gives a constant field, so $|\mathcal{E}(x)|$ rises abruptly at the $\text{p}^{+}$ edge, sits on a near-flat plateau of magnitude $\mathcal{E}_{\max} \approx V/w$ across the intrinsic layer, and falls abruptly at the $\text{n}^{+}$ edge — the shape plotted in panel (iii) above.

When light of photon energy $h\nu \ge E_{g}$ enters through the window, photons are absorbed and generate electron–hole pairs. Pairs created inside the depleted intrinsic layer are separated immediately by the plateau field and swept out at their saturation drift velocities, holes toward the $\text{p}^{+}$ contact and electrons toward the $\text{n}^{+}$ contact; this drift current is the photocurrent. That is exactly why the intrinsic layer is made wide: absorption is exponential with depth ($P(x) = P_{0}e^{-\alpha x}$), so a wide depletion region captures most of the light where the field is strong. Carriers generated outside the depletion region must diffuse before they are collected, which produces the slow tail on the impulse response, so a good design also keeps the $\text{p}^{+}$ entrance layer thin.

(b) Photocurrent equation, operating mode, and the I–V characteristic

Illuminating the junction adds a light-generated current that flows in the reverse direction, so the full diode characteristic becomes

With a reverse bias $V \ll 0$ the exponential collapses and the measured current is simply

where $I_{d}$ is the dark current. The photocurrent is then linear in optical power and independent of bias, and the voltage developed across the load is $v_{\text{out}} = I_{p}R_{L}$. Operating the device in this third quadrant — reverse biased, current proportional to illumination — is called the photoconductive mode (as opposed to the unbiased fourth-quadrant photovoltaic mode used in solar cells). Reverse bias is preferred for communications because it widens the depletion layer, lowers the junction capacitance and so gives both higher speed and better linearity.

(c) Responsivity of a silicon photodetector and its band edge

The responsivity is $R_{0} = I_{p}/P_{\text{in}} = \eta q\lambda /(hc)$. Two effects shape the curve. Because a photon of longer wavelength carries less energy, a fixed optical power delivers more photons per second, so the ideal responsivity rises linearly with $\lambda$. Against that, the quantum efficiency $\eta$ falls at both ends: at short wavelengths the absorption coefficient is very large and most of the light is absorbed in the thin, undepleted $\text{p}^{+}$ surface layer where the carriers recombine before they can be collected; at long wavelengths silicon becomes transparent and $\eta$ collapses. The product therefore peaks near 0.85–0.9 µm at about 0.6 A/W.

The long-wavelength cut-off is set by the band gap — a photon must supply at least $E_{g}$ to lift an electron across it:

Equivalently $\lambda_{g}[\mu\text{m}] = 1.24/E_{g}[\text{eV}] = 1.24/1.11$. This is why silicon detectors serve the 850 nm window but cannot be used at 1310 or 1550 nm, where germanium or InGaAs is required.

(d) The two time constants and the optimum intrinsic width

$\tau_{RC}$ is the electrical time constant of the detector loaded by its circuit: the junction capacitance $C_{j} = \varepsilon A/w$ in parallel with any stray and amplifier capacitance, charging through the load and series resistance, $\tau_{RC} = R_{L}C_{j}$. $\tau_{\text{drift}}$ is the transit time — how long a carrier generated at the far edge of the depletion region takes to cross it at its saturation velocity, $\tau_{\text{drift}} = w/v_{s}$. The two add in quadrature because they are independent broadening mechanisms acting on the same impulse.

The optimum arises because $w$ pulls the two terms in opposite directions. Widening the intrinsic layer reduces the capacitance ($C_{j}\propto 1/w$) and therefore $\tau_{RC}$, but increases the transit time in direct proportion to $w$. Setting $\tau_{RC} = \tau_{\text{drift}}$ minimises the quadrature sum:

Taking a representative silicon detector with $R_{L} = 50\ \Omega$, active area $A = 0.5\ \text{mm}^{2}$, $\varepsilon_{r} = 11.8$ and $v_{s} = 1\times10^{5}\ \text{m/s}$:

at which $\tau_{RC} = \tau_{\text{drift}} = 162\ \text{ps}$, giving $\tau = \sqrt{2}\,(162\ \text{ps}) = 229\ \text{ps}$ and a detector bandwidth $B \approx 0.35/\tau = 1.5\ \text{GHz}$. Note that quantum efficiency argues for a still wider layer, so a real design trades a little responsivity against speed rather than sitting exactly at this point.