Photonics Essentials: Chapter 3 Problems

Source

Thomas P. Pearsall, Photonics Essentials: An Introduction with Experiments (McGraw-Hill, 2003), Chapter 3, Photodiodes, Problems 3.1–3.5, printed pages 58–60.

See also

Photonics Essentials: From Diffusion Current to Equation 3.5 explains how the diffusion current on printed page 38 leads to the second-order spatial derivative in Equation 3.5.

Values read from the book’s plots are estimates. The calculations use

\[E_\gamma(\mathrm{eV})=\frac{1239.84}{\lambda(\mathrm{nm})}, \qquad \mathcal R=\eta\frac{q\lambda}{hc}.\]

Quick results

Problem

Result

3.1

Detection starts near \(0.67\ \mathrm{eV}\) and is suppressed above about \(1.13\ \mathrm{eV}\); the detector is Ge

3.2

\(\mathcal R_{1000}=0.65\ \mathrm{A/W}\), \(\eta\approx0.806\), \(I_{600}\approx0.390\ \mu\mathrm A\)

3.3

Graph estimate: \(I_d\approx1.5\ \mu\mathrm A\)

3.4

The straight semilog segment implies an exponential law; the printed voltage scale gives an unphysical \(n\approx0.19\)

3.5

\(P_D\approx9.21\ \mathrm{nW}\), \(\mathcal R=0.375\ \mathrm{A/W}\), \(I(1\ \mathrm m)=3.46\ \mathrm{nA}\)

Worked solutions

Problem 3.1: Filtered photodiode spectrum

Brief solution

1. Method.

Paraphrase. Interpret a measured spectrum made with an incandescent source, a silicon filter, a monochromator, and an unknown Ge or Si detector.

2. Key step.

The short-wavelength edge is near \(1100\ \mathrm{nm}\), or

3. Answer.

\[E_{\min}\approx\frac{1239.84}{1850} =\boxed{0.67\ \mathrm{eV}}.\]
\[\boxed{\text{germanium}}.\]
\[E_{\max}\approx\frac{1239.84}{1100} =\boxed{1.13\ \mathrm{eV}}.\]

This edge is caused by the silicon filter, not by the Ge detector. Silicon absorbs photons above its band gap and therefore blocks wavelengths shorter than roughly \(1.1\ \mu\mathrm m\).

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Paraphrase. Interpret a measured spectrum made with an incandescent source, a silicon filter, a monochromator, and an unknown Ge or Si detector.

The response first rises at about \(1850\ \mathrm{nm}\). Its photon energy is

\[E_{\min}\approx\frac{1239.84}{1850} =\boxed{0.67\ \mathrm{eV}}.\]

That long-wavelength edge agrees with the room-temperature Ge band gap. A silicon detector would stop responding near \(1100\ \mathrm{nm}\), so the detector must be

\[\boxed{\text{germanium}}.\]

The short-wavelength edge is near \(1100\ \mathrm{nm}\), or

\[E_{\max}\approx\frac{1239.84}{1100} =\boxed{1.13\ \mathrm{eV}}.\]

This edge is caused by the silicon filter, not by the Ge detector. Silicon absorbs photons above its band gap and therefore blocks wavelengths shorter than roughly \(1.1\ \mu\mathrm m\).

A monochromator set to \(\lambda\) can also transmit its second order at \(\lambda/2\). Without the silicon filter, visible second-order light could produce a false infrared response. The silicon filter absorbs most of that visible light, strongly suppressing the artifact.

Problem 3.2: Responsivity and quantum efficiency

Brief solution

1. Method.

At \(1000\ \mathrm{nm}\), divide the measured current by incident power:

2. Key step.

\[\mathcal R_{600} =(0.806)\frac{600}{1239.84} =0.390\ \mathrm{A/W},\]

3. Answer.

\[\mathcal R_{1000} =\frac{0.65\ \mu\mathrm A}{1.00\ \mu\mathrm W} =\boxed{0.65\ \mathrm{A/W}}.\]
\[\eta =\mathcal R\frac{1239.84}{\lambda} =(0.65)\frac{1239.84}{1000} =\boxed{0.806}.\]
\[\boxed{I_{600}=0.390\ \mu\mathrm A}.\]

The trial curve requested in part (d) is therefore

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At \(1000\ \mathrm{nm}\), divide the measured current by incident power:

\[\mathcal R_{1000} =\frac{0.65\ \mu\mathrm A}{1.00\ \mu\mathrm W} =\boxed{0.65\ \mathrm{A/W}}.\]

