Photonics Essentials: Chapter 5 Problems

Source

Thomas P. Pearsall, Photonics Essentials: An Introduction with Experiments (McGraw-Hill, 2003), Chapter 5, Photoconductivity, Problems 5.1–5.6, printed page 100.

Worked solutions

Problem 5.1: Carriers from a laser pulse

Brief solution

1. Method.

The pulse energy is

2. Key step.

\[E_\gamma=\frac{hc}{\lambda} =3.31\times10^{-19}\ \mathrm J.\]

3. Answer.

\[N=\frac{E_{\mathrm{pulse}}}{E_\gamma} =\boxed{3.02\times10^2\ \text{electron-hole pairs}}.\]

The stated \(1\ \mathrm{cm^2}\) area affects the density, not this total.

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The pulse energy is

\[E_{\mathrm{pulse}}=P\Delta t =(10^{-7})(10^{-9})=10^{-16}\ \mathrm J.\]

One \(600\ \mathrm{nm}\) photon has energy

\[E_\gamma=\frac{hc}{\lambda} =3.31\times10^{-19}\ \mathrm J.\]

With complete absorption and one electron-hole pair per photon,

\[N=\frac{E_{\mathrm{pulse}}}{E_\gamma} =\boxed{3.02\times10^2\ \text{electron-hole pairs}}.\]

The stated \(1\ \mathrm{cm^2}\) area affects the density, not this total.

Problem 5.2: Why photodiode gain cannot exceed one

Brief solution

1. Method.

The ideal energy sequence is

2. Key step.

\[\text{valence electron}+\gamma \longrightarrow \text{one conduction electron}+\text{one hole} \longrightarrow \text{one collected electron charge}.\]

3. Answer.

\[G_{\mathrm{photodiode}} =\frac{\text{collected electrons}}{\text{absorbed photons}} \leq\boxed{1}.\]

This statement excludes avalanche multiplication, which is a different high-field gain mechanism.

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The ideal energy sequence is

\[\text{valence electron}+\gamma \longrightarrow \text{one conduction electron}+\text{one hole} \longrightarrow \text{one collected electron charge}.\]

The depletion field separates and collects the pair, but it does not return the same carrier to the absorber to circulate again. Therefore one absorbed photon supplies at most one elementary charge to the external circuit:

\[G_{\mathrm{photodiode}} =\frac{\text{collected electrons}}{\text{absorbed photons}} \leq\boxed{1}.\]

This statement excludes avalanche multiplication, which is a different high-field gain mechanism.

Problem 5.3: Steady-state carrier concentration

Brief solution

1. Method.

The photon generation rate is

2. Key step.

\[\mathcal V=(1\ \mathrm{cm^2})(2\times10^{-4}\ \mathrm{cm}) =2\times10^{-4}\ \mathrm{cm^3}.\]

3. Answer.

\[\Delta n=\Delta p =\frac{\dot N\tau}{\mathcal V} =\frac{(3.12\times10^{12})(10^{-6})}{2\times10^{-4}} =\boxed{1.56\times10^{10}\ \mathrm{cm^{-3}}}.\]

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The photon generation rate is

\[\dot N =\frac{P}{E_\gamma} =\frac{10^{-6}}{2(1.602\times10^{-19})} =3.12\times10^{12}\ \mathrm{s^{-1}}.\]

The illuminated volume is

\[\mathcal V=(1\ \mathrm{cm^2})(2\times10^{-4}\ \mathrm{cm}) =2\times10^{-4}\ \mathrm{cm^3}.\]

At steady state, excess population equals generation rate times lifetime:

\[\Delta n=\Delta p =\frac{\dot N\tau}{\mathcal V} =\frac{(3.12\times10^{12})(10^{-6})}{2\times10^{-4}} =\boxed{1.56\times10^{10}\ \mathrm{cm^{-3}}}.\]

Problem 5.4: Photographic latent image

Brief solution

1. Method.

The three energy-diagram stages can be summarized as follows.

2. Key step.

\[\mathrm{Ag^+}+e^-\longrightarrow\mathrm{Ag^0}.\]

3. Answer.

Development and fixing. Developer preferentially reduces an exposed grain around its silver seed, amplifying the small cluster into an opaque metallic-silver grain. Fixer then dissolves the unexposed AgBr. The remaining silver distribution is the visible negative.

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The three energy-diagram stages can be summarized as follows.

Exposure. A photon excites an electron across the AgBr energy gap:

\[\mathrm{AgBr}+\gamma \longrightarrow e^-_{\mathrm{CB}}+h^+_{\mathrm{VB}}.\]

The mobile electron is trapped at a sensitization site.

