McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 101. |
For the circuit shown below express v0 as a function of v1 and v2. |
| A. | v0 = v1 + v2 |
| B. | v0 = v2 – v1 |
| C. | v0 = v1 – v2 |
| D. | v0 = -v1 – v2 |
| Answer» B. v0 = v2 – v1 | |
| Explanation: considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result. | |
| 102. |
Determine Ad and Ac for the given circuit. |
| A. | ac = 0 and ad = 1 |
| B. | ac ≠ 0 and ad = 1 |
| C. | ac = 0 and ad ≠ 1 |
| D. | ac ≠ 0 and ad ≠ 1 |
| Answer» A. ac = 0 and ad = 1 | |
| Explanation: consider the fact that the potential at the input terminals are identical and obtain the values of v1 and v2. thus obtain the value of vd and vc. | |
| 103. |
Determine the voltage gain for the given circuit known that R1 = R3 = 10kΩ abd R2 = R4 = 100kΩ. |
| A. | 1 |
| B. | 10 |
| C. | 100 |
| D. | 1000 |
| Answer» B. 10 | |
| Explanation: voltage gain is 100/10. | |
| 104. |
What is trans-conductance? |
| A. | ratio of change in drain current to change in collector current |
| B. | ratio of change in drain current to change in gate to source voltage |
| C. | ratio of change in collector current to change in drain current |
| D. | ratio of change in collector current to change in gate to source voltage |
| Answer» B. ratio of change in drain current to change in gate to source voltage | |
| Explanation: the change in drain current which is resulted due to change in gate to source voltage in a fet is measured by trans- conductance. this is termed as trans because it provides relationship between input and output quantity. | |
| 105. |
What is the value of trans conductance? |
| A. | 1 |
| B. | 2 |
| C. | 0.1 |
| D. | 0.01 |
| Answer» C. 0.1 | |
| Explanation: trans-conductance= change in drain current/ change in gate to source voltage | |
| 106. |
002. What is the value ofgm? |
| A. | 1 |
| B. | 2 |
| C. | 0.002 |
| D. | 0 |
| Answer» C. 0.002 | |
| Explanation: gm = change in drain current/ change in gate to source voltage gm = slope of vgs vs id gm = 0.002. | |
| 107. |
Find the maximum value of gm for FET with IDSS=10mA, Vp=-2V, VGS=5V? |
| A. | 10ms |
| B. | 20ms |
| C. | 1ms |
| D. | 0 |
| Answer» A. 10ms | |
| Explanation: gm0=2idss/ | |
| 108. |
A FET has IDSS=4ID and gm0 = 10mS then gm = |
| A. | 10ms |
| B. | 20ms |
| C. | 5ms |
| D. | 14ms |
| Answer» C. 5ms | |
| 109. |
Determine the value of output impedance for JFET, if the value of gm =1mS? |
| A. | 1kohm |
| B. | 0 |
| C. | 100kohm |
| D. | 5kohm |
| Answer» A. 1kohm | |
| Explanation: output impedance=inverse of trans conductance | |
| 110. |
In a small signal equivalent model of an FET, What does gm VGS stand for? |
| A. | a pure resistor |
| B. | voltage controlled current source |
| C. | current controlled current source |
| D. | voltage controlled voltage source |
| Answer» B. voltage controlled current source | |
| Explanation: for fet, the voltage is applied across the gate and source to control the drain current, hence while writing small signal model of an fet, on the output side gm vgs represents a current source which can be controlled by the input voltage vgs. | |
| 111. |
Given yfs = 3.6mS and yos = 0.02mS, determine r0? |
| A. | 100kohm |
| B. | 50mohm |
| C. | 50kohm |
| D. | 20kohm |
| Answer» C. 50kohm | |
| Explanation: r0=1/yos | |
| 112. |
Which of the following statement is incorrect? |
| A. | output of ce amplifier is out of phase with respect to its input |
| B. | cc amplifier is a voltage buffer |
| C. | cb amplifier is a voltage buffer |
| D. | ce amplifier is used as an audio (low frequency) amplifier |
| Answer» C. cb amplifier is a voltage buffer | |
| Explanation: the output of the ce amplifier has a phase shift of 180o with respect to the input. the cc amplifier has av≅1, thus it is a voltage buffer. however, the cb amplifier has a large voltage gain, and its current gain ai≅1, thus it is a current buffer. ce amplifier has an application has an audio amplifier. | |
| 113. |
Given hfe = 60, hie=1000Ω, hoe = 20μ Ω–, hre = 2 * 10-4. Find the current gain of the BJT, correct up to two decimal points. |
| A. | – 58.44 |
| B. | -59.21 |
| C. | – 60.10 |
| D. | – 60.00 |
| Answer» A. – 58.44 | |
| Explanation: current gain, ai = – hf / (1 + | |
| 114. |
Consider the circuit. Given hfe = 50, hie = 1200Ω. Find voltage gain. |
| A. | – 278 |
| B. | -277.9 |
| C. | – 300 |
| D. | – 280 |
| Answer» A. – 278 | |
| Explanation: voltage gain = av = - hferl’/hie | |
| 115. |
Given that IB = 5mA and hfe = 55, find load current. |
| A. | 28ma |
