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17-Phys-B2 Electro-Optical Engineering · December 2018

Question 3 of 7: PIN Photodiode — Load Line, Saturation, Output Characteristic and Bandwidth Limit

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 17-Phys-B2 Electro-Optical Engineering, National Examination December 2018 — a three-hour closed-book examination (one 8.5×11 inch double-sided handwritten note sheet permitted). The cover page states any five of the seven questions constitute a complete paper and only the first five as they appear in the answer book are marked; every question is nonetheless answered in full below so the paper remains a complete study resource.

Reference texts. G. Keiser, Optical Fiber Communications, 4th ed. (fiber modes, dispersion and bend loss; laser-diode longitudinal modes and DFB gratings; PIN photodiode responsivity, noise and receiver design; link power and dispersion budgets); B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 2nd ed. (semiconductor optical gain, LED spontaneous-emission linewidth, avalanche-photodiode noise); A. Yariv, Quantum Electronics / E. Hecht, Optics, 5th ed. (Mach–Zehnder interferometry and electro-optic modulators).

Question 3: PIN Photodiode — Load Line, Saturation, Output Characteristic and Bandwidth Limit (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.

Given data
QuantitySymbolValue
Responsivity$R$0.25 A/W
Dark current$I_d$0.5 nA
Depletion width$d$30 μm
Carrier (saturation) velocity$v_d$$5\times10^4$ m/s
Junction capacitance$C_j$0.45 pF
Reverse bias$V_{\text{bias}}$10 V
Load resistance$R_L$2 MΩ

Find. (a) circuit and loop equation, (b) load line, (c) saturation optical power, (d) $V_{\text{out}}(P)$ from 5–50 μW, (e) illuminated I-V and operating mode, (f) bandwidth-limiting mechanism.

Approach. The photocurrent $I_{\text{ph}}=RP+I_d$ flows through $R_L$, so $V_{\text{out}}=I_{\text{ph}}R_L$ traces a load line that clamps once the diode voltage runs out of headroom; bandwidth is then set by whichever of the carrier transit time or the $R_LC_j$ time constant is slower.

  1. Part (a) — circuit and loop equation.
    Vbias optical power P PIN diode RL Vout Loop: Vbias = I(RL) . RL + VD(reverse)
    PIN detector bias circuit and loop equation.
    The reverse-biased PIN diode acts as a light-controlled current source in series with the bias supply and the load resistor. Kirchhoff's voltage law around the loop gives $$V_{\text{bias}}=I(R_L)\,R_L+V_D,\qquad I(R_L)=I_{\text{ph}}+I_d=RP+I_d,$$ where $V_D$ is the (reverse) voltage left across the diode itself.
  2. Part (b) — load line. Solving the loop equation for $I$ in terms of $V_D$ gives the straight line $I=(V_{\text{bias}}-V_D)/R_L$, running from $(V_D,I)=(V_{\text{bias}},0)=(10\ \text{V},0)$ down to $(0,V_{\text{bias}}/R_L)=(0,5.0\ \mu\text{A})$, slope $-1/R_L$.
  3. Part (c) — saturation optical power. The diode saturates when all of $V_{\text{bias}}$ has been dropped across $R_L$ and the diode itself is left with essentially zero reverse voltage — the load-line endpoint on the current axis: $$I_{\text{sat}}=\frac{V_{\text{bias}}}{R_L}=\frac{10\ \text{V}}{2\times10^{6}\ \Omega} =5.0\ \mu\text{A},$$ $$P_{\text{sat}}=\frac{I_{\text{sat}}}{R}=\frac{5.0\ \mu\text{A}}{0.25\ \text{A/W}} =\boxed{20.0\ \mu\text{W}}.$$
  4. Part (d) — output voltage vs. optical power. Below saturation $V_{\text{out}}=I_{\text{ph}}R_L=RPR_L$, a straight line of slope $RR_L=0.25\times2\times10^{6}=5\times10^{5}\ \text{V/W}=0.5\ \text{V}/\mu\text{W}$; above $P_{\text{sat}}=20.0\ \mu\text{W}$ it clamps at $V_{\text{bias}}=10\ \text{V}$.
$P$ (μW)$I_{\text{ph}}$ (μA)$V_{\text{out}}$ (V)
51.252.5
102.505.0
153.757.5
20 (=$P_{\text{sat}}$)5.0010.0 (clamped)
25–506.25–12.510.0 (clamped, saturated)
  1. Part (e) — illuminated I-V and operating mode.
    V (volt) I (μA) 10 5 load line (slope -1/RL) illuminated I-V (I ≈ Iph, reverse bias) Q-point (saturation, 5 μA, 20 μW)
    I-V load line (red) and illuminated diode characteristic (blue, reverse bias); Q-point at saturation.
    Under constant illumination and reverse bias, the diode's I-V characteristic is approximately a horizontal line at $I\approx I_{\text{ph}}$ across the whole reverse-bias range (the photocurrent is essentially independent of $V_D$ once fully depleted), intersecting the load line at the operating (Q-) point. Because the diode is operated under an applied reverse bias that sweeps out photo-generated carriers rather than at zero bias with an open-circuit photovoltage, it is operating in photoconductive mode (as opposed to the zero-bias photovoltaic mode used e.g. in solar cells).
  2. Part (f) — bandwidth-limiting mechanism. The transit-time-limited bandwidth follows from the depletion-region crossing time, $$\tau_{\text{tr}}=\frac{d}{v_d}=\frac{30\times10^{-6}\ \text{m}}{5\times10^{4}\ \text{m/s}} =0.60\ \text{ns},\qquad f_{\text{tr}}\approx\frac{0.44}{\tau_{\text{tr}}}=\boxed{0.733\ \text{GHz}}.$$ The $R_LC_j$ time constant, using the given $2\ \text{M}\Omega$ load, is $$\tau_{RC}=R_LC_j=(2\times10^{6}\ \Omega)(0.45\times10^{-12}\ \text{F})=900\ \text{ns}, \qquad f_{RC}=\frac{1}{2\pi\tau_{RC}}=\boxed{176.8\ \text{kHz}}.$$ Since $f_{RC}\ (176.8\ \text{kHz})$ is more than three decades below $f_{\text{tr}}\ (0.733\ \text{GHz})$, the huge $2\ \text{M}\Omega$ load resistor — not the depletion transit time — is what limits the usable bandwidth: this detector is firmly RC-limited.