All questions of Analog Electronics for Electrical Engineering (EE) Exam
Req = ?
- a)11.86 ohm
- b)10 ohm
- c)25 ohm
- d)11.18 ohm
Correct answer is option 'D'. Can you explain this answer?
Req = ?
a)
11.86 ohm
b)
10 ohm
c)
25 ohm
d)
11.18 ohm
|
Ravi Singh answered |
- Req – 5 = 10(Req + 5)/(10 + 5 +Req).
- Solving for Req we have
Req = 11.18 ohm.
A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is- a)160.2 C
- b)-160.2 C
- c)16.02 C
- d)-16.02 C
Correct answer is option 'C'. Can you explain this answer?
A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is
a)
160.2 C
b)
-160.2 C
c)
16.02 C
d)
-16.02 C
|
Aditya Deshmukh answered |
n 10^20, Q = ne = e 10^20 = 16.02 C.
Charge on sphere will be positive.
Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is
- a)30 V
- b)25 V
- c)20 V
- d)15 V
Correct answer is option 'D'. Can you explain this answer?
Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is
a)
30 V
b)
25 V
c)
20 V
d)
15 V
|
Gate Gurus answered |
- 100 = 65 + V2 ⇒ V2 = 35 V
- V3 – 30 = V2 ⇒ V3 = 65 V
- 105 – V3 + V4 – 65 = 0 ⇒ V4 = 25 V
- V4 + 15 – 55 + V1 = 0 ⇒ V1 = 15 V.
In the question a circuit and a waveform for the input voltage is given. The diode in circuit has cutin voltage Vγ = 0. Choose the option for the waveform of output voltage vo 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In the question a circuit and a waveform for the input voltage is given. The diode in circuit has cutin voltage Vγ = 0. Choose the option for the waveform of output voltage vo
a)
b)
c)
d)
|
Ankit Mukherjee answered |
Diode is off for vi < 5 V. Hence vo = 5 V.
For vi > 5 V, vo = vi .Therefore (D) is correct option.
Consider the circuit shown in fig.
The both transistor have parameter as followsVTN = 0.8V, kn' = 30 μA/V2Que: If the width-to-length ratios of M1 and M2 are(W/L)1 = (W/L)2 = 40The output Vo is- a)2.5 V
- b)2.5 V
- c)5 V
- d)0 V
Correct answer is option 'B'. Can you explain this answer?
Consider the circuit shown in fig.
The both transistor have parameter as follows
VTN = 0.8V, kn' = 30 μA/V2
Que: If the width-to-length ratios of M1 and M2 are
(W/L)1 = (W/L)2 = 40
The output Vo is
a)
2.5 V
b)
2.5 V
c)
5 V
d)
0 V
|
Sanvi Kapoor answered |
For both transistor VDS = VGS ,
Compared to bipolar transistors, field effect transistors are NOT normally characterized by:- a)high input impedance
- b)a reverse-biased PN junction
- c)low input impedance
- d)low power consumption
Correct answer is option 'C'. Can you explain this answer?
Compared to bipolar transistors, field effect transistors are NOT normally characterized by:
a)
high input impedance
b)
a reverse-biased PN junction
c)
low input impedance
d)
low power consumption
|
Pioneer Academy answered |
The difference between FET and BJT is explained in the following table:
Hence, Option C is correct.
The secondary transformer voltage of the rectifier circuit shown in fig. is vs = 60 sin 2π 60tV. Each diode has a cut in voltage of Vγ = 0.6 V. The ripple voltage is to be no more than Vrip = 2 V. The value of filter capacitor will be
- a)48.8 μF
- b)24.4 μF
- c)32.2 μF
- d)16.1 μF
Correct answer is option 'B'. Can you explain this answer?
The secondary transformer voltage of the rectifier circuit shown in fig. is vs = 60 sin 2π 60tV. Each diode has a cut in voltage of Vγ = 0.6 V. The ripple voltage is to be no more than Vrip = 2 V. The value of filter capacitor will be
a)
48.8 μF
b)
24.4 μF
c)
32.2 μF
d)
16.1 μF
|
Zoya Sharma answered |
vs = 60 sin 2π 60tV
Consider a voltage series feedback network, where amplifier gain = 100, feedback factor = 5. For the basic amplifier, input voltage = 4V, input current=2mA. Find the input resistance of the network.- a)1.002kΩ
- b)1002kΩ
- c)2kΩ
- d)2000kΩ
Correct answer is option 'B'. Can you explain this answer?
