InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The small ozone layer on top of the stratosphere is crucial for human survival. Why? |
|
Answer» The ozone layer absorbs ultraviolet and other low wavelength radiations which are harmful to living cells of human bodies and plants; hence ozone layer is crucial for human survival. |
|
| 2. |
Why is photodiode used in reverse bias? |
|
Answer» When a photo- diode is reverse biased, the width of depletion layer increases as compared to forward biased and a small reverse current (dark current) flows through the diode. Now, when the light is incident on the junction, electron-hole pairs are generated in depletion layer in a big amount (due to broad depletion layer) and these charge carriers can easily cross the barrier, hence contribute to current across the diode. We can say that in reverse bias, diode changes the incident light to current, more significantly due to broad depletion layer i.e. photo current is significant in reverse bias as compared to the forward bias current. |
|
| 3. |
The width of depletion region in a p–n junction diode is 500 nm and an intense electric field of 6×105 V/m is also found to exist. The height of potential barrier is (1) 0.30 V (2) 0.40 V (3) 3 V(4) 4 V |
|
Answer» Correct option: (1) 0.30 V Explanation: Let V be potential barrier across depletion layer of width d. E the electric field in the |E| = V/d or V = |E|.d ∴ V = 6 × 105 × 500 × 10–9 = 0.30 V |
|
| 4. |
How is forward biasing different from reverse biasing in a p-n junction diode? |
|
Answer» 1. Forward Bias: (i) Within the junction diode the direction of applied voltage is opposite to that of built-in potential. (ii) The current is due to diffusion of majority charge carriers through the junction and is of the order of milliamperes. (iii) The diode offers very small resistance in the forward bias. 2. Reverse Bias: (i) The direction of applied voltage and barrier potential is same. (ii) The current is due to leakage of minority charge carriers through the junction and is very small of the order of μ A. (iii) The diode offers very large resistance in reverse bias. |
|
| 5. |
Intensity of light emitted by LED depends on (1) Forward bias voltage (2) Forward current (3) Forward resistance (4) Nature of semiconductor crystal |
|
Answer» Correct option: (2) Forward current Explanation: Intensity of light emitted by LED depended on the forward current flowing in LED, with increase in forward current, intensity of light emitted increases. |
|
| 6. |
Colour of the light emitted by LED depends on (1) Forward bias voltage (2) Reverse bias voltage(3) Nature of semiconductor crystal(4) Forward resistance |
|
Answer» Correct option : (3) Nature of semiconductor crystal Explanation: Colour of the light emitted by LED depends on the energy gap of the semiconductor crystal. Hence the nature of the semiconductor crystal. |
|
| 7. |
An intrinsic semiconductor is at a temperature of 27°C. The probability of an electron jumping from valence band to conduction band (1) does not vary as temperature rises to 40°C (2) decreases linearly with rise in temperature (3) increases exponentially as the energy gap increases (4) decreases exponentially as the energy gap increases |
|
Answer» Correct option: (4) decreases exponentially as the energy gap increases Explanation: As energy gap increases the probability of electron jump from conduction to valency band decreases exponentially. |
|
| 8. |
Most of currently available semiconductor are elemental semiconductors Si or Ge and compound inorganic semiconductors. From 1990 on wards devices using organic semiconductors and semiconducting polymers have been developed. These are futuristic technology in gradients. The (a) in organic (b) organic and (c) organic polymer semiconductors are: (1) (a) GaAs (b) CdSe (c) anthracene (2) (a) GaAs (b) anthracene (c) polyanline(3) (a) anthracene (b) GaAs (c) polyanline (4) (a) CdSe (b) polyanline (a) anthracene |
|
Answer» Correct option : (2) (a) GaAs (b) anthracene (c) polyanline |
|
| 9. |
Are the mobilities of electron and hole equal in semiconductor. |
|
Answer» No, the electron mobility is higher than hole mobility because hole mobility is due to bounded electron movement in valence band. |
|
| 10. |
