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All questions of Operational Amplifier for GATE Physics Exam
A non-inverting Op-Amp summer is shown in the figure. The output voltage V0 is
- a)Sin 100t
- b)2 Sin 100t
- c)3 Sin 100t
- d)3/2 Sin 100t
Correct answer is option 'C'. Can you explain this answer?
A non-inverting Op-Amp summer is shown in the figure. The output voltage V0 is
a)
Sin 100t
b)
2 Sin 100t
c)
3 Sin 100t
d)
3/2 Sin 100t
|
Vedika Singh answered |
V0
= -2(1 + 2R/R) + (2 + sin 100 t)(1 + 2R/R)= – 6 + 6 + 3 sin 100 t
= 3 sin 100 t
For the amplifier shown in figure
- a)The average of the input values is the same magnitude as Vout but of opposite sign
- b)The average of the input values is the different magnitude as Vout with opposite sign
- c)The average of the input values is of the different magnitude as Vout with same sign
- d)The average of the input values is the same magnitudes as Vout of the same sign
Correct answer is option 'A'. Can you explain this answer?
For the amplifier shown in figure
a)
The average of the input values is the same magnitude as Vout but of opposite sign
b)
The average of the input values is the different magnitude as Vout with opposite sign
c)
The average of the input values is of the different magnitude as Vout with same sign
d)
The average of the input values is the same magnitudes as Vout of the same sign
|
Jayant Mishra answered |
Since the input resistors are equal;
R= 100/kΩ
The output voltage is
Whereas, the input average
Shows that average of the input value is the same magnitude as Vout but of opposite sign.
The correct answer is: The average of the input values is the same magnitude as Vout but of opposite sign
R= 100/kΩ
The output voltage is
Whereas, the input average
Shows that average of the input value is the same magnitude as Vout but of opposite sign.
The correct answer is: The average of the input values is the same magnitude as Vout but of opposite sign
In the differentiating circuit given in the figure, the function of R1 is to:
- a)Present oscillations at high frequency.
- b)Eliminate high frequency noise spikes
- c)Maintain high input impedance
- d)Enable circuit to approach ideal differentiator
Correct answer is option 'C'. Can you explain this answer?
In the differentiating circuit given in the figure, the function of R1 is to:
a)
Present oscillations at high frequency.
b)
Eliminate high frequency noise spikes
c)
Maintain high input impedance
d)
Enable circuit to approach ideal differentiator
|
|
Vedika Singh answered |
R1 maintains the high input impedance.
The correct answer is: Maintain high input impedance
The correct answer is: Maintain high input impedance
If a certain Op-Amp has a closed loop gain of 20 and an upper critical frequency of 10 MHz, the gain bandwidth product is :- a)50 MHz
- b)20 MHz
- c)10 MHz
- d)unity gain frequency
Correct answer is option 'B,D'. Can you explain this answer?
If a certain Op-Amp has a closed loop gain of 20 and an upper critical frequency of 10 MHz, the gain bandwidth product is :
a)
50 MHz
b)
20 MHz
c)
10 MHz
d)
unity gain frequency
|
|
Vedika Singh answered |
GBW= 20 x 10 = 200MHz and gain bandwidth product is also defined as the frequency with unity gain.
The correct answers are: 20 MHz, unity gain frequency
The correct answers are: 20 MHz, unity gain frequency
If Av(d) = 3500 and ACM = 0.35, then CMRR is :- a)1225
- b)10000
- c)40 dB
- d)80 dB
Correct answer is option 'B,D'. Can you explain this answer?
If Av(d) = 3500 and ACM = 0.35, then CMRR is :
a)
1225
b)
10000
c)
40 dB
d)
80 dB
|
|
Jayant Mishra answered |
The solution to your question is:
Hence, The correct answers are: 10000, 80 dB
In an integrator, the feedback element is a :- a)zener diode
- b)voltage divider
- c)register
- d)capacitor
Correct answer is option 'D'. Can you explain this answer?
