CBSE Class 12 Physics HOTs Semiconductor Electronics Materials and Devices and Simple Circuits

Refer to CBSE Class 12 Physics HOTs Semiconductor Electronics Materials and Devices and Simple Circuits. We have provided exhaustive High Order Thinking Skills (HOTS) questions and answers for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits. Designed for the 2026-27 exam session, these expert-curated analytical questions help students master important concepts and stay aligned with the latest CBSE, NCERT, and KVS curriculum.

Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Physics HOTS with Solutions

Practicing Class 12 Physics HOTS Questions is important for scoring high in Physics. Use the detailed answers provided below to improve your problem-solving speed and Class 12 exam readiness.

Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits HOTS Solutions for Class 12 Physics

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SEMICONDUCTOR DEVICES

SEMICONDUCTOR DEVICES
 

 

ONE MARKS QUESTIONS

Question. What type of biasing is used to operate a photo diode?
Answer : Reverse bias

Question. Why is n-p-n transistor preferred over p-n-p transistor?
Answer : Electrons are the majority charge carriers in n-p-n transistor and electrons have higher mobility than holes.

Question. How does the energy gap in an intrinsic semiconductor vary, when doped with a trivalent impurity?
Answer : decreases.

Question. How does conductivity of a semiconductor change with rise in its temperature?
Answer : increases

Question. What is the order of energy gap in a semiconductor?
Answer : Approx. 1eV

 

TWO MARKS QUESTIONS

Question. How does the collector current change in a junction transistor, if the base region has larger width?
Answer : As we know
If we increase base width region there is more recombination in the base region. So base current(Ib) will increase. Hence, the collector current decreases if emitter current (Ie) is fixed for a transistor

Question. What is meant by the term, doping of an intrinsic semiconductor? How does it affect the conductivity of a semiconductor?
Answer : Adding impurities-trivalent or pentavalent, conductivity increases

Question. If the output of NAND gate is fed into the input to a NOT gate (i) name the new logic gate obtained and (ii) write down its truth table.
Answer : 

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Question. How does the width of the depletion region of a p-n junction vary, when diode is (i) forward biased (ii) Reverse biased?
Answer : (i) decreases (ii) increases:-

THREE MARKS QUESTIONS

Question.Draw the circuit diagram of a common-emitter amplifier, with appropriate biasing. What is the phase difference between the input and output signals? State two reasons why a common-emitter-amplifier is preferred to a common base amplifier?
Answer : In this case, as shown in Fig.(b), a large band gap Egexists (Eg> 3 eV). There are no electrons in the conduction band, and therefore no electrical conduction is possible. Note that the energy gap is so large that electrons cannot be excited from the valence band to the conduction band by thermal excitation. This is the case of insulators.

Question. With a circuit diagram, briefly explain how a zener diode can be used as a voltage regulator.
Answer : Answer included in 10 years question answer

Question. The V-I characteristic of a silicon diode is shown in the Fig. Calculate the resistance of the diode at (a) ID= 15 mA and (b) VD= –10 V.
Answer : Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
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Question. From the output characteristics shown in Fig, calculate the values ofac and dc of the transistor when VCE is10 V and IC = 4.0 mA.
Answer : 
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Question. What is the order of energy gap in a conductor, semi conductor, and insulator?.
Answer : Conductor - no energy gap
Semi Conductor - It is of the order of 1 ev. Insulator - 6 ev (or) more than 6 ev.
 
Question. Zener effect and avalanche  effect  are the two possible  break down mechanisms  that helps the external current . Why
Answer : Zener effect : The electric field in the depletion  layer reach a point that it can break the covalent bonds and generate electron–hole pairs.
Avalanche breakdown : The minority carriers that  across the depletion layer under the influence of the electric field gain sufficient kinetic energy to be able to break covalent bond in atoms which they collide.
 
Question. Light emission in semiconductor is the process of creation of a photon by means of the annihilation of an electron-hole pair. How can we create them. Give the necessary  condition for the light emission?
Answer : hω = Eeh   greater than or equal to Eg.
Condition for light emission is  excess carriers. 
 
