Photonics Essentials: Chapter 4 Problems

Source

Thomas P. Pearsall, Photonics Essentials: An Introduction with Experiments (McGraw-Hill, 2003), Chapter 4, Electrical Response Time of Diodes, Problems 4.1–4.4, printed pages 75–76.

Worked solutions

Problem 4.1: A 4 MHz fiber receiver

Brief solution

1. Method.

Silicon loses band-to-band response near \(1.1\ \mu\mathrm m\), whereas germanium still absorbs at \(1.3\ \mu\mathrm m\). Of the two choices,

2. Key step.

\[\tau_{\max} =\frac{1}{\pi(4.0\times10^6)} =79.6\ \mathrm{ns}.\]

3. Answer.

\[\boxed{\text{use the Ge photodiode}}.\]

\[C_{\mathrm{total,max}} =\frac{\tau_{\max}}{R_L} =\frac{79.6\ \mathrm{ns}}{50\ \Omega} =\boxed{1.59\ \mathrm{nF}}.\]

The printed C-V graph shows approximately \(3.5\ \mathrm{nF}\) at zero bias, \(2.0\ \mathrm{nF}\) at \(-2\ \mathrm V\), and \(1.55\ \mathrm{nF}\) at \(-3\ \mathrm V\). Therefore choose at least about \(3\ \mathrm V\) reverse bias, preferably with margin for cable and amplifier input capacitance:

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Silicon loses band-to-band response near \(1.1\ \mu\mathrm m\), whereas germanium still absorbs at \(1.3\ \mu\mathrm m\). Of the two choices,

\[\boxed{\text{use the Ge photodiode}}.\]

The specified relation gives the largest permissible time constant:

\[\tau_{\max} =\frac{1}{\pi(4.0\times10^6)} =79.6\ \mathrm{ns}.\]

With a \(50\ \Omega\) load,

\[C_{\mathrm{total,max}} =\frac{\tau_{\max}}{R_L} =\frac{79.6\ \mathrm{ns}}{50\ \Omega} =\boxed{1.59\ \mathrm{nF}}.\]

The printed C-V graph shows approximately \(3.5\ \mathrm{nF}\) at zero bias, \(2.0\ \mathrm{nF}\) at \(-2\ \mathrm V\), and \(1.55\ \mathrm{nF}\) at \(-3\ \mathrm V\). Therefore choose at least about \(3\ \mathrm V\) reverse bias, preferably with margin for cable and amplifier input capacitance:

\[C_D+C_{\mathrm{stray}}\leq1.59\ \mathrm{nF}.\]

For example, \(-4\ \mathrm V\) gives roughly \(C_D=1.35\ \mathrm{nF}\) and leaves about \(0.24\ \mathrm{nF}\) for parasitics.

Problem 4.2: Junction capacitance

Brief solution

1. Method.

For a one-sided abrupt silicon junction under reverse-bias magnitude \(V_R\),

2. Key step.

\[\log(C/A)=\frac12\log N_D+\text{constant}.\]

3. Answer.

Thus every curve has slope \(1/2\); reverse bias shifts a curve downward by widening the depletion region.

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For a one-sided abrupt silicon junction under reverse-bias magnitude \(V_R\),

\[\frac{C}{A} =\sqrt{\frac{\epsilon_s qN_D} {2(V_{\mathrm{bi}}+V_R)}}.\]

Use \(\epsilon_s=11.8\epsilon_0\), \(\epsilon_0=8.854\times10^{-14}\ \mathrm{F/cm}\), and the representative silicon value \(V_{\mathrm{bi}}=0.8\ \mathrm V\). The results are:

\(N_D\) (\(\mathrm{cm^{-3}}\))	0 V	1 V	5 V	10 V
\(10^{15}\)	\(1.02\times10^{-8}\)	\(6.82\times10^{-9}\)	\(3.80\times10^{-9}\)	\(2.78\times10^{-9}\)
\(10^{16}\)	\(3.23\times10^{-8}\)	\(2.16\times10^{-8}\)	\(1.20\times10^{-8}\)	\(8.80\times10^{-9}\)
\(10^{17}\)	\(1.02\times10^{-7}\)	\(6.82\times10^{-8}\)	\(3.80\times10^{-8}\)	\(2.78\times10^{-8}\)
\(10^{18}\)	\(3.23\times10^{-7}\)	\(2.16\times10^{-7}\)	\(1.20\times10^{-7}\)	\(8.80\times10^{-8}\)

All entries are in \(\mathrm{F/cm^2}\). On log-log axes each voltage curve is a straight line:

Calculated junction capacitance per area. Each curve has log-log slope \(1/2\).

\[\log(C/A)=\frac12\log N_D+\text{constant}.\]

Thus every curve has slope \(1/2\); reverse bias shifts a curve downward by widening the depletion region.

Problem 4.3: Built-in voltage from Table 4.1

Brief solution

1. Method.

The area is constant, so multiplying \(1/C^2\) by \(A^2\) changes only the vertical scale and not the voltage-axis intercept. A least-squares fit to all 21 tabulated points gives, with \(C\) in pF and applied voltage \(V\) in volts,

2. Key step.

\[V_{\mathrm{intercept}} =-\frac{4.253\times10^{-4}}{-5.586\times10^{-4}} =0.761\ \mathrm V.\]

3. Answer.

\[\boxed{V_{\mathrm{bi}}\approx0.76\ \mathrm V}.\]

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The area is constant, so multiplying \(1/C^2\) by \(A^2\) changes only the vertical scale and not the voltage-axis intercept. A least-squares fit to all 21 tabulated points gives, with \(C\) in pF and applied voltage \(V\) in volts,

\[\frac{1}{C^2} =(-5.586\times10^{-4})V +4.253\times10^{-4}\ \mathrm{pF^{-2}}.\]

Set the fitted ordinate to zero:

\[V_{\mathrm{intercept}} =-\frac{4.253\times10^{-4}}{-5.586\times10^{-4}} =0.761\ \mathrm V.\]

Therefore,

\[\boxed{V_{\mathrm{bi}}\approx0.76\ \mathrm V}.\]

Problem 4.4: Area and response time

Brief solution

1. Method.

For a device of thickness \(d\), resistivity \(\rho\), permittivity \(\epsilon\), and area \(A\),

2. Key step.

\[R_D=\rho\frac{d}{A}, \qquad C_D=\epsilon\frac{A}{d}.\]

3. Answer.

\[\boxed{R_DC_D=\rho\epsilon},\]

which is independent of area.

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For a device of thickness \(d\), resistivity \(\rho\), permittivity \(\epsilon\), and area \(A\),

\[R_D=\rho\frac{d}{A}, \qquad C_D=\epsilon\frac{A}{d}.\]

Their product is

\[\boxed{R_DC_D=\rho\epsilon},\]

which is independent of area.

The complete circuit also has an external resistance \(R_L\):

\[\tau=(R_D\parallel R_L)C_D.\]

In the usual reverse-biased photodiode, \(R_D\gg R_L\), so

\[\tau\approx R_LC_D\propto A.\]

The fixed load breaks the internal area cancellation. A smaller junction has less capacitance and is therefore faster, until transit time or another parasitic becomes the dominant limit.