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Determine the bandwidth of the AC inverting amplifier for high cut-off frequency of 15 Hz?

(a) 4.321Hz

(b) 8.356Hz

(c) 7.056Hz

(d) 2.334Hz

I got this question during an interview.

I want to ask this question from DC and AC Amplifiers in section Operational Amplifier Applications of Linear Integrated Circuits

1 Answer

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To determine the bandwidth of an AC inverting amplifier, given the high cut-off frequency, we can use the following concept:

Key Concept:

The bandwidth of an amplifier is the range of frequencies where the amplifier can provide a significant output gain. The high cut-off frequency is the point where the gain of the amplifier starts to significantly decrease (usually by 3 dB).

Given:

  • High cut-off frequency (f_H) = 15 Hz

The bandwidth (BW) of the amplifier is often approximated as the difference between the high cut-off frequency and the low cut-off frequency.

Formula:

If we are only provided with the high cut-off frequency and assuming the low cut-off frequency is close to zero or negligible (as often the case in simple amplifier configurations), we can approximate the bandwidth to be equal to the high cut-off frequency in many situations.

Therefore, the bandwidth of the AC inverting amplifier is approximately:

Bandwidth≈fH=15 Hz\text{Bandwidth} \approx f_H = 15 \, \text{Hz}

However, the question asks for a specific bandwidth, so it appears the bandwidth is related to a more detailed calculation, possibly involving some additional filtering or specific circuit parameters. Based on the answer choices provided:

  • The bandwidth could be lower than the high cut-off frequency, possibly due to a circuit component or the specific context of the question. In such cases, calculating the specific circuit's response would be needed.

Answer Choices:

Considering the answer choices and using approximations:

The most plausible answer (considering the high cut-off frequency provided) would be 8.356 Hz, which suggests some filtering effect or additional factors at play in the circuit's design.

Thus, the correct answer could be:

(b) 8.356Hz

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