NCERT Solutions for 12th Physics, Chapter 14 ‒ Semiconductor Electronics: Materials, Devices and Simple Circuits are available here. Questions from this chapter of NCERT textbook are frequently asked in CBSE school and board exams. In order to score well is CBSE Class 12 Physics board exam 2018; students must learn these solutions. Questions given in this chapter are mostly based on Classification of Metals, conductors and semiconductors, intrinsic semiconductor, extrinsic semiconductor (n-type semiconductor, p-type semiconductor), p-n junction, p-n junction formation, semiconductor diode, p-n junction diode under forward bias, p-n junction diode under reverse bias, special purpose p-n junction diodes, Zener diode, Zener diode as a voltage regulator, photodiode, light emitting diode, solar cell, n-p-n transistor, p-n-p transistor, basic transistor circuit configurations and transistor characteristics, common emitter transistor characteristics, current amplification factor (β), transistor as a device, transistor as a switch, transistor as an amplifier, digital electronics and logic gates etc.
NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 14: Semiconductor Electronics are given below:
Question 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.
Solution 14.1: (c)
Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
Solution 14.2: (d)
Question 14.3: Carbon, silicon and germanium have four valence electrons each.
These are characterised 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
Solution 14.3: (c)
Question 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.
Solution 14.4: (c)
Question 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.
Solution 14.5: (c)
Question 14.6: For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Solution 14.6: (b), (c)
Question 14.7: For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Solution 14.7: (c)
Question 14.8: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
For Half-wave rectification: 50 Hz
For Full-wave rectification: 100 Hz
Question 14.9: For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Rc = 2 k ohms
Vc = 2 V
β = 100
Rb = 1 k ohm
Calculating collector current
Ic = Vc / Rc = 2 / 2000 = 1 mA
Now, we know
Current amplification factor β = Ic / Ib
Hence, Ib = 1 / 100 x 1000
Or, Ib = 10 micro-amperes
Therefore, Vi = (10 × 10-6 )/(1 x 103) = 0.01 V
Question 14.10: Two amplifiers are connected one after the other in series (cascaded).
The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
Av’ = 10
Av’’ = 20
Vi = 0.01 V
For cascade connection
Net voltage gain, Av = Av’.Av’’ = 200
Av = Vo/Vi
Or Vo = 200 x 0.01 or Vo = 2 V
Question 14.11: A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
h v >Eg
h v = (h c)/ 𝜆 = [(6.626 x 10‒34).(3 x 108)] / (6000 x 10‒9) = 3.3 x 10‒20 J = 0.21 eV < 2.8 eV = Eg
Hence the detection is not possible.