Electronic Devices Test 2

Concept:

The resistivity of a doped semiconductor is calculated as:

\(\rho = \frac{1}{{q\;\mu \;N}}\)

Where,

μ = mobility of the carrier

N = number of concentration of carriers.

For GaAs, the mobility of electrons is greater than the mobility of holes i.e.

μ_n > μ_p

Calculation:

Given doping concentration of two wafers as:

N_D (wafer 1) = N_A (wafer 2) ≫ n_i

The resistivity of the wafers will be:

\({\rho _1} = \frac{1}{{q{\mu _n}{N_{D1}}}}\) for n type wafer 1

\({\rho _2} = \frac{1}{{q{\mu _p}{N_{A2}}}}\) for p-type wafer – 2

With N_D1 = N_A2, the resistivity of the wafer with less mobility of charge carriers will be more.

Since for GaAs, μ_n > μ_p, we conclude that:

ρ₁ < ρ₂

ρ (wafer 1) < ρ (wafer 2)

Concept:

The electric field in a semiconductor for a given applied voltage V is given by:

\(E = \frac{V}{l}\)

l = length of the semiconductor

Also, the drift velocity of electronics for a given applied field is calculated as:

vd = μnE

μ_n = Mobility of the electron

Calculation:

The electric filed developed in the semiconductor will be uniform with value:

\(E = \frac{V}{l} = 10\;V/cm\)

With μ_n = 1000 cm²/V-sec, the drift velocity will be:

v_d = μ_nE

\(v_d= 1000\frac{{c{m^2}}}{{V- sec}} \times 10\frac{V}{{cm}}\)

\(V_d= {10^4}\frac{{{\rm{cm}}}}{{{\rm{sec}}}}\)

Time taken by the electron to move from 0 to 0.75 cm is t1, i.e.

\(t_1 = \frac{{0.75}}{{{{10}^4}}} s\)

Time taken by the electron to move from 0.75 to 1 cm is t2, i.e.

\(t_2 = \frac{{0.25}}{{{{10}^4}}}s\)

The ratio of t₁ to t₂ will therefore be:

\(\frac{{{{\rm{t}}_1}}}{{{{\rm{t}}_2}}} = 3\)

Note: Please note that we do not even require to evaluate the value of v_d (Drift velocity), as it will get cancelled eventually.

Concept:

According to Einstein's relation:

\(\frac{D}{μ} = \frac{{kT}}{q}\)

D = Diffusion coefficient of charge carrier

μ = Mobility of the charge carrier

\(\frac{kT}{q}=\) Thermal voltage

Calculation:

\(\frac{D}{1900} = 0.025\)

D = 0.025 × 1900 cm²/sec

D = 47.5 cm²/sec

Concept:

The current density in a semiconductor for an applied electric field is given by:

J = σ.E     …1)

Where σ = conductivity of the semiconductor defined as:

σ = qnμ_n + qpμ_p

n = Electron concentration

p = Holes concentration

μ_n = Electron mobility

μ_p = Hole mobility

Calculation:

Given, the semiconductor is doped with Boron (p-type impurity).

So, N_a = 2 × 10¹⁶ cm^-3

Since N_a ≫ n_i, the semiconductor will be dominated by N_a only:

So, σ = qnμ_n + qpμ_p ≈ qpμ_p = qN_aμ_p

Given, J = 120 A/cm²

Using Equation 1),

⇒ 120 = (qN_aμ_p).E

\(E = \frac{{120}}{{q{N_a}{\mu _p}}} \)

\(E= \frac{{120}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times {{10}^{16}}} \right)\left( {500} \right)}}\)

\(E = 75\;V/cm\)

Concept:

The conductivity for a semiconductor at thermal equilibrium in given as:

σ = e(μ_n n + μ_p p)       ---(1)

n = Concentration of electrons

p= Concentration of holes

The amount of exess carriers generated will increase the conductivity by increasing the number of charge carriers, i.e.

