NCERT Solution Physics Class 12 Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
NCERT Solution Physics Class 12 Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits all questions and answers. Physics Class 12 14th Chapter Semiconductor Electronics: Materials, Devices and Simple Circuits exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 14. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits.
14.1.) In an n-type silicon, which of the following statement is true?
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
ANSWER-
We know that n type semiconductor is formed when pentavalent atoms are doped with silicon atoms. And n type means the semiconductors having electrons as a majority charge carrier. Hence option c is correct option.
14.2.) Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
ANSWER-
We know that for a p-type semiconductor is formed when trivalent atom is doped with silicon atoms where holes are majority charge carrier. Hence option d is correct option.
14.3.) Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to (Eg )C , (Eg ) Si and (Eg )Ge. Which of the following statements is true?
(a) (Eg ) Si < (Eg )Ge< (Eg )C
(b) (Eg )C< (Eg )Ge> (Eg ) Si
(c) (Eg )C> (Eg ) Si > (Eg )Ge
(d) (Eg )C = (Eg ) Si = (Eg )Ge
ANSWER-
Out of three elements present the energy band gap of germanium is lowest and energy band gap of carbon is highest hence option c is the correct option for giving relation of energy gaps.
14.4.) In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.
ANSWER-
In an unbiased p-n junction, holes diffuse from the p-region to n-region because they move across the junction by the potential difference. Here p region has higher concentration of holes than the n region hence holes diffuses from higher potential to lower potential.
14.5.) When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) Reduces the majority carrier current to zero. (c) Lowers the potential barrier. (d) None of the above.
ANSWER-
In forward bias positive terminal is connected to the p region while negative terminal is connected to the n region. Due to positive terminal majority charge carriers holes get repelled and potential barrier reduces same happens for n region. Hence, the potential barrier across the junction gets reduced. Hence option c is correct option.
14.6)
14.10)
14.11 You are given the two circuits as shown in Fig. 14.36. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
(a) Here in figure a we have A and B as a input while Y as a output. Given figure is the combination of the NOR gate and NOT gate. And NOR gate is the input to the NOT gate. Hence output of the NOR gate is given by (A+B)̅ which will be input for NOT gate. Output of NOT gate is given by (A+B) ̿ = A+B. hence output of the given diagram is Y =A+B . Which is similar to the output of the OR gat when input is A and B. Hence circuit in figure a acts as OR gate.
(b) From figure b we have two NOT gates connected to the input of the NOR gate . the input to the NOR gate is and . The output from this NOR gate is
(A ̅+B ̅ ) ̅ = A ̿ . B ̿ = A.B
14.12 Write the truth table for a NAND gate connected as given in Fig. 14.37. Hence identify the exact logic operation carried out by this circuit.
ANSWER-
A acts as the two inputs of the NAND gate and Y is the output hence output is given by
Y = (A.B) ̅ = A ̅+A ̅ = A ̅
∴ Truth table is same as NOT gate and given by
INPUT(A) |
OUTPUT ( A ̅) |
0 | 1 |
1 | 0 |
14.13.) You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.
ANSWER-
(A) From figure (a) we can say that it is the combination of two NAND gates. Here A and B are inputs and Y is output and given by
From output you can say that this circuit acts as OR get.
14.14.) Write the truth table for circuit given in Fig. 14.39 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.)
ANSWER-
In given circuit A and B are the inputs while output is Y. The output of the first NOR get is input to second NOR gate.
Output of the first NOR gate is (A+B) ̅ . This (A+B) ̅
This will perform the logic of OR gate. And truth table is given by
A |
B |
Y=A+B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
14.15)
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