Electronic Devices Test 1

• The material which is not a good conductor, or a good insulator is called as a semiconductor.For example, silicon, germanium, etc.
• The charge carriers which are present in more quantity in a semiconductor compared to other particles are called the majority charge carrier.
• The impurity atoms added are called dopants and semiconductors doped with the impurity atoms are called extrinsic or doped semiconductors.
• An ideal, perfectly pure semiconductor (with no impurities) is called an intrinsic semiconductor.
• At absolute zero temperature valence band is full of electrons and the conduction band is empty, hence there are no free electrons in the conduction band and holes in the valence band.
• The charge carrier concentration is zero at zero absolute temperature.
• Hence, at zero absolute temperature the intrinsic semiconductor behaves like insulator.

There are two types of semiconductors:

P-type semiconductor:

• The semiconductor having holes as majority charge carriers and electrons as a minority charge carrier is called a P-type semiconductor.

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

Under equilibrium:

• The fermi level across the entire material will be sum and does not vary with distance.

• If there is any disturbance in the material, like junction contact, injection of impurities at any point, the charge carriers redistribute themselves such that the fermi-potential is same in entire material.

Under non-equilibrium:

• The fermi level is uneven with gradient of charge distribution across distance, in material.
• It can be studied using quasi-fermi states, related to charge distribution.

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

Property of Semiconductors:

• Semiconductors are the materials that have a conductivity between conductors (generally metals) and non-conductors or insulators (such as ceramics).
• Semiconductors can be compounds such as gallium arsenide or pure elements, such as germanium or silicon.
• Semiconductors have an almost empty conduction band and an almost filled valence band.
• In a semiconductor, the mobility of electrons is higher than that of the holes.
• Its Resistivity lies between 10^-5 to 10⁶ Ωm
• Conductivity lies between 10⁵ to 10^-6 mho/m
• The temperature coefficient of resistance for semiconductors is Negative.
• The current Flow in the semiconductor is mainly due to both electrons and holes.

As 25% of donor atoms are ionised, the occupation probability of donor level is 0.75.

Given ND = 2.25 × 1015 cm^-3

Partial ionization of 55%, number of electrons is:

⇒ n = 0.55 × 2.25 × 1015 cm^-3

= 1.2375 × 1015 cm-3

Electron mobility μn = 1000 cm2/v-s

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:

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

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

• Atoms having 5 valence electrons acts as donor impurities 
• Antimony is a donor impurity as it has five valence band electrons.
• Atoms having 3 valence electrons acts as acceptor impurities
• Boron and Indium are acceptor impurities.

Important Points: When we add n-type or donor impurities to the semiconductor, the width of the forbidden energy gap in the lattice structure is reduced. Due to an addition of donor atoms, allowable energy levels are introduced a small distance below the conduction band. These new allowable levels are discrete because the added impurity atoms are placed far apart and hence their interaction is small.

Concept:

• The material which is not a good conductor or a good insulator is called as a semiconductor.
• For example: Silicon
• The charge carriers which are present in more quantity in a semiconductor compared to other particles are called the majority charge carrier.

There are two types of semiconductors:

P-type semiconductor:

• The semiconductor having holes as majority charge carriers and electrons as a minority charge carrier is called a P-type semiconductor.

N-type semiconductor:

• The semiconductor having electrons as majority charge carriers and holes as a minority charge carrier are called N-type semiconductor.

Semiconductor: Semiconductors are the materials that have a conductivity between conductors and insulators.

• Semiconductors are made of compounds such as gallium arsenide or pure elements, such as germanium or silicon.
• There are two types of semiconductor:

1. ​N-type Semiconductor
2. P-type Semiconductor

• In an n-type semiconductor, pentavalent impurity from the V group is added to the pure semiconductor.
• Electrons are the majority charge carriers in the n-type semiconductor.
• So when a pentavalent impurity is mixed into the semiconductor, the number of electrons becomes more than the number of holes.

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

N-type Semiconductor:

• An n-type semiconductor is an intrinsic semiconductor doped with phosphorus (P), arsenic (As), or antimony (Sb) as an impurity.
• Silicon of Group IV has four valence electrons and phosphorus of Group V has five valence electrons.
• If a small amount of phosphorus is added to a pure silicon crystal, one of the valence electrons of phosphorus becomes free to move around (free-electron*) as a surplus electron. When this free electron is attracted to the “+” electrode and moves, current flows.

P-type Semiconductor:

• Impurities such as Aluminium, Gallium, and Indium (elements having three valence electrons) are added to P-type semiconductors.
• Impurities are added to P-type semiconductors in order to increase the number of holes (also known as Acceptor atoms).
• In a P-type semiconductor, the majority of charge carriers are holes whereas the free electrons are in minority.

A semiconductor is a solid substance that has a conductivity between an insulator and metals. Semiconductors have conductivity due to many factors like the addition of an impurity or because of temperature effects.

N-type semiconductors: An extrinsic semiconductor where the dopant atoms provide extra conduction electrons to the host material like Phosphorus (P) in Silicon Si called donor impurities.

This creates an excess of negative (n-type) electron charge carriers that are able to move freely.

In the donor replacement atom, the outermost shell ( valance shell) contains 5 electrons.

Example: Phosphorus (P) etc.

Important Points

P-type semiconductors: A semiconductor, when the impurity (acceptor) with tri valency is added to pure semiconductors, then it is known as a p-type semiconductor.

Impurities with tri valency such as Boron (B), Gallium (Ga), Indium(In), Aluminium(Al), etc are called acceptor impurity.

In the acceptor replacement atom, the outermost shell (valance shell) contains 3 electrons.

Example: Boron (B), Gallium (Ga), Indium (In), Aluminium (Al), etc.

So, the value of the valance electron is 3.

Donor impurities (e.g., phosphorus) in n-type semiconductors add extra electrons by creating donor energy levels just below the conduction band. These electrons are easily excited to the conduction band, increasing the number of free electrons available for conduction compared to an intrinsic semiconductor, which relies on thermal excitation across the bandgap.

• Option A: Incorrect, as donors don’t create holes.

• Option B: Correct, as donors add conduction electrons.

• Option C: Incorrect, as donors reduce the effective excitation energy, not widen the bandgap.

• Option D: Incorrect, as donors don’t reduce mobility.

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