Question.Why does the conductivity of a semi conductor change with the rise in temperature ?
Answer : When a semi conductor  is heated more & more electrons get enough energy to jump across  the forbidden energy gap from valence band to the conduction band, where they are free to conduct electricity. Thereby increasing the conductivity of a semi conductor.
 
Question. On the basis of energy  level diagram,  identify the  P-type semi conductor & N-type semi conductor
Answer :  
C.B
  _ _ _ _ _ _ _ _ _ _ _ Donor  energy level
V.B
 
C.B 
_ _ _ _ _ _ _ _ _ _ _ _Acceptor  energy level
 
V.B
Question. A student has to study the  characteristics of a P-N Junction diode. What kind of a circuit arrangement should she use for this purpose?
Answer : 
CBSE_Class_12_Physics_Semiconductor_Devices_1
 
Question.   Draw the typical shape of input characteristics likely to be obtainedby a student . What do we understand by the forward bias and reverse bias of the diode? In which of these states does the diode being used as a rectifier.?
Answer : 
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Question.  To achieve light amplification, We should provide positive feed back in LASER .Why?
Answer :  Laser is not an amplifier but an oscillator! However, any oscillator is an amplifier with a positive feedback. To obtain lasing , it is necessary to achieve light amplification  and provide positive optical feedback.
 
Question. Self supporting stimulated emission is the  principle of a Laser .why?
Answer :  Probability of stimulated emission is proportional to the density of excess electrons and holes , and to the density of photons. Under the normal circumstances it is negligible. Therefore if  we provide a positive feedback , the stimulated emission  can become self supporting.
 
Question. If  the active layer in a P-N photo diode is the thick  depletion layer  then P-N Photodiode  act very effective . Why?
Answer : In p-I-n photodiode ,due to thick  i – layer  the sensitivity  improves dramatically.. Large reverse biases (< 100V for  Si ) may be required.
 
Question. Modern technology use poly silicon  instead of metal to form the gate. Why?
Answer : Poly silicon  has high conductivity  compare to metal.
 
Question. Digiatal signal is preferred for communication  Why not analog signal?Give its waveform
Answer : A continuous time varying wave form of a current  (or) voltage is called analogue signal.
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A digital signal in a two level voltage signal (ie) “0” and “1” . Digital signals are in the form of pulse of equal level.
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Question. In a given diagram ,is the diode reverse (or) forward biased?.
Answer : 
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Question.Identify the logic  gate, Give its truth table and output wave form?.
Answer : 
NAND GATE.
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Question.The ratio of number of free electrons to holes ne/nh for two different materials A and B are 1 and <1 respectively. Name  the type of semi conductor to which A and B belongs.
Answer : If  ne/nh =1 . Hence A is intrinsic semi conductor. If ne/nh <1 , ne<nh  hence B is P-type.
 
19.Can you identify waveform. How will  you convert A.C To D.C?
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Question. Under what conditions an amplifier can be converted in to an oscillator.Draw a suitable diagram .
Answer : 1. When feed back is positive.        2. When feed factor k is equal to l/AV
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21.Using basic logic gates (NOT, OR, AND )  construct the circuit for the   given logic gate &  give its truth table and wave form.?
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22. Convert a Fraction from Decimal (0.625)10  to Binary? Hint:(0.625)10 =  (0.101)2
 
23. Convert  an octal number (4536) into decimal number? (4536)8 = 4x83 + 5x82 + 3x81 + 6x80 = (1362)10
 
24. Convert Hexa decimal (3A9F) and (2D3.5) into decimal number?
 
(3A9F)16 = 3x163 + 10x162 + 9x161 + 15x160 = 1499910 (2D3.5)16 = 2x162 + 13x161 + 3x160 + 5x16-1 = 723.312510
 
25. In the circuit  diagram  given , a volt meter  is connected  across  a lamp , what  changes would occur at lamp”L” and voltmeter “V:” , if the resistor R  is reduce in value? Give reason for your answer?
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In the given circuit  emitter –base junction of N-P-N  transistor  is forward biased
, with battery “B”. When “r” decreases IE   increases . Because
IC   = IE – IB   .
Therefore  IC will also increase. Hence  bulb will glow with more brightness and
voltmeter reading will increase.
 
