Compensated Semiconductors MCQ Quiz - Objective Question with Answer for Compensated Semiconductors - Download Free PDF

The relationship between carrier concentrations in a semiconductor is given by the mass action law, which states:

n × p = n_i²

Where:

• n = concentration of electrons (minority carriers in p-type material, in cm^-3).
• p = concentration of holes (majority carriers in p-type material, in cm^-3).
• n_i = intrinsic carrier concentration of silicon (in cm^-3).

Given data:

• Intrinsic carrier concentration, n_i = 1.5 × 10¹⁰ cm^-3.
• Acceptor concentration (boron doping), N_A = 1.5 × 10¹⁴ cm^-3.
• In a p-type semiconductor, the majority carrier concentration (holes) is approximately equal to the acceptor concentration: p ≈ N_A.

Step 1: Apply the mass action law

Using the mass action law:

n × p = n_i²

Substitute p ≈ N_A:

n × N_A = n_i²

Solve for the electron concentration, n:

n = n_i² / N_A

Step 2: Substitute the given values

Intrinsic carrier concentration: n_i = 1.5 × 10¹⁰ cm^-3

Acceptor concentration: N_A = 1.5 × 10¹⁴ cm^-3

Substitute these values into the equation:

n = (1.5 × 10¹⁰)² / (1.5 × 10¹⁴)

First, calculate n_i²:

n_i² = (1.5 × 10¹⁰) × (1.5 × 10¹⁰) = 2.25 × 10²⁰

Now, divide by N_A:

n = (2.25 × 10²⁰) / (1.5 × 10¹⁴)

n = 1.5 × 10⁶ cm^-3

The concentration of electrons in the silicon sample is 1.5 × 10⁶ cm^-3.

Calculation:

Initial (without light) hole concentration under thermal equilibrium is

\(\rm p_0 = \frac{n_i^2}{n_0} = \frac{n_i^2}{N_D} = \frac{(10^{10})^2}{10^{16}} = 10^4 cm^{-3}\)

After light incident the excess hole concentration is

p_n(0) = p_n (∞) + [p_n (0) - p_n (∞)]e^-x/L_p

pn(x) = p0 + [pn (0) - p0]e-x/Lp

p_n(x) = 10⁴ + 10⁸ \(\rm e^{-x/100 \times 10^{-4}}\) cm^-3

\(\rm p_n(x) = 10^4 + 10^8 e^{-10^{2}x} cm^{-3}\)

i) For the same dopant concentration, the conductivity of an n-type is higher than p-type since

μ­n > μp and σ ∝ μ

μn = mobility of electron

μp = mobility of hole

ii) Under high electric fields (due to the voltage applied) the mobility of electrons decreases due to lattice collisions. 

iii) When the doping concentration is high either electrons or holes are in high concentration, hence drift current is significant compared to the diffusion current. While at low concentration diffusion current is significant.

iv) Direct bandgap semiconductors:

We can see that there is no change in momentum during recombination.

(i.e., when electrons move from conduction to valence)

Indirect bandgap semiconductor:

We can see that there is a finite change in momentum ΔP during recombination.

The momentum of various particles:

i) Photon → h / λ = E / C

ii) Phonon → E / v

As v ≪ C

∴ The momentum of the photon is negligible.

And also the change in momentum in the case of GaAs is also negligible.

∴ That momentum can be given to a photon.

∴ Radiative recombination is possible in the case of GaAs. 

Concept:

For a compensated semiconductor with acceptor concentration greater than the donor concentration, the majority carrier hole concentration is calculated as:

\({p_0} = \frac{{{N_a} - {N_d}}}{2} + \sqrt {{{\left( {\frac{{{N_a} - {N_d}}}{2}} \right)}^2} + n_i^2} \)

With N_a - N_d ≫ n_i, the above equation becomes:

p₀ ≅ N_a - N_d

Also, the charge carriers at thermal equilibrium follow mass-action law, i.e.

\({n_0}{p_0} = n_i^2\)

Calculation:

Boron is an acceptor atom (i.e a p-type impurity)

N_a = 1.5 × 10¹⁵ cm^-3

Arsenic is a donor atom (i.e. an n-type impurity)

N_d = 8 × 10¹⁴ cm^-3

Since, N_a > N_d, the given semiconductor is a p-type semiconductor.

