In a p-n junction at equilibrium, i.e. no bias
The hole electrons flow (drift, diffusion) directions are as follows (convention based on the above PN diagram)
Particle type and flow |
Directions of flow |
Current |
1) Hole diffusion |
→ |
→ |
2) Hole drift |
← |
← |
3) Electron diffusion |
← |
→ |
4) Electron drift |
→ |
← |
Option (d) is not true.
Drift Current:
Drift current Idrift = Jdrift × A
Where, Jdrift = Drift current density
A = Area of semiconductor
Now, Jdrift = σ ⋅ E
= (nqμn + p.qμp)
σ = Conductivity
E = Electric field
⇒ Idr = (nq.μn + p.qμp) ⋅ E A,
n = Number of electrons in n region
p = Number of holes in p-region
q = Charge on electrons and holes
μp = Mobility of holes
μn = 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:
\(\frac{dp}{dx}\) for holes and
\(\frac{dn}{dx}\) for electrons.
Diffusion current is calculated by:
\(I=-qD_p\frac{dp}{dx}\) -----(1)
where,
Dp is hole diffusion constant in cm2/sec
and q is charge in Coulomb.
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:
\(μ=\frac{V_d}{E}\) cm^{2}/Vs ------(1)
Where,
V_{d} = Drift velocity
E = field intensity
Calculation:
E = 10 V/cm
V_{d} = 70 m/s = 7000 cm/s
From equation (1):
\(μ=\frac{7000}{10}\)
μ = 700 cm^{2}/Vs
Note:
Electron mobility is always greater than hole mobility,
Hence electron can travel faster and contributes more current than a hole.
According to Einstein’s relationship for a semiconductor, the ratio of the diffusion constant to the mobility of the charge carriers is
According to Einstein's relation, the ratio of Diffusion constant to mobility remains constant, at fixed
Temperature, i.e.
\(\frac{{{D_n}}}{{{\mu _n}}} = \frac{{{D_p}}}{{{\mu _p}}} = {V_T} = \frac{{kT}}{q} = \frac{T}{{11,600}}\)
V_{T} – Temperature equivalent
K – Boltzman’s constant
T – Temperature in Kelvin
Explanation:
Current in a material can be expressed as:
I = NqAvd
Where
N = Carrier concentration
q = electrons charge (magnitude)
A = cross-section area
The current density (J) will be:
\(J = \frac{I}{A} = Nq{v_d}\) ---(1)
Also, J = σ E ----(2)
σ = conductivity
E = Electric field
J = Current density
vd = μ E ----(3)
μ = mobility of carriers
E = Electric field
Analysis:
On using equation (1), (2) & (3) we get,
J = NqVd = 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 (4) is the correct answer.
Concept:
The hall voltage VA is given by:
\( {V_H} = \frac{{BI}}{{ρ_cW}} = \frac{R_HBI}{W}\)
ρc: Charge density
RH: Hall coefficient
\(R_H = \frac{1}{- nq} = \frac{1}{pq}\)
W is the side across which the magnetic field is applied.
Calculation:
W = Thickness = 2.5 mm, I = 4A, RH = 5 x 10^{-7}, B = 0.8 Wb/m2
\( {V_H} = \frac{R_HBI}{W}\)
\(V_H=\frac{5\times10^{-7}\times0.8\times4}{2.5\times10^{-3}}\)
V_{H} = 6.4 × 10-4 V
Concept:
Mobility of charge carriers (μ) defines how fast the charge carriers or travel from one place to another.
It is drift velocity (saturation velocity) per unit electric field. i.e
\(\mu = \frac{{{v_d}}}{E}\) -----(1)
vd = Drift velocity (Saturation velocity)
E = Electric field given by:
\(E = \frac{V}{L}\) -----(2)
Calculation:
L = 4 cm
V = 8 Volts
v_{d} = 104 cm/s
from Equation (2):
\(E = \frac{V}{L}\\ =\frac{8}{4}=2 \ V/cm\)
From equation (1):
\(\mu = \frac{{{v_d}}}{E}\\=\frac{104}{2} = 52 \ cm^2/Vs\)
Note: The term impurity in an intrinsic semiconductor:
Impurity scattering is caused by crystal defects such as ionized impurities.
