# how can I calculate the cutoff frequency for this low pass filter?

simulate this circuit – Schematic created using CircuitLab

• Add a source to your circuit and click where it says "simulate this circuit". Mar 26, 2017 at 0:27

While there is not a compact analytic solution, some insight may derived from an open circuit time constant analysis.

In this method, we first compute the time constant associated with each capacitance:

$$\tau_1 = 68 ~\text{pF} \cdot 0 = 0~s\\ \tau_2 = 2.2 ~\mu\text{F} \cdot \left(500\Omega || \left(5\text{k}\Omega+10\text{k}\Omega\right)\right) \approx 1.06~\text{ms} \\ \tau_3 = 100 ~\mu\text{F} \cdot \left(\left(500\Omega+5\text{k}\Omega\right)||10\text{k}\Omega\right) = 0.355~\text{s}$$

The overall time constant is the sum of these values:

$$\tau_{OCT} = \tau_1+\tau_2+\tau_3 = 0.356~\text{s}$$

The predicted -3dB frequency is therefore:

$$f_{OCT} = \frac{1}{2\pi\tau_{OCT}} = 0.447~\text{Hz}$$

This is within 0.2% of the value simulated in CircuitLab (0.446 Hz). Note that -3dB frequency is almost entirely set by the time constant of C3. In other words, there is a dominant pole.

• I think you just gave him the answer to a homework question. It probably would have been better to provide an explanation and formulas, but left actual numbers out of it Mar 26, 2017 at 1:46