Also, since N_a - N_d ≫ n_i the majority carrier hole concentration will be:

p₀ = N_a - N_d

p₀ = 1.5 × 10¹⁵ – 8 × 10¹⁴ cm^-3

p₀ = (15 - 8) × 10¹⁴ cm_-3

p₀ = 7 × 10¹⁴ cm^-3

Using mass action law

\({n_0} = \frac{{n_i^2}}{{{p_0}}} = \frac{{{{10}^{20}}}}{{7 \times {{10}^{14}}}}\)

n₀ = 1.42 × 10⁵ cm^-3

Concept:

According to law of mass action \(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

Application:

The newly created semiconductor is a compensated semiconductor with doping profile, \(\rm N_A = 10^{18} cm^{-3}\) and \(\rm N_D = 10^{15} cm^{-3}\).

The hole concentration due to N_A is very large than Nd

 here, \(\rm N_A\gg N_D\)

So, we can neglect \(\rm N_D\) and the electron concentration can be found as,

using law of mass action 

\(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

\(\rm = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{10}^{18}}}}\)

\(\rm = 2.25 × 10^2\)

Concept:

According to law of mass action \(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

Application:

The newly created semiconductor is a compensated semiconductor with doping profile, \(\rm N_A = 10^{18} cm^{-3}\) and \(\rm N_D = 10^{15} cm^{-3}\).

The hole concentration due to N_A is very large than Nd

 here, \(\rm N_A\gg N_D\)

So, we can neglect \(\rm N_D\) and the electron concentration can be found as,

using law of mass action 

\(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

\(\rm = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{10}^{18}}}}\)

\(\rm = 2.25 × 10^2\)

Calculation:

Initial (without light) hole concentration under thermal equilibrium is

\(\rm p_0 = \frac{n_i^2}{n_0} = \frac{n_i^2}{N_D} = \frac{(10^{10})^2}{10^{16}} = 10^4 cm^{-3}\)

After light incident the excess hole concentration is

p_n(0) = p_n (∞) + [p_n (0) - p_n (∞)]e^-x/L_p

pn(x) = p0 + [pn (0) - p0]e-x/Lp

p_n(x) = 10⁴ + 10⁸ \(\rm e^{-x/100 \times 10^{-4}}\) cm^-3

\(\rm p_n(x) = 10^4 + 10^8 e^{-10^{2}x} cm^{-3}\)

i) For the same dopant concentration, the conductivity of an n-type is higher than p-type since

μ­n > μp and σ ∝ μ

μn = mobility of electron

μp = mobility of hole

ii) Under high electric fields (due to the voltage applied) the mobility of electrons decreases due to lattice collisions. 

iii) When the doping concentration is high either electrons or holes are in high concentration, hence drift current is significant compared to the diffusion current. While at low concentration diffusion current is significant.

iv) Direct bandgap semiconductors:

We can see that there is no change in momentum during recombination.

(i.e., when electrons move from conduction to valence)

Indirect bandgap semiconductor:

We can see that there is a finite change in momentum ΔP during recombination.

The momentum of various particles:

i) Photon → h / λ = E / C

ii) Phonon → E / v

As v ≪ C

∴ The momentum of the photon is negligible.

And also the change in momentum in the case of GaAs is also negligible.

∴ That momentum can be given to a photon.

∴ Radiative recombination is possible in the case of GaAs. 

Concept:

According to law of mass action \(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

Application:

The newly created semiconductor is a compensated semiconductor with doping profile, \(\rm N_A = 10^{18} cm^{-3}\) and \(\rm N_D = 10^{15} cm^{-3}\).

The hole concentration due to N_A is very large than Nd

 here, \(\rm N_A\gg N_D\)

So, we can neglect \(\rm N_D\) and the electron concentration can be found as,

using law of mass action 

\(\rm n = \frac{{n_i^2}}{{{N_A}}}\)

\(\rm = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{10}^{18}}}}\)

\(\rm = 2.25 × 10^2\)

At equilibrium recombination rate R = carrier generation rate G

i.e. R = G = α_Tnp = α_Tn_i²

Even in extrinsic semiconductors, the equilibrium concentration of product np is the same hence α_T will remain constant.

\({\alpha _T} = \frac{R}{{np}} = \frac{R}{{n_i^2}} \)

\(= \frac{{4.3 \times {{10}^5}}}{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}} = 1.911 \times {10^{ - 15}}{s^{ - 1}}\)

Concept:

For a compensated semiconductor with acceptor concentration greater than the donor concentration, the majority carrier hole concentration is calculated as:

\({p_0} = \frac{{{N_a} - {N_d}}}{2} + \sqrt {{{\left( {\frac{{{N_a} - {N_d}}}{2}} \right)}^2} + n_i^2} \)

With Na - Nd ≫ ni, the above equation becomes:

p0 ≅ Na - Nd

Also, the charge carriers at thermal equilibrium follow mass-action law, i.e.