| B. | 280ma |
| C. | 2.5a |
| D. | 2a |
| Answer» B. 280ma | |
| Explanation: in given circuit, which is an emitter follower, current gain = 1 + hfe | |
| 116. |
Consider the following circuit, where source current = 10mA, hfe = 50, hie = 1100Ω, then for the transistor circuit, find output resistance RO and input resistance RI. |
| A. | ro = 0, ri = 21Ω |
| B. | ro = ∞, ri = 0Ω |
| C. | ro = ∞, ri = 21Ω |
| D. | ro = 10, ri = 21Ω |
| Answer» C. ro = ∞, ri = 21Ω | |
| Explanation: since hoe is not given, we can consider it to be small; i.e 1/hoe is neglected, open circuited. hence output resistance ro = | |
| 117. |
For the given circuit, input resistance RI = 20Ω, hfe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit. |
| A. | ri = 20Ω, ro = ∞ |
| B. | ri = 20Ω, ro = 2kΩ |
| C. | ri = 59Ω, ro = ∞ |
| D. | ri = 59Ω, ro = 2kΩ |
| Answer» C. ri = 59Ω, ro = ∞ | |
| Explanation: ri = 20k = hie/(1+hfe) = hie/51 hie =1020 Ω | |
| 118. |
If source resistance in an amplifier circuit is zero, then voltage gain (output to input voltage ratio) and source voltage gain (output to source voltage ratio) are the same. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: when a source resistance rs is present, the voltage gain with respect to source becomes | |
| 119. |
The Differential output of the difference amplifier is the amplification of |
| A. | difference between the voltages of input signals |
| B. | difference between the output of the each transistor |
| C. | difference between the supply and the output of the each transistor |
| D. | all of the mentioned |
| Answer» A. difference between the voltages of input signals | |
| Explanation: none. | |
| 120. |
The inputs to the differential amplifier are applied at |
| A. | at x and y |
| B. | at the gates of m1 and m2 |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. at the gates of m1 and m2 | |
| Explanation: none. | |
| 121. |
In Common Mode Differential Amplifier, the outputs Vout1 and Vout2 are related as: |
| A. | vout2 is in out of phase with vout1 with same amplitude |
| B. | vout2 and vout1 have same amplitude but the phase difference is 90 degrees |
| C. | vout1 and vout2 have same amplitude and are in phase with each other and their respective inputs |
| D. | vout1 and vout2 have same amplitude and are in phase with each other but out of phase with their respective inputs |
| Answer» D. vout1 and vout2 have same amplitude and are in phase with each other but out of phase with their respective inputs | |
| Explanation: none. | |
| 122. |
In a small signal differential gain vs input CM level graph, the gain decreases after V2 due to: |
| A. | as the input voltage increases, the output will be clipped |
| B. | when the input voltage to the transistors are high, the transistor enters saturation region and increases the current, which inturn decreases the output voltage = vdd – rd.iss |
| C. | when common mode voltage is greater than or equal to v2, the input transistors enter triode region, the gain begins to fall |
| D. | increasing the input voltage beyond v2 causes the gate oxide to conduct and the gain is reduced |
| Answer» C. when common mode voltage is greater than or equal to v2, the input transistors enter triode region, the gain begins to fall | |
| Explanation: none. | |
| 123. |
Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at |
| A. | f = 10 hz is 55 db |
| B. | f = 10 khz is 45 db |
| C. | f = 100 khz is 25 db |
| D. | f = 1mhz is 0 db |
| Answer» D. f = 1mhz is 0 db | |
| Explanation: use standard formulas for frequency response and voltage gain. | |
| 124. |
STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true? |
| A. | hp network passes dc and low frequencies and attenuate high frequency and opposite for lp network |
| B. | lp network passes dc and low frequencies and attenuate high frequency and opposite for hp network |
| C. | hp network passes dc and high frequencies and attenuate low frequency and opposite for lp network |
| D. | lp network passes low frequencies only and attenuate high frequency and opposite for hp network |
| Answer» B. lp network passes dc and low frequencies and attenuate high frequency and opposite for hp network | |
| Explanation: by definition a lp network allows dc current (or low frequency current) and an lp network does the opposite, that is, allows high frequency ac current. | |
| 125. |
The signal whose waveform is not effected by a linear circuit is |
| A. | triangular waveform signal |
| B. | rectangular waveform signal |
