InterviewSolution
Saved Bookmarks
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.
| 201. |
The purpose of connecting a coupling capacitor in the output circuit of an amplifier is |
| Answer» Capacitor acts for ac or dc. | |
| 202. |
Generally the gain of a transistor amplifier falls at high frequencies due to the |
| Answer» At high frequency, internal capacitance comes into consideration. | |
| 203. |
The open loop gain of an actual op-amp is about 100. |
| Answer» It is very high. | |
| 204. |
If the Q of a single stage single tuned amplifier is doubled, the bandwidth will |
| Answer» . If Q is doubles, bandwidth becomes half. | |
| 205. |
In full wave rectification, if the input frequency is 50 Hz, then frequency at the output of filter is |
| Answer» Filter output will give d.c. signal and d.c. signal Passed zero frequency. | |
| 206. |
A CB amplifier has = 0.98 and R = 600 Ω. If I = 3.5 mA, the current gain is |
| Answer» Current gain of CB amplifier is a. | |
| 207. |
of an op-amp is about 1 kHz. |
| Answer» It is about 1 MHz. | |
| 208. |
In figure, D turns on when |
| Answer» D1 is forward biased when vi > V1. | |
| 209. |
In a full wave rectifier circuit using centre tapped transformer, the peak voltage across half of the secondary winding is 30 V. Then PIV is |
| Answer» In full wave circuit using centre tapped transformer PIV = 2Vm . | |
| 210. |
A voltage tripler circuit and voltage quadrupler circuit use identical components. Then |
| Answer» Since voltage quadrupler circuit uses more components, voltage drop will be more and voltage regulation poorer. | |
| 211. |
A Ge diode operated at a junction temperature of 27°C. For a forward current of 10 mA, V is found to be 0.3 V. If V = 0.4 V then forward current will be |
| Answer» iD2 = (47.73) x 10 mA = 466 mA. | |
| 212. |
The turn ratio of a transformer is 20:1, if a load of 10Ω is connected across the secondary, what will be the effective resistance seen looking into the Primary? |
| Answer» | |
| 213. |
In a CE amplifier circuit the ac voltage between emitter and ground |
| Answer» Common emitter means emitter at ground potential. | |
| 214. |
An amplifier has input impedance of 4 kΩ and output impedance of 80 kΩ. It is used in negative feedback circuit with 10% feedback. If open loop gain is 90, the closed loop input and output impedances are |
| Answer» Closed loop input impedance = 4 (1 + 0.1 x 90) = 40 kΩ, Closed loop Output impedance = = 8 kΩ. | |
| 215. |
In figure the current through resistor R |
| Answer» . | |
| 216. |
Voltage V in the circuit when V < 0 where D is an ideal diode. (Take R = R = R = 1 Ω) |
| Answer» if Vs ≥ 0, then id ≥ 0 if Vs < 0 then id < 0, Diode D is replaced by and open ckt and id = 0, hence VL = id . RL = 0. | |
| 217. |
A bridge rectifier circuit has a dc load current of 10 mA and a filter capacitance of 1000 μF. The peak to peak ripple voltage is |
| Answer» Peak to peak ripple voltage = . | |
| 218. |
In the circuit of figure, the value of is (neglecting ) |
| Answer» . | |
| 219. |
Figure uses 10 V zener diode. The minimum and maximum current through series resistance are |
| Answer» Minimum current = = 10 mA. Minimum current = = 30 mA. | |
| 220. |
In a power amplifier the collector current flows for 270° of the input cycle. The operations is |
| Answer» In class A operation IC exists for 360°. In class B operation IC exists for 180°. In between there is the class AB. | |
| 221. |
When load is coupled to class A amplifier through transformer, efficiency decreases. |
| Answer» Transformer coupling increases efficiency. | |
| 222. |
The gain of an FET amplifier can be changed by changing |
| Answer» Gain is proportional to gm . | |
| 223. |
To prevent a DC return between source and load, it is necessary to use |
| Answer» Capacitor offers infinite impedance to DC. | |
| 224. |
For a base current of 10 μA, what is the value of collector current in common emitter if β = 100 |
| Answer» IC = 10 x 100 μA = 1 mA. | |
| 225. |