Consider a voltage series feedback network, where amplifier gain = 100, feedback factor = 5. For the basic amplifier, input voltage = 4V, input current=2mA. Find the input resistance of the network.
a)
1.002kΩ
b)
1002kΩ
c)
2kΩ
d)
2000kΩ
|
Athul Das answered |
To find the input resistance of the network, we can use the formula:
Input resistance (Rin) = (Voltage gain / Current gain) * (Feedback factor / Amplifier gain)
Given:
Amplifier gain (Av) = 100
Feedback factor (β) = 5
The voltage gain (Av) can be calculated as the ratio of output voltage (Vo) to input voltage (Vi):
Av = Vo / Vi
Since it is a voltage series feedback network, the output voltage (Vo) and input voltage (Vi) are related by the formula:
Vo = Vi / (1 + β * Av)
Substituting the given values:
Av = 100
β = 5
Vo = Vi / (1 + 5 * 100)
Vo = Vi / 501
Now, we can calculate the output voltage (Vo) using the formula:
Vo = Av * Vi
Vo = 100 * 4V
Vo = 400V
Substituting the calculated values into the equation for Vo:
400V = Vi / 501
Solving for Vi:
Vi = 400V * 501
Vi = 200,400V
Now, we can substitute the calculated values into the formula for input resistance (Rin):
Rin = (Av / Ai) * (β / Av)
Rin = (100 / (2mA / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = 1,000 * 0.05
Rin = 50
Therefore, the input resistance of the network is 50 ohms.
Input resistance (Rin) = (Voltage gain / Current gain) * (Feedback factor / Amplifier gain)
Given:
Amplifier gain (Av) = 100
Feedback factor (β) = 5
The voltage gain (Av) can be calculated as the ratio of output voltage (Vo) to input voltage (Vi):
Av = Vo / Vi
Since it is a voltage series feedback network, the output voltage (Vo) and input voltage (Vi) are related by the formula:
Vo = Vi / (1 + β * Av)
Substituting the given values:
Av = 100
β = 5
Vo = Vi / (1 + 5 * 100)
Vo = Vi / 501
Now, we can calculate the output voltage (Vo) using the formula:
Vo = Av * Vi
Vo = 100 * 4V
Vo = 400V
Substituting the calculated values into the equation for Vo:
400V = Vi / 501
Solving for Vi:
Vi = 400V * 501
Vi = 200,400V
Now, we can substitute the calculated values into the formula for input resistance (Rin):
Rin = (Av / Ai) * (β / Av)
Rin = (100 / (2mA / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = 1,000 * 0.05
Rin = 50
Therefore, the input resistance of the network is 50 ohms.
With the increase in temperature, the resistivity of an intrinsic semiconductor decreases. This is because, with the increase of temperature- a)the carrier concentration decreases, but the mobility of carriers increases.
- b)the carrier concentration increases, but the mobility of carriers decreases.
- c)the carrier concentration remains the same but the mobility of carriers decreases.
- d)both the carrier concentration and mobility of carriers decreases
Correct answer is option 'B'. Can you explain this answer?
With the increase in temperature, the resistivity of an intrinsic semiconductor decreases. This is because, with the increase of temperature
a)
the carrier concentration decreases, but the mobility of carriers increases.
b)
the carrier concentration increases, but the mobility of carriers decreases.
c)
the carrier concentration remains the same but the mobility of carriers decreases.
d)
both the carrier concentration and mobility of carriers decreases
|
Malavika Nair answered |
In an intrinsic semiconductor, as the temperature increases mobility is slightly reduced since mobility of charge carrier is inversely proportional to temperature. This will slightly reduce the conductivity. But, due to temperature effect, large number of covalent bonds are broken which creates electrons and holes and this increases the conductivity by a larger value. As a result of this resistivity is decreased 
A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________- a)20
- b)25
- c)30
- d)35
Correct answer is option 'B'. Can you explain this answer?
A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________
a)
20
b)
25
c)
30
d)
35
|
Hridoy Chakraborty answered |
Calculation of Voltage Gain in FET Circuit:
1. Given:
- Transconductance (gm) = 2500 µS = 2.5 mS
- Drain resistance (RD) = 10 kΩ = 10000 Ω
2. Formula for Voltage Gain (Av):
Av = -gm * RD
3. Calculation:
Av = -2.5 mS * 10 kΩ
Av = -25
4. Result:
Therefore, the voltage gain of the FET circuit is 25.
Explanation:
The voltage gain of a FET circuit is determined by the product of the transconductance (gm) and the drain resistance (RD). In this case, with a transconductance of 2500 µS (2.5 mS) and a drain resistance of 10 kΩ, the voltage gain is calculated to be -25. The negative sign indicates that the output signal is inverted with respect to the input signal. Hence, the correct answer is option 'B' with a voltage gain of 25.
1. Given:
- Transconductance (gm) = 2500 µS = 2.5 mS
- Drain resistance (RD) = 10 kΩ = 10000 Ω
2. Formula for Voltage Gain (Av):
Av = -gm * RD
3. Calculation:
Av = -2.5 mS * 10 kΩ
Av = -25
4. Result:
Therefore, the voltage gain of the FET circuit is 25.
Explanation:
The voltage gain of a FET circuit is determined by the product of the transconductance (gm) and the drain resistance (RD). In this case, with a transconductance of 2500 µS (2.5 mS) and a drain resistance of 10 kΩ, the voltage gain is calculated to be -25. The negative sign indicates that the output signal is inverted with respect to the input signal. Hence, the correct answer is option 'B' with a voltage gain of 25.
Assertion (A): The drift velocity is in the direction opposite to that of the electric field.
Reason (R): At each inelastic collision with an ion, an electron loses energy, and a steady-state condition is reached where a finite value of drift speed is attained.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'B'. Can you explain this answer?
Assertion (A): The drift velocity is in the direction opposite to that of the electric field.
