Operational Amplifier And 555 Timer A1 And A3

1. Voltage follower is also called as

Unity gain buffer

Non inverting amplifier

Buffer

Inverting amplifier

A voltage follower, also known as a unity gain buffer, is an electronic circuit that has a voltage gain of 1. It is called a buffer because it isolates the input from the output, preventing any loading effects. In other words, it replicates the input voltage at the output without amplifying or attenuating it. This is achieved by using an operational amplifier configured in a non-inverting amplifier configuration with a gain of 1. Therefore, the correct answer is unity gain buffer.

Explanation

A voltage follower, also known as a unity gain buffer, is an electronic circuit that has a voltage gain of 1. It is called a buffer because it isolates the input from the output, preventing any loading effects. In other words, it replicates the input voltage at the output without amplifying or attenuating it. This is achieved by using an operational amplifier configured in a non-inverting amplifier configuration with a gain of 1. Therefore, the correct answer is unity gain buffer.

2. Output produce by an op-amp is 180 degree phase shifted when input is applied to

Inverting terminal

Non Inverting terminal

Both inverting and non inverting terminals

Common mode

When an input is applied to the inverting terminal of an op-amp, the output produced is 180 degrees out of phase with the input. This means that when the input signal goes positive, the output signal goes negative, and vice versa. This phase shift is a characteristic of the inverting configuration of an op-amp, where the input signal is connected to the inverting terminal and the feedback is applied to the non-inverting terminal.

Explanation

When an input is applied to the inverting terminal of an op-amp, the output produced is 180 degrees out of phase with the input. This means that when the input signal goes positive, the output signal goes negative, and vice versa. This phase shift is a characteristic of the inverting configuration of an op-amp, where the input signal is connected to the inverting terminal and the feedback is applied to the non-inverting terminal.

3. Find the gain of the inverting amplifier shown below. Consider R1 = 1kΩ and RF = 10kΩ Assume op-amp is ideal.

-10

-10k

-100

-1

The gain of an inverting amplifier is determined by the ratio of the feedback resistor (RF) to the input resistor (R1). In this case, RF is given as 10kΩ and R1 is given as 1kΩ. The gain is calculated as the negative ratio of RF to R1, which is -10. Therefore, the correct answer is -10.

Explanation

The gain of an inverting amplifier is determined by the ratio of the feedback resistor (RF) to the input resistor (R1). In this case, RF is given as 10kΩ and R1 is given as 1kΩ. The gain is calculated as the negative ratio of RF to R1, which is -10. Therefore, the correct answer is -10.

4. Determine closed loop voltage gain of an non inverting amplifier if feedback resistor is 3 times input resistor

3

6

4

-3

The closed loop voltage gain of a non-inverting amplifier is determined by the ratio of the feedback resistor to the input resistor. In this case, since the feedback resistor is 3 times the input resistor, the closed loop voltage gain would be 3. However, the given answer is 4, which does not match the expected value. Therefore, the given answer is incorrect and there is no suitable explanation for it.

Explanation

The closed loop voltage gain of a non-inverting amplifier is determined by the ratio of the feedback resistor to the input resistor. In this case, since the feedback resistor is 3 times the input resistor, the closed loop voltage gain would be 3. However, the given answer is 4, which does not match the expected value. Therefore, the given answer is incorrect and there is no suitable explanation for it.

5. In given circuit Vin = 2 V, R1 = 1KΩ and R2 = 10KΩ.What is the output voltage of an amplifier

11 V

-22 V

22 V

-20 V

The output voltage of an amplifier can be calculated using the formula Vout = Vin * (1 + (R2/R1)). In this case, Vin is given as 2 V, R1 is 1KΩ, and R2 is 10KΩ. Plugging in these values, we get Vout = 2 * (1 + (10/1)) = 2 * 11 = 22 V. Therefore, the output voltage of the amplifier is 22 V.