At lower temperatures, carriers move more slowly, so there is more time for them to interact with charged impurities.
As a result, as the temperature decreases, impurity scattering increases, and mobility decreases.
The effect is just the opposite of the effect of lattice scattering.
Hall effect can be used to measure
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:
\({V_H} = Ed = \frac{{BI}}{{ρ W}}\)
ρ = Charge density
Hall coefficient can be written as:
\({{\rm{R}}_{\rm{H}}} = \frac{1}{{\rm{ρ }}} = \frac{1}{{{\rm{ne}}}}\)
Where,
ρ = charge density = σ / μ
n = charge concentration
σ = conductivity
μ = mobility constant
Hence, the Hall coefficient becomes
\({{\rm{R}}_{\rm{H}}} = \frac{1}{{\rm{ne }}} = \frac{\mu}{{{\rm{\sigma }}}}\)
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:
Ohms’ law: At constant temperature, the current through a resistance is directly proportional to the potential difference across the resistance.V = R I
Where V is the potential difference, R is resistance and I is current flowing
Current density (J): The electric current per unit area is called current density.
\(Current~density~\left( J \right)=\frac{Current~\left( I \right)}{Area~\left( A \right)}\)
Conductivity (σ): The property of a conductor due to which the current flows through it is called the conductivity of that conductor.
\(Resistance~\left( R \right)=\frac{\rho ~l}{A}\)
Where ρ is resistivity = 1/σ, l is the length and A is the area of the conductor
Electric field (E) = potential difference (V)/length (l)
Analysis:
As, V = R I
\(Resistance~\left( R \right)=\frac{\rho ~l}{A}\)
\(V=\frac{\rho ~l}{A}\times I=\left( \frac{1}{σ } \right)\times l\times \frac{I}{A}=\left( \frac{1}{σ } \right)\times l\times J\)
\(J=σ \times \frac{V}{l}=σ ~E\)
σ = J/E
The Hall Effect may be used to
1. Determine whether the semiconductor is p-type or n - type.
2. Determine the carrier concentration.
3. Calculate the mobilty.
Which of the above statements are correct?
Hall Effect:
It states that if a specimen (metal or semiconductor) carrying a current (I) is placed in a transverse magnetic field (B), an electric field is induced in the direction perpendicular to both I and B.
Hall Voltage is given by:
\({V_H} = Ed = \frac{{BI}}{{ρ W}}\)
The hall coefficient is given as:
\({{\rm{R}}_{\rm{H}}} = \frac{1}{{\rm{ρ }}} = \frac{1}{{{\rm{ne}}}}\)
Where,
ρ = charge density = σ / μ
n = charge concentration
σ = conductivity
μ = mobility constant
Hence Hall coefficient becomes
\({{\rm{R}}_{\rm{H}}} = \frac{1}{{\rm{ne }}} = \frac{\mu}{{{\rm{\sigma }}}}\)
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 HALL Experiment, perpendicular MAGNETIC field and electric field are applied on the material and other parameters are measured.
Select the correct statement about a semiconductor
Effective mass |
Germanium |
Silicon |
Gallium Arsenide |
m_{n} |
0.55m |
1.08m |
0.067m |
m_{p} |
0.37m |
0.56m |
0.48m |
Where m: mass
Observations:
1) The effective mass of the hole is greater than that of the electron.
2) The relaxation times are often of the same order of magnitude for electrons and holes and therefore, they do not make too much difference.
As the effective mass is less for electron it moves faster than that of the hole.