Since

\[\mathcal R=\eta\frac{\lambda(\mathrm{nm})}{1239.84},\]

the quantum efficiency is

\[\eta =\mathcal R\frac{1239.84}{\lambda} =(0.65)\frac{1239.84}{1000} =\boxed{0.806}.\]

Assuming this internal efficiency remains constant at \(600\ \mathrm{nm}\),

\[\mathcal R_{600} =(0.806)\frac{600}{1239.84} =0.390\ \mathrm{A/W},\]

and a \(1\ \mu\mathrm W\) signal produces

\[\boxed{I_{600}=0.390\ \mu\mathrm A}.\]

The trial curve requested in part (d) is therefore

\[\begin{split}\mathcal R(\lambda)\approx \begin{cases} 0.806\,\lambda/1239.84\ \mathrm{A/W}, &400\leq\lambda\lesssim1100\ \mathrm{nm},\\ 0,&\lambda\gtrsim1100\ \mathrm{nm}. \end{cases}\end{split}\]

\(\lambda\) (nm)

400

600

800

1000

1100

1200

1400

\(\mathcal R\) (A/W)

0.260

0.390

0.520

0.650

0.715

0

0

The abrupt cutoff is an idealization. A measured silicon response rolls off as absorption becomes weak near the indirect band edge.

Problem 3.3: Germanium dark current

Brief solution

1. Method.

In reverse bias the curve is nearly horizontal about three vertical divisions below zero. With \(5\times10^{-7}\ \mathrm{A/div}\),

2. Answer.

\[|I_d|\approx3(5\times10^{-7}) =\boxed{1.5\times10^{-6}\ \mathrm A}.\]

The reading is only accurate to roughly half a graph division. It is larger than the dark current normally measured from a comparable silicon diode. Three features increase it:

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In reverse bias the curve is nearly horizontal about three vertical divisions below zero. With \(5\times10^{-7}\ \mathrm{A/div}\),

\[|I_d|\approx3(5\times10^{-7}) =\boxed{1.5\times10^{-6}\ \mathrm A}.\]

The reading is only accurate to roughly half a graph division. It is larger than the dark current normally measured from a comparable silicon diode. Three features increase it:

  • Ge has a smaller band gap, so thermal generation is much stronger.

  • The area, \(8\times10^{-3}\ \mathrm{cm^2}\), provides appreciable bulk and junction volume.

  • Surface leakage, defects, and the measurement temperature add to the generation current.

Problem 3.4: Forward characteristic and ideality factor

Brief solution

1. Method.

A straight line on a plot of \(\log_{10}I\) against \(V\) means

2. Key step.

\[\Delta V_{\mathrm{dec}} =n\frac{k_BT}{q}\ln 10 \approx n(59.6\ \mathrm{mV})\]

3. Answer.

\[\Delta V_{\mathrm{dec}}\approx11\ \mathrm{mV}, \qquad n\approx\frac{11}{59.6}=\boxed{0.19}.\]

This is not physically credible for an ordinary p-n diode, whose ideality factor is normally at least one in this model. The likely explanation is a factor-of-ten error in the printed voltage axis. If the intended interval were \(0.50\ \mathrm V\), the same construction would give \(n\approx1.9\). The defensible result is therefore to report both the literal graph result and the apparent scale error.

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A straight line on a plot of \(\log_{10}I\) against \(V\) means

\[I=I_s\exp\left(\frac{qV}{nk_BT}\right).\]

For one decade of current,

\[\Delta V_{\mathrm{dec}} =n\frac{k_BT}{q}\ln 10 \approx n(59.6\ \mathrm{mV})\]

at \(300\ \mathrm K\).

The dashed segment in the printed graph rises by about 4.5 decades over \(0.050\ \mathrm V\), giving

\[\Delta V_{\mathrm{dec}}\approx11\ \mathrm{mV}, \qquad n\approx\frac{11}{59.6}=\boxed{0.19}.\]

This is not physically credible for an ordinary p-n diode, whose ideality factor is normally at least one in this model. The likely explanation is a factor-of-ten error in the printed voltage axis. If the intended interval were \(0.50\ \mathrm V\), the same construction would give \(n\approx1.9\). The defensible result is therefore to report both the literal graph result and the apparent scale error.