InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
The most widely used diode rectifier circuit is |
| Answer» Bridge rectifier has the advantage of lower PIV. Moreover it does not require centre tapped transformer. | |
| 102. |
In figure base current is 10 μA and β = 100. Then V = |
| Answer» IC = 1 mA and VCE = 20 - 5 x 103 x 1 x 10-3 = 15 V. | |
| 103. |
It has been found that in a rectifier circuit with RC filter one RC section reduces ripple by 15%. Two RC sections are used in cascade the reduction in ripple would be |
| Answer» One filter reduces ripple to 0.15 of initial value. The second filter reduces ripple to 0.15 x 0.15 of initial value. | |
| 104. |
A class C amplifier has input frequency of 200 kHz. If width of collector pulses is 0.1 μ, duty cycle is |
| Answer» . | |
| 105. |
Output of a phase splitter is |
| Answer» A phase shifter gives two voltages which have equal magnitudes but are out of phase by 180°. | |
| 106. |
In a transistor the minimum and maximum values of β are 100 and 121. The proper value of β to be used to locate Q point is |
| Answer» Lowest value of β should be used to locate Q point. | |
| 107. |
In figure the zener has a resistance of 5 ohms. As the load resistance is varied, the output voltage |
| Answer» V0 = 12 + Iz x 5, Max Iz = 130 mA, minimum Iz = 10 mA. | |
| 108. |
Percentage increase in the reverse saturation current of a diode if the temperature is increased from 25°C to 50°C |
| Answer» = 2(T2 - T1)/10 = 2(50-25)/10 x 100% 565.7% . | |
| 109. |
The input to an op-amp integrating amplifier is a constant voltage. The output will be |
| Answer» Integral of a constant is t. | |
| 110. |
FET amplifier is more difficult to analyse than BJT amplifier. |
| Answer» FET amplifier is easier to analyse since gate current is negligible. | |
| 111. |
To bypass a 100 ohm emitter resistor at |
| Answer» XC < 0.1 x 100 or XC < 10 Ω or 10 = or C 1000 mF. | |
| 112. |
In a CE amplifier the dc load line is the same as ac load line when |
| Answer» When ac collector resistance is very high, the net resistance is the same as dc collector resistance. | |
| 113. |
If the output current wave shape of class C circuit has a period of 1 μs and a pulse width of 0.006 μs, the duty cycle is |
| Answer» . | |
| 114. |
A full wave rectifier circuit using centre tapped transformer, the input frequency 50 Hz. The frequency of output is |
| Answer» There are two output waves in one cycle of input wave. | |
| 115. |
If the midband gain of an amplifier is 40 dB, the gain at half power frequency is |
| Answer» Half power frequency is -3 dB frequency. | |
| 116. |
Which configuration is suitable for impedance matching? |
| Answer» CC has high input impedance and low output impedance. Hence suitable for impedance matching. | |
| 117. |
A BJT is said to be operating in saturation region if |
| Answer» For saturation both junctions have to be forward biased. | |
| 118. |
It is desired to reduce distortion in an amplifier from 8% to 2% by using 5% negative feedback. The open and closed loop gains respectively are |
| Answer» . | |
| 119. |
An ideal op-amp requires |
| Answer» An ideal Op-amp has identical transistors and hence does not require input offset voltage. | |
| 120. |
The JFET in the circuit shown in figure has an I = 10 mA and V = -5V. The value of the resistance R for a drain current I = 6.4 mA is |
| Answer» VGS = - 1 V IDS.RS = - VGS . | |
| 121. |
The value of parameter used in transistor model |
| Answer» . | |
| 122. |
An op-amp has zero gain for common mode inputs. Then CMRR = |
| Answer» Since common mode signal is reduced to zero, CMRR is infinite. | |
| 123. |
An op-amp has |
| Answer» High input impedance ensures minimum loading of source. Low output impedance is required for impedance matching at output. | |
| 124. |
In the figure, assume the op-amp is to be ideal. The output V if the circuit is |
| Answer» KCL at mode one. ... (1) Applying KCL at node (3) ... (2) From (1) V2 = 100 L sin ωt Put value of V2 in equation (2) and solve for V0 . | |
| 125. |
As the ratio R/R increases the efficiency of a rectifier increases. |
| Answer» As increases efficiency decreases. | |
| 126. |
If an amplifier with gain of - 100 and feedback of β = - 0.1 has a gain change of 20% due to temperature the change in gain of the feedback amplifier will be |
| Answer» . | |
| 127. |
In figure the cut in voltage of diode is 0.7 V and its bulk resistance is 20 Ω. The peak value of circuit current of during positive and negative half cycles are |
