An op-amp has an open loop gain of 105 and an open-loop upper cut-off frequency of 10 KHZ. If this op-amp is connected as an amplifier with a closed-loop gain of 100, calculate the new upper cut off frequency(in Hz)?Correct answer is '10000'. Can you explain this answer?

Question

Answer

Analysis:
To determine the new upper cut-off frequency, we need to consider the closed-loop gain of the amplifier and the open-loop gain and cut-off frequency of the op-amp.

Step 1: Calculate the closed-loop gain:
Given that the closed-loop gain is 100, we can use the formula for an inverting amplifier:

$$
A_{\text{closed-loop}} = -\frac{R_f}{R_{\text{in}}}
$$

where $R_f$ is the feedback resistor and $R_{\text{in}}$ is the input resistor.

Using the given closed-loop gain of 100, we can rearrange the formula to solve for $R_f$ as:

$$
R_f = -A_{\text{closed-loop}} \times R_{\text{in}}
$$

Assuming $R_{\text{in}}$ is 1 (for simplicity), we have:

$$
R_f = -100 \times 1 = -100
$$

Step 2: Calculate the new upper cut-off frequency:
The open-loop gain of the op-amp is given as 105, and the open-loop upper cut-off frequency is 10 KHz.

The gain-bandwidth product (GBW) of an op-amp is a constant and is given by:

$$
\text{GBW} = A_{\text{open-loop}} \times f_{\text{cut-off}}
$$

where $A_{\text{open-loop}}$ is the open-loop gain and $f_{\text{cut-off}}$ is the open-loop cut-off frequency.

From the given values, we can rearrange the formula to solve for the GBW as:

$$
\text{GBW} = 105 \times 10 \times 10^3 = 1.05 \times 10^6 \text{ Hz}
$$

Step 3: Calculate the new upper cut-off frequency:
To find the new upper cut-off frequency, we divide the GBW by the new closed-loop gain:

$$
f_{\text{cut-off, new}} = \frac{\text{GBW}}{A_{\text{closed-loop}}} = \frac{1.05 \times 10^6}{100} = 10^4 \text{ Hz} = 10000 \text{ Hz}
$$

Therefore, the new upper cut-off frequency is 10000 Hz.

Answer

Answer

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