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The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is the material n-type or p-type?
Number of silicon atoms, N = 5 × 1028 atoms/m3
Number of arsenic atoms, nAs = 5 × 1022 atoms/m3
Number of indium atoms, nIn = 5 × 1020 atoms/m3
Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3
Number of electrons, ne = 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022
Number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
nenh = ni2
Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.
In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by
where n0 is a constant.
Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:
Where,
kB = Boltzmann constant = 8.62 × 10−5 eV/K
T = Temperature
n0 = Constant
Initial temperature, T1 = 300 K
The intrinsic carrier-concentration at this temperature can be written as:
… (1)
Final temperature, T2 = 600 K
The intrinsic carrier-concentration at this temperature can be written as:
… (2)
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.
Therefore, the ratio between the conductivities is 1.09 × 105.
In a p-n junction diode, the current I can be expressed as
where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kBis the Boltzmann constant (8.6×10−5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10−12 A and T = 300 K, then
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
In a p-n junction diode, the expression for current is given as:
Where,
I0 = Reverse saturation current = 5 × 10−12 A
T = Absolute temperature = 300 K
kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1
V = Voltage across the diode
(a) Forward voltage, V = 0.6 V
∴Current, I 
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, V’ = 0.7 V, we can write:
Hence, the increase in current, ΔI = I' − I
= 1.257 − 0.0256 = 1.23 A
(c) Dynamic resistance 
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.
You are given the two circuits as shown in the Figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure.
Hence, the output of the NOR Gate =
This will be the input for the NOT Gate. Its output will be
= A + B
∴Y = A + B
Hence, this circuit functions as an OR Gate.
(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, the output of the given circuit can be written as:
Hence, this circuit functions as an AND Gate.
Write the truth table for a NAND gate connected as given in Fig. 14.45.
Hence identify the exact logic operation carried out by this circuit.
A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.
Hence, the output can be written as:
The truth table for equation (i) can be drawn as:
|
A |
Y |
|
0 |
1 |
|
1 |
0 |
This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:
You are given two circuits as shown in the Figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits.
In both the given circuits, A and B are the inputs and Y is the output.
(a) The output of the left NAND gate will be
, as shown in the following figure.
Hence, the output of the combination of the two NAND gates is given as:
Hence, this circuit functions as an AND gate.
(b) 
is the output of the upper left of the NAND gate and 
is the output of the lower half of the NAND gate, as shown in the following figure.
Hence, the output of the combination of the NAND gates will be given as:
Hence, this circuit functions as an OR gate.
Write the truth table for circuit given in Figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
A and B are the inputs of the given circuit. The output of the first NOR gate is
. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.
Hence, the output of the combination is given as:
The truth table for this operation is given as:
|
A |
B |
Y (=A + B) |
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
1 |
This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energy of a signal is given by the relation:
E = 
Where,
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
E 
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
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