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Question Bank (3 marks) · Q9

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

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

With a capacitor in the feedback path, Vo=−1R1C∫Vin dtV_o = -\dfrac{1}{R_1 C}\int V_{in}\,dt.

Circuit: the input VinV_{in} is applied through resistor R1R_1 to the inverting (−) input, a feedback capacitor CC connects output to that node, and the (+) input is grounded.

Derivation: The inverting node is a virtual ground, so the input current is i=Vin/R1i = V_{in}/R_1. This same current charges the capacitor, whose current is i=−C dVodti = -C\,\dfrac{dV_o}{dt}. Equating,

VinR1=−CdVodt\frac{V_{in}}{R_1} = -C\frac{dV_o}{dt}

Integrating, …

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