Electronic Devices Test 2

II-VI semiconductors are generally p-type semiconductors except for ZnO and ZnTe. II-VI semiconductors are those which contain atoms of materials that have 2 valence electrons and 6 valence electrons.

The number of negative charges per unit area in the p-region is equal to the number of positive charges per unit area in the n-region.

i.e. N_ax_p = N_dx_n

where, x_p and x_n­ are the width of the depletion region extending to the ‘p’ and ‘n’-side, respectively.

Calculation:

Given, N_d = 3 × 10¹⁶ cm^-3, N_a = 4 × 10¹⁵ cm^-3

x_p = 0.447 μm

with ⇒ N_ax_p = N_dx_n

\({x_n} = \frac{{{N_a}{x_p}}}{{{N_d}}}\)

\({x_n} = \frac{{4 \times {{10}^{15}} \times 0.447 \times {{10}^{ - 4}}}}{{3 \times {{10}^{16}}}}\)

= 5.96 × 10^-6

= 0.0596 μm

Semiconductor: 

• Semiconductors are 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 semiconductors:

• ​N-type Semiconductor
• P-type Semiconductor

In a P-type semiconductor material, holes are generated because a trivalent impurity atom has one less electron than the surrounding silicon atom. Thus, leaving a vacancy in a covalent bond that acts as a hole. The presence of a hole does not make the semiconductor material positively charge because in an impurity atom, no. of electron and proton is equal before and after the doping, and the same applies for the silicon atom.

Thus, p-type semiconductor crystal remains neutral.

Important Points:

• Extrinsic P-type Semiconductor is formed when a trivalent impurity is added to a pure semiconductor.
• Example of trivalent impurity is Boron, Gallium, and Indium.
• Trivalent impurity like boron has 3 valence electrons.
• Each atom of the impurity fits in the silicon crystal by forming covalent bonds with the surrounding silicon atoms.
• The dopant boron atom has one less electron than surrounding silicon and thus vacancy is generated that acts as a hole.

• Extrinsic p-Type Semiconductor is formed when a trivalent impurity is added to a pure semiconductor.
• Example of trivalent impurity are Boron, Gallium, and Indium
• Trivalent impurity like boron, have 3 valence electrons.
• Each atom of the impurity fits in the silicon crystal by forming covalent bonds with the surrounding silicon atoms.
• The dopant boron atom has one less electron than surrounding silicon and thus vacancy is generated that acts as a hole.

Drift:

• Directed motion of charge carriers in semiconductors occurs through two mechanisms:
• Charge drift under the influence of applied electric field and
• Diffusion of charge from a region of high charge density to one of low charge density.

Phenomenon:

• When no electric field is applied to the semiconductor which is above 0º K, the conduction electrons move within the crystal with random motion and repeatedly collide with each other and the fixed ions.
• Due to the randomness of their motion, the net average velocity of these charge carriers in any given direction is zero. Hence, no current exists in the crystal under this condition of no field.
• Now, consider the case when an electric field is applied to the crystal. Under the influence of this field, the charge carriers attain a directed motion which is superimposed on their random thermal motion.
• This results in a net average velocity called drift velocity in the direction of the applied electric field.
• Electrons and holes move in opposite directions but because of their opposite charges, both produce current in the same direction.
• In extrinsic semiconductors, this current is essentially a majority carrier flow.
• The drift velocity is proportional to electric field strength E, the constant of proportionality being called mobility µ.

Diffusion:

• It is the gradual flow of charge from a region of high density to a region of low density.
• It is a force-free process based on the non-uniform distribution of charge carriers in a semiconductor crystal.
• It leads to an electric current without the benefit of an applied field.
• This flow or diffusion of carriers is proportional to the carrier density gradient, the constant of proportionality being called diffusion constant or diffusion coefficient (D) which has a unit of m2/s.

Recombination:

• It is the collision of an electron with a hole.
• The process is essentially the return of a free conduction electron to the valence band and is accompanied by the emission of energy.
• The recombination rate is directly proportional to the carrier concentration for the simple reason that the larger the number of carriers, the more likely is the occurrence of electron-hole recombination.
• This phenomenon is important in describing minority carrier flow.
• In a semiconductor, the thermal generation of electron-hole pairs also takes place continuously.
• Hence, there is a net recombination rate given by the difference between the recombination and generation rates.

Concept:
Internal quantum efficiency η is the ratio of the number of electrons injected into the depletion region to the number of photons produced.