| C. | sine/cosine wave signal |
| D. | sawtooth waveform signal |
| Answer» C. sine/cosine wave signal | |
| Explanation: only sine/cosine wave are not affected by a linear circuit while all other waveforms are affected by a linear circuit. | |
| 126. |
Which of the following is not a classification of amplifiers on the basis of their frequency response? |
| A. | capacitively coupled amplifier |
| B. | direct coupled amplifier |
| C. | bandpass amplifier |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| Explanation: none of the options provided are correct. | |
| 127. |
General representation of the frequency response curve is called |
| A. | bode plot |
| B. | miller plot |
| C. | thevenin plot |
| D. | bandwidth plot |
| Answer» A. bode plot | |
| Explanation: general representation of frequency response curves are called bode plot. bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes. | |
| 128. |
Under what condition can the circuit shown be called a compensated attenuator. |
| A. | c1r1 = c2r2 |
| B. | c1r2 = c2r1 |
| C. | c1c2 = r1r2 |
| D. | r1 = 0 |
| Answer» A. c1r1 = c2r2 | |
| Explanation: standard condition of a compensated attenuator. here is the derivation for the same. | |
| 129. |
Which of the following is true? |
| A. | coupling capacitors causes the gain to fall off at high frequencies |
| B. | internal capacitor of a device causes the gain to fall off at low frequencies |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| Explanation: both the statements are false. | |
| 130. |
Which of the following is true? |
| A. | monolithic ic amplifiers are directly coupled or dc amplifiers |
| B. | televisions and radios use tuned amplifiers |
| C. | audio amplifiers have coupling capacitor amplifier |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: these all are practical applications of different types of amplifiers. | |
| 131. |
During high frequency applications of a B.J.T., which parasitic capacitors arise between the base and the emitter? |
| A. | cje and cb |
| B. | ccs |
| C. | cb |
| D. | ccs and cb |
| Answer» A. cje and cb | |
| Explanation: there are two capacitors which arise between bases and emitter. one is cje due to depletion region associated between base and emitter. cb is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor. | |
| 132. |
During high frequency applications of a B.J.T., which parasitic capacitors arise between the collector and the emitter? |
| A. | no capacitor arises |
| B. | ccs |
| C. | cb |
| D. | ccs and cb |
| Answer» A. no capacitor arises | |
| Explanation: the emitter and the collector are far away from each other when the b.j.t. is being constructed. hence, we find that they don’t share a common junction where charges can accumulate. thus, no such parasitic capacitors appear. | |
| 133. |
During high frequency applications of a B.J.T, which parasitic capacitors arise between the collector and the base? |
| A. | cje and cb |
| B. | ccs |
| C. | cπ |
| D. | cµ |
| Answer» D. cµ | |
| Explanation: only one capacitor up between the base and the collector. this is due to the depletion region present between the base and the collector region. | |
| 134. |
3 BJT FREQUENCY RESPONSE |
| A. | cje and cb |
| B. | ccs and cµ |
| C. | cb |
| D. | ccs and cb |
| Answer» B. ccs and cµ | |
| Explanation: there are two capacitors attached to the collector terminal. the collector-base junction provides a depletion capacitance (cµ) while the collector substrate junction provides a certain capacitance (ccs). | |
| 135. |
Which parasitic capacitors do not affect the frequency response of the C.E. stage, of the B.J.T.? |
| A. | cje and cb |
| B. | ccs and cµ |
| C. | cb and cµ |
| D. | no parasitic capacitor gets deactivated |
| Answer» D. no parasitic capacitor gets deactivated | |
| Explanation: while observing the frequency response of a c.e. stage, we find that all the parasitic capacitances of the b.j.t. end up slowing the speed of the b.j.t. the frequency response of this stage is affected by all the parasitic capacitors. | |
| 136. |
Which parasitic capacitors don’t affect the frequency response of the C.B. stage of the B.J.T.? |
| A. | none of the parasitic capacitances |
| B. | all the parasitic capacitances |
| C. | some of the coupling capacitors |
| D. | ccs and cb |
| Answer» B. all the parasitic capacitances | |
| Explanation: all the parasitic capacitors of a | |
| 137. |
Which parasitic capacitors don’t affect the frequency response of the C.C. stage of the B.J.T.? |
| A. | ccs |