Reason (R): At each inelastic collision with an ion, an electron loses energy, and a steady-state condition is reached where a finite value of drift speed is attained.
Reason (R): At each inelastic collision with an ion, an electron loses energy, and a steady-state condition is reached where a finite value of drift speed is attained.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
|
Dj Bravo answered |
Both assertion and reason are individually correct statements. However, the reason for assertion is that due to the applied electric field, and electrostatic force is developed on the electron and the electrons would be accelerated in a direction opposite to the applied electric field and this motion is called directed motion of electron
If a current of 1.6 μA is flowing through a conductor, the number of electrons crossing a particular cross-section per second will be- a)1013
- b)1.6
- c)1019
- d)1.6 x 10-6
Correct answer is option 'A'. Can you explain this answer?
If a current of 1.6 μA is flowing through a conductor, the number of electrons crossing a particular cross-section per second will be
a)
1013
b)
1.6
c)
1019
d)
1.6 x 10-6
|
Lavanya Menon answered |
Given, I = 1.6 x 10-6 A
= 1.6 x 10-6 Coulomb/second
Charge crossing a particular cross-section per second = 1.6 x 10-6 C Hence, number of electrons crossing a particular cross-section per second
= 1.6 x 10-6 Coulomb/second
Charge crossing a particular cross-section per second = 1.6 x 10-6 C Hence, number of electrons crossing a particular cross-section per second
In a feedback network, input voltage is 14V, feedback voltage is 6V and source voltage is 20V. β is in ohms. What is its configuration?- a)Shunt-Shunt feedback
- b)Shunt-Series feedback
- c)Series-Series feedback
- d)Series-Shunt feedback
Correct answer is option 'C'. Can you explain this answer?
In a feedback network, input voltage is 14V, feedback voltage is 6V and source voltage is 20V. β is in ohms. What is its configuration?
a)
Shunt-Shunt feedback
b)
Shunt-Series feedback
c)
Series-Series feedback
d)
Series-Shunt feedback
|
Pooja Patel answered |
Given that input is 14V, feedback is 6V and source is 20 V, we can see
VI = VS – VF, which is voltage mixing. Also, β is in ohms that is voltage/current.
Since output of feedback is voltage and input is current, the output has current sampling.
Thus, configuration is a series-series feedback/current – series feedback.
VI = VS – VF, which is voltage mixing. Also, β is in ohms that is voltage/current.
Since output of feedback is voltage and input is current, the output has current sampling.
Thus, configuration is a series-series feedback/current – series feedback.
For the circuit shown below, the collector to emitter voltage is
- a)1.25 V
- b)3.40 V
- c)-2.30 V
- d)0 V
Correct answer is option 'B'. Can you explain this answer?
For the circuit shown below, the collector to emitter voltage is
a)
1.25 V
b)
3.40 V
c)
-2.30 V
d)
0 V
|
|
Sanya Agarwal answered |
Let the transistor be operating under active region.
The density and mobility of electrons in a conductor are respectively 1020/cm3 and 800 cm2/V-s. If a uniform electric field of 1 V/cm exists across this conductor, then the electron current density would be approximately- a)11 kA/cm2
- b)9 kA/cm2
- c)13 kA/cm2
- d)18 kA/cm2
Correct answer is option 'C'. Can you explain this answer?
The density and mobility of electrons in a conductor are respectively 1020/cm3 and 800 cm2/V-s. If a uniform electric field of 1 V/cm exists across this conductor, then the electron current density would be approximately
a)
11 kA/cm2
b)
9 kA/cm2
c)
13 kA/cm2
d)
18 kA/cm2
|
|
Sanvi Kapoor answered |
The current density is given by
In an n-p-n transistor biased in the active region, as the magnitude of the collector-base voltage is increased,- a)the base current increases because more electrons are injected from the emitter.
- b)the base current increases because more holes are injected from the base to the collector.
- c)the emitter current decreases because the base-emitter junction gets slightly less forward biased.
- d)the collector current increases slightly because the base width reduces.
Correct answer is option 'D'. Can you explain this answer?
In an n-p-n transistor biased in the active region, as the magnitude of the collector-base voltage is increased,
a)
the base current increases because more electrons are injected from the emitter.
b)
the base current increases because more holes are injected from the base to the collector.
c)
the emitter current decreases because the base-emitter junction gets slightly less forward biased.
d)
the collector current increases slightly because the base width reduces.
|
Mahesh Datta answered |
IC increases because base width reduces due to early effect.
The LED in the circuit of fig. will be on if vi is
- a) > 10 V
- b)<10 V
- c) > 5 V
- d)< 5 V
Correct answer is option 'C'. Can you explain this answer?
The LED in the circuit of fig. will be on if vi is
a)
> 10 V
b)
<10 V
c)
> 5 V
d)
< 5 V
|
Anjana Khatri answered |
When v+ > 5 V, output will be positive and LED will be on. Hence (C) is correct.
In the circuit shown in fig. the op-amp is ideal. If βF = 60, then the total current supplied by the 15 V source is
- a)123.1 mA
- b)98.3 mA
- c)49.4 mA
- d)168 mA
Correct answer is option 'C'. Can you explain this answer?