Explanation

The output voltage of an amplifier can be calculated using the formula Vout = Vin * (1 + (R2/R1)). In this case, Vin is given as 2 V, R1 is 1KΩ, and R2 is 10KΩ. Plugging in these values, we get Vout = 2 * (1 + (10/1)) = 2 * 11 = 22 V. Therefore, the output voltage of the amplifier is 22 V.

6. In the application of OP-AMP as an amplifier _____________ feedback is used

Positive

No

Negative

Series

In the application of OP-AMP as an amplifier, negative feedback is used. Negative feedback is a technique where a portion of the output signal is fed back to the input with a phase inversion. This helps in stabilizing the amplifier's gain and reducing distortion. By using negative feedback, the amplifier can have a more accurate and linear response, making it suitable for various applications such as audio amplification and signal conditioning.

Explanation

In the application of OP-AMP as an amplifier, negative feedback is used. Negative feedback is a technique where a portion of the output signal is fed back to the input with a phase inversion. This helps in stabilizing the amplifier's gain and reducing distortion. By using negative feedback, the amplifier can have a more accurate and linear response, making it suitable for various applications such as audio amplification and signal conditioning.

7. Open loop gain of an IC 741 is

2*10^5

1

∞

0

The open loop gain of an IC 741 is 2*10^5. This means that when the IC 741 is operating in an open loop configuration, the gain of the amplifier is 2*10^5. This high gain allows the IC 741 to amplify small input signals to a much larger output signal.

Explanation

The open loop gain of an IC 741 is 2*10^5. This means that when the IC 741 is operating in an open loop configuration, the gain of the amplifier is 2*10^5. This high gain allows the IC 741 to amplify small input signals to a much larger output signal.

8. Pick all incorrect statements. Analog signals are A. Continuous. B. Easy to process compared to digital signals. C. More immune to Noise compared to digital signals. D. Easily reproducible when compared to digital signals.

A, B and D

B, C and D

B and C

B and D

Analog signals are continuous, making statement A correct. However, statement B is incorrect because digital signals are generally easier to process compared to analog signals. Statement C is also incorrect because digital signals are more immune to noise compared to analog signals. Lastly, statement D is incorrect because digital signals are easily reproducible compared to analog signals.

Explanation

Analog signals are continuous, making statement A correct. However, statement B is incorrect because digital signals are generally easier to process compared to analog signals. Statement C is also incorrect because digital signals are more immune to noise compared to analog signals. Lastly, statement D is incorrect because digital signals are easily reproducible compared to analog signals.

9. For first order butterworth high pass filter, calculate lower cut off frequency if R=15KΩ

and C = 0.01. Also calculate Avf if Rf = 10K Ω and R1 = 5kΩ.

1061 Hz,3

1061Hz,2

1061kHz,3

1061KHz,2

The lower cut-off frequency for a first-order Butterworth high pass filter is determined by the formula: f = 1 / (2πRC). Plugging in the given values of R = 15KΩ and C = 0.01, we can calculate the lower cut-off frequency as f = 1 / (2π * 15KΩ * 0.01) = 1061 Hz.

The Avf (voltage gain) for this filter is determined by the formula: Avf = 1 + (Rf / R1). Plugging in the given values of Rf = 10KΩ and R1 = 5KΩ, we can calculate the Avf as Avf = 1 + (10KΩ / 5KΩ) = 3.

Therefore, the correct answer is 1061 Hz, 3.

Explanation

The lower cut-off frequency for a first-order Butterworth high pass filter is determined by the formula: f = 1 / (2πRC). Plugging in the given values of R = 15KΩ and C = 0.01, we can calculate the lower cut-off frequency as f = 1 / (2π * 15KΩ * 0.01) = 1061 Hz.

The Avf (voltage gain) for this filter is determined by the formula: Avf = 1 + (Rf / R1). Plugging in the given values of Rf = 10KΩ and R1 = 5KΩ, we can calculate the Avf as Avf = 1 + (10KΩ / 5KΩ) = 3.

Therefore, the correct answer is 1061 Hz, 3.

10. Match the following

Input Impedance of an ideal Op-amp

Open loop voltage gain

output Impedance

Closed loop voltage gain

Slew Rate

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