Question
Download Solution PDFFor an op-amp having a slew rate SR = 5V/ms, what is the maximum closed-loop voltage gain that can be used when the input signal varies by 0.2V in 10 ms.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Slew rate
It is the maximum rate of change of output voltage for all possible input signals.
It is defined as:
\(SR = {\left| {\frac{{d{V_0}}}{{dt}}} \right|_{max}}\frac{{volts}}{{\mu sec}}\)
Calculation:
Given slew rate is 5 V/ms and the rate of change of the input signal is 0.2 V in 10 ms
\(\frac{{d{V_0}}}{{dt}} = \frac{{5V}}{{ms}}\)
\(\frac{{d{V_{in}}}}{{dt}} = \frac{{0.2}}{{10\;ms}}\)
Required voltage gain is
\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{\frac{{d{V_0}}}{{dt}}}}{{\frac{{d{V_{in}}}}{{dt}}}}\)
\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{d{V_0}}}{{dt}} \times \frac{{dt}}{{d{V_{in}}}}\)
\(\frac{{{V_o}}}{{{V_{in}}}} = \frac{5}{{1\;ms}} \times \frac{{10\;ms}}{{0.2}}\)
\(\frac{{{V_0}}}{{{V_{in}}}} = \frac{{50}}{{0.2}}\)
Last updated on May 28, 2025
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