TYPICAL QUESTIONS & ANSWERS

 

PART - I

OBJECTIVE TYPE QUESTIONS

 

Each Question carries 2 marks.

 

Choose correct or the best alternative in the following:

 

Q.1       The breakdown mechanism in a lightly doped p-n junction under reverse biased condition is called

                    (A)  avalanche breakdown.      

                   (B)  zener breakdown.

(C)   breakdown by tunnelling.             

(D)  high voltage breakdown.

 

             Ans: A

 

Q.2       In a CE – connected transistor amplifier with voltage – gain Av, the capacitance Cbc is amplified by a factor

(A)                                                  (B)

(C)                                       (D)

 

             Ans: B

 

Q.3       For a large values of |VDS|, a FET – behaves as

(A)    Voltage controlled resistor.          

(B)    Current controlled current source.

(C)   Voltage controlled current source.

(D)  Current controlled resistor.

 

             Ans: C

 

Q.4       Removing bypass capacitor across the emitter-leg resistor in a CE amplifier causes

(A)  increase in current gain.                   (B) decrease in current gain.

(C)  increase in voltage gain.                  (D)decrease in voltage gain.           

 

             Ans: D

 

Q.5       For an op-amp having differential gain Av and common-mode gain Ac the CMRR is given by

(A)                                            (B)

(C)                                            (D)       

 

             Ans: B

Q.6       When a step-input is given to an op-amp integrator, the output will be

(A)     a ramp.                                       

(B)     a sinusoidal wave.

(C)     a rectangular wave.                     

(D)    a triangular wave with dc bias.

 

             Ans: A

 

Q.7       Hysteresis is desirable in Schmitt-trigger, because

(A)     energy is to be stored/discharged in parasitic capacitances.      

(B)     effects of temperature would be compensated.

(C)     devices in the circuit should be allowed time for saturation and desaturation.

(D)    it would prevent noise from causing false triggering.    

 

             Ans: C

 

Q.8       In a full-wave rectifier without filter, the ripple factor is

(A)     0.482                                          (B) 1.21

(C) 1.79                                              (D) 2.05

 

             Ans: A

 

Q.9       A minterm of the Boolean-function, f(x, y, x) is

                    (A)                                     (B)              

                   (C) x z                                                 (D) (y +z) x           

 

             Ans: B

 

Q.10                                                                      The minimum number of flip-flops required to construct a mod-75 counter is

                   (A) 5                                                   (B)  6

(C) 7                                                   (D) 8

 

             Ans: C

 

Q.11                                                                      Space charge region around a p-n junction

(A)    does not contain mobile carriers  

(B)    contains both free electrons and holes

(C)   contains one type of mobile carriers depending on the level of doping of the p or n regions

(D)  contains electrons only as free carriers

            

             Ans: A

 

Q.12                                                                      The important characteristic of emitter-follower is

(A)  high input impedance and high output impedance

(B)  high input impedance and low output impedance

(C)  low input impedance and low output impedance

(D)  low input impedance and high output impedance

             Ans: B

 

Q.13                                                                      In a JFET, at pinch-off voltage applied on the gate

(A) the drain current becomes almost zero

(B)     the drain current begins to decrease

(C)     the drain current is almost at saturation value.

(D)  the drain-to-source voltage is close to zero volts.

 

             Ans: C

 

Q.14                                                                      When an amplifier is provided with current series feedback, its

(A)  input impedance increases and output impedance decreases    

(B)  input and output impedances both decrease

(C)  input impedance decreases and output impedance increases    

(D)  input and output impedances both increase  

 

             Ans: D

 

Q.15                                                                      The frequency of oscillation of a tunnel-collector oscillator having L= 30μH and C = 300pf is nearby                           

(A)  267 kHz                                        (B)  1677 kHz

(C) 1.68 kHz                                       (D)  2.67 MHz

 

             Ans: B          

 

Q.16                                                                      The open-loop gain of an op-amp available in the market may be around.

(A)   10–1                                              (B)  10

(C)  105                                               (D)  102

       

             Ans: C

 

Q.17                                                                      The control terminal (pin5) of 555 timer IC is normally connected to ground through a capacitor    (~ 0.01μF). This is to

(A)  protect the IC from inadvertent application of high voltage

(B)  prevent false triggering by noise coupled onto the pin

(C)  convert the trigger input to sharp pulse by differentiation            

(D) suppress any negative triggering pulse

 

             Ans: B

 

Q.18      The value of ripple factor of a half-wave rectifier without filter is approximately

(A)  1.2                                               (B)  0.2

(C)  2.2                                               (D)  2.0

 

              Ans: A

Q.19      The three variable Boolean expression xy + xyz + y + xz

(A)                                            (B) 

(C)                                           (D)

 

              Ans: C

                          

 

Q.20      The fan-out of a MOS-logic gate is higher than that of TTL gates because of its

(A)     low input impedance                     (B) high output impedance

(C)  low output impedance                   (D) high input impedance

 

              Ans: D

 

Q.21      In an intrinsic semiconductor, the Fermi-level is

                   (A)   closer to the valence band

                   (B)   midway between conduction and valence band

(C)   closer to the conduction band      

(D)  within the valence band

 

              Ans: C

 

Q.22      The reverse – saturation current of a silicon diode

(A)  doubles for every 10°C increase in temperature

(B)  does not change with temperature

(C)  halves for every 1°C decrease in temperature

(D)  increases by 1.5 times for every 2°C increment in temperature

 

              Ans: A

 

Q.23      The common collector amplifier is also known as

(A)  collector follower                          (B)  Base follower

(C)  Emitter follower                            (D)  Source follower

 

              Ans: C

 

Q.24      In class–A amplifier, the output current flows for

 

(A)  a part of the cycle or the input signal.

(B)  the full cycle of the input signal.

(C)  half the cycle of the input signal.   

(D)  3/4th of the cycle of the input signal.             

 

              Ans: B

 

Q.25      In an amplifier with negative feedback

(A)  only the gain of the amplifier is affected         

(B)  only the gain and bandwidth of the amplifier are affected

(C)  only the input and output impedances are affected      

(D)  All of the four parameters mentioned above would be affected

 

              Ans: D

 

Q.26      Wien bridge oscillator can typically generate frequencies in the range of

(A)  1KHz – 1MHz                            

(B)  1 MHz – 10MHz

(C)  10MHz – 100MHz                     

(D)  100MHz – 150MHz

 

              Ans: A

 

Q.27      A differential amplifier, amplifies

(A)  and mathematically differentiates the average of the voltages on the two input lines

(B)     and differentiates the input waveform on one line when the other line is grounded

(C)     the difference of voltages between the two input lines

(D)    and differentiates the sum of the two input waveforms

 

              Ans: C

 

Q.28      The transformer utilization factor of a half-wave rectifier is approximately

(A)  0.6                                               (B)  0.3

(C)  0.9                                               (D)  1.1

 

              Ans: B

                            0.286 0.3

 

Q.29      The dual of the Boolean expression: x + y + z is

(A)                                         (B) 

(C)                                          (D)

 

              Ans: C

                          

 

Q.30      It is required to construct a counter to count upto 100(decimal). The minimum number of flip-flops required to construct the counter is

             

(A)   8                                                  (B)  7

(C)  6                                                  (D)  5

 

              Ans: A

Q.31     The power conversion efficiency of an output stage is defined as_______.         

                   (A) (Load power + supply power) / supply power

                   (B) (Load power + supply power) / (load power-supply power)

                   (C) Load power / supply power

                   (D) Supply power / load power

 

             Ans. (C)

             Power gain is defined as the ratio of output signal power to that of input signal power.

 

Q.32     A highly stable resonance characteristic is the property of a ____ oscillator.

                   (A) Hartley                                          (B) Colpitts                                                          

                   (C) Crystal                                          (D) Weinbridge

 

             Ans. (C)

 

Q.33     The gate that assumes the 1 state, if and only if the input does not take a 1   state is called________.

                   (A) AND gate                                     (B) NOT gate

                   (C) NOR gate                                     (D) Both (B) & (C)

 

             Ans. (D)

             Y=   therefore output is high only when the values of both A and B are 0.

 

Q.34      The width of depleted region of a PN junction is of the order of a few tenths of a ___________.

 

                   (A) millimeter                                       (B) micrometer

                   (C) meter                                            (D) nanometer

 

             Ans. (B)

 

Q.35           For NOR circuit SR flip flop the not allowed condition is ________.

                   (A) S=0, R=0.                                     (B) S=0, R=1.

                   (C) S=1, R=1.                                     (D) S=1, R=0.

 

             Ans. (C)

             When S=R=1 the output is subject to unpredictable behaviour when S and R return  to 0 simultaneously.

 

Q.36     In negative feedback the return ratio is __________.

                   (A) 0                                                   (B) 1

                   (C) greater than 0                                (D) greater than 1

 

             Ans. (C)

             In a negative feed back circuit, always the return ratio will be in the range of 0 to 1.