Drift velocity
v_{d} = μ × E
μ: Mobility of charge carrier
\(\mu = \frac{{{V_d}}}{E}\frac{{c{m^2}}}{{volt - Sec}}\;or\frac{{{m^2}}}{{volt - sec}}\)
Drift velocities of Hole and Electron
V_{n} = μ_{n} × E
V_{p} = μ_{p} × E
Charge carriers can move in a semiconductor because of two phenomena:
Drift Mechanism:
Diffusion Mechanism:
Analysis:
Drift velocity is directly proportional to the electric field, i.e.
Vd ∝ E
Vd = μE
The proportionality constant 'μ' is called as the mobility and is given by:
\(\mu = \frac{{{V_d}}}{E} = \frac{{Drift\;velocity}}{{electric\;field}}\)
Substituting the respective units, we get:
\( \mu= \;\frac{{m/s}}{{V/m}}\)
∴ SI unit of μ is \(\frac{{{m^2}}}{{V - sec}}\) or m2V-1s-1
A piece of silicon is doped uniformly with phosphorus with a doping concentration of 10^{16} /cm^{3} The expected value of mobility versus doping concentration for silicon full dopant ionization is shown below. The Charge of an electron is 1.6 × 10^{-19} C. The conductivity (in S cm^{-1}) of the silicon sample at 300 K is_________
Concept:
For n-type SC conductivity is given by σ ≈ q.n.μn
Application:
We see that at a doping concentration of \({10^{16}}\), the mobility values are:
μ_{n} = 1200 cm^{2}V^{-1}sec^{-1}
μ_{p} = 400 cm^{2}V^{-1}sec^{-1}
Assuming complete ionization. We have electron concentration:
np = 10^{16} cm^{-3}
Now, since intrinsic carrier concentration of silicon of the order 10^{10}, we have hole concentration p of the order of:
\(\frac{{{{\left( {{{10}^{10}}} \right)}^2}}}{{{{10}^{16}}}} ≈ {10^4}\)
Thus we can neglect hole concentration while calculating conductivity.
Thus, we have conductivity:
σ ≈ q.n.μ_{n}
σ = 1.6 × 10^{-19} × 10^{16} × 1200
σ = 1.92 S cm^{-1}
Explanation:
Hall Effect: It states that if a “specimen (meta or semiconductor) carrying the current I is placed in traverse magnetic field B, then Electric field Intensity E is induced in a direction perpendicular to both I and B”.
From Hall Experiment, we can determine the following properties:
From the above properties, we can say that,
Option (2) is correct.
Concept:
The formula for the conductivity (σ) of the semi-conductor is
\(\sigma = {n_i}q\left( {{\mu _n} + {\mu _p}} \right)\)
Where,
q is the charge on the moving particle
ni is intrinsic carrier density
μn is electron mobility
μp is hole mobility
Calculation:
Given that,
σ = ?
ni at 300°K = 2.5 × 1013 /cm
μn = 3800 cm2/V-s
μp = 1800 cm2/V-s
q = 1.6 × 10-19 C
Then, putting the value in the above formula
\(\sigma = {2.5 × 10^{13}} × 1.6 × {10^{ - 19}} × \left( {3800 + 1800} \right)\)
= 22400 × 10^{-6 }S/cm = 0.0224 S/cm
Hall coefficient is defined as:
\({R_H} = \frac{{P\mu _p^2 - n\mu _n^2}}{{q{{\left( {p{\mu _p} + n{\mu _n}} \right)}^2}}}\)
\({R_H} = \frac{1}{{{\rho _c}}} = \frac{{ - 1}}{{nq}}\)
\(n = \frac{1}{{{R_H}q}}\)
We can find the electron concentration as:
\({R_H} = \frac{1}{{{\rho _c}}} = \frac{1}{{pq}}\)
\(P = \frac{1}{{{R_H}q}}\)
ρc = charge density
σ_{n} = nμ_{n}q and σ_{p} = pμ_{p}q
Concept:
The hall voltage V_{A} is given by:
\({V_{12}} = {V_H} = \frac{{BI}}{{ρ_cW}} = \frac{R_HBI}{W}\)
ρ_{c}: Charge density
R_{H}: Hall coefficient
\(R_H = \frac{1}{- nq} = \frac{1}{pq}\)
W is the side across which the magnetic field is applied.
