# NCERT 12 Physics Semiconductor Electronics Materials, Devices And Simple Circuits Chapter 14 Exercise

Q.1. In n- type 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 carries and trivalent atoms are the dopants

Sol. n-type is obtained by doping the Ge or Si with pentavalent atoms. Is n-type semiconductor, electrons are majority carriers and holes and holes are minority carriers, hence option (c) is True.
Q.2. Which of the statements given in Q.1 is true for p-type semiconductors.
Sol. P-type semiconductor is obtained by doping Go or Si with trivalent atoms. In p-type semiconductor holes are majority carriers and holes are minority carriers. Hence option (d) is correct.

Q.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 (Eg)c, (Eg)sir  and (Eg)Ge.

(a) (Eg)Sir < (Eg)Ge. < (Eg)C

(b) (Eg)C < (Eg)Ge. < (Eg)sir

(c) (Eg)C < (Eg)sir. < (Eg)Ge

(d) (Eg)C < (Eg)sir. < (Eg)Ge

Sol. The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given tree elements. Hence option (c) is correct.

Q.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.

Sol. In an unbiased p-n junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration. Thus option (c) is correct.

Q.5. When a forward bias is applied to p-n junctions, it

(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.

Sol. When a forward bias is applied across the p-n junction, the applied voltage opposes the barrier voltage. Due to it, the potential barrier across the junction is lowered. Hence option (c) is correct.

Q.6. For a transistor action, which of the following statements are current.

(a) collector current is equal to the sum of base current and emitter current.

(b) The input resistance depends upon the current lc in the transistor.

(c) The emitter junction is forward biased and collector is reverse biased.

(d) Both the emitter junction as well as the collector junction are forward biased. current Ic for a transistor action, the emitter junction is forward biased and collector junction is reverse biased.

Q.7. For 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.

Sol.  (c) the voltage gain is low at high and low frequencies and constant and high at mid-frequencies.
Q.8. In half wave rectifier, 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.
Sol. The output ripple frequency is 50 Hz for half wave rectifier and 100 Hz for full-wave rectifier.
Q.9. For a common emitter amplifier, the audio signal voltage across the collector 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 k Ω. Q.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 V, calculate the output a. c. signal. Q.11. A p-n junction is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 600 nm ? Q.12. 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 ? Q.13 In an intrinsic semiconductor the energy gap  is 1.2 e V. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and 300 K ? Assume that the temperature dependence intrinsic concentration  is given by  Q.14. In a p-n junction diode, the current I can be expressed 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,  is the Boltzmann constant (8.6× 10-5e V/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 dynamic resistance ?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V ?  Q.15. You are given two circuits as shown in (a) and (b), Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Sol. (a) In (a), the out put of NOR gate is made input for NOT gate. Then Boolean expression is Q.16 Write the truth table for a NAND gate as given in. Hence identify the exact logic operation carried out by this circuit.

Sol. In (a), we have NAND gate, with one input A and one output Y. The Boolean expression is Q.17. You are given two circuits as sown in (a) and (b) which consists of NAND gates. Identify the logic operation carried out by the two circuits. Q.18. Write the truth table for circuit given in (a) below of NOR gates and identify the logic operation (OR, and NOT) which this circuit is performing.

Sol.  The Boolean expression for the given circuit is It means OR gate is created. Truth table of this gate is shown in adjoining table (b).

Q.19. Write the truth table for the circuit given in (a) and (b), consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by two circuits. Updated: October 14, 2020 — 1:15 pm