 

Q.37           A phase shift oscillator uses __________________.

                   (A) LC tuning                                      (B) Piezoelectric crystal

                   (C) Balanced bridge                            (D) Variable frequency operation

 

             Ans. (C)                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

 

Q.38     The voltage gain of basic CMOS is approximately _________.

                   (A) (gmro)/2                                         (B) 2gmro

                   (C) 1 / (2gmro)                                     (D) 2ro / gm

 

             Ans. (A)

 

Q.39     Transistor is a

                            (A) Current controlled current device.

                   (B) Current controlled voltage device.

                   (C) Voltage controlled current device.

                   (D) Voltage controlled voltage device.

 

             Ans. (A)

                   The output current depends on the input current.

 

Q.40     A bistable multivibrator is a

                   (A) Free running oscillator.                   (B) Triggered oscillator.

                   (C) Saw tooth wave generator.            (D) Crystal oscillator.

 

             Ans. (B)

                   The transistors would change their state of operation from ON to OFF and vice versa depending on the external trigger provided.

 

 

Q.41     If the output voltage of a bridge rectifier is 100V, the PIV of diode will be

                   (A) 100√2V                                        (B) 200/π V

                   (C) 100πV                                          (D) 100π/2 V

 

             Ans. (D)

                   Peak inverse voltage = max secondary voltage

                   Vdc=2Vm/ π =100

                   Vm=100 π /2

 

Q.42           In the voltage regulator shown below, if the current through the load decreases,

                   (A) The current through R1 will increase.

                   (B) The current through R1 will decrease.

                   (C) zener diode current will increase.

                   (D) zener diode current will decrease.

 

             Ans. (C)                                                  

                                                  

Q.43     In Boolean algebra A + AB

                   (A)  A + B                                         

                   (B)  A + B

                   (C)  A + B                                         

                   (D)  A +B                                          

 

             Ans. (A)                                                                                                                               

                   A.1+A B= A (1+B) +A B = A + AB +A B = A+B (A +A) = A+B

 

 

Q.44     For a JFET, when VDS is increased beyond the pinch off voltage, the drain current

                   (A) Increases                                      

                   (B) decreases

                   (C) remains constant.                          

                   (D) First decreases and then increases.

 

             Ans. (C)

                   At pinch off voltage drain current reaches its maximum off. Now if we   further increase VDS above Vp the depletion layer expands at the top of the channel. The channel acts as a current limiter & holds drain current constant

 

Q.45           The type of power amplifier which exhibits crossover distortion in its output is

                            (A)  Class A                                        (B) Class B

                   (C)  Class AB                                     (D) Class C

 

             Ans. (B)

                   The transistors do not conduct until the input signal is more than cut-in voltage of the B-A junction. In class B, the devices being biased at cut-off, one device stops conducting before the other device starts conducting leaving to Cross-over distortion.

 

Q.46           The main advantage of a crystal oscillator  is that its output is

                            (A) 50Hz to 60Hz                               (B) variable frequency

                   (C) a constant frequency.                     (D) d.c

 

             Ans. (C)

                   The quality factor (Q) of a crystal as a resonating element is very high, of  the order of thousands. Hence frequency of a crystal oscillator is highly stable.

 

Q.47     The lowest output impedance is obtained in case of BJT amplifiers for         

                   (A) CB configuration.                         

                   (B) CE configuration.

                   (C) CC configuration.                         

                   (D) CE with RE configuration.

       

             Ans. (C)

                   The output impedance in case of CC configuration is on the order of a few ohms.

                   (In case of CB ≈ 450kΩ and in case of CE ≈ 45kΩ)

 

Q.48     N-channel FETs are superior to P-channel FETs, because

                   (A) They have higher input impedance

                   (B) They have high switching time

                   (C) They consume less power

                   (D) Mobility of electrons is greater than that of holes

 

             Ans. (D)

 

Q.49     The upper cutoff frequency of an RC coupled amplifier mainly depends upon

                   (A) Coupling capacitor

                   (B) Emitter bypass capacitor

                   (C) Output capacitance of signal source

                   (D) Inter-electrode capacitance and stray shunt capacitance

   

             Ans. (D)

 

Q.50     Just as a voltage amplifier amplifies signal-voltage, a power amplifier

                   (A) Amplifies power

                   (B) Amplifies signal current

                   (C) Merely converts the signal ac power into the dc power                                                      

                   (D) Merely converts the dc power into useful ac power

 

             Ans. (D)

 

Q.51     A radio frequency signal contains three frequency components, 870 KHz,  875 KHz and 880 KHz. The signal needs to be amplified. The amplifier used should be

                   (A) audio frequency amplifier               (B) wide band amplifier

                   (C) tuned voltage amplifier                   (D) push-pull amplifier

 

             Ans. (C)

                   We need to amplify 3 signal frequencies i.e., 870 kHz, 875 kHz and 880 kHz.

                   These frequencies lie in a bandwidth of 10 kHz and we should use only tuned voltage amplifiers  to amplify them.

 

Q.52     An oscillator of the LC type that has a split capacitor in the circuit is

                   (A) Hartley oscillator                          

                   (B) Colpitts oscillator                          

                   (C) Weinbridge oscillator                    

                   (D) R-C phase shift oscillator

 

             Ans. (B)

                   We have two capacitors in the tank circuit, which serve as a simple ac voltage divider.

 

Q.53           The function of a bleeder resistor in a power supply is

                   (A) the same as that of load resistor

                   (B) to ensure a minimum current drain in the circuit

                   (C) to increase the output dc voltage                                                                                

                   (D) to increase the output current

 

             Ans. (B)                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              

Q.54     In a bistable multivibrator circuit, commutating capacitor is used

                   (A) to increase the base storage charge

                   (B) to provide ac coupling

                   (C) to increase the speed of response                                                                              

                   (D) to provide the speed of oscillations

 

             Ans. (C)

                   The commutating capacitor is used for the speedy transition of the state of the bistable.

                   

Q.55     n-type silicon is obtained by

                           (A) Doping with tetravalent element

                   (B) Doping with pentavalent element

                   (C) Doping with trivalent element

                                                                              (D) Doping with a mixture of trivalent and tetravalent element

 

             Ans: (B)        

                   The pentavalent atom provides an excess electron while the other four form the       covalent bond with the neighbouring atoms. This excess free electron provides the n type conductivity.

 

Q.56     The forward characteristic of a diode has a slope of approximately 50mA/V at   a desired

             point. The approximate incremental resistance of the diode is

                        (A) 50Ω                                              (B) 35Ω

                   (C) 20Ω                                              (D) 10Ω

            

             Ans: (C)

                   Resistance at any point in the forward characteristics is given by R= ΔV/ ΔI =      1/50mA = 20Ω

 

Q.57     Two stages of BJT amplifiers are cascaded by RC coupling. The voltage gain   of the first

             stage is 10 and that of the second stage is 20. The overall gain of   the coupled amplifier is 

 

                   (A) 10x20                                          

                   (B) 10+20

                   (C) (10+20)2                                      

                   (D) (10x20)/2

          

             Ans: (A)

                   The voltage gain of a multistage amplifier is equal to the product of the gains of    the individual stages.

Q.58     In the voltage range, Vp < VDS < BVDSS of an ideal JFET or MOSFET

                            (A) The drain current varies linearly with VDS.

                   (B) The drain current is constant.

                                                                              (C) The drain current varies nonlinearly with VDS.

                   (D) The drain current is cut off.

            

             Ans: (B)

                   It is the saturation region or pinch off region, and drain current remains almost constant   at its maximum value, provided VGS is kept constant. 

 

Q.59     In a voltage shunt negative feedback amplifier system, the input resistance Ri and the output

             resistance Ro of the basic amplifier are modified as follows:

                   (A) Ri is decreased and Ro increased.

                   (B) Both Ri and Ro are decreased.

                   (C) Both Ri and Ro are increased

                                                                              (D) Ri is increased and Ro is decreased.

            

             Ans: (B)

                   Here, a fraction of output voltage obtained by parallel sampling is applied in parallel with the input voltage through feedback and both input and output resistance decrease by a factor equal to (1+ βAv).

       

Q.60           The use of crystal in a tunable oscillator

                                                                              (A) Improves frequency stability.

                  (B) Increases the gain of the oscillator.

                   (C) Helps to obtain optimum output impedance.

                                                                              (D) Facilitates generation of wide range of frequencies.

 

                   Ans: (A)

                   Piezoelectric crystal is used as a resonant tank circuit. The crystal is made of quartz material and provides a high degree of frequency stability.

 

Q.61     The large signal bandwidth of an opamp is limited by its

                   (A) Loop gain                                      (B) slew rate

                                                                              (C) output impedance              (D) input frequency

 

                   Ans: (B)

 

Q.62     Rectification efficiency of a full wave rectifier without filter is nearly equal to

 

                   (A) 51%                                              (B) 61%

                  (C) 71%                                              (D) 81% 

      

             Ans: (D)

                   Efficiency of a full wave rectifier is given by

                                                                              [(2Im / π) 2 x  RL] / [(Im / √2) 2 x (Rf + RL)] = 81%, when Rg is zero.

