Topic: Analog Circuits
The very low frequency gain of the low pass filter shown in the given figure is (Op-amp with input resistor 1 k$\Omega$ and feedback loop of 100 k$\Omega$ parallel with 0.01 $\mu$F capacitor)
- A. 10 dB
- B. 20 dB
- C. 30 dB
- D. 40 dB
Answer: D. 40 dB
Explanation: At very low frequencies, the feedback capacitor acts as an open circuit. The gain magnitude of this inverting amplifier is $A_{f} = |-\frac{R_{f}}{R_{1}}| = |\frac{100\text{k}}{1\text{k}}| = 100$. Converting to decibels: $20 \log_{10}(100) = 40 \text{ dB}$.