In an integrator, the feedback element is a :
a)
zener diode
b)
voltage divider
c)
register
d)
capacitor
|
|
Jayant Mishra answered |
Integrator circuit is as shown below
The correct answer is: capacitor
The correct answer is: capacitor
Ideal op-amp has infinite voltage gain because - a)To control the output voltage
- b)To obtain finite output voltage
- c)To receive zero noise output voltage
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
Ideal op-amp has infinite voltage gain because
a)
To control the output voltage
b)
To obtain finite output voltage
c)
To receive zero noise output voltage
d)
None of the mentioned
|
|
Vedika Singh answered |
As the voltage gain is infinite, the voltage between the inverting and non-inverting terminal (i.e. differential input voltage) is essentially zero for finite output voltage.
When negative feedback is used, then the gain bandwidth product of an Op-Amp.- a)fluctuates
- b)increases
- c)decreases
- d)stays the same
Correct answer is option 'D'. Can you explain this answer?
When negative feedback is used, then the gain bandwidth product of an Op-Amp.
a)
fluctuates
b)
increases
c)
decreases
d)
stays the same
|
|
Jayant Mishra answered |
Applying the feedback will reduce the gain but increase the bandwidth hence, the gain bandwidth product = Av x f. he remains constant for any voltage feedback amplifier.
The correct answer is: stays the same
The correct answer is: stays the same
A differential amplifier.- a)is a part of an Op-Amp
- b)has one input only
- c)has two output
- d)has one input and one output
Correct answer is option 'A,C'. Can you explain this answer?
A differential amplifier.
a)
is a part of an Op-Amp
b)
has one input only
c)
has two output
d)
has one input and one output
|
|
Jayant Mishra answered |
Any differential amplifier has two inputs and two outputs (one is Vout and other is ground potential) The correct answers are: is a part of an Op-Amp, has two output
The circuit below represent a non-inverting integrator. Find the expression for the output voltage.
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The circuit below represent a non-inverting integrator. Find the expression for the output voltage.
a)
b)
c)
d)
|
|
Vedika Singh answered |
Let the inverting terminal of Op-Amp is at potential ‘V and hence non-inverting terminal also appears to have the same potential ‘V due to virtual ground concept.
Input current to op-amp is zero. Hence at non-inverting terminal mode, we have
Input current to Op-Amp is zero. Hence at inverting terminal mode we have
Substituting the values and solving all the above equations, we get
The circuit connection of Op-Amp given in the figure (i) and (ii) represent
- a)Detectors for figure (i) and logarithmic amplifier for figure (ii)
- b)Logarithmic amplifier for figure (i) and detector for figure (ii)
- c)Logarithmic amplifier for both figure (i) and (ii)
- d)Detectors for both figure (i) and (iii)
Correct answer is option 'B'. Can you explain this answer?
The circuit connection of Op-Amp given in the figure (i) and (ii) represent
a)
Detectors for figure (i) and logarithmic amplifier for figure (ii)
b)
Logarithmic amplifier for figure (i) and detector for figure (ii)
c)
Logarithmic amplifier for both figure (i) and (ii)
d)
Detectors for both figure (i) and (iii)
|
Pie Academy answered |
The circuit connection of Op-Amp given in figure (i) and (ii) represents logarithmic amplifier for figure (i) and detector for figure (ii)
The correct answer is: Logarithmic amplifier for figure (i) and detector for figure (ii)
The correct answer is: Logarithmic amplifier for figure (i) and detector for figure (ii)
Given the circuit below, calculate the output voltage of the non-inverting Op-Amp summer:
- a)9V
- b)4V
- c)8V
- d)6V
Correct answer is option 'A'. Can you explain this answer?
Given the circuit below, calculate the output voltage of the non-inverting Op-Amp summer:
a)
9V
b)
4V
c)
8V
d)
6V
|
|
Vedika Singh answered |
The output voltage of a non-inverting Op-Amp is.
= 9V
The correct answer is: 9V
= 9V
The correct answer is: 9V
A certain Op-Amp has bias currents of 50μA and 49.3μA. The input offset current is (in nA).