If the input to the ideal comparator shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of |
| Answer» | |
| 226. |
In figure The minimum and maximum load currents are |
| Answer» When RL = ∞, IL = 0, When RL = 100 Ω, or 120 mA. | |
| 227. |
In figure, V = 0.6 V, β = 99. Then V and I are |
| Answer» VC = 20 - 1.98 x 10-3 x 5.4 x 103 9.3 V. | |
| 228. |
The input impedance of op-amp circuit of figure is |
| Answer» Due to the presence of virtual ground at input, the resistance in the series path of input of inverting amplifier is input impedance. | |
| 229. |
In a BJT circuit a transistor is replaced by transistor. To analyse the new circuit |
| Answer» All voltages and currents have reverse polarity. | |
| 230. |
To protect the diodes in a rectifier and capacitor input filter circuit it is necessary to use |
| Answer» Resistor reduces surge current. | |
| 231. |
In the amplifier circuit of figure = 100 and = 1000 Ω. The voltage gain of amplifier is about |
| Answer» . | |
| 232. |
The efficiency of a full wave rectifier using centre tapped transformer is twice that in full wave bridge rectifier. |
| Answer» Efficiency of full wave rectifier with centre tapped transformer is slightly higher than that of bridge rectifier. | |
| 233. |
Negative feedback reduces noise originating at the amplifier input. |
| Answer» It has no effect on noise originating at amplifier input. | |
| 234. |
The output V in figure is |
| Answer» Input to non-inverting op-amp is -10 x 10-6 x 103 = -10 mV. Therefore output = = -100 mV. | |
| 235. |
In a CE amplifier the input impedance is equal to the ratio of |
| Answer» Input is applied to base with emitter grounded. The input impedance is the ratio of ac base voltage to ac base current. | |
| 236. |
For a system to work, as oscillator the total phase shift of the loop gain must be equal to |
| Answer» Gain of system with + ve feedback = for oscillation but V0 ≠ 0 so, that 1 - AB = 0 AB = 1 ∠0° or 360°. | |
| 237. |
An amplifier has a large ac input signal. The clipping occurs on both the peaks. The output voltage will be nearly a |
| Answer» When a sinusoidal voltage is clipped on both sides it resembles a square wave. | |
| 238. |
A CB amplifier has = 6 Ω, R = 600 Ω and = 0.98. The voltage gain is |
| Answer» | |
| 239. |
A bridge rectifier circuit has input of 50 Hz frequency. The load resistance is R and filter capacitance is C. For good output wave shape, the time constant RC should be at least equal to |
| Answer» Time constant RLC must be about 5 x 20 ms. | |
| 240. |
Maximum efficiency of class B power amplifier is 50%. |
| Answer» Maximum efficiency for class B-operation can be 78.5% . | |
| 241. |
In figure what is the base current if V = 0.7 V |
| Answer» . | |
| 242. |
In figure I = 4 mA. Then V = |
| Answer» VS = ISRS = 4 x 10-3 x 1.1 x 103 = 4.4 V. | |
| 243. |
The self bias provides |
| Answer» Stable Q point is the advantage of self bias circuit. | |
| 244. |
In figure what is value of I if β = 100. Neglect V |
| Answer» Since IB = 100 μA, IC = βIB = 10 mA. | |
| 245. |
In figure the saturation collector current is |
| Answer» Since IE IC, . | |
| 246. |
In figure = 8 V and = 8 V. Which diode will conduct? |
| Answer» Both D1 and D2 are forward biased. | |
| 247. |
A forward voltage of 9 V is applied to a diode in series with a 1 kΩ load resistor. The voltage across load resistor is zero. It indicates that |
| Answer» Since load voltage is zero, the current is zero and diode is open-circuited. If load was open-circuited, load voltage would be about 9 V. | |
| 248. |
Which power amplifier can deliver maximum load power? |
| Answer» Since efficiency of class C operation is high, maximum load power is delivered. | |
| 249. |
In class C operation of an amplifier circuit, the collector current exists for |
| Answer» In class C operation output is in the form of short duration pulses. | |
| 250. |
The '' parameters of the circuit shown in the figure are = 25 Ω, = 0.999 and = 10Ω The Voltage gain is |
| Answer» Zi = 25 + 10-6 x 0.99 x 104 25 . | |