Electronic Devices Test 1

Conceptual Approach:

Fermi level is the level where the probability of finding an electron equals the probability of finding a hole.

Given that the probability of a state being filled with an electron at E = EF + KT is equal to the probability that the state is empty at E = EF + KT, we can conclude that EF + KT in the Fermi-level.

Conventional Method:

Carriers in a semiconductor follow Fermi-Dirac distribution.

The probability that an electron occupies an energy level E is given by:

\(f\left( E \right) = \frac{1}{{1 + e\frac{{\left( {E - {E_F}} \right)}}{{KT}}}}\)

Where EF = Fermi energy level.

The probability of a state being empty is given by:

\(1 - f\left( E \right) = 1 - \frac{1}{{1 + e\frac{{\left( {E - {E_F}} \right)}}{{KT}}}}\)

Now, the probability that an energy state at E_C + kT is filled will be given by:

\(f\left( {{E_C} + KT} \right) = \frac{1}{{1 + {e^{\frac{{\left( {{E_C} + KT - {E_F}} \right)}}{{KT}}}}}}\)

Similarly, the probability that a state at E_C + KT is empty will be given by:

\(1 - f\left( {{E_C} + KT} \right) = 1 - \frac{1}{{1 + {e^{\frac{{\left( {{E_C} + KT - {E_F}} \right)}}{{KT}}}}}}\)

Given the two probabilities as equal, we can write:

f(E_C + KT) = 1 – f (E_C + KT)

\(\frac{1}{{1 + {e^{\frac{{\left( {{E_C} + KT - {E_F}} \right)}}{{KT}}}}}} = 1 - \frac{1}{{1 + {e^{\left( {\frac{{{E_C} + KT - {E_F}}}{{KT}}} \right)}}}}\)

\(\frac{1}{{1 + {e^{\frac{{\left( {{E_C} + KT - {E_F}} \right)}}{{KT}}}}}} = \frac{{{e^{\frac{{\left( {{E_C} + KT - {E_F}} \right)}}{{KT}}}}}}{{1 + {e^{\left( {\frac{{{E_C} + KT - {E_F}}}{{KT}}} \right)}}}}\)

\({e^{\left( {\frac{{{E_C} + KT - {E_F}}}{{KT}}} \right)}} = 1\)

Taking the natural log on both sides, we get:

\(\frac{{{E_C} + KT - {E_F}}}{{KT}} = 0\)

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.

\({n_0}{p_0} = n_i^2\) 

Application:

Given dopant concentration for Ga As as:

N_d = 1 × 10¹³ cm^-3

N_a = 2.5 × 10¹³ cm^-3

Since the acceptor concentration is greater than the donor concentration, the material will be p-type.

The majority carrier hole concentration will be:

p₀ ≅ N_a - N_d

p₀ = 2.5 × 10¹³ – 1 × 10¹³

= 1.5 × 10¹³ cm^-3

Electron concentration as obtained using mass action law as:

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

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

\({n_0} = \frac{{{{\left( {2.8 \times {{10}^6}} \right)}^2}}}{{1.5 \times {{10}^{13}}}}\)

n₀ = 0.216 cm^-3

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:

The maximum power delivered by a solar cell is given by:

P_m = V_m . I_m = f.f × V_OC­ I_SC

Calculation:

Putting on the respective values, we get:

= 0.8 × 0.9 × 10 mA

In the accumulation region, the MOS capacitance is due to the accumulation of charges at the gate surface and substrate.

Hence the Capacitance is defined as:

C_MOS = C_ox = \(\frac{{{\varepsilon _{ox}}}}{{{t_{ox}}}}\)

It is independent of doping concentration. Hence, it remains unchanged.