 

 

Q.63    When the temperature of a doped semiconductor is increased, its conductivity

                  (A) decreases.

                  (B) increases.

                  (C) does not change.

                  (D) increases or decreases depending on whether it is p- or n-type.

 

            Ans: B

Q.64    The main characteristics of a Darlington Amplifier are  

(A)   High input impedance, high output impedance and high current gain.

(B)   Low input impedance, low output impedance and low voltage gain.

(C)   High input impedance, low output impedance and high current gain.

(D)  Low input impedance, low output impedance and high current gain.

 

            Ans: C

 

Q.65    The transconductance, gm, of a JFET is computed at constant VDS, by the following:

                  (A)                                                  (B)

                  (C)                                         (D)

            Ans: A

 

Q.66    The feedback factor β at the frequency of oscillation of a Wien bridge oscillator is

                  (A) 3                                                          (B)

                  (C)                                                       (D)

           

            Ans: B

 

Q.67    In an amplifier with negative feedback, the bandwidth is

(A)   increased by a factor of β

(B)   decreased by a factor of β

(C)   increased by a factor of (1+Aβ)

(D)  not affected at all by the feedback

where A = gain of the basic amplifier and β = feedback factor

 

            Ans: C

 

Q.68    The ‘slew rate’ of an operational amplifier indicates

(A)   how fast its output current can change

(B)   how fast its output impedance can change

(C)   how fast its output power can change

(D)  how fast its output voltage can change

when a step input signal is given.

           

            Ans: D

 

Q.69    In a clamping circuit, the peak-to peak voltage of the waveform being clamped is

(A)  affected by the clamping

(B)  not affected by the clamping

(C)  determined by the clamping voltage value

(D)  determined by the ratio of rms voltage of the waveform and the clamping voltage

            Ans: B

Q.70    Regulation of a.d.c. power supply is given by

(A)  product of no-load output voltage and full-load current

(B)  ratio of full-load output voltage and full-load current

(C)  change in output voltage from no-load to full-load

(D)  change in output impedance from no-load to full-load

 

            Ans: D

 

Q.71    A ‘literal’ in Boolean Algebra means

(A)  a variable inn its uncomplemented form only

(B)  a variable ORed with its complement

(C)  a variable in its complemented form only

(D)  a variable in its complemented or uncomplemented form

 

            Ans: D

 

Q.72    In an unclocked R-S flip-flop made of NOR gates, the forbidden input condition is

                  (A) R = 0, S = 0                                         (B) R = 1, S = 0

                  (C) R = 0, S = 1                                         (D) R = 1, S = 1

 

            Ans: D

 

Q.73    The current amplification factor in CE configuration is

                  (A) α                                                          (B) β + 1

                  (C)                                                       (D) β

 

            Ans: D

 

Q.74    A zener diode

(A)   Has a high forward voltage rating.

(B)   Has a sharp breakdown at low reverse voltage.

(C)   Is useful as an amplifier.

(D)  Has a negative resistance.

 

            Ans: B

 

Q.75    N-channel FETs are superior to P-channel FETs, because

(A)   They have a higher input impedance.

(B)   They have high switching time.

(C)   They consume less power.

(D)  Mobility of electrons is greater than that of holes.

 

            Ans:

 

Q.76    The maximum possible collector circuit efficiency of an ideal class A power amplifier is

                  (A) 15%                                                     (B) 25%

                  (C) 50%                                                     (D) 75%

 

            Ans: C

 

Q.77    Negative feedback in an amplifier

(A)   Reduces the voltage gain.

(B)   Increases the voltage gain.

(C)   Does not affect the voltage gain.

(D)  Converts the amplifier into an oscillator.

 

            Ans: A

 

Q.78    For generating 1 kHz signal, the most suitable circuit is

                  (A) Colpitts oscillator.                                 (B) Hartley oscillator.

                  (C) Tuned collector oscillator.                     (D) Wien bridge oscillator.

 

            Ans: D

 

Q.79    Phe output stage of an op-amp is usually a

(A)              (A) Complementary emitter follower.

(B)              (B)  Transformer coupled class B amplifier.

(C)              (C ) Class A power amplifier.

(D)             (D)  Class B amplifier.

           

            Ans: A

 

Q.80    When a sinusoidal voltage wave is fed to a Schmitt trigger, the output will be

                  (A) triangular wave.                                    (B) asymmetric square wave.

                  (C) rectangular wave.                                 (D) trapezoidal wave.

 

            Ans: B

 

Q.81    If the peak value of  the input voltage to a half wave rectifier is 28.28 volts and no filter is use, the maximum dc voltage across the load will be

                  (A) .                                              (B) 15 V.

                  (C) 9 V.                                                     (D) 14.14 V.

 

            Ans: C

 

Q.82    The logic gate which detects equality of two bits is

                  (A) EX-OR                                                (B) EX-NOR

                  (C) NOR                                                   (D) NAND

 

            Ans: B

 

Q.83    The electron relaxation time of metal A is s, that of B is s. The ratio of resistivity of B to resistivity of A will be

                  (A) 4.                                                         (B) 2.0.

                  (C) 0.5.                                                      (D) 0.25.

 

            Ans: B

 

Q.84    The overall bandwidth of two identical voltage amplifiers connected in cascade will

(A)   Remain the same as that of a single stage.

(B)   Be worse than that of a single stage.

(C)   Be better than that of a single stage.

(D)  Be better if stage gain is low and worse if stage gain is high.

 

            Ans: B

 

Q.85    Field effect transistor has

                  (A) large input impedance.                          (B) large output impedance.

                  (C) large power gain.                                  (D) large votage gain.

 

            Ans: A

 

Q.86    Which of the following parameters is used for distinguishing between a small signal and a large-signal amplifier?

                  (A) Voltage gain                                         (B) Frequency response

                  (C) Harmonic Distortion                             (D) Input/output impedances

 

            Ans: D

 

Q.87    Which of  the following parameters is used for distinguishing between a small signal and a large-signal amplifier?

                  (A) Instability                                              (B) Bandwidth

                  (C) Overall gain                                          (D) Distortion

 

            Ans: B

 

Q.88    If the feedback signal is returned to the input in series with the applied voltage, the input impedance ______.

                  (A) decreases                                             (B) increases

                  (C) does not change                                   (D) becomes infinity

 

            Ans: B

 

Q.89    Most of linear ICs are based on the two transistor differential amplifier because of its

(A)   input voltage dependent linear transfer characteristic.

(B)   High voltage gain.

(C)   High input resistance.

(D)  High CMRR

 

            Ans: D   

 

Q.90    The waveform of the output voltage for the circuit shown in Fig.1 (RC >> 1) is a

                  (A) sinusoidal wave                                    (B) square wave

                  (C) series of spikes                                     (D) triangular wave.

 

            Ans: D

 

 

Q.91    A single phase diode bridge rectifier supplies a highly inductive load. The load current can be assumed to be ripple free. The ac supply side current waveform will be

                  (A) sinusoidal                                             (B) constant dc.

                  (C) square                                                  (D) triangular

 

            Ans: C

 

 

Q.92    Which of  the following Boolean rules is correct?

                  (A) A + 0 = 0                                             (B) A + 1 = 1

                  (C)                                       (D)

 

            Ans: B

 

Q.93    A single phase diode bridge rectifier supplies a highly inductive load. The load current can be assumed to be ripple free. The ac supply side current waveform will be

                  (A) sinusoidal                                             (B) constant dc.

                  (C) square                                                  (D) triangular

 

            Ans: C

 

Q.94    Which of  the following Boolean rules is correct?

                  (A) A + 0 = 0                                             (B) A + 1 = 1

                  (C)                                       (D)

 

            Ans: B


PART – II

 

NUMERICALS

 

Q.1      In a transformer, give the relationship between

             (i)  turns ratio and its primary and secondary impedances

             (ii) turns ratio and primary/secondary voltage.

             In a transformer-coupled amplifier, the transformer used has a turns ratio N1:N2 = 10:1. If the   source impedance is 8 KΩ what should be the value of load impedance for maximum power transfer to the load? Also find the load voltage if the source voltage is 10 volts.      (6)

 

             Ans: p. 439 – 40

 

                                                                              Given: ;  - load side;  - amplifier side or source side.

                         

             Turns ratio and impedances:

             Turns ration and voltages:

             ; i.e.

                                                                              Load voltage = V2 ;  .