Correct answer is '700'. Can you explain this answer?
A certain Op-Amp has bias currents of 50μA and 49.3μA. The input offset current is (in nA).
|
Akanksha Choudhary answered |
The input offset current is the different of input bias currents i.e.
The correct answer is: 700
The correct answer is: 700
If a certain Op-Amp has a mid range open-loop gain of 200000 and a unity-gain frequency of 5MHz, the gain bandwidth product in Hz is :
Correct answer is '5000000'. Can you explain this answer?
If a certain Op-Amp has a mid range open-loop gain of 200000 and a unity-gain frequency of 5MHz, the gain bandwidth product in Hz is :
|
|
Saanvi Roy answered |
gain bandwidth product = frequency at which gain is unity
= 5 MHz
= 5000000 Hz
The correct answer is: 5000000
= 5 MHz
= 5000000 Hz
The correct answer is: 5000000
The circuit shown in the figure can be used as a
- a)high pass filter
- b)low pass filter
- c)differentiator
- d)Integrator
Correct answer is option 'B,D'. Can you explain this answer?
The circuit shown in the figure can be used as a
a)
high pass filter
b)
low pass filter
c)
differentiator
d)
Integrator
|
|
Jayant Mishra answered |
The given circuit is
It is an integrator and low-pass filter.
The correct answers are: low pass filter, Integrator
It is an integrator and low-pass filter.
The correct answers are: low pass filter, Integrator
The certain inverting amplifier has a closed loop gain of 25. The Op-Amp has an open-loop gain of 100000. If another Op-Amp with an open loop gain of 200000 is substituted in the configuration, the closed loop gain is?
Correct answer is '25'. Can you explain this answer?
The certain inverting amplifier has a closed loop gain of 25. The Op-Amp has an open-loop gain of 100000. If another Op-Amp with an open loop gain of 200000 is substituted in the configuration, the closed loop gain is?
|
Hrishikesh Verma answered |
The closed loop gain depends on the configuration and open loop gain is completely independent of it. Hence even if open loop gain is changed, the closed loop gain remains at 25 with the same configuration.
The correct answer is: 25
The correct answer is: 25
The Common Mode Rejection Ratio (CMRR) of a differential amplifier using an operational amplifier is 100 dB. The output voltage for a differential input of is 2V. The common mode gain is (in dB)
Correct answer is '0.1'. Can you explain this answer?
The Common Mode Rejection Ratio (CMRR) of a differential amplifier using an operational amplifier is 100 dB. The output voltage for a differential input of is 2V. The common mode gain is (in dB)
|
|
Niti Mukherjee answered |
Where Ad = differential voltage gain
= large signal voltage gains
= Common mode voltage gain
The CMRR is generally expressed is dB and denoted by
The correct answer is: 0.1
The input and output impedance of the amplifier in given figure, is, (The Op-amp data sheet gives Zin = 2MΩ, ZOut = 75Ω and Aol = 200000).
- a)17.4GΩ; 8.6mΩ
- b)8.6GGΩ; 17.4mΩ
- c)7.4GΩ; 3.6mΩ
- d)7.4GΩ; 8.6mΩ
Correct answer is option 'A'. Can you explain this answer?
The input and output impedance of the amplifier in given figure, is, (The Op-amp data sheet gives Zin = 2MΩ, ZOut = 75Ω and Aol = 200000).
a)
17.4GΩ; 8.6mΩ
b)
8.6GGΩ; 17.4mΩ
c)
7.4GΩ; 3.6mΩ
d)
7.4GΩ; 8.6mΩ
|
|
Jayant Mishra answered |
The attenuation B, of the feedback circuit is
= [1 + 200000(0.0435)](2MΩ)
= 17.4GΩ
= 8.6Ω
The correct answer is: 17.4GΩ; 8.6/mΩ
= [1 + 200000(0.0435)](2MΩ)
= 17.4GΩ
= 8.6Ω
The correct answer is: 17.4GΩ; 8.6/mΩ
A certain Op-Amp has an open loop voltage gain of 1,00,000 and a common mode gain of 0.2. The CMRR is :- a)500000
- b)114 dB
- c)22.8 dB
- d)100000
Correct answer is option 'A,B'. Can you explain this answer?