A “Hole” in a semiconductor is produced when an electron breaks away from a covalent bond in a semiconductors. Which of the following statements regarding a hole is incorrect? (1) A hole has same magnitude of charge as on an electron (2) In an external electric field a hole moves in a direction opposite to that of free electron (3) The energy of a hole is less as compared to that of an electron(4) In same external applied electric field the drift speed acquired by a hole is less than the drift speed acquired by a free electron |
|
Answer» Correct option: (3) The energy of a hole is less as compared to that of an electron Explanation: The energy of hole is less than the energy of an electron. |
|
| 11. |
Why does the electrical conductivity of a pure semiconductor increases on heating? |
|
Answer» As the temperature increases, more and more of the electrons in the valence band gain energy to cross the energy gap and enter the conduction band. This results in an increase in the number of charge carriers. Hence conductivity increases. |
|
| 12. |
Figure shows combination of gates. For Y = 1 which of the following options is correct.(1) A = 0, B = 1(2) A = 1, B = 1(3) A = 1, B = 0(4) A = 0, B = 0 |
|
Answer» Correct option: (3) A = 1, B = 0 Explanation: For A = 1, B = 0; the output of gate 1 (i.e. OR gate) C1 = 0. The output of gate C2 (i.e. NAND gate) C2 = 0. C1 = 0 and C2 = 0 are input of gate 3 which is NOR gate, hence output Y = 1. |
|
| 13. |
In binary system the decimal numbers are represented in terms of 0 and 1. We use a low voltage plus to represent binary 0 and a high voltage plus to represent binary 1. Fig. (a) and (b) give the plus representation of decimal number X and Y respectively. The plus representation decimal number (X + Y) is: |
|
Answer» Correct option: (3) Explanation: The binary representation of X is 1001. The decimal number X = 1(2)3 + 0(2)1 + 1(2)0 = 9 The binary representation of Y is 10101 The decimal number Y = 1(2)4 + 0(2)3 + 1(2)2 + 0(2)1 + 1(2)0 = 21 The decimal number X + Y = 9 + 21 = 30. The binary representation of (X+Y) is 11110. The pluse representation of this number is as shown in (3). |
|
| 14. |
To get the output Y = 1 from the circuit given below, the input must be: |
|
Answer» Correct option: (3) Explanation: Gate 1 is OR gate. For A = 1, B = 0 the output C1 = 1. C1 = 1 and C = 1 are input of gate 2 which is an AND gate. The output Y = 1. |
|
| 15. |
Why does the reverse current in pn-junction show a sudden increase at the critical voltage? Name any semiconductor device which operates under the reverse bias in the breakdown region. |
|
Answer» At critical voltage/breakdown voltage, a large number of covalent bonds break, resulting in availability of large number of charge carriers. Zener diode. |
|
| 16. |
A silicon PN diode has a reverse saturation current of 20 μA at a temperature of 20°C. The reverse saturation current of the same diode at a temperature of 40°C will be, (1) 10 µA(2) 20 µA(3) 40 µA (4) 80 µA |
|
Answer» Correct option : (4) 80 µA Explanation: For a silicon p-n junction, for every 10°C rise in temperature, the reverse saturation current gets doubled. |
|
| 17. |
An a.c voltage of peak value 10 V is connected in series with a silicon diode a load resistance of 100 Ω. The forward resistance of the diode to 20 Ω. The peak output voltage across the load resistance will be: (1) 5 V(2) 7.75 V (3) 9.3 V (4) 10 V |
|
Answer» Correct option: (2) 7.75 V Explanation: Barrier potential of silicon diode = 0.7 V Peak current through the diode in the circuit = 10 - 0.7/100+20 = 77.5 mA Peak output voltage = 77.5 mA × 100 Ω = 7.75 V |
|
| 18. |
Why are NAND and NOR gates called digital building blocks? |
|
Answer» The repeated use of NAND or NOR gate can produce all the three basic gates (OR AND and NOT) whose different combinations provide us large number of digital circuits. Hence NAND and NOR are called digital building blocks. |
|
| 19. |
Fig. shows VCE vs Vi characteristics of a transistor, known as transfer characteristics. The curve is divided into three regions as shown in Fig. The transistor is to be used as (a) an amplifier (b) a switch. The part of transfer–characteristics used is(1) (a) and (c) any of I, II or II(2) (a) – region I, (b) – region II(3) (a) – region II; (b) region I and III (4) (b) region I and II (b) region II |
|