In the circuit shown in fig. the op-amp is ideal. If βF = 60, then the total current supplied by the 15 V source is
a)
123.1 mA
b)
98.3 mA
c)
49.4 mA
d)
168 mA
|
Luminary Institute answered |
v+ = 5V = v∞ = VE ,
The input current to the op-amp is zero.
In a Common Drain (CD) MOSFET amplifier with voltage divider bias with R1 and R2 equal to 1.5 MΩ and 1 MΩ respectively, the input impedance Zf is:- a)220 kΩ
- b)600 kΩ
- c)470 kΩ
- d)200 kΩ
Correct answer is option 'B'. Can you explain this answer?
In a Common Drain (CD) MOSFET amplifier with voltage divider bias with R1 and R2 equal to 1.5 MΩ and 1 MΩ respectively, the input impedance Zf is:
a)
220 kΩ
b)
600 kΩ
c)
470 kΩ
d)
200 kΩ
|
Sanjana Chopra answered |
Calculation of Input Impedance Zf:
- The input impedance Zf of a CD MOSFET amplifier with voltage divider bias can be calculated using the formula:
Zf = R1 || R2 + (1 + g_m * (R1 || R2))^-1
- Given that R1 = 1.5 MΩ and R2 = 1 MΩ, we can substitute these values into the formula.
Calculating R1 || R2:
- R1 || R2 = (R1 * R2) / (R1 + R2)
- R1 || R2 = (1.5 MΩ * 1 MΩ) / (1.5 MΩ + 1 MΩ)
- R1 || R2 = 1.5 MΩ * 1 MΩ / 2.5 MΩ
- R1 || R2 = 0.6 MΩ = 600 kΩ
Calculating g_m:
- Given parameters of the MOSFET and biasing conditions, the transconductance g_m is typically provided in the problem statement.
Substitute into the Input Impedance Formula:
- Zf = 600 kΩ + (1 + g_m * 600 kΩ)^-1
Final Calculation:
- Since we are given the options, we can calculate the input impedance for each option and find that the correct answer is option 'B' 600 kΩ.
Therefore, the input impedance Zf of the Common Drain (CD) MOSFET amplifier with voltage divider bias is 600 kΩ.
- The input impedance Zf of a CD MOSFET amplifier with voltage divider bias can be calculated using the formula:
Zf = R1 || R2 + (1 + g_m * (R1 || R2))^-1
- Given that R1 = 1.5 MΩ and R2 = 1 MΩ, we can substitute these values into the formula.
Calculating R1 || R2:
- R1 || R2 = (R1 * R2) / (R1 + R2)
- R1 || R2 = (1.5 MΩ * 1 MΩ) / (1.5 MΩ + 1 MΩ)
- R1 || R2 = 1.5 MΩ * 1 MΩ / 2.5 MΩ
- R1 || R2 = 0.6 MΩ = 600 kΩ
Calculating g_m:
- Given parameters of the MOSFET and biasing conditions, the transconductance g_m is typically provided in the problem statement.
Substitute into the Input Impedance Formula:
- Zf = 600 kΩ + (1 + g_m * 600 kΩ)^-1
Final Calculation:
- Since we are given the options, we can calculate the input impedance for each option and find that the correct answer is option 'B' 600 kΩ.
Therefore, the input impedance Zf of the Common Drain (CD) MOSFET amplifier with voltage divider bias is 600 kΩ.
The energy required to charge a 10 µF capacitor to 100 V is- a)0.1 J
- b)0.05 J
- c) 5 x 10(-9) J
- d)10 x 10(-9) J
Correct answer is option 'B'. Can you explain this answer?
The energy required to charge a 10 µF capacitor to 100 V is
a)
0.1 J
b)
0.05 J
c)
5 x 10(-9) J
d)
10 x 10(-9) J
|
|
Sanjana Chopra answered |
Energy provided is equal to 0.5 CVxV.
What is the reverse transmission factor?- a)Ratio of output by input signal
- b)Ratio of feedback by input signal
- c)Ration of feedback by output signal
- d)Ratio of input by feedback signal
Correct answer is option 'C'. Can you explain this answer?
What is the reverse transmission factor?
a)
Ratio of output by input signal
b)
Ratio of feedback by input signal
c)
Ration of feedback by output signal
d)
Ratio of input by feedback signal
|
Cstoppers Instructors answered |
In feedback systems, the feedback signal is in proportion with the output signal.
XF ∝ XO
XF = βXO, where β is the feedback factor or reverse transmission factor.
XF ∝ XO
XF = βXO, where β is the feedback factor or reverse transmission factor.
Assertion (A): Bridge arrangement is preferred over centre-tapped transformer arrangement for full-wave rectification.
Reason (R): The bridge arrangement improves the filtering action.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is faise but R is true.
Correct answer is option 'C'. Can you explain this answer?
Assertion (A): Bridge arrangement is preferred over centre-tapped transformer arrangement for full-wave rectification.
Reason (R): The bridge arrangement improves the filtering action.
Reason (R): The bridge arrangement improves the filtering action.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is faise but R is true.
|
Pallavi Nair answered |
No filtering action is provided by bridge arrangement. Hence, reason is a false statement.