\({n_0}{p_0} = n_i^2\)

Calculation:

Boron is an acceptor atom (i.e a p-type impurity)

Na = 1.5 × 1015 cm-3

Arsenic is a donor atom (i.e. an n-type impurity)

Nd = 8 × 1014 cm-3

Since, Na > Nd, the given semiconductor is a p-type semiconductor.

Also, since Na - Nd ≫ ni the majority carrier hole concentration will be:

p₀ = N_a - N_d

p₀ = 1.5 × 10¹⁵ – 8 × 10¹⁴ cm^-3

p₀ = (15 - 8) × 1014 cm-3

p₀ = 7 × 10¹⁴ cm^-3

Using mass action law

\({n_0} = \frac{{n_i^2}}{{{p_0}}} = \frac{{{{10}^{20}}}}{{7 \times {{10}^{14}}}}\)

n₀ = 1.42 × 10⁵ cm^-3

The relationship between carrier concentrations in a semiconductor is given by the mass action law, which states:

n × p = n_i²

Where:

• n = concentration of electrons (minority carriers in p-type material, in cm^-3).
• p = concentration of holes (majority carriers in p-type material, in cm^-3).
• n_i = intrinsic carrier concentration of silicon (in cm^-3).

Given data:

• Intrinsic carrier concentration, n_i = 1.5 × 10¹⁰ cm^-3.
• Acceptor concentration (boron doping), N_A = 1.5 × 10¹⁴ cm^-3.
• In a p-type semiconductor, the majority carrier concentration (holes) is approximately equal to the acceptor concentration: p ≈ N_A.

Step 1: Apply the mass action law

Using the mass action law:

n × p = n_i²

Substitute p ≈ N_A:

n × N_A = n_i²

Solve for the electron concentration, n:

n = n_i² / N_A

Step 2: Substitute the given values

Intrinsic carrier concentration: n_i = 1.5 × 10¹⁰ cm^-3

Acceptor concentration: N_A = 1.5 × 10¹⁴ cm^-3

Substitute these values into the equation:

n = (1.5 × 10¹⁰)² / (1.5 × 10¹⁴)

First, calculate n_i²:

n_i² = (1.5 × 10¹⁰) × (1.5 × 10¹⁰) = 2.25 × 10²⁰

Now, divide by N_A:

n = (2.25 × 10²⁰) / (1.5 × 10¹⁴)

n = 1.5 × 10⁶ cm^-3

The concentration of electrons in the silicon sample is 1.5 × 10⁶ cm^-3.

Concept:

For a compensated semiconductor with acceptor concentration greater than the donor concentration, the majority carrier hole concentration is calculated as:

\({p_0} = \frac{{{N_a} - {N_d}}}{2} + \sqrt {{{\left( {\frac{{{N_a} - {N_d}}}{2}} \right)}^2} + n_i^2} \)

With N_a - N_d ≫ n_i, the above equation becomes:

p₀ ≅ N_a - N_d

Also, the charge carriers at thermal equilibrium follow mass-action law, i.e.

\({n_0}{p_0} = n_i^2\)

Calculation:

Boron is an acceptor atom (i.e a p-type impurity)

N_a = 1.5 × 10¹⁵ cm^-3

Arsenic is a donor atom (i.e. an n-type impurity)

N_d = 8 × 10¹⁴ cm^-3

Since, N_a > N_d, the given semiconductor is a p-type semiconductor.

Also, since N_a - N_d ≫ n_i the majority carrier hole concentration will be:

p₀ = N_a - N_d

p₀ = 1.5 × 10¹⁵ – 8 × 10¹⁴ cm^-3

p₀ = (15 - 8) × 10¹⁴ cm_-3

p₀ = 7 × 10¹⁴ cm^-3

Using mass action law

\({n_0} = \frac{{n_i^2}}{{{p_0}}} = \frac{{{{10}^{20}}}}{{7 \times {{10}^{14}}}}\)

n₀ = 1.42 × 10⁵ cm^-3

\({\delta _p}\left( x \right) = {{\rm{\Delta }}_p}{e^{ - \frac{x}{{{L_p}}}}}\)

\(= {L_p} = \sqrt {{D_p}{\tau _p}}\)

\({{\rm{\Delta }}_p} = {{\rm{\Delta }}_{po}}\exp \left( { - \frac{L}{{\sqrt {{D_p}{\tau _p}} }}} \right)\)

= 1.6 × 10¹³/cm³