Concentration of electron = n + δn

Concentration of holes = p + δp

The conductivity now will be:

σ_new = e(μ_n (n + δn) + μ_p (p + δp))      ---(2)

The change (increase) in conductivity (Δσ) is calculated as:

\({\rm{\Delta }}\sigma = e\left( {{\mu _n}\left( {n + δ n} \right) + {δ _p}\left( {p + δ p} \right)} \right) - e\;\left( {{\mu _n}n + {\mu _p}p} \right)\)

Δσ = eμ_nδ_n + eμ_pδ_p

Since the excess electron and holes are always generated in pairs, i.e. δp = δn, the conductivity can be written as:

Δσ = e(μ_n + μ_p) δp 'or'

Δσ = e(μn + μp) δn

Calculation:

The generation rate is the ratio of the excess carriers generated with the minority carrier lifetime, i.e.

\(g' = \frac{{δ p}}{{{\tau _p}}}\)

The excess hole concentration will be:    

δp = g’ τ_p 

\(δ p = 2 \times {10^{21}} \times 2 \times {10^{ - 7}}\)

\(δ p = 4 \times {10^{14}}\;c{m^{ - 3}}\)

The change in conductivity ‘or’ the conductivity is increased by:

\(\therefore {\rm{\Delta }}\sigma = 1.6 \times {10^{ - 19}}\left( {8500 + 400} \right) \times 4 \times {10^{14}}\)

= 0.57 (Ω cm)^-1

Concept:

Mass-action law states that:

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

n₀ = concentration electrons

p₀ = concentration of holes

n_i = Intrinsic carrier concentration

Since the electrons and holes recombine in pairs, the minority carrier recombination rate will be equal to the majority carrier recombination rate, i.e.

\(\frac{{{p_0}}}{{{\tau _{p0}}}} = \frac{{{n_0}}}{{{\tau _{n0}}}}\)

τ = Recombination lifetime

Calculation:

Given, \({n_i} = 1.5 \times {10^{10}}\;c{m^{ - 3}}\)

N_d = 10¹⁶ cm^-3

N_d ≫ n_i

The electron concentration is:

n₀ = N_d = 10¹⁶ cm^-3

Using mass-action law, the hole concentration (minority concentration) is obtained as:

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

\({p_0} = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{\left( {10} \right)}^{16}}}}\)

\({p_0} = 2.25 \times {10^4}\;c{m^{ - 3}}\)

Now, the hole recombination rate in thermal equilibrium is:

\({R_{p0}} = \frac{{{p_0}}}{{{\tau _{p0}}}}\)

The recombination rate of majority carriers will be the same as that of minority carrier holes i.e.

\(R_p'=R_n'\)

\(\frac{{{p_0}}}{{{\tau _{p0}}}} = \frac{{{n_0}}}{{{\tau _{n0}}}}\)

∴ The lifetime of the majority carrier electron will be:

\({\tau _{n0}} = \frac{{{n_0}}}{{{p_0}}}\;{\tau _{p0}}\)

\({\tau _{n0}} = \frac{{{{10}^{16}}}}{{2.25 \times {{10}^4}}} \times 20 \times {10^{ - 6}}\)

τ_n0 = 8.89 × 10⁶ sec

Concept:

The total current density in a semiconductor due to electrons is the sum of the diffusion current density (due to concentration gradient) and drift current density (due to electric field) i.e.

\(J = e{\mu _n}nE + e\;{D_n}\frac{{dn}}{{dx}}\;\)       ---(1)

n = Concentration of electron

E = Electric field

According to Einstein’s relation:

\(\frac{{{D_n}}}{{{\mu _n}}} = \frac{{kT}}{q} = \frac{{kT}}{e}\)

Calculation:

Given that the donor concentration varies as:

\({N_d}\left( x \right) = {N_{d0\;}}{e^{ - \alpha x}}\) for \(0 \le x \le \frac{1}{\alpha }\) 

Given that the semiconductor is in thermal equilibrium (no current), i.e.

J = 0

Equation (1) can now be written as:

\(e{\mu _n}E{N_{d0}}{e^{ - \alpha x}} + e\;{D_n}\;{N_{d0}}\left( { - \alpha } \right){e^{ - \alpha x}} = 0\)

\(E = \frac{{{D_n}}}{{{\mu _n}}}\alpha \)       ---(2)

Using Einstein’s relation, the above can be written as:

\(E = \alpha \left( {\frac{{kT}}{e}} \right)\)

∴ The electric field in the given range \(0 \le x \le \;\frac{1}{\alpha }\) is independent of ‘x’.