Semiconductor Physical Properties Assignment Questions 
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Question 1  In an experiment, the current through and the voltage across the opposite faces of three specimen blocks of material, each 1 cm^{3}, was measured using the circuit of FIGURE 1. Use the results of the experiment, given in the table below, to classify each material as insulator, conductor or semiconductor.
Answer  It is given in the question the voltage and current values of three specimens as follows
Specimen

V

I

A

10V

40µA

B

10V

0.2A

C

1000V

10nA

The length breadth and width of the material are 1cm each.
To check the given materials are insulators, conductors and semiconductors we can calculate the resistivity.
Resistivity ρ = RA/l
We know R = V/I
A = 1cm^{2} and length = 1cm
So
Specimen

V

I

R = V/I

ρ = RA/l

A

10V

40µA

10/(40 x 10^{6}) = 250kΩ

=(250 x 10^{3} x 1)/1 = 250 x 10^{3} Ωcm

B

10V

0.2A

10/0.2 = 50Ω

(50 x 1)/1= 50 Ωcm

C

1000V

10nA

1000/(10 x 10^{9}) = 100 x 10^{9} Ω

= (100 x 10^{9} x 1)/1 = 100 x 10^{9} Ωcm

Based on these values
Specimen

Materials

A

Semiconductor

B

Conductor

C

Insulator

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Question 2 (i)  Sketch and label energyband diagrams of a metal, an insulator and a semiconductor and give approximate values of the energy gap where appropriate.
Answer  Energy band diagram
1 eV = 1.6 × 10^{19}J
Materials

Energy band diagram

Band gap values

Insulator


Energy gap value is greater than 5eV

Semiconductor


Energy gap less than 5e V

Metal


Band gap will overlap.

Question 2 (ii)  The three graphs of FIGURE 2 show how the resistivities of metals, insulators and intrinsic semiconductors vary over a small temperature range. Explain the shape of the graphs in terms of the energyband theory of electrons in solids.
Answer  We can explain as follows
Materials

Explanation

Metal

Metals are having low resistivity. When the temperature is increased the resistivity is also increased. Metals have positive temperature coefficient.

Insulator

Insulators are having high resistivity. When temperature is increased the resistivity will decrease, showing the negative temperature coefficient.

Intrinsic Semiconductor

Intrinsic semiconductors are having the greatest resistivity. When the temperature is increased the resistivity will decrease, showing negative temperature coefficient.

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Question 3 (i)  Determine the current flowing through a 1 cm cube of pure silicon when a potential of 1 V is applied across opposite faces (FIGURE 3(a)) if the number density of holeelectron pairs in the cube 1.56 x 10^{16} m^{3}.
Answer  We need to determine the current flowing
As we know the resitivity is ρ = RA/l
R = ρl/A
V/I = ρl/A = l/Aσ = l/Aqdesity
= (1×10^{2})/(1×10^{6} × 1.6×10^{19} × 1.56×10^{16})
I = (1×10^{6} × 1.6×10^{19} × 1.56×10^{16})/(1×10^{2})
I = 2.496×10^{7}A
= 0.2496µA
Question 3 (ii)  A lamp is now allowed to shine upon one face of the cube as shown in FIGURE 3(b). This has the effect of generating holeelectron pairs at the surface of the exposed face and so raising the number densities of the carriers at that surface to a value of 1 x 10^{20} carriers m^{3}. For the polarities of external connections as shown, calculate the change in hole current due to the effects of the light.
Answer  We need to determine the current flowing
As we know the resitivity is ρ = RA/l
R = ρl/A
V/I = ρl/A = l/Aσ = l/Aqdesity
= (1×10^{2})/(1×10^{6} × 1.6×10^{19 }× 1×10^{20})
I = (1×10^{6} × 1.6×10^{19} × 1×10^{20})/(1×10^{2})
I = 1.6×10^{3 }A
= 1.6mA
Question 3 (iii)  If the polarity of the supply is now reversed, calculate the change in electron current due to the light. Again, assume a linear density gradient.
Answer  If the polarity is reversed then the current will change its direction as 1.6mA.
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Question 4  Use the FermiDirac probability function
p(W) = (1+ exp((WW_{F})/kT))^{1}
to calculate, for pure silicon at a temperature of 27^{o}C, the probability of:
(a) an electron just being at the bottom of the conduction band
Answer  Electron at the bottom of conduction band
ww_{F} = 0
so exp[(ww_{F})/KT] = 1
p(w) = (1+1)^{1} = ½5
(b) A hole just being at the top of the valence band.
Answer  Hole at the top of conduction band
ww_{F} = 0
so exp[(ww_{F})/KT] = 1
p(w) = (1+1)^{1} = 1/2
Question 5  A silicon 'chip' has the dimensions 8 x 2 x 2 mm.
(i) Calculate its resistance as measured between the square faces, assuming it is of intrinsic material. Take the carrier number density of holeelectron pairs as 1.5 x 10^{16} m^{3} at room temperature.
Answer  We need to calculate the resistance
Given that a pure si at absolute temperature
So σ = q(μn + μp)ni
We know the values of μn + μp = 0.15 + 0.05 = 0.2
Now σ = 1.6×10^{19} × 0.2×1.5×10^{16}
= 0.48×10^{3}
Now ρ = (σ)^{1}
ρ = 2083
R = ρl/A = (2083×8×10^{3})/(16×10^{6}) = 10.4MΩ
(ii) Estimate its resistance at 10^{o}C above room temperature.
Answer  Resistance at 10^{o}C above room temperature
Almost same as 10.4MΩ because the si sample is not doped at this temperature.
(iii) Calculate the resistance, again across the square faces and at room temperature, if the 'chip' is now uniformly doped with aluminium, so that one in every 10^{7} atoms is an impurity atom (a cubic metre of silicon contains about 5 x 10^{28} atoms). Also apply the 'law of mass action' to calculate the minority carrier density.
Answer  σ = q(μnno + μppn)
Given that n_{0} = 5×10^{28}
Now p_{o} = (n_{i}^{2})/n_{0} = 5120
Now σ = 1.6×10^{19}[0.15×5×10^{28} + 0.5×5120]
= 12×10^{8}
Now ρ = (σ)^{1}
ρ = 0.0833×10^{8}
R = ρl/A = (0.0833×108×8×10^{3})/(16×10^{6}) = 0.042×10^{5}Ω
(iv) Estimate the resistance of the doped 'chip' at 10^{o}C above room temperature.
Answer  When temperature is greater than room temperature the resistance will have slight variations.
(v) Sketch, on the same axes, the energy/probability distribution curve for ntype silicon at two temperatures T_{1} and T_{2} (where T_{2} > T_{1}) to show the effect of temperature upon the position of the Fermi level.
Answer  The diagram is as below
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