A certain Op-Amp has an open loop voltage gain of 1,00,000 and a common mode gain of 0.2. The CMRR is :
a)
500000
b)
114 dB
c)
22.8 dB
d)
100000
|
|
Jayant Mishra answered |
Aol = 100000
Acm = 0.2
Expressed in decibels
CMRR =
= 114dB
The correct answers are: 500000, 114 dB
Acm = 0.2
Expressed in decibels
CMRR =
= 114dB
The correct answers are: 500000, 114 dB
Common-mode rejection ratio is given by- a)
- b)
- c)
- d)
Correct answer is option 'A,B'. Can you explain this answer?
Common-mode rejection ratio is given by
a)
b)
c)
d)
|
|
Jayant Mishra answered |
The correct answers are:
How many PN junctions is/are present in a bipolar junction transistor?
Correct answer is '2'. Can you explain this answer?
How many PN junctions is/are present in a bipolar junction transistor?
|
Pooja Choudhury answered |
A bipolar junction transistor (BJT) contains 2 PN junctions. If the transistor is NPN type then it contains a P-type semiconductor sandwiched between two N-type semiconductor. If the transistor is PNP type then it contains a N-type semiconductor sandwiched between two P-type semiconductor. In both the cases there are two PN junctions.
The output V02 (at the end of third Op-Amp) of the below circuit is :
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The output V02 (at the end of third Op-Amp) of the below circuit is :
a)
b)
c)
d)
|
|
Vedika Singh answered |
Similarly
The correct answer is:
The midrange open-loop gain of an Op-Amp :- a)extends from the lower critical frequency to the upper critical frequency
- b)extends from 0Hz to the upper critical frequency
- c)rolls off at 20 dB/decade beginning at 0Hz
- d)both (b) and (c)
Correct answer is option 'B'. Can you explain this answer?
The midrange open-loop gain of an Op-Amp :
a)
extends from the lower critical frequency to the upper critical frequency
b)
extends from 0Hz to the upper critical frequency
c)
rolls off at 20 dB/decade beginning at 0Hz
d)
both (b) and (c)
|
|
Jayant Mishra answered |
It extends from 0Hz to the upper critical frequency.
The correct answer is: extends from 0Hz to the upper critical frequency
The correct answer is: extends from 0Hz to the upper critical frequency
The Op-Amp has the following parameters; Aol = 50000; Zin = 4MΩ and Zout = 50Ω. The values of input impedance, output impedance and close loop voltage gains are :
- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
The Op-Amp has the following parameters; Aol = 50000; Zin = 4MΩ and Zout = 50Ω. The values of input impedance, output impedance and close loop voltage gains are :
a)
b)
c)
d)
|
|
Vedika Singh answered |
The feedback attenuation B is
zero for all practical purposes. The closed loop voltage gain is.
The correct answers are: Zout = 980mΩ, Acl(t) = -100
In order to obtain a solution of the differential equation
involving voltage V(t) and V1 an Op-Amp circuit would require at least.- a)one Op-Amp adder
- b)two Op-Amp adder
- c)one Op-Amp integrator
- d)two Op-Amp integrator
Correct answer is option 'A,D'. Can you explain this answer?
In order to obtain a solution of the differential equation
involving voltage V(t) and V1 an Op-Amp circuit would require at least.
involving voltage V(t) and V1 an Op-Amp circuit would require at least.
a)
one Op-Amp adder
b)
two Op-Amp adder
c)
one Op-Amp integrator
d)
two Op-Amp integrator
|
|
Jayant Mishra answered |
Since, it is a two degree linear differential equation, hence, two integrators are required, also one adder to calculate the sum of outputs.
