simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Add a source to your circuit and click where it says "simulate this circuit". \$\endgroup\$
    – The Photon
    Mar 26, 2017 at 0:27

1 Answer 1


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.

  • \$\begingroup\$ 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 \$\endgroup\$
    – DerStrom8
    Mar 26, 2017 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.