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GATE 2017 EC – Question 52

Analog Circuits · Op-amp Circuits · 2 marks · Numerical answer

The amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, $A_0 = 10^5$ V/V and an open-loop cut-off frequency, $f_c = 8$ Hz. The voltage gain of the amplifier at 15 kHz, in V/V, is ________.

A non-inverting amplifier with the input $V_i$ at the non-inverting terminal, $R_1 = 1\ \text{k}\Omega$ from the inverting terminal to ground, and a feedback resistor $R_2 = 79\ \text{k}\Omega$.

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Show answer and explanation

Correct answer: 44 to 45

Explanation

The closed-loop gain at low frequency is $1 + \frac{79}{1} = 80$. The gain-bandwidth product of the op-amp is $10^5 \times 8 = 8 \times 10^5$ Hz, so the closed-loop bandwidth is $\frac{8 \times 10^5}{80} = 10$ kHz. At 15 kHz the gain is $\frac{80}{\sqrt{1 + (15/10)^2}} = \frac{80}{1.803} = 44.4$.