HOTS (ELECTRONIC DEVICES) 1.Determine the current through resistance R in each circuit.

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1 HOTS (ELECTRONIC DEVICES) 1.Determine the current through resistance R in each circuit. Diodes D1 and D2 are identical and ideal. Sol. In circuit (i) Both D1 and D2 are forward baiased hence both will conduct current and resistance of each diode is 0..Therefore I = 3/15 = 0.2 A (ii) Diode D1 is forward bias and D2 is reverse bias, therefore resistance D1 is 0 and resistance D2 is infinite. Hence D1 will conduct and D2 do not conduct. No current flows in the circuit. 2. In only one of the circuits given below the lamp L light glow. Identify which circuit is it? Give reason for your answer? Sol.In fir (i) emitter base junction has no source of emf. Therefore Ic =0, bulb will not glow. In figure (ii) emitter base junction is forward biased, therefore lamp L will glow. 3. semiconductors obey OHM S law for only low fields.give reason? Sol.The drift velocity of a charge carrier is proportional to electric E. Therefore V = eet/m ie. V 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. 4. Relate input frequency and the output frequencies of a half wave rectifier and a full wave rectifier? Sol.In half-wave rectification, we get pulsating output for half cycle only. Therefore

2 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. 5. 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 thetransistor? In which of these states does the transistor not remain when being used as a switch? Sol. When both emitter and collector are reverse biased, no current flows through the transistor. This condition is known as cut-off state. When both Emitter and collector are reverse biased, no current flows through the transistor. This condition is known as saturation state. 6. Construct AND gate using NAND GATE and give its truth table?. AND Gate using NAND GATE:- Sol. 7. Why should we use Zener diode in the output stage of the electronic circuit? Explain the working of a Zener diode as a voltage regulator. Sol. Zener effect : The electric field in the depletion layer reach a point that it can break the covalent bonds and generate electron hole pairs. If the output voltage of a DC power supply does not change with load it is called regulated power supply.

3 Zener diode as a voltage regulator 8. In NPN transistor circuit, the collector current is 5mA. If 95% of the electrons emitted reach the collector region, what is the base current? Sol. Ic=95% of Ie = (95 / 100 ) Ie Ie = (100 / 95) 5 ma = 5.26mA Ie= Ic+ Ib Ib = 0.25 ma 9.The base current of a transistor is 105 A and collector current is 2.05 ma. Determine the value of, Ie, and A change of 27 A in the base current produces a change of 0.65 ma in the collector current. Find a.c. SOL. How does the collector current change in junction transistor if the base region has larger width? Sol. Current decreases.

Questions for practice: 1.The V-I characteristic of a silicon diode is shown in the Fig. below. Calculate the resistance of the diode at (a) ID = 15 ma and (b) VD = 10 V. 2.

4 Questions for practice: 1.The V-I characteristic of a silicon diode is shown in the Fig. below. Calculate the resistance of the diode at (a) ID = 15 ma and (b) VD = 10 V. 2. The current in the forward bias is known to be more (~ma) than the current in the reverse bias (~ A). What is the reason then to operate the photodiodes in reverse bias? 3. Why are Si and GaAs are preferred materials for solar cells? 4.From the output characteristics shown in Fig. below, calculate the values of and of the transistor when VCE is 10 V and IC = 4.0 ma. 5. Write the truth table for circuit given in Fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

5 6. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 ev. Can it detect a wavelength of 6000 nm? 7. 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. 1. (a) 10 Ω (b) Ω 2. See ncert text book 3. See ncert text book Answers: 4., 5. (Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.) 6. No (h has to be greater than Eg ) V

6 Hots on Communication System 1. Line of sight communication (LOS) is good for only 30-50Km and loosing required Sol. LOS connection. Why? How to solve the problem in microwave communication. Microwave system can travel only a limited distance before problems occur. Signals may grow too weak because of attenuation. The receiver might be so far away that target falls below the horizon. Losing the LOS or their may be obstacles between dishes. Microwave systems use repeaters to solve this problem. They capture and regenerate the signal to remove the propagation effects before passing it to the receiving antenna. 2. How do we make television broadcasts for larger coverage and for long distance? Sol.By using (i) tall antennas which is familiar landmark in many cities and (ii) using artificial satellites called geostationary satellites. Since television signals are of high frequency and are not reflected by ionosphere so we use satellites to get them reflected & transmission of TV signals can be used for larger coverage as well as for long distance. 3.In TV transmission which is used either A.M or F.M? Sol.In TV transmission FM is used for Sound signal and AM for picture signal. 4.In demodulation stage, how the RF signal is removed from AF? Sol.Rectifier is used to rectify the modulated wave (i.e) negative halt of the modulated wave is eliminated. It separates the audio signal from the carrier wave. 5. In the transmission of television signals sky waves are not used Give reason? Sol.The television signals have frequencies in MHz range. As ionosphere cannot reflect radio waves of frequency greater than 40 M back to earth, the sky waves cannot be used in the transmission of TV signals. 6. What would be the modulation index for an amplitude modulated wave for which the maximum amplitude is a while the minimum amplitude is b? Sol. Modulation index, am = Em/Ec (1) Maximum amplitude of modulated wave a=ec + Em...(2) Minimum amplitude of modulated wave b = Ec - Em (3) From (2) and (3), Ec = a+b/2, Em = a-b/2 From (1), modulation index, am = Em/Ec = (a-b)/2 / (a+b)/2 = a-b/ a+b 7. A carrier wave of peak voltage 20 V is used to transmit a message signal. What should be the peak voltage of the modulating signal, in order to have a modulation index of 80%? SOL. Modulation index, ma = Em / Ec Em = ma x Ec = 0.80 x 20 V = 16 V 8. A message signal of frequency 10 khz and peak value of 8 volts is used to modulate a carrier of frequency 1MHz and peak voltage of 20 volts. Calculate: (i) Modulation index

7 (ii) The side bands produced. (2) Sol (i) Modulation index, ma = Em / Ec = 8/20 = 0.4 (ii) Side bands frequencies = fc ± fm Thus the side bands are at 1010KHz and 990 khz. 9.How will you find the range of the receiving antenna using line of sight communication. SOL. The transmitted wave traveling in a straight line, directly reach the receiver and are then picked up by the receiving antennae. It can be seen that due to the final curvature of the earth such a wave cannot be seen beyond the tangent point R1 and R2 Suppose height of the is h and the radius of earth is r (that is OR1 = OR2 = OP = r). In the right-angled triangles OQR1, QPR1 we have 10. A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below: Give two drawbacks from which this arrangement suffers. Describe briefly with the help of a block diagram the alternative arrangement for the transmission and reception of the message signal. SOL.

8 Two drawbacks of the given arrangement are: (1) Audio signals cannot be efficiently radiated and do not propagate well in space. (2) Simultaneous transmission of signals by different transmitters can cause confusion at the receiver end due to overlapping of frequencies.

9 Questions for practice: 1.A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth m. 2. A message signal of frequency 10 khz and peak voltageof 10 volts is used to modulate a carrier of frequency 1 MHz and peakvoltage of 20 volts. Determine (a) modulation index, (b) the side bandsproduced. 3.Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level? 4. A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? 5. For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. Determine the modulation index, km 2. a) 0.5 b) 990 k Hz No 3258 km V 5. m = 2/3 Answers

Section:A Very short answer question

Section:A Very short answer question 1.What is the order of energy gap in a conductor, semi conductor, and insulator?. Conductor - no energy gap Semi Conductor - It is of the order of 1 ev. Insulator -