Electron-Hole Pair Generation MCQ Quiz - Objective Question with Answer for Electron-Hole Pair Generation - Download Free PDF

Ans.(3)

A. Correct: In semiconductor diodes, carriers can indeed be generated by photo-excitation, especially in devices like photodiodes or solar cells where light generates electron-hole pairs.

B. Correct: When a photodiode is exposed to light whose energy is greater than the band gap (E > E_g) of the semiconductor, electron-hole pairs are created due to photon absorption. This is a standard principle in photodetector operation.

C. Incorrect: Photovoltaic devices do not convert electricity into optical radiation. Instead, they convert optical radiation (light) into electricity, as seen in solar cells. Therefore, this statement is false.

D. Correct: A photodiode can be used as a photodetector, which is a device that detects optical signals (light) and converts them into electrical signals. This is a well-established application.

E. Incorrect: While Zener diode specifications do relate to voltage regulation, this statement is outside the context of photonic devices discussed in the other options and is not relevant to the comparison here.

Hence, the correct statements are A, B, D, and E, which corresponds to Option 3.

Concept:

Extrinsic p-type semiconductors are formed when a trivalent impurity is added to a pure semiconductor. Examples of trivalent impurity are Boron, Gallium, and Indium.

Also, the law of mass action is used to determine the minority carriers in a doped semiconductor.

According to law:

n₀.p₀ = ni2

n₀ = concentration of electrons

p₀ = concentration of holes

For a steady-state carrier injection, the excess carrier concentration dies out exponentially for a distance ‘x’ due to recombination.

Mathematically this is defined as:

δp(x) = δp(0) e-x/Lp

δn(x) = δn(0) e-x/Ln

Ln and Lp are the diffusion length of electrons and holes respectively.

Calculation:

Given, P = 10¹⁷ cm^-3 = P₀

\(\rm n_{0}=\frac{n_{i}^{2}}{P_{0}}=\frac{10^{20}}{10^{17}}=10^{3} \mathrm{~cm}^{-3}\)

L_n = 2 μm

n'_PO = 10¹⁴ cm^-3

Excess electron concentration at any distance x is

δn_p(x) = n'_P0 e^-x/L_n

= 10¹⁴ e^-2/2

= 10¹⁴ e^-1

= 3.67 × 10¹³ cm^-3

= 0.367 × 10¹⁴ cm^-3

The current due to electron:

I_n = nq μ_n EA

The current due to holes:

I_p = pq μ_p EA

Given:

The total current is given as I

 \(\rm I_n=\dfrac{3}{4}I\)

\(\rm I_p=\dfrac{1}{4}I\)

\(\dfrac{I_n}{I_p}=3=\dfrac{nqμ_nEA}{pqμ_pEA}=\dfrac{nμ_n}{pμ_p}\)

\(\dfrac{n}{p}=\dfrac{3μ_p}{μ_n}=3\left[\dfrac{V_{d_p}}{V_{d_n}}\right]\) [∵ V_d = μE]

\(\dfrac{n}{p}=3\times\dfrac{1}{2}=1.5\)

Concept:

If we optically generate excess charge carries in semiconductor

then generation rate

\({g_{op}} = \frac{{{\delta _n}}}{{{\tau _n}}}\)

Calculation

10¹² cm^-3 are generated in 3 μsec

\({g_{op}} = \frac{{{{10}^{12}}}}{{3\;\mu sec}}\)

From given Relation

\(\frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} = \frac{{{\delta _n}}}{{{\tau _n}}}\)

\({\delta _n} = \frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} \times 1\;\mu sec\)

\(= \frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} \times 1 \times {10^{ - 6}}\)c

⇒ 3.33 × 10¹¹ cm^-3

≈ 3 × 10¹¹ /cm³

Illumination of a semiconductor produces EHP i.e. electrons and holes are created equally irrespective of concentration of electrons and holes in the semiconductor.

i.e. Δn = Δp

Δn × Δp ≠ n_i²

The semiconductor is not in equilibrium, due to generation of EHP due to light illumination.

Concept:

Extrinsic p-type semiconductors are formed when a trivalent impurity is added to a pure semiconductor. Examples of trivalent impurity are Boron, Gallium, and Indium.

Also, the law of mass action is used to determine the minority carriers in a doped semiconductor.

According to law:

n₀.p₀ = ni2

n₀ = concentration of electrons

p₀ = concentration of holes

For a steady-state carrier injection, the excess carrier concentration dies out exponentially for a distance ‘x’ due to recombination.