Ep = Energy in photon (in eV) given by:

\(E_p=\frac{hc}{λ.e}\)

c = Velocity of light 

λ = wavelength of LED

Calculation:

Given, the wavelength of LED light λ = 900 nm 

Forward current through LED I = 20 mA

η = 0.1

Output optical power will be:

P = η Ep I

The Energy is calculated as:

\(= \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 9 \times {{10}^{ - 7}}}}\)

= 1.38 eV

∴ P = 0.1 × 1.38 × 20 × 10-3

= 2.76 mW

Explanation:

The minority current is given by:

\({I_0} = Aqn_i^2\left[ {\frac{{{D_p}}}{{{L_p}{N_D}}} + \frac{{{D_n}}}{{{L_n}{N_A}}}} \right]\)

σ ∝ Doping concentration

Hence For diffusion current

\({I_p} ∝ \frac{1}{L_p\sigma_p}\)

\({I_n} ∝ \frac{1}{L_n\sigma_n}\)

\(\frac{{{I_p}}}{{{I_n}}} = \left( {\frac{{{L_n}}}{{{L_p}}}} \right)\left( {\frac{{{σ _p}}}{{{σ _n}}}} \right)\)

\(4 = \left( {\frac{{{L_n}}}{{{L_p}}}} \right)\left( {\frac{{1.4}}{{2.8}}} \right)\)

Concept:

The conductivity of the silicon sample is given by σ 

⇒ σ = n q μ_n,

Here,

• n is the impurity concentration
• μ is the mobility of electrons
• q is the electronic charge in coulomb

The resistance of the silicon sample is given by

⇒ R = ρL/A   

Here, 

• L is the length of the sample
• A is the cross-sectional area of the sample
• ρ is the resistivity of the sample

Calculation

Given:

n = 5 × 10²⁰ atom/m³, μ_n = 0.125 m²/v-sec, q = 1.6 × 10^-19 C

⇒ σ = ( 5 × 10²⁰) × (1.6 × 10^-19) × (0.125)

⇒ σ = 10 

Therefore Resistivity (ρ) = 1/σ = 0.1

Also given: L = 10^–3 m, A = 10^–10 m²

⇒ R = 10⁶ Ω = 1MΩ 

Hence the resistance of the silicon sample is 1MΩ 

Therefore the correct answer is option b. 

V_H = (B * I) / (n * q * t)

where

• B = 8 T (in SI units),

• I = 4 A,

• n is carrier density,

• q = 1.6 × 10⁻¹⁹ C,

• t = 5 m,

we get:

V_H = (8 * 4) / (n * (1.6 × 10⁻¹⁹) * 5)
V_H = 32 / (n * 8 × 10⁻¹⁹)

 n = 10²⁰ m⁻³ ,

V_H = 32 / (10²⁰ * 8 × 10⁻¹⁹)
V_H = 32 / 80
V_H = 0.4 V

Hence the closest answer is (d) 0.4 V.

Drift Current:

• Drift current in the diode is the combined effect of movement of the minority charge carrier and majority charge carriers and also on the electric field applied to the diode.
• Drift current is due to the motion of charge carriers when a force exerted on them by an electric field.
• In the p-n junction diode, electrons are the majority charge carriers in the n-region and holes are the majority charge carriers in the p-region.
• When an electric field is applied to the diode there is more number of covalent bonds break and the concentration of charge carriers also increases in both region (p-type, n-type), and hence they affect the drift current in the diode.

Drift current I_drift = J_drift × A

Where, J_drift = Drift current density

A = Area of semiconductor

Now, J_drift = σ ⋅ E

= (nqμ_n + p.qμ_p)

σ = Conductivity

E = Electric field

⇒ I_dr = (nq.μ_n + p.qμ_p) ⋅ E A,

n = Number of electrons in n region

p = Number of holes in p-region

q = Charge on electrons and holes

μ_p = Mobility of holes

μ_n = Mobility of electrons

Additional Information

Diffusion

1) Diffusion is a natural phenomenon.

2) The migration of charge carriers from higher concentration to lower concentration or from higher density to lower density is called diffusion.

3) Diffusion is mainly due to the concentration gradient and is always negative.

It is given by:

dp/dx for holes and

dn/dx for electrons.

Diffusion current is calculated by:

where,

D_p is hole diffusion constant in cm²/sec

and q is charge in Coulomb.

According to Einstein's relation, the ratio of Diffusion constant to mobility remains constant, at fixed

Temperature, i.e.