 

Q.2       The FET circuit given below in Fig.1, has R1=3.5MΩ, R2 = 1.5MΩ, RS = 2KΩ, RL = 20KΩ and gm = 2.5mS. Find its input impedance, output impedance and voltage gain.                                                   (8)

 

       

            

 

 

 

 

 

             Ans:

            

             (i)

             (ii)

                                                                              (iii) Voltage gain

 

Q.3       In an amplifier with negative feedback, the gain of the basic amplifier is 100 and it employs a feedback factor of 0.02. If the input signal is 40mV, determine

                                                                                   (i)  voltage gain with feedback and

                                                                                   (ii) value of output voltage.                                       (3)

 

             Ans:

                                                                              (i)

                                                                              (ii)

 

Q.4       In the circuit shown below in Fig.2, R1=12KΩ, R2 = 5KΩ, R3 = 8KΩ, RF = 12KΩ. The inputs are: V1 = 9V, V2 = -3V and V3 = -1V. Compute the output voltage.                                                                      (4)

                                                                              Ans:

            

                                                                                    = - 12(0.75 – 0.6 – 0.125) = - 0.3 V.

 

 

 

 

Q.5       A half-wave rectifier has a load resistance of 3.5 KΩ. If the diode and secondary of the transformer have a total resistance of 800KΩ and the ac input voltage has 240 V (peak value), determine:

                                                                                         (i)        peak, rms and average values of current through load

                        (ii)       DC power output

(iii)            AC power input

(iv)            rectification efficiency                                                                                        (7)

 

             Ans:

 

             (i)  Im   =    Vm        =         240               =   55.8mA =  peak current.

                            RL + Rf        3.5X 103 + 800     

                = average current.

                   .

             (ii) DC power output =  watts.

             (iii) AC power input =  watts.

             (iv) Rectification Effect =.

 

 

Q.6       A BJT has a base current of 250 μA and emitter current of 15mA Determine the collector current gain and β.                                                                  (2)

 

             Ans:

The current relationship in a transistor is given by

                                                                                 i.e.

                                                            Given:

                                                                         IE = 15 mA

                                                                    

                                                            And     .

 

Q.7      In the circuit shown in Fig.1, [IDSS] = 4mA, Vp = 4V. Find the quiescent values of 1D, VGS and VDS of the FET.                                                                  (6)

Fig.1

            Ans:            

           

                  The potential divider bias circuit can be replaced by Thevenin equivalent as shown, where

            

Applying Kirchoff’s Voltage law to gate–source circuit gives  as there is no gate current flow.

Hence

 

Given, IDSS = 4 mA and VP = 4 V. substituting these values in the equation for IP

                             

Forming the quadratic equation in VGS.

                             

Solving,  VGS = 3 V(Solution VGS = 4V is not realistic).                                                      

Solving for ID, ID = (1 – ¾) = 0.25 mA

Hence, VDS = 30 – 0.25(18 + 4) = 24.5 Volts.

                             

Q.8      In the transformer coupled class A amplifier shown in Fig.2 below, the transistor has hFE = β = 40 and hie = 25Ω. Assume that the transformer is ideal and that  and also . Determine values of R1 and R2 to obtain quiescent current ICQ = 100mA. If the collector current swing , find the peak values of load current and load voltage. Transformer turns ratio is primary 1:6 secondary.                                            (10)

 

           

            Ans:

                 The equivalent circuit replacing R1, R2 as part of thevenin source is shown

 

                      ……(1)                                     ………(2)

                  Applying KVL around base-emitter circuit,

                 

                  i.e.                              …….(3) 

 

                  Any value of RB by which is acceptable. Hence, choosing a reasonable value of 2.2K for RB,      volts.

                  Use equations (1) and (2) to find R1 and R2

                  From (1)

                            i.e.                         ……..(4)

       

                  substituting the value of R1 in equation (2)

                     ….(5)

                  simplifying,                   ………(6)

                  From (4) and (6), we get

                             ………(7)

                  using equation (6),

                  using equation (7), 

                  The load Rac at the collector leg of the transistor is reflected load of RL as per the turns ratio (a) of the transformer.

                  i.e.  

                  As ic swings ± 80 mA either side of ICQ = 100 mA on the Rac load line, and VCEQ = 12V (DC load line being almost vertical)

                 

                                                                       = 12 + 5.55 = 17.55 Volts.

                  Hence, Maximum Load voltage(secondary):

       

           

                  Hence, Maximum Load current:

                      .

 

Q.9      A negative feedback of β = 2.5 x 10-3 is applied to an amplifier of open loop gain 1000. Calculate the change in overall gain of the feedback amplifier if the gain of the internal amplifier is reduce by 20%.                (4)

 

            Ans:

If A is the gain of the basic amplifier, the overall gain Af of the amplifier with negative f.b.  is

 given A = 1000 and β = 2.5 x 10-3,

When A is reduced by 20%, the new A, say An = 1000 – 0.2 x 103 = 800                   

The new voltage gain with f.b. is

.

 

Q.10     A full wave rectifier is fed with a voltage, 50 sin 100 πt. Its load resistance is 400Ω. The diodes used in the rectifier have an average forward resistance of 30Ω. Compute the

                       (i)    average and rms values of load current,

                       (ii)    ripple factor and

                       (iii)   efficiency of rectification.                                                                                        (6)

 

             Ans:

                  The maximum value of load current

                  ; given: Vmax = 50volts; RL = 400Ω, RF = 30 Ω                    

 

                  Thus

                 

               Average current Iavg  =  2 Imax   for a FWR

                                                      π                                                        

                  i.e. Iavg = 2 X 0.1163  = 0.074A or 74 mA = Idc

                                     λ

                  The r.m.s value of load current is

                 

       

                  Ripple factor

                  Efficiency of rectification in a FWR is given by               

 

                    i.e.  or 75.5%.

 

Q.11                                                                      For the circuit shown in Fig.2, find thee maximum and minimum values of zener diode current.                                                                                                        (8)

Fig.2

 

                  Ans: 9mA, 1.0mA.

 

Q.12                                                                      The parameters of the transistor in the circuit shown in Fig.3 are  Find

                  (i) midband gain

                  (ii) the value of Cb necessary to give a lower 3 dB frequency of 20Hz

                  (iii) the value of Cb necessary to ensure less than 10% tilt for a 100 Hz square wave input.

                                                                                                                                                            (8)

                                                                              Fig.3

 

                  Ans:

                         (i) 32.26              (ii) 2.567 Mf            (iii) 16.13 Mf

 

 

Q.13                                                                      For the circuit shown in Fig.4, determine voltage gain, input impedance, output impedance, common-mode gain and CMRR if , ,  and transistors Q1 and Q2 are identical with . Determine output voltage when  and when .                                                                                                                    (8)

Fig.4

                  Ans:

                  A = 150, Zin = 0.66MΩ, Zout = 1MΩ

                  Acm = 0.5, CMRR = 300

                  Vout = 7.5V and 0.15V

 

Q.14                                                                      A class B push-pull amplifier is supplied with . The minimum voltage reached by the collector due to signal swing is . The dissipation in both the transistors totals 30 W. What is the conversion efficiency of the amplifier?                 (7)

            

             Ans:

             DC supply voltage, Vcc = 40v

              Vmin = 8v, Pd = 30w.

             As Pd = Pin(dc) – Pout(ac)

                    

                    

                         

                         

                                                                                         

            

                

                

                

             Conversion Efficiency,

                                                                                                                   

 

Q.15                                                                      The input to an op-amp differentiator circuit is a sinusoidal voltage of peak value  and frequency of 2 kHz. If the values of differentiating components are given as R = 40 k and C =, determine the output voltage.                                                                                   (4)

 

             Ans:

            

            

             Scale factor = CR =

            

            

            

            

 

Q.16                                                                      Draw the circuit of a monostable multivibrator using two transistors. Use the following data in your circuit:, , B,  for both the transistors. The resistor and capacitor connected to the base of  have values  and C = 0.1  respectively. Determine the monostable pulse width.                                                         (8)

            

             Ans:

             ,         Input pulse width,

                 

                                          

                                           T = 1.4 714 m sec.

Monostable Multivibrator Circuit

Q.17                                                                      For the series regulator given below, , , the transistor , =1.2 K, =10V and . Calculate (i) output voltage (ii) load current (iii) the base current in the transistor (iv) zener current.                                                    (8)

 

 
 

 

             Ans:  

            

            

            

            

            

 

Q.18            Find the values of collector and emitter currents in a transistor having ICBO=3μA, and αdc=0.98 when its base current is 60μA.                                                                                                              (6)

             

              Ans: 

              =

              We know that

             

             

             

             

             

             

             

              Emitter current,

               =

                    .

 

Q.19                                                                      The h-parameters of the transistor in the amplifier circuit shown below are: hie=2.2 KΩ , hfe=52, hoe=and hre is negligible. The output load resistor dissipates a signal power of 9 mW. Determine the power gain of the amplifier using its equivalent circuit. The reactances of the capacitors may be neglected.                     (8)

 

12KΩ

 
                                                                                                                  + Vcc

 

 

 

 

 

 

 

 

 


             Ans:

            

            

            

             Input impedance to E,

                                          

             Power drawn from the source, .

             Base current,

                                        =

            

             Output impedance,

              

            

            

            

                                                                              AC-load resistance,

            

            

                                                                                                                           

             Output voltage,

            

            

            

            

            

            

             E = 0.2715

             Power gain, Ap =

            

            

            

             .

