
The slew rate of an op-amp is found to be 10V/microsecond. A voltage change of 100V will produce a time delay of
(A) \[10micro\sec onds\]
(B) \[1micro\sec ond\]
(C) \[5micro\sec onds\]
(D) \[1\sec \]
Answer
233.1k+ views
Hint: In the given question, we have been given the slew rate of an op-amp. An op-amp is an operational amplifier; it is a linear device that serves the purpose of direct current amplification and conditions, filters and operates on the signal. We have been given the slew rate, which is the maximum rate of change of an operational amplifier’s output voltage with time. We have been given a value of the input voltage and have been asked to find out the corresponding time delay produced. Let us see the detailed solution given below.
Formula Used: \[time-delay=\dfrac{input-voltage}{slew-rate}\]
Complete step by step solution:
We have been given that
The input voltage to the operational amplifier \[\left( input-voltage \right)=100V\]
The slew-rate of the given operational amplifier \[\left( slew-rate \right)=10V/micro\sec ond\]
Now, the relation between the time delay, the input voltage and the slew rate of an operational amplifier is given below.
\[time-delay=\dfrac{input-voltage}{slew-rate}\]
Substituting the values of the terms given to us in the question, we get
\[\begin{align}
& time-delay=\dfrac{100V}{10V/micro\sec ond} \\
& \Rightarrow time-delay=10micro\sec onds \\
\end{align}\]
Hence we can say that the time delay produced by the operational amplifier will be \[10micro\sec onds\] .
Therefore, option (A) is the correct answer
Note:
If any student is not aware of the relation between the time delay, the input voltage and the slew rate of an amplifier, he/she can still find the correct answer by simply looking at the units of the quantities given to us in the question. The unit of the input voltage is volt and the unit of the slew rate is volt per microsecond. Hence if we divide the input voltage by the slew rate, the volt will get cancelled in the division and the microsecond will come to the numerator and hence we will get the answer in microseconds which is a unit of time or time-difference or delay.
Formula Used: \[time-delay=\dfrac{input-voltage}{slew-rate}\]
Complete step by step solution:
We have been given that
The input voltage to the operational amplifier \[\left( input-voltage \right)=100V\]
The slew-rate of the given operational amplifier \[\left( slew-rate \right)=10V/micro\sec ond\]
Now, the relation between the time delay, the input voltage and the slew rate of an operational amplifier is given below.
\[time-delay=\dfrac{input-voltage}{slew-rate}\]
Substituting the values of the terms given to us in the question, we get
\[\begin{align}
& time-delay=\dfrac{100V}{10V/micro\sec ond} \\
& \Rightarrow time-delay=10micro\sec onds \\
\end{align}\]
Hence we can say that the time delay produced by the operational amplifier will be \[10micro\sec onds\] .
Therefore, option (A) is the correct answer
Note:
If any student is not aware of the relation between the time delay, the input voltage and the slew rate of an amplifier, he/she can still find the correct answer by simply looking at the units of the quantities given to us in the question. The unit of the input voltage is volt and the unit of the slew rate is volt per microsecond. Hence if we divide the input voltage by the slew rate, the volt will get cancelled in the division and the microsecond will come to the numerator and hence we will get the answer in microseconds which is a unit of time or time-difference or delay.
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