Given: V_EC = 5V. This means that the pnp transistor is operating in the active node

\({I_{{E_1}}} = 10\;A\)

\({I_{{B_1}}} = \frac{{10}}{{1 + \beta }} = \frac{{10}}{{16}} = 0.625\;A\)

\({I_{{C_1}}} = \beta {I_{{B_1}}} = 15 \times \frac{{10}}{{16}} = 9.375\;A\)

\({I_{{C_1}}} = {I_{{S_1}}}{e^{{V_{EB}}/{V_T}}}\)

\(\Rightarrow {I_{S1}} = {I_{C1}}{e^{ - {V_{EB}}/{V_T}}}\)

\({I_{{S_1}}} = 9.375 \times {e^{ - \frac{{0.85}}{{26 \times {{10}^{ - 3}}}}}} \)

\(I_{S1}= 59.4 \times {10^{ - 15}}\;A\)

\(\frac{{{I_{{S_1}}}}}{{{I_{{S_2}}}}} = \frac{{{A_{EBJ1}}}}{{{A_{EBJ2}}}} = \frac{{59.4 \times {{10}^{ - 15}}A}}{{{I_{{C_2}}}{e^{ - {V_{BE}}/{V_T}}}}} \)

\(\frac{I_{S_{1}}}{I_{S_{2}}}= \frac{{59.4 \times {{10}^{ - 15}}}}{{2.03 \times {{10}^{ - 15}}}}= 29.3\)

Drift velocity is directly proportional to the electric field, i.e.

V_d ∝ E

V_d = μE

The proportionality constant 'μ' is called as the mobility and is given by:

The Relation between bandgap and intrinsic carrier concentration is

\(n_i^2 \propto {e^{ - {E_g}/KT}}\)

When E_g increase there is decrease in \(n_i^2\)

Now factor affecting BJT gain are

1) Emitter injection efficiency

2) Recombination in base

Due to decrease inn \(n_i^2\)( hence n_i ) the emitter injection efficiency increase. Hence there is decrease in current gain.

Current in a material can be expressed as:

I = NqAv_d

Where

N = Carrier concentration

q = electrons charge (magnitude)

A = cross-section area

The current density (J) will be:

Also, J = σ E     ----(2)

σ = conductivity

E = Electric field

J = Current density

v_d = μ E    ----(3)

μ = mobility of carriers

E = Electric field

Analysis:

On using equation (1), (2) & (3) we get,

J = NqV_d = NqμE = σE

⇒ σ = Nqμ  

So as conductivity is directly proportional to the mobility of ions and the number of EH pair ions.

The conductivity of an intrinsic semiconductor depends upon the number of hole electron pairs and mobility. The number of hole electron pairs increases with an increase in temperature, while its mobility decreases. However, the increase in hole electron pairs is greater than the decrease in their mobility.

So with an increase in temperature, conductivity in semiconductors also increases.

Hence option (d) is the correct answer.

Hall Voltage states that if a specimen (metal or semiconductor) carrying a current I is placed in transverse magnetic field B, an electric field is induced in the direction perpendicular to both I and B.

Hall Voltage is given by:

ρ = Charge density

Hall coefficient can be written as:

Where,

ρ = charge density = σ / μ 

n = charge concentration

σ = conductivity

μ = mobility constant

Hence, the Hall coefficient becomes

The Hall effect provides information on the sign, concentration, and mobility of charge carriers in the normal state.

A positive sign for the Hall coefficient indicates that the majority carriers are holes and the semiconductor is P-type.

A negative sign for the Hall coefficient indicates that the majority carriers are electrons and the semiconductor is N-type.

Applications of Hall-effect:

Hall effect can be used to find:

1. Carrier concentration

2. Type of semiconductor

3. Conductivity

4. Mobility

It cannot be used to find a magnetic field.

Common Confusion Point:

Looking at the formula one can think that the magnetic field can be calculated but in the HALL Experiment, perpendicular MAGNETIC field and electric field are applied on the material and other parameters are measured.

Concept:

The conductivity for a doped semiconductor is obtained as:

σ = q n0 μn = q p0 μp

where,

ni = Intrinsic carrier concentration.

μn = electron mobility

μp = Hole mobility

When a semiconductor crystal is optically excited, one hole is created for every electron, i.e. electron-hole pass are created.  