 

Q.20                                                                      The collector voltage of a Class B push pull amplifier with VCC=40 Volts swings down to a minimum of 4 volts. The total power dissipation in both the transistors is 36 watts. Compute the total dc power input and conversion efficiency of the amplifier.                     (7)

            

             Ans:           

            

            

            

             =

             36 =

             36 =

            

            

            

            

                                               =

                                     

 

Q.21           The transistor in the feedback circuit shown below has β=200. Determine

(i) feedback factor, (ii) feedback ratio, (iii) voltage gain without feedback and (iv) voltage gain with feedback in the circuit. In the transistor, under the conditions of operation, VBE may be assumed to be negligible.   (6)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


             Ans:

             ,  ,  , 

            

            

             AC Emitter resistance,

                                                  

             Voltage gain without feedback,

            

             A = 1008.16

             Feedback ratio

                                         

             Feedback factor =

           Feedback factor =  48.0076

             Voltage gain with feedback,

            

            

 

Q.22                                                                      In the differential amplifier circuit shown below, the transistors have identical characteristics and their β=100.Determine the (i) output voltage (ii) the base currents and (iii) the base voltages taking into account the effect of the RB and VBE. Take VBE=0.7 Volts.                              (8)

 

 

 

 

 

 

 

 

 

 

 

 

 


            

             Ans:

             Tail current,                

             The collector current in transistor Q2 is half thus tail current (i.e. 0.75mA) because each transistor gets half the tail current.

            

            

             Tail current 

            

            

            

             And Tail current,

                                         

                                         

             And output voltage,

            

                    =

            

             If the results obtained are compared, we find that the results obtained improve with each refinement, but the improvement is not significant.

              The ideal tail current is 1.41mA

            

            

            

 

Q.23                                                                      Design a series voltage regulator to supply 1A to a load at a constant voltage of 9V. The supply voltage to regulator is 15V±10%. The minimum zener current is 12mA. For the transistor to be used, assume VBE=0.6V and β=50.                                                          (7)

            

             Ans:

            

            

            

            

             Voltage drop in resistor 

             Current through resistor R,

            

             .

 

Q.24           Obtain the minterms of the function

                                ƒ(A,B,C) =

                   and draw the K-map of the function.                                                                          (4)

            

             Ans:

            

            

                                =

                                =

 

Q.25                                                                      A load line intersects the forward V-I characteristic of a silicon diode at Q, where the slope of the curve is 40mA/V. Calculate the diode resistance at the point Q.                        (4)

 

             Ans:

             DC or static resistance,

                                                                             

             .                                                

 

VCC=20V

 
Q.26                                                                      The power amplifier shown below is operated in class A, with a base current drive of 8.5mA peak. Calculate the input dc power, the power dissipated in the transistor, the signal power delivered to the load and the overall efficiency of the amplifier, if transistor β=30 and VBE=0.7V.                                                                        (8)

 

 

 

 

 

 

 

 

 

 

 

 

 

 


             Ans:

            

            

             Now dc – load line is drawn joining points (20v, 0) and (0, 1.25A)

             For operating point Q,

            

            

            

            

             .

            

                           = 0.5202 w.

             Power delivered to the transistor,

            

                        = 20(0.579) – (0.579)2.16

                        = 6.216w

             Power lost in transistor =

                                                   = 6.216 – 0.5202

                                                   = 5.6958

             Collector Efficiency,

                                                            =

                                                            = 8.368 %.

             Power rating of transistor = Zero-signal power dissipation

                                                       =

                                                       =

                                                       =  6.216w.

 

 

Q.27           Find the period of the output pulse in the circuit shown below:                             (4)

             Ans:

TIMERS

 

                                                                              The pulse width = .

 

 

Q.28                                                                      Analyze half-wave and full-wave rectifier circuits (without filter) to deduce the values of rectification efficiency assuming ideal diodes.                                     (8)

            

Ans:

             Rectification efficiency of half-wave rectifier, which is defined as the ratio of dc-output power to the ac-input power, is given as

            

                 = .

             Now,  .

             Pac = Power dissipated in diode junction + Power dissipated in load resistance RL

                   =

                   =

            

                     i.e.  40.6 % if RF is neglected.

             Full-wave rectifier:

            

            

             Rectification efficiency,

            

                   if RF is neglected.              

 

Q.29      An intrinsic silicon bar is 4mm long and has a rectangular cross section   60x100

(mm) 2. At 300K, find the electric field intensity in the bar and voltage across the bar

 when a steady state current of 1mA is measured. (Resistivity of intrinsic silicon at

 300K is 2.3 x 103 W-m                                                                                                     (7)

              

Ans:

                           Length=4mm

                           A=60 x 100 (mm) 2

                                    Current I =1mA

                           Resistivity  r = 2.3 x 103 Wm

                                            J = σ E

                                            E = J / σ = (I/A) (1/σ) = (I/A) r

                                            E = (1 x 10-6/ (60 x 10-6 x 100 x 10-6))x 2.3 x103

                                                = 383.33 x 103 V/m

                                            V = EL = 383.33 x 103  x 4mm =1.53 x 103 V

 

Q.30      The resistivity of doped silicon material is 9 x 10-3ohm-m.  The Hall co-efficient is

3.6 x 10-4m3 columb-1.  Assuming single carrier conduction, find the mobility and

density of charge carrier (e = 1.6 x 10-19 coulomb)                                                           (7)

 

Ans:

            RH = 3.6x10-4 m3/columb, ρ = 9x10-3 ohm-m

Mobility = μn = σ RH = (1/ρ)RH = (1/9x10-3)x  3.6x10-4 = 400 cm2/V-s

 

Density of charge carriers = σμ

              = (1/9x10-3) 400 = 44.44m coulomb

 

Q.31      A single tuned amplifier with capacitive coupling consists tuned circuit having R=10, L=20mH and C=0.05 F.  Determine the (i) Resonant frequency  (ii) Q-factor of the tank circuit (iii) Bandwidth of the amplifier.                                                                                                     (7)

 

 Ans:

 (i)  Reasonant frequency: fr = 1 / (2√LC) = 1 / (2√ 20x10-3x0.05x10-6) = 5032 Hz

 (ii) Q-factor of the tank circuit: Q = XL / R  = 2frL / R.

XL=2frL /  = (2  x 5032 x 20x10-3)/10=63.23

Q = 63.23 / 10 =6.32

(iii) Band width of the amplifier: As Q = fr  / BW  

             *6.32 = 5032 / BW 

BW = 769.20Hz

 

Q.32      Calculate the output voltage ‘V0’ for the following non inverting op-amp summer,

with V1 =2V and V2 = -1V                                                                                              (7)

 

 

Ans:

VO = ([R2V1 + R1V2] / [R1 + R2] )* ([R + Rf] / R)

If in the summer circuit the value of resistance are selected as R1 = R2 = R and

             Rf = 2R . Then

VO = - [(2R) V1/R + (2R) V2/R]

             = - [2(V1 + V2)]

      = - [2(2 -1)   ] = -2 V

 

Q.33      The current flowing through a certain P-N junction at room temperature when reverse biased is 0.15μA. Given that volt-equivalent of temperature, VT is 26mV, and the bias voltage being very large in comparison to VT, determine the current flowing through the diode when the applied voltage is 0.12V.                                                                                              (7)

 

 Ans:

The diode current is given by

I = Io (e-(Vp/ηVT)- 1)

             For large reverse bias, the diode current

I = Io = 0.15 x 10-6A.

Given V=0.12V, VT = 0.026 V, assuming Si diode, i.e.,η = 2

I = 0.15 x 10-6 (e (0.12/ (2 x 0.026)) - 1)

  = 1.36 μA

 

Q.34      In a transistor amplifier, change of 0.025V in signal voltage causes the base current to change by 15μA and collector current by 1.2 mA. If collector and load resistances are of 6kΩ and 12kΩ, determine                                                                         

          i) input resistance                (ii) current gain

          iii) ac load                                        (iv) voltage gain

          v) power gain                                                                                                                           (7) 

 

          Ans:

          i) Input resistance = change in input voltage / change in input current = 0.025/15 x 10-6 = 1.67kΩ

          ii) Current gain = β = change in output current / change in input current = 1.2mA/15 x 10-6 = 80

          iii) AC load = Rc || RL = 6k x 12k/(6k+12k) = 4kΩ

          iv) Voltage gain: output voltage = 1.2 x 10-3 x 4 x 103 = 4.8V = Vo

                            input voltage = 0.025V = Vi

               Voltage Gain = Vo / Vi = 4.8 / 0.025 =192

          v) Power gain = voltage gain  x  current gain = 192 x 80 = 15360

 

Q.35      What is a load line and how is it used in the calculation of current and voltage gains for a single stage amplifier?                                                                                                      (7)

         

           Ans: 

            In a transistor, the collector current IC depends on base current IB. The variation of IC for a specific load RC as a function of input voltage is along a straight line. This line for a fixed load is called as dc load line.  The output characteristic of a CE amplifier is plotted in the Fig. 13a. Consider the following specifications of the CE amplifier. 