| B. | ccs and cb |
| C. | cb |
| D. | ccs and cµ |
| Answer» A. ccs | |
| Explanation: in the follower stage, the load is present at the emitter. the parasitic capacitors present between the collector and the substrate i.e. cµ gets deactivated. this is observed from the small signal analysis where both the terminals of this capacitor get shorted to a.c. ground. | |
| 138. |
If the transconductance of the B.J.T increases, the transit frequency |
| A. | increases |
| B. | decreases |
| C. | doesn’t get affected |
| D. | doubles |
| Answer» A. increases | |
| Explanation: the transit frequency is directly proportional to the transconductance of the | |
| 139. |
If the total capacitance between the base and the emitter increases by a factor of 2, the transit frequency |
| A. | reduces by 2 |
| B. | increases by 2 |
| C. | reduces by 4 |
| D. | increases by 4 |
| Answer» A. reduces by 2 | |
| Explanation: the transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the b.j.t. hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2. | |
| 140. |
Which effect plays a critical role in producing changes in the frequency response of the B.J.T.? |
| A. | thevenin’s effect |
| B. | miller effect |
| C. | tellegen’s effect |
| D. | norton’s effect |
| Answer» A. thevenin’s effect | |
| Explanation: the miller effect results in a change in the capacitance seen between the base and the collector. this is why, it affects the frequency response of the b.j.t. deeply by changing the poles and affecting the high frequency voltage gain stage. | |
| 141. |
If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied? |
| A. | 1 + gmrl |
| B. | 1 – gmrl |
| C. | 1 + 2*gmrl |
| D. | 1 – 2*gmrl |
| Answer» A. 1 + gmrl | |
| Explanation: the low frequency gain of the | |
| 142. |
If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side gets multiplied? |
| A. | 1 + 1/gmrl |
| B. | 1 – 1/gmrl |
| C. | 1 + 2/gmrl |
| D. | 1 – 2/gmrl |
| Answer» A. 1 + 1/gmrl | |
| Explanation: the low frequency response of | |
| 143. |
Given that transition capacitance is 5 pico F and diffusion capacitance is 80 pico F, and base emitter dynamic resistance is 1500 Ω, find the β cut-off frequency. |
| A. | 7.8 x 106 rad/s |
| B. | 8.0 x 106 rad/s |
| C. | 49.2 x 106 rad/s |
| D. | 22.7 x 106 rad/s |
| Answer» A. 7.8 x 106 rad/s | |
| Explanation: the frequency in radians is calculated by | |
| 144. |
8 x 106. |
| A. | 200 |
| B. | 100 |
| C. | 141.42 |
| D. | 440.2 |
| Answer» C. 141.42 | |
| Explanation: the current gain for the ce circuit is a = β | |
| 145. |
Gain bandwidth frequency is GBP= 3000 Mhz. The cut-off frequency is f=10Mhz. What is the CE short circuit current gain at the β cutoff frequency? |
| A. | 212 |
| B. | 220 |
| C. | 300 |
| D. | 200 |
| Answer» A. 212 | |
| Explanation: ft = 3000mhz βfβ = 3000mhz | |
| 146. |
Which of the statement is incorrect? |
| A. | at unity gain frequency the ce short circuit current gain becomes 1 |
| B. | unity gain frequency is the same as gain |
| Answer» A. at unity gain frequency the ce short circuit current gain becomes 1 | |
| Explanation: gain bandwidth product for any mosfet is ft = gm/2π(cgs+cgd) thus gbp is approximately 5.9 mhz. | |
| 147. |
In an RC coupled CE amplifier, when the input frequency increases, which of these are incorrect? |
| A. | reactance csh decreases |
| B. | voltage gain increases |
| C. | voltage gain decreases due to shunt capacitance |
| D. | an rc coupled amplifier behaves like a low pass filter |
| Answer» B. voltage gain increases | |
| Explanation: when frequency increases, shunt reactance decreases. the voltage drop across shunt capacitance decreases and net voltage gain decrease. rc coupled amplifier acts as a low pass filter at high frequencies. | |
| 148. |
In Miller’s theorem, what is the constant K? |
| A. | total voltage gain |
| B. | internal voltage gain |
| C. | internal current gain |
| D. | internal power gain |
| Answer» B. internal voltage gain | |
| Explanation: the constant k=v2/v1, which is the internal voltage gain of the network. | |
| 149. |
When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2? |
| A. | c1 |
| B. | c2 |
| C. | both are equal |
| D. | insufficient data |
| Answer» A. c1 | |
| Explanation: given r1>r2 | |
| 150. |
Find net voltage gain, given hfe = 50 and hie = 1kΩ. |
| A. | 27.68 |
| B. | -22 |
| C. | 30.55 |
| D. | -27.68 |
| Answer» D. -27.68 | |
| Explanation: apply millers theorem to resistance between input and output. | |
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