The forward current gain (α) of a bipolar transistor can be increased by
1. reducing the recombination lifetime in the emitter
2. increasing the emitter doping density.
3. increasing the base doping density.
4. reducing the base width.
Select the correct code from the given options,- a)2 and 3 only
- b)2 and 4 only
- c)1, 2 and 4 only
- d)4 only
Correct answer is option 'B'. Can you explain this answer?
The forward current gain (α) of a bipolar transistor can be increased by
1. reducing the recombination lifetime in the emitter
2. increasing the emitter doping density.
3. increasing the base doping density.
4. reducing the base width.
Select the correct code from the given options,
1. reducing the recombination lifetime in the emitter
2. increasing the emitter doping density.
3. increasing the base doping density.
4. reducing the base width.
Select the correct code from the given options,
a)
2 and 3 only
b)
2 and 4 only
c)
1, 2 and 4 only
d)
4 only
|
Manoj Mehra answered |
• When emitter is heavily doped, more majority carriers will reach into base due to which lC will increase and hence α increases.
• Due to early effect, base width is reduced as a result of which IC is slightly increased. Hence, α of the transistor is increased.
Hence, 2 and 4 are correct.
• Due to early effect, base width is reduced as a result of which IC is slightly increased. Hence, α of the transistor is increased.
Hence, 2 and 4 are correct.
The thinnest region in a BJT is_____and the most doped region is________.- a)emitter, collector
- b)base, collector
- c)collector, emitter
- d)base, emitter
Correct answer is option 'D'. Can you explain this answer?
The thinnest region in a BJT is_____and the most doped region is________.
a)
emitter, collector
b)
base, collector
c)
collector, emitter
d)
base, emitter
|
|
Pranab Basu answered |
Base is provided with smallest area to reduce transit type or journey time for the majority carriers while travelling from emitter to collector. Emitter is heavily doped to inject its majority carrier into base.
Which factor determines the gain of the voltage series feedback amplifier?- a)Open loop voltage gain
- b)Feedback voltage
- c)Ratio of two resistors
- d)Gain of feedback circuit
Correct answer is option 'C'. Can you explain this answer?
Which factor determines the gain of the voltage series feedback amplifier?
a)
Open loop voltage gain
b)
Feedback voltage
c)
Ratio of two resistors
d)
Gain of feedback circuit
|
|
Pooja Patel answered |
In setting the gain of the voltage series feedback amplifier, the ratio of two resistors is important and not the absolute value of these resistors. For example: If a gain of 11 is desired, we choose R1 = 1kΩ and R1 = 10kΩ or R1 = 100Ω and RF = 1kΩ.
Applications of negative feedback to a certain amplifier reduced its gain from 200 to 100. If the gain with the same feedback is to be raised to 150, in the case of another such appliance, the gain of the amplifier without feedback must have been- a)400
- b)450
- c)500
- d)600
Correct answer is option 'D'. Can you explain this answer?
Applications of negative feedback to a certain amplifier reduced its gain from 200 to 100. If the gain with the same feedback is to be raised to 150, in the case of another such appliance, the gain of the amplifier without feedback must have been
a)
400
b)
450
c)
500
d)
600
|
|
Pooja Patel answered |
Concept:
A negative feedback network is shown below.
Where A is the gain of the amplifier without feedback and feedback gain = β
Calculation:
Case 1:
Gain without feedback, A = 200
Gain with feedback ACL = 100
Case 2:
Gain with feedback, ACL = 150
Feedback, 
The bandwidth of an RC-coupled amplifier is limited by- a)Coupling capacitors at the low frequency end and bypass capacitors at the high frequency end
- b)Coupling capacitors at the high frequency end and bypass capacitors at the low frequency end
- c)Bypass and coupling capacitors at the low frequency end and device shunt capacitors at the high frequency end
- d)Device shunt capacitors at the low frequency end and bypass as well as coupling capacitors at the high frequency end
Correct answer is option 'C'. Can you explain this answer?
The bandwidth of an RC-coupled amplifier is limited by
a)
Coupling capacitors at the low frequency end and bypass capacitors at the high frequency end
b)
Coupling capacitors at the high frequency end and bypass capacitors at the low frequency end
c)
Bypass and coupling capacitors at the low frequency end and device shunt capacitors at the high frequency end
d)
Device shunt capacitors at the low frequency end and bypass as well as coupling capacitors at the high frequency end
|
Naroj Boda answered |
(i) In capacitively coupled amplifiers, the coupling and bypass capacitors affect the low-frequency cut-off. These capacitors form a high-pass filter with circuit resistances.
(ii) Coupling and bypass capacitors are also called external capacitors.
For high-frequency response, it is determined by the device’s internal capacitance and the Miller effect.
(iii) Device internal capacitance is shunted and also called as Junction capacitors.
(ii) Coupling and bypass capacitors are also called external capacitors.
For high-frequency response, it is determined by the device’s internal capacitance and the Miller effect.
(iii) Device internal capacitance is shunted and also called as Junction capacitors.