In the depletion region, the MOS capacitance is the series sum of C_ox and C_D (depletion layer capacitance)

\(\begin{array}{l} \frac{1}{{{{\rm{C}}_{{\rm{Mos}}}}}} = \frac{1}{{{{\rm{C}}_{{\rm{ox}}}}}} + \frac{1}{{{{\rm{C}}_{\rm{D}}}}}\\ {C_D} = \frac{{{\varepsilon _{ox}}}}{{{W_{dep}}}} \end{array}\)

And \({W_{dep}}\propto \frac{1}{{\sqrt {{N_a}} }}\)

Where N_a is the substrate doping concentration, i.e.

\({C_D}\propto \sqrt {{N_a}}\)

Dry oxidation is used to from thin oxide films

Wet oxidation proceeds at faster rate and is used to form thick oxide films.

This is because water vapor diffuses through SiO₂ faster than oxygen.

The Corresponding Reaction involved are

Si + O₂ → SiO₂

Concept:

According to space charge neutrality, the net positive charge in the ‘n’ region must equal the net negative charge in the ‘p’ region, i.e.

q N_a x_p = q N_d x_n

x_p = space charge region extending to p-side

x_n = space charge region extending to n-side

Also the built-in potential is given by:

\({V_{bi}} = \frac{{kT}}{q}\;\;In\left( {\frac{{{N_a}{N_d}}}{{n_i^2}}} \right)\)

N_a = doping concentration on p-side

N_d = doping concentration on n-side

Application:

At zero bias, only 20% of the total width (W) is to be in the p-region, i.e.

x_p = 0.2 W

∴ x_n = 0.8 W

Since, q N_a x_p = q N_d x_n, we can write:

q N_a (0.2 W) = q N_d (0.8 W)

\({N_d} = \frac{{{N_a}}}{4}\)        ---(1)

Given:

V_bi = 1.2 V, n_i = 1.8 × 10⁶ cm^-2 and V_T = 25.9 mV

The built-in potential will be:

\({V_{bi}} = \frac{{kT}}{q}\left( {\frac{{{N_a}{N_a}}}{{4n_i^2}}} \right)\)

\(1.2 = \left( {25.9 \times {{10}^{ - 3}}} \right)ln\left( {\frac{{{N_a} \cdot {N_a}}}{{4\;n_i^2}}} \right)\)

Taking anti-log on both sides, we get:

\({e^{\left( {\frac{{1.2}}{{25.9 \times {{10}^{ - 3}}}}} \right)}} = \frac{{N_a^2}}{{4n_i^2}}\)

\(1.323 \times {10^{20}} = \frac{{N_a^2}}{{4{{\left( {1.8 \times {{10}^6}} \right)}^2}}}\)

\(N_a^2 = 1.715 \times {10^{33}}\)

Concept:

The external quantum efficiency is given by the relation:

\({η _{ext}} = \frac{{η_{a}^2}}{{4η _s^2}}\)

η_s = Refractive index of the LED

Calculation:

η_s = 3.6

η_a = 1

\({η _{ext}} = \frac{{1}}{{4(3.6)^2}}\)

η_ext = 0.0193

Out of every 100 photons generated in Ga – As, only 1.93 are emitted out of the specimen.

∴ for 500 photons → 9.5 will be emitted.

Taking the greatest integer value, we get:

[9.5] = 9

Concept:

The current equation for an ideal pn junction with V voltage (forward-biased) is given as:

\(I = {I_s}\left( {e^\frac{V}{{{V_T}}} - 1} \right) ≈ {I_s}e^\frac{V}{{{V_T}}}\)

I_S = Saturation current

V_T = Thermal voltage

Calculation:

For a voltage V1, the value of the current i1 is:

\({I_1} = {I_s}e^\frac{{{V_1}}}{{{V_T}}}\)

For a voltage V2, the value of the current I2 is:

\({I_2} = {I_s}e^\frac{{{V_2}}}{{{V_T}}}\)

A factor of 50 increase means, I2 = 50 I1

\( \frac{{{I_2}}}{{{I_1}}} = \frac{{{I_s}e^\frac{{{V_2}}}{{{V_T}}}}}{{{I_s}e^\frac{{{V_1}}}{{{V_T}}}}}\)