The correct answers are: two Op-Amp integrator , one Op-Amp adder
The correct answers are: two Op-Amp integrator , one Op-Amp adder
The natural semiconductor LH0063C has a slew rate of- a)1400V/µs
- b)6000V/µs
- c)500V/µs
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
The natural semiconductor LH0063C has a slew rate of
a)
1400V/µs
b)
6000V/µs
c)
500V/µs
d)
None of the mentioned
|
Chirag Verma answered |
National semiconductor LH0063C has a slew rate of 6000V/µs. Generally, the practical op-amp is available with slew rate from 0.1V/µs to well above 1000V/µs.
The data of Op-Amp shown in figure is.
Zin = 2MΩ, ZOut = 75Ω and AoI = 200000, then the correct matching pairs is.- a)Input impedance = +17.4GΩ
- b)Output impedance = +28.6MΩ
- c)Close loop voltage gain = +125.0
- d)Attenuation of the feedback = +0.0435
Correct answer is option 'A,D'. Can you explain this answer?
The data of Op-Amp shown in figure is.
Zin = 2MΩ, ZOut = 75Ω and AoI = 200000, then the correct matching pairs is.
Zin = 2MΩ, ZOut = 75Ω and AoI = 200000, then the correct matching pairs is.
a)
Input impedance = +17.4GΩ
b)
Output impedance = +28.6MΩ
c)
Close loop voltage gain = +125.0
d)
Attenuation of the feedback = +0.0435
|
|
Pie Academy answered |
The attenuation, B, of the feedback circuit is
= [1 + (200000)(0.0435)](2MΩ)
= 17.4 GΩ
= 8.6mΩ
The correct answers are: Attenuation of the feedback = +0.0435, Input impedance = +17.4 GΩ
= [1 + (200000)(0.0435)](2MΩ)
= 17.4 GΩ
= 8.6mΩ
The correct answers are: Attenuation of the feedback = +0.0435, Input impedance = +17.4 GΩ
Phase shifts through an Op -Amp is caused by.- a)the gain Roll-off
- b)the internal RC circuit
- c)negative feedback
- d)the external RC circuit
Correct answer is option 'B'. Can you explain this answer?
Phase shifts through an Op -Amp is caused by.
a)
the gain Roll-off
b)
the internal RC circuit
c)
negative feedback
d)
the external RC circuit
|
|
Jayant Mishra answered |
The internal RC circuit provides the phase shift through any amplifier
The correct answer is: the internal RC circuit
The correct answer is: the internal RC circuit
If the voltage gain for each input of a summing amplifier with a 4.7 kΩ feedback resistor is unity, the input resistors must have a value of (in kΩ?
Correct answer is '4.7'. Can you explain this answer?
If the voltage gain for each input of a summing amplifier with a 4.7 kΩ feedback resistor is unity, the input resistors must have a value of (in kΩ?
|
|
Pragati Sharma answered |
Voltage gain for each
The correct answer is: 4.7
The correct answer is: 4.7
The inverting input terminal of an operational amplifier (Op-Amp) is shorted with output terminal apart from being grounded. A voltage signal V1 is applied to the noninverting input terminal of the Op-Amp. Under this configuration the Op-Amp function is as :- a)voltage to current converter
- b)an oscillator
- c)an open loop inverter
- d)a voltage follower
Correct answer is option 'D'. Can you explain this answer?
The inverting input terminal of an operational amplifier (Op-Amp) is shorted with output terminal apart from being grounded. A voltage signal V1 is applied to the noninverting input terminal of the Op-Amp. Under this configuration the Op-Amp function is as :
a)
voltage to current converter
b)
an oscillator
c)
an open loop inverter
d)
a voltage follower
|
|
Jayant Mishra answered |
The lowest gain can be obtained from a non-inverting amplifier with feedback 1.