Mathematically this is defined as:

δp(x) = δp(0) e-x/Lp

δn(x) = δn(0) e-x/Ln

Ln and Lp are the diffusion length of electrons and holes respectively.

Calculation:

Given, P = 10¹⁷ cm^-3 = P₀

\(\rm n_{0}=\frac{n_{i}^{2}}{P_{0}}=\frac{10^{20}}{10^{17}}=10^{3} \mathrm{~cm}^{-3}\)

L_n = 2 μm

n'_PO = 10¹⁴ cm^-3

Excess electron concentration at any distance x is

δn_p(x) = n'_P0 e^-x/L_n

= 10¹⁴ e^-2/2

= 10¹⁴ e^-1

= 3.67 × 10¹³ cm^-3

= 0.367 × 10¹⁴ cm^-3

Considering low level injection the excess electron generated can be neglected. The excess holes generated at time \({\rm{t}} = 0\) is \({\rm{\delta p}}\left( 0 \right) = {{\rm{G}}_{{\rm{opt}}}}.{{\rm{\tau }}_{{\rm{po}}}} \Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{20}} \times {\rm{\;}}0.1{\rm{\;\mu sec}}\)

\(\Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{13}}{\rm{holes}}/{\rm{c}}{{\rm{m}}^3}\)

Now hole concentration as a function of time is

\(\begin{array}{l} {\rm{\delta p}}\left( {\rm{t}} \right) = {{\rm{\delta }}_{\rm{p}}}\left( {\rm{O}} \right){{\rm{e}}^{ - {\rm{t}}/{{\rm{\tau }}_{{\rm{po}}}}{\rm{\;}}}}\\ {\rm{\delta p}}\left( {0.3{\rm{\;\mu sec}}} \right) = 1.5 \times {10^{13}}.{{\rm{e}}^{ - \frac{{0.3 \times {{10}^{ - 6}}}}{{0.1 \times {{10}^{ - 6}}}}}}\\ = 1.5 \times {10^{13}} \times {{\rm{e}}^{ - 3}}\\ \Rightarrow {{\rm{\delta }}_{\rm{p}}}\left( {0.3{\rm{\mu sec}}} \right) = 7.47 \times {10^{11}}{\rm{\;hole}}/{\rm{c}}{{\rm{m}}^3} \end{array}\)

Given,

NA­ = 1015 cm-3 ⇒ hole concentration p0 = 1 × 1015 cm-3

At, t = 0, 1014 EHP created

Excess holes generated Δp = 1014 cm-3

Excess electrons generated Δn = 1014 cm-3

Carrier recombination life time τn = τP = 10-8 s

A time ‘t’ the carrier contention of electrons

δn(t) = Δn e-t/τP

\(= 10^{14}\left({e^{ - \frac{{20}}{{10}}}}\right)\)

= 1.35 × 10+13 cm-3

Similarly,

δp(t) = 1.35 × 1013 cm-3

Total hole concentration p(t) = p + δP(t) = 100 × 1013 + 1.35 × 1013 cm-3

= 101.35 × 1013 cm-3

Total electron concentration n(t) = n + δn(t) = 0 + 1.35 × 1013

∴ Conductivity = q[p(t) μP + n(t) μn]

= 1.6 × 10-19 [101.35 × 1013 × 800 + 1.35 × 1013 × 1100]

= 0.13 S/cm

Illumination of a semiconductor produces EHP i.e. electrons and holes are created equally irrespective of concentration of electrons and holes in the semiconductor.

i.e. Δn = Δp

Δn × Δp ≠ n_i²

The semiconductor is not in equilibrium, due to generation of EHP due to light illumination.

Concept:

Extrinsic p-type semiconductors are formed when a trivalent impurity is added to a pure semiconductor. Examples of trivalent impurity are Boron, Gallium, and Indium.

Also, the law of mass action is used to determine the minority carriers in a doped semiconductor.

According to law:

n₀.p₀ = ni2

n₀ = concentration of electrons

p₀ = concentration of holes

For a steady-state carrier injection, the excess carrier concentration dies out exponentially for a distance ‘x’ due to recombination.

Mathematically this is defined as:

δp(x) = δp(0) e-x/Lp

δn(x) = δn(0) e-x/Ln

Ln and Lp are the diffusion length of electrons and holes respectively.