26.Determine the current through resistance  “R” in each circuit. Diodes D1 and
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32.Differentiate  the electrical conductivity of both types of extrinsic semiconductors in terms of the energy band picture
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33. semiconductors obey OHM’S law for only low fields .Give reason? The drift velocity of a charge carrier is proportional to electric E.
Therefore V = eET/m ie. V α  E
But V cannot be increased indefinitely by increasing E . At high speed relaxation
time (T) begins to decrease due to increase in collision frequency. S: so drift velocity saturates at thermal velocity (lOms-1). An electric field of 106 V/m causes saturation of drift velocity. Hence semi- conduction obey ohm’s law for low electrical field and above this field ( E < 106 V/m ) current becomes independent of potential.
 
34. Relate  input frequency and the output frequencies of a half wave rectifier and a full wave rectifier?
In half-wave rectification, we get pulsating output for half cycle only. Therefore
output ripple frequency is same as the input frequency. But in full wave rectification we get output for both half cycle. Hence the output  at ripple frequency is full-wave rectification is double the input frequency.
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35. Emitter is heavily doped while collector is moderately doped . Why? There are two main differences between emitter and collector of a transistor construction wise. First emitter is heavily doped while collector is moderately doped. Second the area of the collector is larger than the emitter.So output will be maximum. 
 
36. Key point for its proper functioning of base region in a transistor depends upon the thickness of the base Why ?
Base region in a transistor controls the current. The majority charge carriers supplied by the emitter do not combine in the same region due to its thin size and light doping and most of them pass on to the collector. This is the key point for its proper functioning. 
 
37. For faster action which transistor is used and why?
For faster action NPN Transistor is used .In an NPN transistor, current
conduction is mainly by free electron ,whereas in PNP type transistor .it is mainly holes Since electron are more mobile than holes we prefer NPN for faster action as well as high conduction current.
 
38.A student has to study the input and output characteristics of a n-p-n silicon transistor in the common emitter configuration. What kind of a circuit arrangement should she use for this purpose?
 
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39.  Draw the typical shape of input characteristics likely to be obtained by a student. What do we understand by the cut off, active and saturation of the transistor? In which of these states does the transistor not remain when being used as a switch?
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Rectification — It is the process in which alternating current is converted to direct current.P-N junction Diode as Half wave rectifier
CBSE_Class_12_Physics_Semiconductor_Devices_18
As shown in the figure above the a.c input is applied across the primary of transformer. The secondary is connected to junction diode and load resistance (Rc) in series.
WORKING PRINCIPLE:- when positive half cycle or a.c takes place the diode is forward biased and output is obtained across the load resistance Rc. But when it negative half cycle diode is reverse biased and no current flows, hence there is no output across load resistance Rc. So only half is obtained as output.
Full wave rectification:
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48.Compliment(1’s) the following binary numbers (a) 110010 (b) 111101 (c) 101010 (d)111111 (a) 1’s compliment=001101
(b) 1’s compliment=000010
(c) 1’s compliment=010101 (d) 1’s compliment=000000
 
49. In the circuit, Vs=0.2V, V0 = -10 V. Find Vi & gain Av = V0 / Vi & Av ‘ = V0  / Vs ?
Sol: Vs=0.2 V, V0 = -10 V
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Advanced HOTS Questions with Solutions: Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

About Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits HOTS for Class 12 Physics

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In the 2026 pattern, 50% of the marks are for competency-based questions. Our CBSE Class 12 Physics HOTs Semiconductor Electronics Materials and Devices and Simple Circuits are to apply basic theory to real-world to help Class 12 students to solve case studies and assertion-reasoning questions in Physics.

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