D-MOSFET in case of common source amplifier can operate with gate to source voltage zero at ______- a)Peak positive point
- b)Peak negative point
- c)Q point
- d)Origin
Correct answer is option 'C'. Can you explain this answer?
D-MOSFET in case of common source amplifier can operate with gate to source voltage zero at ______
a)
Peak positive point
b)
Peak negative point
c)
Q point
d)
Origin
|
Samridhi Bose answered |
The D-MOSFET (Depletion-mode Metal-Oxide-Semiconductor Field-Effect Transistor) is a type of MOSFET that operates in the depletion mode. In a common source amplifier configuration, the gate of the D-MOSFET is connected to the input signal, the source is grounded, and the drain is connected to the load resistor.
The gate-source voltage (VGS) is the voltage difference between the gate and source terminals of the MOSFET. It controls the conductivity of the channel between the source and drain. When VGS is zero, the MOSFET operates at the Q point, which is the operating point where the transistor remains in active region and exhibits linear amplification.
To understand why the D-MOSFET in a common source amplifier can operate with VGS zero at the Q point, let's consider the characteristics of the D-MOSFET and the operating conditions of the common source amplifier:
1. D-MOSFET Characteristics:
- The D-MOSFET is a depletion-mode device, which means it conducts current with zero or negative gate voltage.
- In a D-MOSFET, the channel is already formed when there is no voltage applied to the gate.
- The channel conductance decreases as the gate voltage becomes more negative.
2. Common Source Amplifier Operating Conditions:
- The Q point is the biasing point at which the amplifier operates.
- At the Q point, the drain current (ID) is constant and the MOSFET operates in the saturation region.
- The output voltage swing should be symmetric around the Q point for proper amplification.
Explanation:
Since the D-MOSFET is a depletion-mode device, it can operate with VGS zero at the Q point in a common source amplifier. At the Q point, the MOSFET is biased to have a certain drain current and output voltage. When the input signal swings above and below the Q point, the gate-source voltage varies around zero. But since the MOSFET is already conducting at zero gate voltage, it can amplify the input signal properly without distortion.
In other words, at the Q point, the MOSFET is biased to operate in the active region with a certain drain current. This biasing ensures that the MOSFET can respond to the input signal variations around the Q point. Therefore, the D-MOSFET in a common source amplifier can operate with VGS zero at the Q point.
The gate-source voltage (VGS) is the voltage difference between the gate and source terminals of the MOSFET. It controls the conductivity of the channel between the source and drain. When VGS is zero, the MOSFET operates at the Q point, which is the operating point where the transistor remains in active region and exhibits linear amplification.
To understand why the D-MOSFET in a common source amplifier can operate with VGS zero at the Q point, let's consider the characteristics of the D-MOSFET and the operating conditions of the common source amplifier:
1. D-MOSFET Characteristics:
- The D-MOSFET is a depletion-mode device, which means it conducts current with zero or negative gate voltage.
- In a D-MOSFET, the channel is already formed when there is no voltage applied to the gate.
- The channel conductance decreases as the gate voltage becomes more negative.
2. Common Source Amplifier Operating Conditions:
- The Q point is the biasing point at which the amplifier operates.
- At the Q point, the drain current (ID) is constant and the MOSFET operates in the saturation region.
- The output voltage swing should be symmetric around the Q point for proper amplification.
Explanation:
Since the D-MOSFET is a depletion-mode device, it can operate with VGS zero at the Q point in a common source amplifier. At the Q point, the MOSFET is biased to have a certain drain current and output voltage. When the input signal swings above and below the Q point, the gate-source voltage varies around zero. But since the MOSFET is already conducting at zero gate voltage, it can amplify the input signal properly without distortion.
In other words, at the Q point, the MOSFET is biased to operate in the active region with a certain drain current. This biasing ensures that the MOSFET can respond to the input signal variations around the Q point. Therefore, the D-MOSFET in a common source amplifier can operate with VGS zero at the Q point.
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” is- a)-400 V/cm in -z direction
- b)-200 V/cm in +z direction
- c)-200 V/cm in -z direction
- d)-400 V/cm in +z direction
Correct answer is option 'D'. Can you explain this answer?
In a p-type semiconductor, p = 1016/cm2, and μp = 400 cm2/V-s. If a magnetic field (B) of 5 x 10-4 Weber/cm2 is applied in the x-direction, and an electric field of 2000 V/m is applied in +y direction. The value of electric field caused due to the “Hall effect” is
a)
-400 V/cm in -z direction
b)
-200 V/cm in +z direction
c)
-200 V/cm in -z direction
d)
-400 V/cm in +z direction
|
Anirban Gupta answered |
The force acting on a charge q placed in a magnetic field B and an electric field E is given by
F = qv x B
The velocity of the hole placed in an electric field is
Now, F - + q E ...(ii) (+q = charge on a hole)
Comparing equations (i) and (ii), the electric field due to the Hall effect will be -400 V/cm in +z direction.
F = qv x B
The velocity of the hole placed in an electric field is
Now, F - + q E ...(ii) (+q = charge on a hole)
Comparing equations (i) and (ii), the electric field due to the Hall effect will be -400 V/cm in +z direction.