When the non-inverting amplifier is configured for unit gain, it is called a voltage follower because the output voltage is equal to and in phase with the input. In other words, in the voltage follower, the output follows the input. To obtain the voltage follower from non-inverting amplifier of figure 1. simply open R1 and short R
The correct answer is: a voltage follower
When the non-inverting amplifier is configured for unit gain, it is called a voltage follower because the output voltage is equal to and in phase with the input. In other words, in the voltage follower, the output follows the input. To obtain the voltage follower from non-inverting amplifier of figure 1. simply open R1 and short R
f
The resulting circuit is shown is figure 2. In this figure, all the output voltages are feedback into the inverting terminal of the Op-Amp ; consequently the gain of the feedback is 1.The correct answer is: a voltage follower
What is the minimum voltage required to make the PN junction of a real silicon transistor in forward biased?
Correct answer is '0.7'. Can you explain this answer?
What is the minimum voltage required to make the PN junction of a real silicon transistor in forward biased?
|
|
Pooja Choudhury answered |
0.7 volts is the minimum voltage required to make the PN junction of a real silicon transistor in forward biased. This 0.7 volt potential difference makes the PN junction between base and emitter terminal in forward biased.
The frequency at which the open loop gain is equal to 1 is called. - a)The upper critical frequency
- b)The notch frequency
- c)The cut-off frequency
- d)The unity-gain frequency
Correct answer is option 'D'. Can you explain this answer?
The frequency at which the open loop gain is equal to 1 is called.
a)
The upper critical frequency
b)
The notch frequency
c)
The cut-off frequency
d)
The unity-gain frequency
|
Gitanjali Iyer answered |
Unity-gain frequency
The unity-gain frequency, also known as the gain crossover frequency, is the frequency at which the open-loop gain of a feedback system is equal to 1. It is an important parameter in the analysis and design of control systems.
Open-loop gain
Before delving into the unity-gain frequency, it is crucial to understand the concept of open-loop gain. In a control system, the open-loop gain is the gain of the system without any feedback. It represents the amplification or attenuation of the input signal by the system without any corrective action.
Definition
The unity-gain frequency is defined as the frequency at which the magnitude of the open-loop gain is unity, or 1. It is the frequency at which the output signal of the system is equal in magnitude to the input signal, resulting in a gain of 1.
Significance
The unity-gain frequency is a critical parameter in control systems because it determines the bandwidth and stability of the system. It represents the frequency at which the system transitions from a high-gain region to a low-gain region.
Frequency response
The frequency response of a control system is a plot of the gain and phase shift of the open-loop transfer function as a function of frequency. It provides valuable insights into the behavior of the system at different frequencies.
Frequency at gain crossover
At the unity-gain frequency, the open-loop gain of the system is equal to 1. This means that the output signal is equal in magnitude to the input signal. In terms of the frequency response, the gain crossover occurs at this frequency.
Application in stability analysis
The unity-gain frequency is used in stability analysis of control systems. It helps determine the stability margins and the frequency range within which the system remains stable. The gain margin and phase margin are calculated based on the proximity of the crossover frequency to the critical point of -180 degrees phase shift.
Conclusion
In summary, the unity-gain frequency is the frequency at which the open-loop gain of a control system is equal to 1. It is an important parameter in stability analysis and determines the bandwidth of the system. Understanding the unity-gain frequency helps in the design and analysis of control systems.
The unity-gain frequency, also known as the gain crossover frequency, is the frequency at which the open-loop gain of a feedback system is equal to 1. It is an important parameter in the analysis and design of control systems.
Open-loop gain
Before delving into the unity-gain frequency, it is crucial to understand the concept of open-loop gain. In a control system, the open-loop gain is the gain of the system without any feedback. It represents the amplification or attenuation of the input signal by the system without any corrective action.
Definition
The unity-gain frequency is defined as the frequency at which the magnitude of the open-loop gain is unity, or 1. It is the frequency at which the output signal of the system is equal in magnitude to the input signal, resulting in a gain of 1.
Significance
The unity-gain frequency is a critical parameter in control systems because it determines the bandwidth and stability of the system. It represents the frequency at which the system transitions from a high-gain region to a low-gain region.
Frequency response
The frequency response of a control system is a plot of the gain and phase shift of the open-loop transfer function as a function of frequency. It provides valuable insights into the behavior of the system at different frequencies.
