To determine the maximum closed-loop voltage gain for the given op-amp, we first need to understand the relationship between the slew rate and the maximum rate of change of the output voltage.
The slew rate \( \text{SR} \) of an operational amplifier is defined as the maximum rate of change of the output voltage per unit time, typically expressed in \(\text{V}/\mu\text{s}\). The formula for the slew rate is given by:
\(\text{SR} = \frac{\Delta V_{\text{out}}}{\Delta t}\)
In the question, the slew rate \( \text{SR} \) is provided as \(4 \text{ V}/\mu\text{s}\). The change in input voltage \( \Delta V_{\text{in}} \) is given as \(0.1 \text{ V}\) over a time interval \( \Delta t \) of \(10 \, \mu\text{s}\).
We are tasked with finding the maximum closed-loop voltage gain \( A_v \). The voltage gain affects how the input voltage change is amplified at the output. Therefore,
\(\Delta V_{\text{out}} = A_v \times \Delta V_{\text{in}}\)
Substituting the known values into the slew rate formula, we have:
\(4 = \frac{A_v \times 0.1}{10}\)
Solving for the voltage gain \( A_v \):
\(4 = \frac{A_v \times 0.1}{10} \\[5pt] A_v = \frac{4 \times 10}{0.1} \\[5pt] A_v = 400\)
The maximum closed-loop voltage gain that can be used under the given conditions is therefore 400.
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