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Question Bank (3 marks) · Q10

Q.Explain how an operational amplifier can be used as a differentiator?

Karnataka PUCTextbookLong· 3mImportance★★★★★est
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[!TLDR]

With a capacitor at the input and a feedback resistor, Vo=−RfC dVindtV_o = -R_f C\,\dfrac{dV_{in}}{dt}.

Circuit: the input VinV_{in} is applied through a capacitor CC to the inverting (−) input, a feedback resistor RfR_f connects output to that node, and the (+) input is grounded.

Derivation: The inverting node is a virtual ground. The capacitor current is i=C dVindti = C\,\dfrac{dV_{in}}{dt}; this current flows through RfR_f, so Vo=−iRfV_o = -i R_f. Therefore

Vo=−RfC dVindtV_o = -R_f C\,\frac{dV_{in}}{dt} …

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