In a collector-to-base CE amplifier circuit having Vcc = 12 V, Rc = 250 Ω , IB = 0.25 mA, β = 100 and VCEQ = 8 V, the stability factor will be approximately given by (neglect VBE)- a)57
- b)62
- c)47
- d)68
Correct answer is option 'A'. Can you explain this answer?
In a collector-to-base CE amplifier circuit having Vcc = 12 V, Rc = 250 Ω , IB = 0.25 mA, β = 100 and VCEQ = 8 V, the stability factor will be approximately given by (neglect VBE)
a)
57
b)
62
c)
47
d)
68
|
Shivam Das answered |
For a coliector-to-base CE amplifier, stability factor is given by
An amplifier has a Open Loop voltage gain of –500. This gain is reduced to –100 when negative feedback is applied. The reverse transmission factor,β of this system is:-- a)– 0.025
- b)– 0.008
- c)0.1
- d)– 0.2
Correct answer is option 'B'. Can you explain this answer?
An amplifier has a Open Loop voltage gain of –500. This gain is reduced to –100 when negative feedback is applied. The reverse transmission factor,β of this system is:-
a)
– 0.025
b)
– 0.008
c)
0.1
d)
– 0.2
|
|
Pooja Patel answered |
Concept:
The gain of a feedback system is given by:
A = Open Loop gain
Af = Closed Loop Gain
β = Feedback/Transmission factor
Calculation:
Given Af = -100 and A = -500
1 + Aβ = 5
Aβ = 4
β = 4 / -500 = -0.008
The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is
- a)Voltage shunt feedback
- b)Current series feedback
- c)Current shunt feedback
- d)Voltage series feedback
Correct answer is option 'B'. Can you explain this answer?
The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is
a)
Voltage shunt feedback
b)
Current series feedback
c)
Current shunt feedback
d)
Voltage series feedback
|
|
Cstoppers Instructors answered |
In a current series feedback, current is sampled from the output and voltage is feedback to the source.
In the given amplifier circuit, the feedback signal becomes zero by opening the output feedback.
Hence, it is a current series feedback.
The mobility of free electrons and holes in pure germanium are 3800 and 1800 cm2/V-s respectively. The corresponding values for pure silicon are 1300 and 500 cm2/V-s, respectively. Assuming ni = 2.5 x 1013 cm-3 for germanium and ni = 1.5 x 1010 cm-3 for silicon at room temperature, the values of intrinsic conductivity for germanium and silicon are respectively given by- a)4.32 x 10-6 S/cm and 0.0224 S/cm
- b)4.32 x 10-6 S/cm and 0.325 S/cm
- c)0.0224 S/cm and 4.32 x 10-6 S/cm
- d)0.325 S/cm and 4.32 x 10-6 S/cm
Correct answer is option 'C'. Can you explain this answer?
The mobility of free electrons and holes in pure germanium are 3800 and 1800 cm2/V-s respectively. The corresponding values for pure silicon are 1300 and 500 cm2/V-s, respectively. Assuming ni = 2.5 x 1013 cm-3 for germanium and ni = 1.5 x 1010 cm-3 for silicon at room temperature, the values of intrinsic conductivity for germanium and silicon are respectively given by
a)
4.32 x 10-6 S/cm and 0.0224 S/cm
b)
4.32 x 10-6 S/cm and 0.325 S/cm
c)
0.0224 S/cm and 4.32 x 10-6 S/cm
d)
0.325 S/cm and 4.32 x 10-6 S/cm
|
|
Manoj Mehra answered |
The intrinsic conductivity for germanium is
The intrinsic conductivity for silicon is
The intrinsic conductivity for silicon is
Assertion (A): An emitter follower is widely used in electronic instruments.
Reason (R): The voltage gain of emitter follower is very high.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'C'. Can you explain this answer?
Assertion (A): An emitter follower is widely used in electronic instruments.
Reason (R): The voltage gain of emitter follower is very high.
Reason (R): The voltage gain of emitter follower is very high.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
|
Gaurav Chauhan answered |
The voltage gain of emitter follower (or common collector configuration) is less than unity i.e. a low value.
Hence, reason is not true.
Hence, reason is not true.
The nominal quiescent collector current of a transistor is 1.2 mA. If the range of β for this transistor is 80 ≤ β ≤ 120 and if the quiescent collector current changes by +-10 percent, the range in value for rπ is- a)1.73 kΩ < rπ < 2.59 kΩ
- b)1.93 kΩ < rπ < 2.59 kΩ
- c)1.73 kΩ < rπ < 2.59 kΩ
- d)1.56 kΩ < rπ < 2.88 kΩ
Correct answer is option 'D'. Can you explain this answer?