\(\frac{{50\;{I_1}}}{{{I_1}}} = e\frac{{{V_2} - {V_1}}}{{{V_T}}}\)

\( \ln \left( {50} \right) = \frac{{{V_2} - {V_1}}}{{{V_T}}}\)

\( {V_2} - {V_1} = {V_T}\ln \left( {50} \right) \)

With V_T = 25 mV, we get:

V₂ - V₁ = 25m × 4 = 97.8 mV

Concept:

Drain current for a MOSFET in saturation is given by:

\({{I}_{D}}=\frac{1}{2}{{\mu }_{n}}{{C}_{ox}}\left( \frac{W}{L} \right){{\left( {{V}_{Gs}}-{{V}_{T}} \right)}^{2}}\left( 1+λ {{V}_{DS}} \right)\)

V_GS = Gate to source voltage

λ = Channel length modulation parameter

Calculation:

\(\frac{{{I}_{D1}}}{{{I}_{D2}}}=\frac{1+λ {{V}_{DS}}_{1}}{1+λ {{V}_{DS}}_{2}}~\)

\({{I}_{D2}}=\frac{\left( 1+λ {{V}_{DS}}_{2} \right)}{\left( 1+λ {{V}_{DS}}_{1} \right)}{{I}_{D1}}\)

Putting on the respective values, we get:

\({{I}_{D2}}=\left[ \frac{1+\left( 0.2 \right)\left( 1 \right)}{1+\left( 0.2 \right)\left( 0.5 \right)} \right]1~mA\)

\(\Rightarrow \frac{1.2}{1.1}=\frac{12}{11}~mA\)

Concept:

The depletion region extending to the base is defined as:

\({x_b} = {\left( {\frac{{2\epsilon{_s}\left( {{V_{bi}} + {V_{R}}} \right)}}{q}\frac{{{N_c}}}{{{N_B}}}\frac{1}{{\left( {{N_C} + {N_B}} \right)}}} \right)^{1/2}}\)

N_C, N_B = Carrier and base doping concentrations

V_R = Applied reverse bias voltage

Since punch thorough voltage (V_pt) >> V_bi, we can neglect V_bi , the above equation can be rearranged as:

\({V_{pt}} = \frac{{ex_{{B_o}}^2}}{{2\epsilon{_s}}}\frac{{{N_B}\left( {{N_C} + {N_B}} \right)}}{{{N_C}}}\)

X_B0 = Metallurgical base width

Calculation:

Putting on the respective values, we get:

\( 25 = \frac{{\left( {1.6 \times {{10}^{ - 19}}} \right){{\left( {0.5 \times {{10}^{ - 4}}} \right)}^2}\left( {{{10}^{16}}} \right)\left( {{N_C} + {{10}^{16}}} \right)}}{{2\left( {11.7} \right)\left( {8.85 \times {{10}^{ - 14}}} \right){N_C}}}\)

\(12.94 = 1 + \frac{{{{10}^{16}}}}{{{N_C}}}\)

N_C = 8.38 × 10¹⁴ cm^-3

Concept:

Excess carriers once generated decay exponentially due to recombination at a rate given by:

\(\delta n\left( t \right) = \delta n\left( 0 \right){e^{ - \frac{t}{\tau }}}\)

Whre τ = Minority Carrier Recombination Lifetime

Calculation:

Given, δn(0) = 10¹⁵cm^-3.

τ_no = 10^-6sec = 1 μsec

\(\Rightarrow \delta n\left( t \right) = \delta n\left( 0 \right){e^{ - \frac{t}{\tau }}}\)

⇒ At t = 10μsec

\(\Rightarrow \delta n\left( {10\mu } \right) = \delta n\left( 0 \right){e^{ - \frac{{10\mu }}{{1\mu }}}}\)

⇒ δn(10μ) = 10¹⁵ × e^-10

= 4.54 × 10¹⁰cm^-3