Frequency at gain crossover
At the unity-gain frequency, the open-loop gain of the system is equal to 1. This means that the output signal is equal in magnitude to the input signal. In terms of the frequency response, the gain crossover occurs at this frequency.
Application in stability analysis
The unity-gain frequency is used in stability analysis of control systems. It helps determine the stability margins and the frequency range within which the system remains stable. The gain margin and phase margin are calculated based on the proximity of the crossover frequency to the critical point of -180 degrees phase shift.
Conclusion
In summary, the unity-gain frequency is the frequency at which the open-loop gain of a control system is equal to 1. It is an important parameter in stability analysis and determines the bandwidth of the system. Understanding the unity-gain frequency helps in the design and analysis of control systems.
What should be the values of the component R and R2 such the frequency of the Wien bridge oscillator is 300Hz?
(given C = 0.01 μf and R1 = 12kΩ.
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
What should be the values of the component R and R2 such the frequency of the Wien bridge oscillator is 300Hz?
(given C = 0.01 μf and R1 = 12kΩ.
(given C = 0.01 μf and R1 = 12kΩ.
a)
b)
c)
d)
|
|
Vedika Singh answered |
The frequency of oscillation f0 is exactly the resonant frequency of balanced Wien bridge and is given by 
Assuming that the resistor are equal in value, and the capacitors are equal is value in the reactive lag of the wien bridge. At the frequency, the gain required for sustained oscillation is given by
The correct answer is:
Assuming that the resistor are equal in value, and the capacitors are equal is value in the reactive lag of the wien bridge. At the frequency, the gain required for sustained oscillation is given by
The correct answer is:
Figure shows a practical integrator with 
If a step (dc) voltage of +3V is applied as input for 
( f in second), the output voltage is.
- a)a ramp function of - 6V
- b)a step function of -12V
- c)a ramp function of - 4V
- d)a ramp function of -1 5V
Correct answer is option 'C'. Can you explain this answer?
Figure shows a practical integrator with If a step (dc) voltage of +3V is applied as input for ( f in second), the output voltage is.
If a step (dc) voltage of +3V is applied as input for ( f in second), the output voltage is.a)
a ramp function of - 6V
b)
a step function of -12V
c)
a ramp function of - 4V
d)
a ramp function of -1 5V
|
|
Pie Academy answered |
= -4V
The correct answer is: a ramp function of -4V
The low-pass active filter shown in the figure has a cut-off frequency of 2kHz and a pass band gain of 1.5. The values of the resistors are :
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The low-pass active filter shown in the figure has a cut-off frequency of 2kHz and a pass band gain of 1.5. The values of the resistors are :
a)
b)
c)
d)
|
|
Vedika Singh answered |
At f = fcc (cut o ff frequency at which gain falls to 
of its low frequency value)
(i) Pass-Band gain = Amax = 1.5.
Hence,
The correct answer is:
of its low frequency value)(i) Pass-Band gain = Amax = 1.5.
Hence,
The correct answer is:
The circuit shown is based on ideal operational amplifiers. It acts as a
- a)subtractor
- b)buffer amplifier
- c)adder
- d)divider
Correct answer is option 'B'. Can you explain this answer?
The circuit shown is based on ideal operational amplifiers. It acts as a
a)
subtractor
b)
buffer amplifier
c)
adder
d)
divider
|
|
Pie Academy answered |
A buffer amplifier (sometimes simply called a buffer) is one that provides electrical impedance transformation from one circuit to another, with the aim of the signal source being unaffected by (“buffered from”) whatever currents that the load may produce.
The correct answer is: buffer amplifier
The correct answer is: buffer amplifier
In one of the following circuits, negative feedback does not operate for a negative input. Which one is it ? The Op-Amps are running from supplies.- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In one of the following circuits, negative feedback does not operate for a negative input. Which one is it ? The Op-Amps are running from supplies.
a)
b)
c)
d)
|
|
Jayant Mishra answered |
Since diode does not conduct in reverse bias hence, the circuit given in option (c) does not operate for a negative feedback.
