Electronic Devices Test 1

Concept:

For a given position of Fermi-level, the concentration of electrons and holes are given by:

\({n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_F}} \right)}}{{kT}}}}\)

\({p_0} = {N_v}{e^{ - \frac{{\left( {{E_F} - {E_V}} \right)}}{{kT}}}}\)

N­_c and N_v are the effective density of states.

We can now calculate the intrinsic Fermi level position.

For the intrinsic semiconductor, the electron and hole concentrations are equal.

∴ We can write:

\({N_c}\;{e^{ - \left( {\frac{{{E_c} - {E_{Fi}}}}{{kT}}} \right)}} = {N_v}\;{e^{ - \left( {\frac{{{E_{Fi}} - {E_v}}}{{kT}}} \right)}}\)

Taking natural lag of both sides of the above equation, we get:

\({E_{{F_i}}} = \frac{1}{2}\left( {{E_c} + {E_v}} \right) + \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

Since, \(\frac{1}{2}\left( {{E_c} + {E_v}} \right) = {E_{midgap}}\)

\({E_{{F_i}}} = {E_{midgap}} + \frac{1}{2}kT\;In\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

\({E_{Fi}} - {E_{midgap}} = \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)    ---(1)

Also, since \({N_c} \propto {\left( {m_n^*} \right)^{3/2}}\)

And \({N_v} \propto {\left( {m_p^*} \right)^{3/2}}\)

Equation (1) becomes:

\({E_{{F_i}}} - {E_{mid}} = \frac{3}{4}\;ln\;\left( {\frac{{m_p^*}}{{m_n^*}}} \right)\)      ---(2)

m^*_n = Effective mass of electron

m^*_p = Effective mass of holes

Calculation:

Given, m^*n = 1.08 m₀

m^*_p = 0.56 m₀

\({E_F} - {E_{mid}} = \frac{3}{4}kT\;In\left( {\frac{{0.56\;{m_0}}}{{1.08\;{m_0}}}} \right)\)

\(= \frac{3}{4} \times 0.026 \times \left( { - 0.656} \right)\)

=  -0.0128 eV

The negative sign indicates that the Fermi level is 0.0128 below the centre of the bandgap.

Concept:

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

\({n_0} = \frac{{{N_D} - {N_A}}}{2} + \sqrt {{{\left( {\frac{{{N_A} - {N_D}}}{2}} \right)}^2} + n_i^2} \)

n_i = Intrinsic concentration

Note: If the donor concentration is very very greater than the intrinsic concentration, then we simply take the majority concentration as the donor concentration, i.e.

For N_D - N_A >> ni

n₀ = N_D - N_A

Calculation:

Given: ND = 5.00 × 1016 cm3 , NA = 0 and ni = 2.865 × 1016 cm-3

Since ND - NA is comparable to the the intrinsic concentration, we cannot take n₀ = N_D - N_A.

\({n_0} = \frac{{{N_D} - {N_A}}}{2} + \sqrt {{{\left( {\frac{{{N_A} - {N_D}}}{2}} \right)}^2} + n_i^2} \)

\(\frac{{{N_D} - {N_A}}}{2} = \frac{{5 \times {{10}^{16}}}}{2} \)

\(= 2.5 \times {10^{16}}\)

\(n = \left( {2.5 \times {{10}^{16}}} \right) + \sqrt {{{\left( {2.5 \times {{10}^{16}}} \right)}^2} + {{\left( {2.865 \times {{10}^{16}}} \right)}^2}} \)

\(n = 2.5 \times {10^{16}} + \sqrt {6.25 \times {{10}^{32}} + 8.21 \times {{10}^{32}}} \)

n = (2.5 + 3.8) × 10¹⁶

n = 6.3 × 10¹⁶ cm^-3

Concept:

The majority carrier electron concentration is given by:

\({n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_F}} \right)}}{{kT}}}}\)  ---(1)

For an intrinsic semiconductor, the intrinsic carrier concentration will be:

\({n_i} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_{{F_i}}}} \right)}}{{kT}}}}\)   ---(2)

Equation (1) can be written as:

\({n_0} = {N_c}{e^{ - \frac{{\left[ { - \left( {{E_c} - {E_{{F_i}}}} \right) + \left( {{E_F} - {E_F}} \right)} \right]}}{{kT}}}}\)

\({n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_{{F_i}}}} \right)}}{{kT}}}}{e^{\frac{{\left( {{E_F} - {E_{{F_i}}}} \right)}}{{kT}}}}\)

Using Equation (2), the above equation can be written as:

\({n_0} = {n_i}\;{e^{\frac{{\left( {{E_F} - {E_{{F_i}}}} \right)}}{{kT}}}}\)   ---(3)

Application:

With N_d – N­_a ≫ n_i, the majority carrier electron concentration will be:

n₀ = N_d - N_a

n₀ = 10¹⁴ cm^-3

Taking natural l0g on both sides of the equation (3), we get:

\(ln\left( {\frac{{{n_0}}}{{{n_i}}}} \right) = \frac{{{E_F} - {E_{{F_i}}}}}{{kT}}\)

\({E_F} - {E_{{F_i}}} = kT\;ln\left( {\frac{{{n_0}}}{{{n_i}}}} \right)\)

Putting on the respective values, we get:

\({E_F} - {E_{{F_i}}} = 0.026\;ln\left( {\frac{{{{10}^{14}}}}{{{{10}^{10}}}}} \right)\)

\({E_F} - {E_{{F_i}}} = 0.239\;eV\)

Concept:

The hole concentration for a p-type semiconductor is given by:

\({p_0} = {n_i}{e^{\frac{{\left( {{E_{{Fi}}} - {E_F}} \right)}}{{kT}}}}\)

E_Fi = Intrinsic Fermi Level

n_i = Intrinsic carrier concentration

Application:

Given doping concentration is:

N_a = 10¹⁶ cm^-3

Required Fermi level position is 0.45 eV below the intrinsic i.e.

\({E_{{F_i}}} - {E_F} = 0.45\;eV\)

So, we can calculate the required concentration of holes as:

\({p_0} = {n_i}{e^{\frac{{\left( {{E_F}_i - {E_F}} \right)}}{{kT}}}}\)

\({p_0} = \left( {1.8 \times {{10}^6}} \right){e^{\frac{{0.45}}{{0.026}}}}\)

p₀ = 6.32 × 10¹³ cm^-3

Since p₀ (Required) < N_a (Given)

∴ Donor atoms must be added to decrease this hole concentration to the required level.

Since,

p₀ = N_a - N_d

With p₀ = Required concentration

N_a = Already given concentration

N_d = N_a – p₀

Nd = 10¹⁵ – 6.32 × 10¹³

∴  9.368 × 1014 cm-3 additional donor impurity atoms must be added so that the Fermi level is 0.45 eV below the intrinsic level.

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

When photon with energy less than 2.5 eV is incident on the semiconductor the recombination of hole electron doesn't take place, hence it cannot be detected 

The wavelength corresponding to 2.5 eV band gap is

\(\frac{{hc}}{\lambda } = 2.5 \times 1.6 \times {10^{ - 19}}\)

\(\frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^{ + 8}}}}{\lambda } = 2.5 \times 1.6 \times {10^{ - 19}}\)

\(\lambda = 4.97 \times {10^{ - 7}}m\)

≃ 497 nm

For λ > 497 nm

\(E = \frac{{hc}}{\lambda } < 2.5\;eV\)

Hence the photon cannot be absorbed by the semiconductor.