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F.Sc ICS Notes: Physics XII: Chapter 18 Electronics Exercise Short Questions:

F.Sc ICS Notes: Physics XII: Chapter 18 Electronics Exercise Short Questions:

Question 18.1 How does the motion of an electron in a n-type substance differ from the motion of holes in a p-type substance?
Answer 18.1 The majority carriers in n-type substance are the free electrons and majority carriers in p-type substances are holes. The electrons move from lower to higher potential where as hole moves from higher to lower potential. The motion of electrons in n-type substance is much rapid than the motion of holes in p-type substance.

Question 18.2 What is the net charge on a n-type or p-type substance?
Answer 18.2 p-type and n-type substances are electrically neutral. Since they are made as a result of combination of atoms of intrinsic semi conductors and atoms of impurity which are neutral.

Question 18.3 The anode of a diode is 0.2 V positive with respect to its cathode. Is it forward biased?
Answer 18.3 The anode of diode (p-type) is at higher potential (.2 volt) with respect to cathode (n-type). Therefore the p-side of diode is positive respect to n side. Therefore p-n junction is forward biased.

Question 18.4 Why charge carriers are not present in the depletion region?
Answer 18.4 When p-n junction is formed, some of the electron of n-type diffuse into p-type substance to neutralize the holes. As a result the flow of electron stops due to the formation of potential barrier. The region on both sides of junction is called depletion region because all the mobile carriers are depleted off from this region.

Question 18.5 What is the effect of forward and reverse biasing of a diode on the width of depletion region?
Answer 18.5 When p-n junction is forward biased, the width of the depletion region decreases. However on reverse biasing the junction, width of depletion region is increased.

Question 18.6 Why ordinary silicon diodes do not emit light?
Answer 18.6 In case of silicon energy gap between is 1ev between valence and conduction band. Therefore when electron jump from conduction band to valence band it will give off photon of energy 1ev. This photon of energy approximately 1ev will not lie in visible range of electromagnetic radiation. Hence light is not emitted by using diode made up of silicon.

Question 18.7 Why a photo diode is operated in reverse biased state?
Answer 18.7 Photodiode is basically the device for detecting light. That is why the photodiode is operated in reverse biased state. In reverse biasing, the current due to voltage is almost zero, so when light is allowed to fall on depletion region electron hole pairs are created which give rise to size-able current and this current is the detection of light. The reverse current is directly proportional to the intensity of light.

Question 18.8 Why is the current in a transistor very small?
Answer 18.8 As base of the transistor is thin and is lightly dopes, therefore very few of the charge carriers coming from emitter get neutralized in base. This will give rise to very small base current. Due to reverse biasing of collector-base junction, almost all the electrons enter into collector. A few charge carriers pass through the base circuit and a negligible current flow into through it.

Question 18.9 What is the biasing requirement of the junctions of a transistor for its normal operation? Explain how these requirements are met in a common emitter amplifier?
Answer 18.9 For normal operation of transistor, the emitter to base junction is forward biased and collector to base junction is reversed biased. In common emitter amplifier, both the requirements can be met by connecting the two batteries  batteries VBB and VCC to be connected forward biased for emitter-base junction and collector-base junction is reverse biased.

Q 18.10 What is the principle of virtual ground? Apply it to find the gain of an inverting amplifier.
Answer 18.10 For an op. amplifier AOL is open loop gain which is very high of the order 10(5). Thus for given Vo, (V+ - V-) = 0
V+ = V-
If V+ is grounded then V- becomes almost zero that is V- = 0. Hence inverting input will be virtually grounded.
From the fig
                  I1 = I2
                                          (V- -- Vi) / R1 = (V- --Vo)/ R2
                                               (0-Vi )/ R1 = (0- Vo)/R2
                                                       Vo/Vi = -R2 / R1
                                                            G = - R2 /R1

Question 18.11 The inputs of a gate are 1 and 0, Identify the gate if its output is (a) 0, (b) 1.
Answer 18.11
a, Inputs of given gate 1 and 0, output is 0 so in  first case the gates may be And, NOR, XNOR Gate.
b, Inputs of given gate 1 and 0, output is 1 so the Gates may be OR, NAND, XOR Gates.

Question 18.12 Tick the correct answer.

(i) A diode characteristic curve is a plot between
  1. current and time
  2. voltage and time
  3. voltage and current
  4. forward voltage and reverse voltage

(ii) The color of light emitted by a LED depends on
  1. its forward bias
  2. its reverse bias
  3. the amount of
  4. the type of semi-conductor current material used

(iii) In a half-wave rectifier the diode conducts during
  1. both halves of the input cycle
  2. a portion of the positive half of the input cycle
  3. a portion of the negative half of the input cycle
  4. one half of the input cycle

(iv) In a bridge rectifier of Fig. Q.18.1 when V is positive at point B with respect to point A, which diodes are ON.
  1. D2 and D4
  2. D1 and D3
  3. D2 and D3
  4. D1 and D4

(v) The common emitter current amplification factor β is given by
  1. a. IC / IE
  2. b. IC / IB
  3. c. IE / IB
  4. d. IB / IE

(vi) Truth table of logic function
  1. summarizes its output values
  2. tabulates all its input conditions only
  3. display all its input/output possibilities
  4. is not based on logic algebra

(vii) The output of a two inputs OR gate is 0 only when its
  1. both inputs are 0
  2. either input is 1
  3. both inputs are 1
  4. either input is 0

(viii) A two inputs NAND gate with inputs A and B has an output 0 if
  1. A is 0
  2. B is 0
  3. both A and B are zero
  4. both A and B are 1

(ix) The truth table shown below is for
  1. XNOR gate
  2. OR gate
  3. AND gate
  4. NAND gate

Written By: Asad Hussain.