Calculation:

Before Illumination:

NA = 1017 cm-3 and ni = 2.2 × 106 cm-3

Since NA ≫ ni, the excess hole concentration at thermal equilibrium will be:

p0 = NA = 1017 cm-3

The electron concentration using mass-action law is obtained as:

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

\({n_0} = \frac{{{{\left( {2.2 \times {{10}^6}} \right)}^2}}}{{{{10}^{17}}}} = 4.8 \times {10^{ - 5}}c{m^{ - 3}}\)

The electron concentration before illumination is negligible.

Assuming room temperature, all dopants are ionized.

∴ p_o = N_A = 10¹⁷ cm^-3

Therefore, the electron concentration will be:

\({n_o} = \frac{{n_i^2}}{{{N_A}}} = \frac{{{{\left( {2.2 \times {{10}^6}} \right)}^2}}}{{{{10}^{17}}}} = 4.8 \times {10^{ - 5}}c{m^{ - 3}}\)

Here, we may notice that the electron contribution is negligible. So, we obtain the conductivity of the sample (in dark) as:

σ_o = q (μ_n n_o + μ_p p_o) ≅ qμ_pp_o

= (1.6 × 10^-19)(230)(10¹⁷)

σ₀ = 3.68 (Ω-cm)^-1

When electrons are generated optically, one hole is also created with the generation of an electron. Therefore, the excess hole concentration and excess electron concentration are the same for this case, i.e.

Δp = Δn = 10¹⁶ cm^-3

Now, the electron and hole concentration after illumination will be:

n = n0 + Δn

n = 4.8 × 10-5 + 1016

n = 1016

Similarly,

p = p0 + Δp

p = 1017 + 1016 cm-3

p = 11 × 1016 cm-3

∴ the conductivity of the given sample when the light is ON will be:

σ = q(n0 + Δn)μn + q(p0 + Δp)μp

σ = q[(1016)μn + (11 × 1016)μp]

= (1.6 × 10-19) (1016 × 5300 + (11 × 1016) (230))

= 1.6 × 10-19 (5.3 × 1019 + 2.53 × 1019)

= 1.6 × (5.3 + 2.53)

σ = 12.528 (Ω-cm)-1      

Concept:

For a field-effect transistor (FET) under certain operating conditions, the resistance of the drain-source channel is a function of the gate-source voltage alone and the JFET will behave as an almost pure ohmic resistor

For a MOSFET in the linear region:

VGS > Vth and

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

\({{I}_{D}}=\mu C_{ox} \frac{W}{2L}\left\{\left( {{V}_{GS}}-{{V}_{th}} \right)^2\right\}\)

Taking the derivative of the above w.r.t. V_DS, we get:

\(\frac{\partial I_D}{\partial V_{DS}}=\mu C_{ox}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)\)

Voltage-controlled Resistor (r_DS) is defined as:

\({r_{DS}} = \frac{{{V_{DS}}}}{{{I_{DS}}}} = \frac{1}{{{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)}}\)

Calculation:

Given:

V_GS = 2.0 V,

V_TH = -0.5 V,

\(\frac{W}{L} = 100\)

C_ox = 10^-8 f/cm²

μ_n = 800 cm²/V-s,

V_DS = 5 V

Putting on the respective values in Equation (1), we get:

\({r_{DS}} = \frac{1}{{800 \times {{10}^{ - 8}} \times 100 \times \left( {2 - \left( { - 0.5} \right)} \right)}}\)

\({r_{DS}} = 500\;{\rm{\Omega }}\)

Concept:

Mobility: It denotes how fast is the charge carrier is moving from one place to another.

It is demoted by μ.

Mobility is also defined as:

Where,

V_d = Drift velocity

E =  field intensity

Calculation:

E = 10 V/cm

V_d = 70 m/s = 7000 cm/s

From equation (1):

μ = 7000/10

μ = 700 cm²/Vs

Note: 

Electron mobility is always greater than hole mobility, 

Hence electron can travel faster and contributes more current than a hole.

The diffusion capacitance of a PN junction \({C_D} = \frac{{\tau I}}{{n{V_T}}}\)

τ = Carrier life time

I = Diode current

\(I = {I_0}\left( {{e^{\frac{v}{{\eta {V_T}}}}} - 1} \right)\)

\({I_1} = {I_0}\left( {{e^{\frac{{0.5}}{{{V_T}}}}} - 1} \right)\)

\({I_2} = {I_0}\left( {{e^{\frac{1}{{{V_T}}}}} - 1} \right)\)

\(\Rightarrow \frac{{{C_D}_1}}{{{C_D}_2}} = \frac{{{I_1}}}{{{I_2}}}\)

\({C_D}_2 = \frac{{{I_2}}}{{{I_1}}}{C_D}_1\)