 

                            

                                                                         Fig. 13a

             IBQ = 40μA

             ICQ=8mA,            VCEQ = 6V

             VS = Vm sinωt

             The amplitude Vm is chosen to provide a signal component of base current

             Ib = Ibm sinωt

             Where Ibm = 20μA

Total instantaneous base current iB is the superposition of the dc level and the signal current.

Therefore            iB = IBQ + Ib = 40 + 20 sinωt μA

From the figure, we see that variation in IB causes both IC and VCE to vary sinusoidally about their quiescent levels. These quantities are expressed as

iC  = ICQ+ iC = ICQ + Icm sinωt

vCE = VCEQ + vce = VCEQ + Vcm sinωt V

From figure 3a, Icm  = 4mA and Vcm= 2V

 

Q.36    The transconductance of a FET used in an amplifier circuit is 4000 micro-siemens. The load resistance is 15kΩ and drain circuit resistance is 10 MΩ. Calculate the voltage gain of the amplifier circuit            .                                

 

          Ans:

          Given: gm = 4000, RL = 15kΩ, rd = 10MΩ                                                                  (7) 

            Vo = -gm  x Vgs (rd || RL) 

          Vo / Vin = -(gm  x Vgs (rd || RL)) /  Vin

          But Vgs = Vin

          Therefore voltage gain Vo / Vin = -(gm  (rd || RL))

          Vo / Vin = 4000x10-6 x ( ( 106x15x103) / (106 + 15x103 ) )

            = (60x106) / (103(103 + 1)

            ≈ 60

 

Q.37  An output waveform displayed on an oscilloscope provided the following measured values

          i) VCE min=1.2V, VCE max=22V, VCEQ=10V

          ii) VCE min=2V, VCE max=18V, VCEQ=10V

          Determine the percent second harmonic distortion in each case.                    (14)

 

          Ans:      

          D2 = (|B2| / |B1|) x 100%

          i) B2 = (Imax + I min – 2IcQ) / 4

                  = (VCE max + VCE min – 2VCEQ) / 4

                  = (22+1.2-20) / 4 = 3.2 / 4 = 0.8

             B1 = (I max- Imin) / 2         = (VCE max – VCE min) / 2 = (22-1.2) / 2 = 10.4

             D2 = (0.8 / 10.4)x100 = 7.69%

          ii) B2 = (2+18-20) / 4 = 0

               B1 = (18-2) / 2 = 8

               D2 = 2.5/8 x 100 = 0% 

 

Q.38    A sample of pure silicon has electrical resistivity of 3000Ωm. The free carrier density in it is 1.1x1016/m3. If the electron mobility is three times that of hole mobility, find electron mobility and hole mobility. The electronic charge is equal to 1.6x10-19coulomb.                          (6)

Ans:

For pure Silicon,   ni = n = p = 1.1x1016/ m3, and mn= 3mp

s = ni (mn + mp) e = 1.1x1016 (mp + 3mp) 1.6x10–19= 7.04 x 10-3 mp

   = (3000 Wm)-1

Thus mpx7.04x10-3 = 1/3x10-3

i.e., mp = .047 m2 V-1 S-1 and mn = 3 mp =0.141 m2 V-1 S-1

 

Q.39    Explain Zener breakdown’. The zener diode in the circuit shown below regulates at 50V, over a range of diode currents from 5 to 40mA. The supply voltage V = 150V. Compute the value of R to allow voltage regulation from a zero load current to a maximum load current Imax. What is Imax?                                                                                                           (8)

           

Ans:

            Zener break down takes place in diodes having heavily doped p and n regions with essentially narrow depletion region. Considerable reverse bias gives rise to intense electric field in the narrow depletion region causing breakdown of covalent bonds and so creating a number of electron-hole pairs which substantially add to the reverse current which may sustain at a constant voltage across the junction. This breakdown is reversible.

 

 

Problem:

               For IL = 0, VL= 50 volts and

               Iz = Is = (150 – 50) / R £ 40 mA

              Hence R ³ 100/40 KW, i.e 2.5 KW

              For IL = Imax, Iz ³ 5mA

              But for R = 2.5 KW, Is = 40 mA.

  Hence Imax = 40 – 5 = 35 mA

 

Q.40    Draw a figure to show the output V-I characteristic curves of a BJT in CE   configuration. Indicate thereon, the saturation, active and cut off regions. Explain how, using these characteristics, one can determine the value of hfe or βF.                                                 (6)

  Ans:

Fig. 18b

 

Fig. 18b. shows the characteristics of BJT in CE configuration. To find hfe, draw a constant VCE line (vertical) going through desired Q point. Choose constant IB lines suitably, which cut the constant Vce line at X and Y.

                 hfe = ΔIC/ ΔIB. From fig hfe = (IC2-IC1) / (IB4-IB2)

          = (6-2) mA / (60-20) mA =100.

 

Q.41      Draw a small signal h-parameter equivalent circuit for the CE amplifier shown

in fig below.

 

Find an expression for voltage gain of the amplifier.Compute the value of voltage gain, if RC = RL = 800Ω, R1 = 1.5kΩ, R2 = 3kΩ, hre≈ 0, hoe = 100μS, hfe = 90 and hie = 150Ω.                        (9)

 

Ans:

Fig. 19 shows the small signal h-parameter model for CE amplifier

Fig.19

 

Voltage gain of amplifier is Av = VL / Vi      

Rb = R1R2 / (R1 + R2) in the equivalent circuit.

By current division in output circuit,

Rb = (R1R2) / (R1+R2) = (1.5x103x3x103) / (1.5+3)103 = 1kΩ

          Av = (-90x800x800) / [150(800+800+100x10-6x800x800) = -230.7

 

Q.42    The circuit of a common source FET amplifier is shown in the figure below. Find expressions for voltage gain Av and current gain Ai for the circuit in mid frequency region where Rs is bypassed by Cs. Find also the input resistance Rin for the amplifier. If RD = 3kΩ, RG = 500kΩ, μ = 60, rds = 30kΩ, compute the value of Av, Ai, and Rin.                                      (8)

 

Ans:

Using voltage source model of the FET, the equivalent circuit is as in Fig. 20a

Fig. 20 a

 

Vo = (-RDμVgs) / (RD+rds)

Vgs = Vi

Av = Vo / Vi = (-μRD) / (RD+rds)

id = (μVgs) / (rds+RD)

Vgs = iiRG

             Av  = (-60x3x103) / (3x103+30x103) = -5.45

Ai = id/ii ; i.e Ai = μRG / (RD+rds) = [60x500x103] / [30x103+3x103] = 909

It is obvious that Rin = RG = 500kΩ

 

Q.43    State Barkhausen criterion for sustained oscillations in a sinusoidal oscillator. The capacitance values of the two capacitors C1 and C2 of the resonant circuit of a colpitt oscillator are C1 = 20pF and C2 = 70pF.The inductor has a value of 22μH. What is the operating frequency of oscillator?                                                                                                                         (5)

 

            Ans:

            Consider a basic inverting amplifier with an open loop gain ‘A’. With feedback network attenuation factor ‘ß’ less than unity. The basic amplifier produces a phase shift of 180o between input and output. The feedback network must introduce 180o phase shift. This ensures positive feedback.

             Barkhausen criterion states that for sustained oscillation,

1. The total phase shift around the loop, as the signal proceeds from input through the amplifier, feedback network and back to input again, is precisely 0o or 360o.

2. The magnitude of the product of the open loop gain of the amplifier, ‘A’ and the feedback factor ‘ß’ is unity, i.e. |Aß| = 1.

Operating frequency of Colpitts oscillator is given by

f = 1 / (2π√ LCeq)

Where Ceq = (C1C2) / (C1 + C2)

      = (20 x 70) / (20 + 70) = 15.56pF.

                f = 1 / {2 x π √ 22 x 10-6 x 15.56 x 10-9} = 272.02 KHz

 

Q.44     Suggest modification in the given circuit of Opamp to make it (i) inverting (ii)  

non inverting.                                                                                                         (7)

 

Ans:

Fig 24 a (i) shows an inverting amplifier. For an inverting amplifier, the input is to be applied to the inverting terminal. Therefore point P2 is connected to the signal to be amplified.

Fig 24 a (i)                                                               Fig 24 a (ii)

 

Fig 24 a (ii) shows a non-inverting amplifier. For a non-inverting amplifier, the input is to be applied to the non-inverting terminal. Therefore point P1 is connected to the source.

  

Q.45      In the circuit of Q44, if input offset voltage is 0,

 (i) Find the output voltage Vo due to input bias current, when IB=100nA     (3)

 (ii) How can, the effect of bias current be eliminated so that output voltage is   zero?     (4)

 

 Ans:

(i) When the voltage gain of op-amp is very large, no current flows into the op-amp. Therefore IB flows into R2

Vo = IB x R2 = 100 x 10-9 x 106 =100 mV

      (ii) If Vo = 0, then R1 || R2.