Calculation:

Given, P = 10¹⁷ cm^-3 = P₀

\(\rm n_{0}=\frac{n_{i}^{2}}{P_{0}}=\frac{10^{20}}{10^{17}}=10^{3} \mathrm{~cm}^{-3}\)

L_n = 2 μm

n'_PO = 10¹⁴ cm^-3

Excess electron concentration at any distance x is

δn_p(x) = n'_P0 e^-x/L_n

= 10¹⁴ e^-2/2

= 10¹⁴ e^-1

= 3.67 × 10¹³ cm^-3

= 0.367 × 10¹⁴ cm^-3

The current due to electron:

I_n = nq μ_n EA

The current due to holes:

I_p = pq μ_p EA

Given:

The total current is given as I

 \(\rm I_n=\dfrac{3}{4}I\)

\(\rm I_p=\dfrac{1}{4}I\)

\(\dfrac{I_n}{I_p}=3=\dfrac{nqμ_nEA}{pqμ_pEA}=\dfrac{nμ_n}{pμ_p}\)

\(\dfrac{n}{p}=\dfrac{3μ_p}{μ_n}=3\left[\dfrac{V_{d_p}}{V_{d_n}}\right]\) [∵ V_d = μE]

\(\dfrac{n}{p}=3\times\dfrac{1}{2}=1.5\)

Considering low level injection the excess electron generated can be neglected. The excess holes generated at time \({\rm{t}} = 0\) is \({\rm{\delta p}}\left( 0 \right) = {{\rm{G}}_{{\rm{opt}}}}.{{\rm{\tau }}_{{\rm{po}}}} \Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{20}} \times {\rm{\;}}0.1{\rm{\;\mu sec}}\)

\(\Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{13}}{\rm{holes}}/{\rm{c}}{{\rm{m}}^3}\)

Now hole concentration as a function of time is

\(\begin{array}{l} {\rm{\delta p}}\left( {\rm{t}} \right) = {{\rm{\delta }}_{\rm{p}}}\left( {\rm{O}} \right){{\rm{e}}^{ - {\rm{t}}/{{\rm{\tau }}_{{\rm{po}}}}{\rm{\;}}}}\\ {\rm{\delta p}}\left( {0.3{\rm{\;\mu sec}}} \right) = 1.5 \times {10^{13}}.{{\rm{e}}^{ - \frac{{0.3 \times {{10}^{ - 6}}}}{{0.1 \times {{10}^{ - 6}}}}}}\\ = 1.5 \times {10^{13}} \times {{\rm{e}}^{ - 3}}\\ \Rightarrow {{\rm{\delta }}_{\rm{p}}}\left( {0.3{\rm{\mu sec}}} \right) = 7.47 \times {10^{11}}{\rm{\;hole}}/{\rm{c}}{{\rm{m}}^3} \end{array}\)

Concept:

Charge carriers in presence of illumination

n = n₀ + δ_n

p = p₀ + δ_p

where δ_n and δ_p are excess charge carriers given by:

δ_n = δ_p = Gτ_p

Conductivity

σ = e(nu_n + pu_p)

σ = e((n₀ + δ_n)u_n + (p₀ + δ_p)μ_p)

σ = e(n₀u_n + p₀u_p) + e(μ_n + μ_p)δ_n

δ_n = Gτ_p

σ = e(n₀u_n + p₀u_p) + e(u_n + μ_p) Gτ_p

Change in conductivity

= e(u_n + μ_p)Gτ_p

Calculations:

Δσ = 1.6 × 10^-19(1300 + 500) × 10^-4 × 2 × 10¹⁹ × 10^-6Ω^-1

= 1.6(1800)(2) × 10^-10Ω^-1

⇒ 5760 × 10^-10Ω^-1

Concept:

The electron and hole generation rate is given by

\(G = \frac{{{\rm{\Delta }}n}}{\tau }\) 

Where Δn = excess electron-hole pair generated and τ = minority carrier lifetime.

For n-type Si, hole is minority carrier

Hence

\(G = \frac{{{\rm{\Delta }}n}}{{\tau p}}\)    …(1)

Calculation:

\(G = \frac{{2 \times {{10}^{13}}}}{{{{10}^{ - 6}}}}\) 

Concept:

If we optically generate excess charge carries in semiconductor

then generation rate

\({g_{op}} = \frac{{{\delta _n}}}{{{\tau _n}}}\)

Calculation

10¹² cm^-3 are generated in 3 μsec

\({g_{op}} = \frac{{{{10}^{12}}}}{{3\;\mu sec}}\)

From given Relation

\(\frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} = \frac{{{\delta _n}}}{{{\tau _n}}}\)

\({\delta _n} = \frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} \times 1\;\mu sec\)

\(= \frac{{{{10}^{12}}}}{{3 \times {{10}^{ - 6}}}} \times 1 \times {10^{ - 6}}\)c

⇒ 3.33 × 10¹¹ cm^-3

≈ 3 × 10¹¹ /cm³