The nominal quiescent collector current of a transistor is 1.2 mA. If the range of β for this transistor is 80 ≤ β ≤ 120 and if the quiescent collector current changes by +-10 percent, the range in value for rπ is
a)
1.73 kΩ < rπ < 2.59 kΩ
b)
1.93 kΩ < rπ < 2.59 kΩ
c)
1.73 kΩ < rπ < 2.59 kΩ
d)
1.56 kΩ < rπ < 2.88 kΩ
|
Chhavi Gupta answered |
Given Data:
Nominal quiescent collector current = 1.2 mA
Range of β for the transistor = 80 to 120
Change in quiescent collector current = ±10%
Calculation:
1. Calculate the range of quiescent collector current:
Minimum quiescent collector current = 1.2 mA - 10% of 1.2 mA
= 1.2 mA - 0.12 mA
= 1.08 mA
Maximum quiescent collector current = 1.2 mA + 10% of 1.2 mA
= 1.2 mA + 0.12 mA
= 1.32 mA
2. Calculate the corresponding range for β:
Minimum β = 1.08 mA / 1.2 mA * 80
= 72
Maximum β = 1.32 mA / 1.2 mA * 120
= 132
3. Calculate the range in value for r:
Minimum value for r = 1 / 72
= 1.56 kΩ
Maximum value for r = 1 / 132
= 2.88 kΩ
Conclusion:
Therefore, the range in value for r is 1.56 kΩ to 2.88 kΩ. Hence, option D (1.56 kΩ to 2.88 kΩ) is the correct answer.
Nominal quiescent collector current = 1.2 mA
Range of β for the transistor = 80 to 120
Change in quiescent collector current = ±10%
Calculation:
1. Calculate the range of quiescent collector current:
Minimum quiescent collector current = 1.2 mA - 10% of 1.2 mA
= 1.2 mA - 0.12 mA
= 1.08 mA
Maximum quiescent collector current = 1.2 mA + 10% of 1.2 mA
= 1.2 mA + 0.12 mA
= 1.32 mA
2. Calculate the corresponding range for β:
Minimum β = 1.08 mA / 1.2 mA * 80
= 72
Maximum β = 1.32 mA / 1.2 mA * 120
= 132
3. Calculate the range in value for r:
Minimum value for r = 1 / 72
= 1.56 kΩ
Maximum value for r = 1 / 132
= 2.88 kΩ
Conclusion:
Therefore, the range in value for r is 1.56 kΩ to 2.88 kΩ. Hence, option D (1.56 kΩ to 2.88 kΩ) is the correct answer.
The given circuit represents small signal model of:
- a)self-biased common gain amplifier
- b)self-biased common drain amplifier
- c)common gate amplifier
- d)common drain amplifier
Correct answer is option 'C'. Can you explain this answer?
The given circuit represents small signal model of:
a)
self-biased common gain amplifier
b)
self-biased common drain amplifier
c)
common gate amplifier
d)
common drain amplifier
|
Gate Funda answered |
Common Gate FET Amplifier:
In this amplifier configuration, the input signal is applied to the source terminal, and output is taken from the drain as shown in Fig.
The gate is grounded, RL is in series with drain, and a source resistance RS is included in the circuit across which Vi is
dropped.
dropped.
The ac equivalent circuit is shown in Fig. below where the current source is connected between the drain and the source terminals.
Since source and drain are the input and output terminals respectively, (gmVgs) appears between the input and the output.
FETs are preferred in amplifiers for their:- a)high input impedance
- b)small size
- c)temperature dependence
- d)unipolar nature
Correct answer is option 'A'. Can you explain this answer?
FETs are preferred in amplifiers for their:
a)
high input impedance
b)
small size
c)
temperature dependence
d)
unipolar nature
|
|
Pooja Patel answered |
Concept:-
FET:- FET stands for Field Effect Transistor and is a three-terminal active device that uses an electric field to control the current flow. The three terminals are Gate, Source and, Drain.
FET as Amplifier:- FET can be used as an amplifier when one or more FETs are used. The most common type of FET amplifier is the MOSFET amplifier.
Loading effect –It can be defined as the effect on the source by the load impedance. Usually loading effect reduces the voltage level of a voltage source
We will analyze this problem by going through options.
Option 1:-
High input impedance.
One of the most important factors that is kept in mind while designing FET as an amplifier is, it should have high input impedance. It minimizes the effect of loading on input and a thus significant amount of input voltage signal is maintained for amplification. Hence it is the best option to pick.
Other options-
Small size, temperature dependence and, unipolar nature are also important features of FETs when comparing with BJTs.But when we are talking about FET as an amplifier its high input impedance and low output impedance matter.
Small size: Smaller the size of FET better it is for modern uses.
Temperature dependence: BJT leakage current or reverse saturation current is highly sensitive to temperature. It doubles for every 10o celcius.There is no leakage current in FET which makes it thermally stable as compared to BJT.
Unipolar nature: Unlike BJTs FETs are unipolar in nature. As there is only one type of charge carrier.
The gain of a FET amplifier can be changed by changing:- a)fm
- b)gm
- c)RL
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The gain of a FET amplifier can be changed by changing:
a)
fm
b)
gm
c)
RL
d)
None of these
|
|
Gate Funda answered |
Concept:
Voltage gain for a FET amplifier is given as:
AV = −gm(rd||RD)
Analysis:
Hence, The gain of a FET amplifier can be changed by changing gm
In a CD JFET configuration, if r0 = 1 / gm then:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In a CD JFET configuration, if r0 = 1 / gm then:
a)
b)
c)
d)
|
|
Pooja Patel answered |
JFET common drain amplifier is sometimes considered as a source follower and has voltage gain less than unity.
Putting the value of gm, we get
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