Calculation:
W = Thickness = 4 mm, I = 1 mA, V_{H} = 0.005 V, B = 0.1 Wb/m^{2}
\(0.005 = \frac{{{0.1} \times {{10}^{ - 3}}}}{{e\times {p} \times {4\times{10}^{ - 3}}}}\)
Since hall voltage is positive, the majority charge carriers will be holes, with the concentration calculated as:
\(p = \frac{{{0.1} \times {{10}^{ - 3}}}}{{1.6\times 10^{-19}\times 4\times 10^{-3}\times 0.005}} \)
p ≈ 3 × 10^{19} m^{-3}
A dc voltage of \(10{\rm{\;V}}\) is applied across an n – type silicon bar having a rectangular cross - section and a length of \(1{\rm{\;cm}}\) as shown in figure. The donor doping concentration \({{\rm{N}}_{\rm{D}}}\) and the mobility of electrons \({{\rm{\mu }}_{\rm{n}}}\) are \({10^{ - 16}}{\rm{c}}{{\rm{m}}^{ - 3}}\) and \(1000{\rm{c}}{{\rm{m}}^2}{{\rm{V}}^{ - 1}}{{\rm{s}}^{ - 1}},\) respectively. The average time (in μsec) taken by the electrons to move from one end of the bar to other end is____________
Concept:
drift velocity is given by \({{\rm{v}}_{\rm{d}}} = {{\rm{\mu }}_{\rm{n}}} \times {\rm{E}}\)
Application:
We have, field intensity \({\rm{E}} = \frac{{\rm{v}}}{{\rm{d}}} = \frac{{10}}{1} = 10\frac{{\rm{V}}}{{{\rm{cm}}}}\)
Now, drift velocity, \({{\rm{v}}_{\rm{d}}} = {{\rm{\mu }}_{\rm{n}}} \times {\rm{E}}\)
\(\Rightarrow {{\rm{v}}_{\rm{d}}} = 1000 \times 10 = {10^4}\frac{{{\rm{cm}}}}{{{\rm{sec}}}}\)
Thus, average time taken for transit,
\(\begin{array}{l} {\rm{t}} = \frac{{\rm{d}}}{{{{\rm{v}}_{\rm{d}}}}} = \frac{1}{{{{10}^4}}} = {10^{ - 4}}{\rm{sec}}\\ \Rightarrow {\rm{t}} = 100{\rm{\mu sec}} \end{array}\)
Hence, \({\rm{t}}\left( {{\rm{in\;\mu sec}}} \right) = 100\)Concept:
The number of minority carriers generated because of light is given by the generation rate as:
\(G = \frac{{excess\;minority\;carrier\;generated}}{{minority\;carrier\;life\;time}} \)
\(G= \frac{{{\rm{\Delta }}p}}{{\tau_p}} = \frac{{{\rm{\Delta }}n}}{{\tau_n}}\)
Analysis:
N_{A} = 10^{16} / cm^{3}
G = 10^{20} / cm^{3} – sec
τ = 100 μsec
ni = 10^{10} / cm^{3}
before shining light:
Hole concertration (p) ≃ N_{A} = 10^{16} / cm^{3}
Electron concentration (n) \(= \frac{{n{i^2}}}{p} = \frac{{{{10}^{20}}}}{{{{10}^{16}}}} = {10^4}/c{m^3}\)
After illumination of light minority carrier will be generated.
So after light illumination hole conc. (p’) = p + Δp
p’ = 10^{16} + 10^{20} × 10^{-6} × 100
= 2 × 10^{16} / cm^{3}
After illumination e^{-} conc. (n’) = n + Δn
= 10^{4} + 10^{20} × 10^{-6} × 100
≃ 10^{16} / cm^{3}
Product of e^{-} and hole concertration = n’ × p’
= 2 × 10^{32} / cm^{-6}