The correct answer is:
The correct answer is:
The transistor gain of the circuit shown in figure is given by
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The transistor gain of the circuit shown in figure is given by
a)
b)
c)
d)
|
|
Vedika Singh answered |
A and B can be measured at the same potential. From figure;
or
Also
Combing (i) and (ii)
The correct answer is:
or
Also
Combing (i) and (ii)
The correct answer is:
The circuit shown below represent a non-inverting integrator. For high frequency input (ω → 
) impedance in ohms is?
Correct answer is '0'. Can you explain this answer?
The circuit shown below represent a non-inverting integrator. For high frequency input (ω → ) impedance in ohms is?
) impedance in ohms is?|
|
Athul Chaudhary answered |
Input impedance is given as 
For (ω →
) input impedance → 0
The correct answer is: 0
For (ω →
) input impedance → 0The correct answer is: 0
The output voltage of a certain Op-Amp appears as shown in figure in response to a step input. What is the slew rate (in V/μs)
Correct answer is '18'. Can you explain this answer?
The output voltage of a certain Op-Amp appears as shown in figure in response to a step input. What is the slew rate (in V/μs)
|
|
Sinjini Nair answered |
The output goes from the lower to the upper limit. Since, this response is not ideal, the limit are taken at the 90% points, as indicated. So the upper limit is + 9Vand the lower limit is - 9V. The slew rate is 
= 18 V/μs
The correct answer is: 18
= 18 V/μs
The correct answer is: 18
A bistable multi vibrator with a saturation voltage 15Vis shown in the diagram. The magnitude of the positive and negative thresholds at the inverting terminal for which the multi vibrator will switch to the other state are in volts (up to 2 decimal places).
Correct answer is '0.45'. Can you explain this answer?
A bistable multi vibrator with a saturation voltage 15Vis shown in the diagram. The magnitude of the positive and negative thresholds at the inverting terminal for which the multi vibrator will switch to the other state are in volts (up to 2 decimal places).
|
|
Swara Reddy answered |
The threshold voltage is 
where
The correct answer is: 0.45
where
The correct answer is: 0.45
An averaging amplifier has five inputs. Ratio 
must be?
Correct answer is '0.2'. Can you explain this answer?
An averaging amplifier has five inputs. Ratio must be?
must be?|
|
Nilotpal Basu answered |
Now to average 5 inputs, Rf should be
The correct answer is: 0.2
The output voltage for the scaling adder in figure shown below is (in volts)
Correct answer is '-8.84'. Can you explain this answer?
The output voltage for the scaling adder in figure shown below is (in volts)
|
|
Anirban Kapoor answered |
The output voltage is
The correct answer is: -8.84
The difference between the bandwidth of the amplifiers given below when both the Not answered Op-Amps have an open loop gain of 100dB and a unity-gain bandwidth of 3MHz (in KHz) is :
Correct answer is '19.5'. Can you explain this answer?
The difference between the bandwidth of the amplifiers given below when both the Not answered Op-Amps have an open loop gain of 100dB and a unity-gain bandwidth of 3MHz (in KHz) is :
|
|
Lekshmi Deshpande answered |
For non-inverting amplifier, closed loop gain is
For the inverting amplifier, the closed loop gain is.
Using the absolute value of Acl the closed-loop bandwidth is
Therefore difference in their bandwidth is
= 63.8 - 44.3
= 19.5 kHz
The correct answer is: 19.5
For the inverting amplifier, the closed loop gain is.
Using the absolute value of Acl the closed-loop bandwidth is
Therefore difference in their bandwidth is
= 63.8 - 44.3
= 19.5 kHz
The correct answer is: 19.5
The output voltage for the summing amplifier shown in figure is : (in volts)',
Correct answer is '-7'. Can you explain this answer?
The output voltage for the summing amplifier shown in figure is : (in volts)',
|
|
Tarun Singh answered |
The correct answer is: -7
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