             Let Rp=R1 || R2

Then voltage from inverting terminal to ground is

VI = -IB2 x Rp

      Let R1 = Rp = (R1 x R2) / (R1 + R2) = 90.9kΩ

      Add resistor R1 between non-inverting terminal and ground

       Choose the value of R1 = 90.9kΩ to make the output voltage as zero.

Since, VI –VN = 0 or VI = VN

where VI : Voltage at the inverting terminal w.r.t. ground

  and VN : Voltage at the non-inverting terminal w.r.t. ground

Hence, -IB2  x Rp = -IB1 x  R1        

                   For IB1 = IB2

 

Q.46    A differential amplifier has inputs Vs1=10mV, Vs2 = 9mV. It has a differential mode gain of 60 dB and CMRR is 80 dB. Find the percentage error in the output voltage and error voltage. Derive the formulae used.                                                                                                (14)      

         

          Ans:

          In an ideal differential amplifier output Vo is given by

          Vo = Ad (V1-V2)

          Ad = gain of differential amplifier

                    

          But in practical differential amplifiers, the output depends on difference signal Vd as well as on common mode signal VC.

          Vd = V1 – V2 ---------- 1

          VC = (V1 + V2) / 2 --------2

          Therefore output of above linear active device can be given as

          Vo = A1V1 + A2V2

         

          Where A1(A2) is the voltage amplification factor from input 1(2) to output under the condition that input 2(1) is grounded.

          Therefore from 1 and 2

          V1 = Vc + 0.5Vd                 and                      V2=Vc-0.5Vd

          Vo = AdVd + AcVc

            Where Ad = 0.5(A1-A2) and Ac = 0.5(A1+A2)

          the voltage gain of difference signal is Ad and voltage gain of common mode signal is Ac.

          Common mode rejection ratio = ρ = |Ad/Ac|. The equation for output voltage can be written as

          Vo = AdVd (1+ (Ad/Ac) (Vc/Vd))

          Vo = AdVd (1+ (1 / ρ) (Vc/Vd))        

          Vs1=10mV = V1    , Vs2=9mV = V2

          Ad=60dB,             CMRR=80dB   

          Vd = V1-V2=10mV-9mV=1mV

          Ad = 60dB = 20log10 Ad, since Ad =1000

          Vc = (V1+V2)/2 = (10+9) / 2 = 9.5mV

          CMRR = Ad / Ac

          10³ = 1000 / Ac

          Ac = 0.1

          Vo = AdVd + AcVc = 1000 x 10-3 + 0.1 x 9.5 x 10-6 

                = 1.00095 V

 

 

Q.47      Prove the following postulate of Boolean algebra using truth tables

x + y + z = (x + y).(x + z )                                                                                 (3)

 

Ans:

The truth table below demonstrates the equality (x + y + z) = (x + y) (x + z)

 

X

y

z

(x+y)

(x+z)

(x+y)(x+z)

x+y+z

0

0

0

0

0

0

0

0

0

1

0

1

0

0

0

1

0

1

0

0

0

0

1

1

1

1

1

1

1

0

0

1

1

1

1

1

0

1

1

1

1

1

1

1

0

1

1

1

1

1

1

1

1

1

1

1

 

Q.48      Simplify the following Boolean function using K-map:

                _      _          _   

f(a,b,c) = a c + a b + a b c + b c    Give the logic implementation of the simplified function in SOP form using suitable gates.                                                                  (6)

 

Ans:              

a b

 

c

 
                                                                                                                                                             

                                                               _

The simplified function by implementation of K-map is f (a,b,c) = a b + c . Assuming the availability of complements, the logic implementation is as in Fig. 40b.

 

Fig. 40b

 

 


PART – III

 

DESCRIPTIVES

 

Q.1       A triangular wave shown in fig(1) is applied to the circuit in fig(2). Explain the working of the circuit. Sketch the output waveform.

 

             Vin

 

             25v

 

 

                                                                                                                                                t

 

 

             -25v

 

                                      Fig(1)

 

           

                                    R

                                                                                     RL

              Vin                D1                               D2

                                                                                              Vout

 


                                 12v                                 8v

 

 


                                    Fig (2)

 

              Ans:

           Vin

 

          +12v

 

 

                                                                                                                                                t

 

 

         -12v

 

 

 

During the positive half-cycle of the input triangular voltage, the diode D1 remain forward biased as long as input voltage exceeds battery voltage +12v and diode D2 remains reverse biased and acts as open circuit. Thus up to +12v of the applied signal there would be output voltage across the output terminals and the triangular signal will be clipped off above 12v level.

              During negative half-cycle of the input signal voltage, diode D1 remains reverse-biased while D2 remains forward-biased as long as input signal voltage exceeds the battery voltage -8v in magnitude.

 

Q.2       Define ‘diffusion capacitance’ of a pn junction diode. Obtain an expression for the same. Why is the diffusion capacitance negligible for a reverse biased diode?                                                             (9)

 

             Ans:

             When a P-N junction is forward biased, a capacitance which is much larger than the transition capacitance, comes into play. This type of capacitance is called the diffusion capacitance, CD.

             Diffusion capacitance is given by the equation,  where dQ represents the change in number of minority carriers stored outside the depletion region when a change in voltage across diode, dV, is applied.

 

             If is the mean life-time of charge carriers, then a flow of charge Q yields a diode current I given as

                            

             We know that  

 

                            So,      

                                        
                  

               

             So diffusion capacitance

             

 

             For a forward bias,

             

 

             Thus the diffusion capacitance is directly proportional to the forward current through the diode.

             CD: Diffusion capacitance.

             CT: Transition capacitance.

 

 

 

 

                                                                                                                                                                                                                       

                                                                              15

 

                                                                       

 

                                                                                                                                                                                                                        

                                                                              10

 


                                                                         

                                                                               5             

                                                                                                     CD

                                                 CT

 

 

 

 

             In a reverse biased diode both CD and CT are present but CT >> CD. Hence in a reverse biased diode CD is neglected and only CT is considered.

 

Q.3       Draw the circuit of h-parameter equivalent of a CE amplifier with un by-passed emitter resistor.  Derive an expression for (i) its input impedance and (ii) voltage gain, using the equivalent circuit.                        (10)

 

             Ans:

VS

 

Ie

 

Ib

 

             CE – amplifier AC – Equivalent Circuit with an un-bypassed Emitter Resistor (RE).

 

 

 

CE – Amplifier n – parameter Equivalent circuit with RE.

 

             Input Impedance:

                                  Zin (base) or Zb = hie

             With an un-bypassed emitter register RE in the circuit,

                                                                             

                                            = .

                                             But   .

                                 

                                     

                                                                                     

RL)

 
                                                                              Voltage Gain:

                                                                                         

                        

             The minus sign indicates that output voltage Vout is 1800 out of phase with input voltage Vin.

             With an un-bypassed RE in the circuit,

            

            

                                                                             

             Usually ,                                                                   

RL)

 
            

 

Q.4       What is a ‘multistage amplifier’? Give the requirements to be fulfilled for an ideal coupling network.               (6)

 

             Ans:

             The voltage amplification or power gain or frequency response obtained with a single stage of amplification is usually not sufficient to meet the needs of either a composite electronic circuit or load device. Hence, several amplifier stages are usually employed to achieve greater voltage or current amplification or both. A transistor circuit containing more than one stage of amplification is known as a MULTI-STAGE amplifier.

             In a multistage amplifier, the output of first-stage is combined to next stage through a coupling device.

                              For an ideal coupling network the following requirements should be fulfilled.

1.               It should not disturb the dc-bias conditions of the amplifiers being coupled.

2.               The coupling network should transfer ac signal waveform from one amplifier to the next amplifier without any distortion.

3.               Although some voltage loss of signal cannot be avoided in the coupling network but this loss should be minimum, just negligible.

4.               The coupling network should offer equal impedance to the various frequencies of signal wave.

 

Q.5       Draw a neat sketch to illustrate the structure of a N-channel E-MOSFET. Explain its operation.                   (9)

 

             Ans:

                                          N-Channel  E-MOSFET Structure

Operation of N-Channel E-MOSFET

             Operation:

              It does not conduct when VGS = 0. In enhancement mosfet drain (ID) current flows only when VGS exceeds gate-to-source threshold voltage.

             When the gate is made positive with respect to the source and the substrate, negative change carriers within the substrate are attracted to the +ve gate and accumulate close to the surface of the substrate. As the gate voltage increased, more and more electrons accumulate under the gate, these accumulated electrons i.e., minority charge carriers make N-type channel stretching from drain to source.

             Now a drain current starts flowing. The strength of the drain current depends upon the channel resistance which, in turn, depends on the number of charge carriers attracted to the positive gate. Thus drain current is controlled by the gate potential.