Latent-image formation. The negatively charged trap attracts a mobile \(\mathrm{Ag^+}\) ion and reduces it:

\[\mathrm{Ag^+}+e^-\longrightarrow\mathrm{Ag^0}.\]

Repeated events create a small cluster of neutral silver atoms. This cluster stores the invisible latent image.

Development and fixing. Developer preferentially reduces an exposed grain around its silver seed, amplifying the small cluster into an opaque metallic-silver grain. Fixer then dissolves the unexposed AgBr. The remaining silver distribution is the visible negative.

Problem 5.5: Hole lifetime after sensitization

Brief solution

1. Method.

From Table 5.2,

2. Key step.

\[\begin{split}\begin{aligned} \frac1{\tau_p} &=n_{r1}vs_{p1}+n_{r2}vs_{p2}\\ &=(2\times10^{15})(10^7)(10^{-15}) +(2\times10^{16})(10^7)(10^{-15})\\ &=2.2\times10^8\ \mathrm{s^{-1}}, \end{aligned}\end{split}\]

3. Answer.

\[\boxed{\tau_p=4.55\times10^{-9}\ \mathrm s\approx5\ \mathrm{ns}}.\]

Sensitization lengthens the electron lifetime but shortens the hole lifetime.

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From Table 5.2,

\[N_{r1}=2\times10^{15},\quad N_{r2}=2\times10^{16}\ \mathrm{cm^{-3}},\quad v=10^7\ \mathrm{cm/s},\]

and \(s_{p1}=s_{p2}=10^{-15}\ \mathrm{cm^2}\). Under the occupancy approximations used in the example, \(n_{r1}\approx N_{r1}\) and \(n_{r2}\approx N_{r2}\). Hence

\[\begin{split}\begin{aligned} \frac1{\tau_p} &=n_{r1}vs_{p1}+n_{r2}vs_{p2}\\ &=(2\times10^{15})(10^7)(10^{-15}) +(2\times10^{16})(10^7)(10^{-15})\\ &=2.2\times10^8\ \mathrm{s^{-1}}, \end{aligned}\end{split}\]

so

\[\boxed{\tau_p=4.55\times10^{-9}\ \mathrm s\approx5\ \mathrm{ns}}.\]

Sensitization lengthens the electron lifetime but shortens the hole lifetime.

Problem 5.6: Reversed capture preference

Brief solution

1. Method.

Now \(s_{n2}=10^{-15}\ \mathrm{cm^2}\) and \(s_{p2}=10^{-17}\ \mathrm{cm^2}\). The type-2 center captures electrons efficiently but holes slowly, so the sensitization preference reverses: holes become the long-lived carrier.

2. Key step.

whereas electron capture through the same active population gives

3. Answer.

\[\tau_p\sim\frac{1}{N_{r1}vs_{p2}} =\frac{1}{(2\times10^{15})(10^7)(10^{-17})} =\boxed{5\times10^{-6}\ \mathrm s},\]

\[\tau_n\sim\frac{1}{N_{r1}vs_{n2}} =\boxed{5\times10^{-8}\ \mathrm s}.\]

These are order-of-magnitude lifetimes, not an exact trap-occupancy solution. Direct substitution into the occupancy approximation used for Equations 5.18–5.23 predicts \(p_{r1}>N_{r1}\), which is impossible. That signals that the original high-illumination occupancy assumption no longer applies. An exact pair of lifetimes would require the illumination level, charge neutrality condition, and coupled rate equations, none of which the problem specifies.

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Now \(s_{n2}=10^{-15}\ \mathrm{cm^2}\) and \(s_{p2}=10^{-17}\ \mathrm{cm^2}\). The type-2 center captures electrons efficiently but holes slowly, so the sensitization preference reverses: holes become the long-lived carrier.

A useful limiting estimate keeps the example’s available active type-2 population near \(N_{r1}\). Then

\[\tau_p\sim\frac{1}{N_{r1}vs_{p2}} =\frac{1}{(2\times10^{15})(10^7)(10^{-17})} =\boxed{5\times10^{-6}\ \mathrm s},\]

whereas electron capture through the same active population gives

\[\tau_n\sim\frac{1}{N_{r1}vs_{n2}} =\boxed{5\times10^{-8}\ \mathrm s}.\]

These are order-of-magnitude lifetimes, not an exact trap-occupancy solution. Direct substitution into the occupancy approximation used for Equations 5.18–5.23 predicts \(p_{r1}>N_{r1}\), which is impossible. That signals that the original high-illumination occupancy assumption no longer applies. An exact pair of lifetimes would require the illumination level, charge neutrality condition, and coupled rate equations, none of which the problem specifies.