\(= \frac{{{e^{\frac{1}{{{V_T}}}}}}}{{{e^{\frac{{0.5}}{{{V_T}}}}}}}{C_D}_1\left( {\therefore {e^{\frac{1}{{{V_T}}}}} > > 1} \right)\)

\(= {e^{\left( {\frac{{0.5}}{{0.0259}}} \right)}} \times {10^{ - 8}}F\)

= 2.42 F

Concept:

Energy of a photon emitted by a semiconductor is given by:

\(E = \frac{{hc}}{λ}\)

Rearranging the above, we get:

\(λ = \frac{{hc}}{E}\)

Differentiating this with respect to T:

\(\frac{{dλ }}{{dT}} = \frac{{ - hc}}{{E_g^2}}\frac{{d{E_g}}}{{dT}}\)    ----(1)

Calculation:

Given \({E_g} = \sqrt {1.24} ~eV\) , \(\frac{{d{E_g}}}{{dT}} = - 5 × {10^{ - 4}}~eV/K\) and hc = 1240 eV nm.

Putting these values in Equation- (1) we get:

\(\frac{{dλ }}{{dT}} = \frac{{ - 1240\left( {eV\;nm} \right)}}{{{{\left( {\sqrt(1.24)} \right)}^2}e{V^2}}}\left( { -~5 × {{10}^{ - 4}}} \right)\frac{{eV}}{K}\)

\(\frac{{dλ }}{{dT}} = 0.5\;nm/K\)

dλ = 0.5 × dT

Given dT = 10°C or 10 K, the required change in emitted wavelength will be:

dλ = 0.5 × 10 = 5 nm

At T = 300 K, we have:

n_i = 1.5 × 10¹⁰ cm^-3

N_d = 10¹⁶ cm^-3

Since N_d ≫ n_i, the electron concentration will be:

n₀ = N_d

= 10¹⁶ cm^-3

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

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

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

= 2.25 × 10⁴ cm^-3

Hence, the hole recombination rate in thermal equilibrium will be:

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

\( = \frac{{2.25 \times {{10}^4}}}{{20 \times {{10}^{ - 6}}}} = 1.125 \times {10^9}c{m^{ - 3}}{s^{ - 1}}\)

Now, the recombination rate of the majority carrier electrons will be the same as that of the minority carrier holes, i.e.

R’_p = R’_n

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

Hence, the lifetime of the majority carrier electrons will be:

\({\tau _{n0}} = \frac{{{n_0}}}{{{p_0}}}{\tau _{p0}} = \frac{{{{10}^{16}}}}{{2.25 \times {{10}^4}}} \times 20 \times {10^{ - 6}}\)

= 8.89 × 10⁶ sec

Smaller is the wave length used in lithography, better is the resolution.

Among the given options, x-rays have minimum wavelength and are expected to give best resolution.

Note: X-rays are presently not used because of increased cost associated while using them.

Concept:

The built-in potential barrier for a PN junction with given doping concentration Na, Nd, is given by:

\({{V}_{bi}}=\frac{kT}{q}\ln \left( \frac{{{N}_{a}}{{N}_{d}}}{n_{i}^{2}} \right)\)

Also, the conductivity is defined as:

σ = eμn

n = carrier concentration

Calculation:

The conductivity of a p-type semiconductor is defined as:

σ_po = eμ_pp_po

So, the hole (majority) concentration in p-side is:

\({P_{po}} = \frac{{{\sigma _{po}}}}{{e{\mu _p}}}\)

\({P_{po}} = \frac{{\frac{1}{2}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {1800} \right)}}\)

\({P_{po}} = 1.74 \times {10^{15}}c{m^{ - 3}}\)

Similarly, the electron (majority) concentration in n-side is obtained as:

\({n_{no}} = \frac{{{\sigma _{no}}}}{{e{\mu _n}}}\)

\( = \frac{{\frac{1}{1}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {3800} \right)}}\)

= 1.64 × 10¹⁵ cm^-3

Hence, the built-in potential in the germanium diode is:

\({V_{bi}} = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)

\( = \frac{{KT}}{q}\ln \left( {\frac{{{p_{po}} \times {n_{no}}}}{{n_i^2}}} \right)\)

\( = 0.0259\ln \left( {\frac{{1.74 \times {{10}^{15}} \times 1.64 \times {{10}^{15}}}}{{{{\left( {2.25 \times {{10}^{13}}} \right)}^2}}}} \right)\)

= 0.218 V