| Answer» During positive half cycle current I = or I 9.12 mA. During negative half cycle diode is reverse biased and current is zero. | |
| 128. |
A voltage gain of a CE amplifier connected in collector to base configuration is 50. Collector to base resistor R = 1 kΩ, R = 4Ω, the O/P resistance R =? |
| Answer» R0 = RC || RF' and , where RF is Miller resistance. . | |
| 129. |
A voltage multiplier circuit, using diodes and capacitors is suitable for |
| Answer» It is a high voltage device. However current has to be low since diode rated currents are low. | |
| 130. |
The main advantage of a crystal oscillator |
| Answer» Crystal oscillator has the advantage of frequency stability. | |
| 131. |
In a CE amplifier the ac emitter resistance |
| Answer» Emitter is at ground potential for ac signals. | |
| 132. |
A full wave rectifier using centre tapped transformer and a bridge rectifier use similar diodes and have equal no load output voltage. Under equal load conditions |
| Answer» Since two diodes are in series bridge rectifier, voltage drop is more. | |
| 133. |
Assume that op-amp in figure is ideal. If input V is triangular, the output V will be |
| Answer» It is a differentiating amplifier. | |
| 134. |
In an amplifier with a gain of - 1000 and feedback factor β = - 0.1, the change in gain is 20% due to temperature. The change in gain for feedback amplifier will be |
| Answer» . When gain changes, age change in | |
| 135. |
In the diode circuit of figure the diodes are ideal. The average current through ammeter is |
| Answer» . | |
| 136. |
A typical value of for a FET is about 25 μs. |
| Answer» gm is about 0.1 x 10-3 siemens. | |
| 137. |
The circuit shown is |
| Answer» Circuit can be redrawn as. This is an inverting amplifier with inverting terminal virtual ground. Hence, Vo = - iin RF This RF is termed as mutual resistance here, Rm. output voltage with load Where R0 is output Resistance of Op-Amp. if RL >> R0 output terminals become open circuited. and VoL = Rm.Iin where Rm work as mutual resistance. | |
| 138. |
Consider 49 cascaded amplifiers having individual rise time as 2 n sec. 3 n sec. ... 50 n sec. The input waveform rise time is 1 n sec. Then the output signal rise time is given time by (Assume output signal rise time is measured within 10 percent range of the final output signal.) |
| Answer» Output rise time (tr) 1.1ti02 + t12 + ... + t492, where t, t2 ... t49 are the individual rise time. tr 1.1 (1 ns)2 + (2 ns)2 + (50 ns)2 1.1 0.228 msec. | |
| 139. |
The zener diode in the rectangular circuit shown in the figure has a zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this circuit ensuring proper functioning over the input voltage range between 20 and 30 volts, is |
| Answer» It is zener diode, hence 5.8 volt remain constant. By applying KVL 30 = 1000 Imax + 5.8 Imax = 24.2 mA. | |
| 140. |
Assuming V = 0.2 V and β = 50, the minimum base current (I) required to drive the transistor in the given figure to saturation is |
| Answer» or = Ic = βIB and 2.8 mA | |
| 141. |
A negative feedback can be of |
| Answer» Series voltage, series current, shunt voltage and shunt current. | |
| 142. |
In a class C power amplifier the input frequency of ac signal is 1 MHz. If tank circuit has C = 1000 pF, the value of L = |
| Answer» | |
| 143. |
The V of the op-amp circuit shown in the given is |
| Answer» V0 = (Vs2 - Vs1) [Vs1 - Vs1] = 0. | |
| 144. |
In calculating output impedance of an amplifier the source is replaced by an open circuit. |
| Answer» It is replaced by short circuit. | |
| 145. |
For transistor 2 N 338 the manufacturer specifies P = 100 mW at 250°C free air temperature and maximum junction temperature of 125°C. Its thermal resistance is |
| Answer» . | |
| 146. |
A half wave diode rectifier uses a diode having forward resistance of 50 ohms. The load resistance is also 50 ohms. Then the voltage regulation is |
| Answer» No load output voltage = Output voltage at full load = . | |
| 147. |
A Hartley oscillator is used for |
| Answer» LC oscillators are used for radio frequencies. | |
| 148. |
In Class C operation the collector current looks like |
| Answer» Since collector current exists for less than 180°, it looks like narrow pulses. | |
| 149. |
An op-amp integrating circuit uses |
| Answer» Capacitor in feedback path provides integration. | |
| 150. |
In a CE amplifier, the output voltage is equal to product of |
| Answer» Since output is taken from collector terminal, output voltage is equal to collector current multiplied by collector resistance. | |