 

Q.6       Show that in an amplifier, the gain reduces if negative feedback is used.                             (6)

 

             Ans:

             When the feedback voltage (or current) is applied to weaken the input signal, it is called negative feedback

β

 

 

             For an open-loop amplifier,

             Voltage again,

             Let a fraction (say β) of the output voltage Vout, be supplied back to the input and A be the open-loop gain. Now

                                                                               For + ve feedback case

             And    

             For negative feedback case,

             Actual input voltage to amplifier,

            

                        

                                                                             

 

Q.7       In a voltage series feedback amplifier, show that                                                           

a.                the input impedance increases with negative feedback.

b.                the output impedance decreases due to negative feedback.                                   (10)

 

             Ans:

             The input impedance can be determined as follows:

            

                   =

             Or 

            

                  

            

            

             The effect of negative feedback on the output impedance of an amplifier is explained below.

            

            

            

             Or  

             Or  

            

             Thus, series voltage negative feedback reduces the output impedance of an amplifier by a factor (1 + βA).

 

Q.8       List the advantages of a crystal oscillator.   (4)

            

       `     Ans:

             Advantages:

1.               It is very simple circuit as it does not need any tank circuit rather than crystal itself.

2.               Different oscillation frequencies can be had by simply replacing one crystal with another.

3.               The Q-factor, which is a measure of the quality of resonance circuit of a crystal, is very high.

4.               Most crystals will maintain frequency drift to within a few cycles at 250c.

 

Q.9       Explain how the timer IC 555 can be operated as an astable multivibrator, using timing diagrams.                 (8)

            

            

 

 

             Ans:

c = 0.01µF

 

The Timer -555 As An Astable Multivibrator

             An astable multivibrator, often called a free-running multivibrator, is a rectangular-wave generating circuit. The timing during which the output is either high or low is determined by the externally connected two resistors and a capacitor.

Internal Circuitary With External Connections

                                                                              When Q is low, or output Vout is high, the discharging transistor is cut-off and capacitor C begins charging towards Vcc through resistances RA and RB. Because of this, the charging time constant is (RA + RB)C. Eventually, the threshold voltage exceeds + , comparator 1 has a high output and triggers the flip-flop so that its Q is high and the timer output is low. With Q high, the discharge transistor saturates and pin-7 grounds so that the capacitor C discharges through resistance RB, trigger voltage at inverting input of comparator-2 decreases. When it drops below . The output of comparator 2 goes high and this reset the flip-flop so that Q is low and the timer output is high.

 

Q.10                                                                      Establish from first principles, the continuity equation, valid for transport of carriers in a semi-conductor.                                                                       (10)

            

             Ans:

                                                                              The continuity equation states a condition of a dynamic equilibrium for the concentration of mobile carriers in any elementary volume of the semiconductor.

             The carrier concentration in the body of semiconductor is a function of both time and distance. The differential equation governing this functional relationship, called the continuity equation, is based upon the fact that charge can be neither created nor destroyed.

             Consider an infinitesimal element of volume of area a, and length dx, as shown in fig, within which the average hole is p. If τn is the mean lifetime of holes then P/τn equals the holes per second lost by recombination per unit volume. If ‘e’ is the electronic charge, then, because of recombination, the number of coulombs per second decreases within the volume and decrease within the volume =             ------(1)

             If ‘g’ is the thermal rate of generation of electron-hole pairs per unit volume, the number of coulombs per second increases within the volume and increase within the

                                               volume = e.a.dx.g                ------(2)

             In general, the current varies with distance within the semiconductor. If the current entering the volume at x is In and leaving at x + dx is

                  Decrease within the volume = dIn.                    -------(3)          

             Because of three effects enumerated above, the hole concentration must change with time, and the total number of coulombs per second increases within the volume.

                                                                                   Increase within the volume =            ------(4)

             Since the charge must be conserved, so

                                    -------(5)

             The hole current In is the sum of drift current and diffusion current so,

                                                      ------(6)

             If the semiconductor is in thermal equilibrium with its surroundings and is subjected to no applied fields, the hole density will attain a constant volume Po.

             Under these conditions .

             So from the equation (5), we have             ------(7)

             Combining equations (5), (6) and (7) we have the equation of conservation of charge, called the continuity equation,

            

             

Q.11                                                                      What are the important characteristics of a cascade amplifier? Write the circuit of cascade amplifier and determine an expression for its voltage gain in terms of its circuit parameters.                                                                                                                                       (8)

             Ans:

                  Important characteristics:

1.      High input impedance

2.      Low voltage gain

3.      Input Miller capacitance is at a minimum with the common base stage providing good high frequency operation.

 

            

                    = .

             With a stage gain of only 1, no Miller effect occurs at transistor Q1. Voltage gain of     stage-2,

            

                 Over-all voltage gain,  .

 

Q.12                                                                      Write a neat sketch to shown the construction of a depletion-enhancement MOSFET and explain its operation in both the modes.                                              (9)

 

             Ans                                                          

                                                          N-channel DE- MOSFET Structure

            

             DE-MOSFET can be operated with either a positive or a negative gate. When gate is positive with respect to the source it operates in the enhancement. When the gate is negative with respect to the source, it operates in depletion-mode.

             When the gate is made negative w.r.t the substrate, the gate repels some of the negative charge carriers out-of the N-channel. This creates a depletion region in the channel and therefore, increases the channel resistance and reduces the drain current. The more negative the gate, the less the drain current.

 

Q.13                                                                      Draw the circuit of Hartley oscillator and derive an expression for its frequency of oscillation.                                                                                                  (10)

 

            

 

 

 

 

             Ans:

 

             The Hartley oscillator widely used as a local oscillator in  radio receivers.

                               --------(1)

             Here ,       and    

             Substituting these values in equation (1), we get

            

             Equating imaginary parts of above equation to zero we get,                   

While

             Or 

            

 

             Or 

 

Q.14           Write the circuit of current mirror used in a op-amp design and explain its operation.                                                                                         (8)

            

 

 

 

 

 

            

             Ans:

Current Mirror Circuit

 

             In the design of op-amps, current strategies are used that are not practical in discrete amplifiers. The new strategies are prompted by the fact that resistors utilize a great deal of ‘real-estate’, and precise matching of active devices is very practical since they share same piece of silicon. One such approach is the use of the current mirrors to bias differential pairs. The current Ix, set by transistor Q1 and resistor Rx is mirrored in the current I through the transistor Q2.

               and  .

            

             Since 

 

Q.15     Explain, using neat circuit diagram and waveforms, the application of timer IC555 as monostable multivibrator.                                                                                                                                       (9)   

 

             Ans:

 

Trigger Input, Output and Capacitor Voltage Wave Forms

 

Internal Circuitry with external connections

       

             Operation:

             Initially, when the output at pin-3 is low i.e., the circuit is in a stable state, the transistor is on and capacitor C is shorted to ground, when a negative pulse is applied to pin 2, the trigger input false below , the output of comparator goes high which resets the flip-flop and consequently the transistor turns off and output at pin-3 goes high. As the discharge transistor is cut-off, the capacitor C begins charging towards through resistance RA with a time constant equal to RAC. When the increasing capacitor voltage becomes slightly greater than, the output of comparator-1 goes high, which sets the flip-flop. The transistor goes to saturation, thereby discharging the capacitor C and output of the timer goes low.

 

Q.16                                                                      Write the circuit diagram of a square wave generator using an opamp and explain its operation.                                                                                                (7)

 

            

             Ans:

             The circuit’s frequency of oscillation is dependable on the charge and discharge of a capacitor C through feedback resistor Rf. The heart of the oscillator is an inverting op-amp comparator.

             The comparator uses positive feedback that increases the gain of the amplifier. A fraction of the output is feedback to the non-inverting input terminal. Combination of Rf and C acting as a low-pass R-C-Circuit is used to integrate the output voltage Vout and the capacitor voltage Vc is applied to the inverting input terminal in place of external signal.

                      where 

             When Vin is positive,    and

             When Vin is negative,  .

 

Q.17           Distinguish between synchronous and asynchronous counters.

              Show the logic diagram of a 3-bit UP-DOWN synchronous counter using suitable flip-flops, with parallel, carry based on NAND gates and explain its operation drawing wave diagrams.                                       (12)

            

              Ans:            

              Difference between synchronous and asynchronous counter :

1.       In synchronous counters synchronized at the same time. But in the case of asynchronous counter the output of first flip-flop is given as the clock input of the next flip-flop.

2.       In synchronous counter the output occurs after nth clock pulse if number of bits are N. But in asynchronous counter the output is derived by previous one that’s why n+1 step or clock pulse will be required.

 

             Design of 3 bit UP DOWN counter:-

              For M = 0, it acts as an UP counter and for M = 1 as a DOWN counter. The number of

              flip-flop required is 3. The input of the flip-flops are determined in a manner similar to the following table.

 

 

 

 

Truth Tables

 Direction

 

Present

State

Required FlipFlop

M

Q3

Q1

Q0