Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits Solutions
Question - 11 : - A p-n photodiode is fabricated from a semiconductor withband gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Answer - 11 : -
Energy band gap of the given photodiode, Eg =2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energyof a signal is given by the relation:
E =
Where,
h = Planck’sconstant
= 6.626 × 10−34 Js
c = Speedof 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, thephotodiode cannot detect the signal.
The number of silicon atoms per m3 is 5 × 1028. Thisis doped simultaneously with 5 × 1022 atomsper m3 of Arsenic and 5 × 1020 perm3 atoms of Indium. Calculate the number ofelectrons and holes. Given that ni= 1.5 × 1016 m−3. Is thematerial n-type or p-type?
Answer - 12 : -
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 thermalequilibrium, the concentrations of electrons and holes in a semiconductor arerelated as:
nenh = ni2
Therefore, the number of electrons isapproximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than thenumber of holes, the material is an n-type semiconductor.
Question - 13 : - In an intrinsic semiconductor the energy gap Egis1.2 eV. Its hole mobility is much smaller than electron mobility andindependent of temperature. What is the ratio between conductivity at 600K andthat at 300K? Assume that the temperature dependence of intrinsic carrierconcentration niisgiven by
where n0 isa constant.
Answer - 13 : -
Energy gap of the given intrinsicsemiconductor, Eg =1.2 eV
Thetemperature 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
Theintrinsic carrier-concentration at this temperature can be written as:
Final temperature, T2 =600 K
Theintrinsic carrier-concentration at this temperature can be written as:
… (2)
The ratiobetween the conductivities at 600 K and at 300 K is equal to the ratio betweenthe respective intrinsic carrier-concentrations at these temperatures.
Therefore, the ratio between the conductivitiesis 1.09 × 105.
Question - 14 : - In a p-n junction diode, the current I can be expressed as
where I0 iscalled the reverse saturation current, V is the voltage across thediode and is positive for forward bias and negative for reverse bias, and I isthe current through the diode, kBis theBoltzmann constant (8.6×10−5 eV/K) and T is the absolute temperature. Iffor a given diode I0 = 5 × 10−12 Aand T = 300 K, then
(a) What will be the forward current at a forward voltage of0.6 V?
(b) What will be the increase in the current if the voltageacross the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changesfrom 1 V to 2 V?
Answer - 14 : -
In a p-n junction diode, the expression for current isgiven as:
Where,
I0 =Reverse saturation current = 5 × 10−12 A
T = Absolutetemperature = 300 K
kB =Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 JK−1
V = Voltageacross 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 dynamicresistance in the reverse bias will be infinite.
You are given the two circuits as shown in Fig. 14.44.Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
Answer - 15 : -
(a) A and B arethe inputs and Y isthe output of the given circuit. The left half of the given figure acts as theNOR Gate, while the right half acts as the NOT Gate. This is shown in thefollowing figure.
Hence, the output of theNOR Gate =
This will be the input forthe NOT Gate. Its output will be
= A + B
∴Y = A + B
Hence, thiscircuit functions as an OR Gate.
(b) A and B arethe inputs and Y isthe output of the given circuit. It can be observed from the following figurethat the inputs of the right half NOR Gate are the outputs of the two NOTGates.
Hence, the output of the given circuit can be written as:
Hence, this circuit functions as an AND Gate.
Question - 16 : - Write the truth table for a NAND gate connected as givenin Fig. 14.45.
Hence identify the exact logic operation carried out bythis circuit.
Answer - 16 : -
A actsas the two inputs of the NAND gate and Y is the output, as shown inthe 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 thislogic circuit is shown as:
Question - 17 : - You are given two circuits as shown in Fig. 14.46, whichconsist of NAND gates. Identify the logic operation carried out by the twocircuits.
Answer - 17 : -
In both the given circuits, A and B arethe inputs and Y isthe 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 gatesis given as:
Hence, this circuit functions as an AND gate.
(b) 
is the output of the upperleft of the NAND gate and 
is the output of the lower half of the NANDgate, as shown in the following figure.Hence, the output of the combination of the NAND gateswill be given as:
Hence, this circuit functions as an OR gate.
Question - 18 : - Write the truth table for circuit given in Fig. 14.47below 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 gatewill be 0 and hence Y=1. Similarly work out the values of Y for othercombinations of A and B. Compare with the truth table of OR, AND, NOT gates andfind the correct one.)
Answer - 18 : - A and B are the inputs of thegiven circuit. The output of the first NOR gate is. It can be observed from the followingfigure that the inputs of the second NOR gate become the out put of the firstone.
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 circuitfunctions as an OR gate.
Write the truth table for the circuits given in Fig. 14.48consisting of NOR gates only. Identify the logic operations (OR, AND, NOT)performed by the two circuits.
Answer - 19 : - (a) A actsas the two inputs of the NOR gate and Y is the output, as shown inthe following figure. Hence, the output of the circuit is
The truth table for the same is given as:
| A | Y |
| 0 | 1 |
| 1 | 0 |
This is the truth table of a NOT gate. Hence, this circuitfunctions as a NOT gate.
(b) A and B arethe inputs and Y isthe output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs ofthe first two NOR gates are
as shown in the followingfigure.
are the inputs for the last NOR gate. Hence, the outputfor the circuit can be written as:
The truth table for the same can be written as:
| A | B | Y (=A⋅B) |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
This is the truth table of an AND gate. Hence, thiscircuit functions as an AND gate.