Answer» Correct option: (3) (a) – region II; (b) region I and III Explanation: In a transistor if Vi is low so that t emitter–base junction is NOT FORWARD biased V0 is high. If Vi is high the transistor is driven into saturation region and V0 is low. One of these two positions can be used as “off” and other as “on” state of transistor. The transistor works as a switch. Region II is the “active region” of a transistor. In this region transistor is used as an amplifier. A small variation in Vi produces a small change in input current IB. This produces a large change in output current IC and therefore a large change in output voltage V0. This is amplification. |
|
| 20. |
In the circuit diagram shown below; the potential difference between the points A and B will be (assume diode is ideal) (1) 0 V(2) 0.6 V(3) 3 V(4) 6 V |
|
Answer» Correct option: (4) 6 V Explanation: As the diode is in reverse bias, (p–side at lower potential). The reverse biased resistance of an ideal diode is infinite. There is no current in circuit. There is no p.d across resistance R = 2Ω. Hence p.d between points A and B is 6 V. |
|
| 21. |
For a p–n junction diode which of the following statement is correct (1) In forward biasing there is a diffusion and in reverse bias there is drifting of charge carriers across the junction. (2) In forward bias there is drifting and in reverse bias there is diffusion of charge carriers across the junction. (3) In both forward and reverse biasing the current is due to diffusion of charge carriers across the junction.(4) In both forward and reverse bias there is current due to drifting of charge carriers across the junction. |
|
Answer» Correct option: (1) In forward biasing there is a diffusion and in reverse bias there is drifting of charge carriers across the junction. Explanation: Factual knowledge about moment of charge carriers in a p–n junction diode. |
|
| 22. |
Fig. (a) shows a “black–box” with three connecting terminals P, Q and R. Three components two diodes and one resistance is connected across the three terminals in same unknown arrangement. Fig. (b) shows black–box connected as shown in circuit. In I vs V graph obtained when P is negative and Q is positive are shown in Fig. (c). The arrangement of components between P, Q and R is |
|
Answer» Correct option: (1) Explanation: The characteristics curve shown in Fig. (c) is forward bias characteristic of a diode with a knee voltage of 0.7 V. Since P is connected to negative terminal and Q to positive terminal diode D1 is forward biased and has a series resistance R. Due to series resistance R, I vs V graph is a straight line. The forward bias resistance of diode is very small. This is shown in (1). In (2) diode between P and Q is forward biased but there is no series resistance. I vs V graph would be a almost vertical straight line. In (3) diode between P and Q is reverse biased therefore I = 0. Same is true for (4). |
|
| 23. |
The current through an ideal p–n junction shown in the circuit diagram will be (1) zero (2) 1 mA(3) 10 mA(4) 30 mA |
|
Answer» Correct option: (1) zero Explanation: As p is at lower potential with respect to n–side. Hence diode is in reverse bias. In reverse bias diode does not conduct. i.e. current through diode is zero. |
|
| 24. |
The barrier potential of a p–n junction depends on (a) type of semiconductor (b) amount of doping (c) temperature. Which one of the followings is correct? (1) (a) and (b) only (2) (b) only (3) (b) and (c) only (4) (a), (b) and (c) |
|
Answer» Correct option: (4) (a), (b) and (c) Explanation: Factual knowledge about junction diode. |
|
| 25. |
In a p–n junction a potential barrier of 300 meV exists across the junction. A hole with a kinetic energy of 400 meV approaches the junction. The K.E of the hole when it crosses the junction (i) from p to n–side and (ii) from n to p–side will be (1) 100 eV, 700 eV(2) 100 eV, 100 eV(3) 700 eV, 700 eV(4) 700 eV, 100 eV |
|
Answer» Correct option: (1) 100 eV, 700 eV Explanation: The p–region is at a lower potential than n–region in depletion layer when hole crosses depletion layer from n– to p–side; it (hole) is accelerated. Therefore K = Kinetic energy of hole = Ki + |e| V = (400 + 300) meV = 700 meV When hole crosses depletion layer from p– to n–side, it is retarded. Therefore K’ = Ke – |e| V = (400–300) meV = 100 meV |
|
| 26. |
State the reason, why GaAs is most commonly used in making of a solar cell. |
|
Answer» In solar radiations, intensity is maximum near 1.5 eV. In GaAs, Eg ≈ 1.53 eV, so solar cell made of GaAs has high absorption coefficient of solar radiations. |
|