I have a circuit that adds a DC offset to an AC signal. From what I know, C1 and R2 form a high-pass filter. Since the input signal frequency is 50Hz, I would like to reduce the cut-off frequency of the filter as much as possible.

Ceramic capacitors with high capacitance are difficult to find, so instead I'm using high-valued resistors (with a buffer just before the ADC connected to the output).

I want to know whether R1 will affect the cut-off frequency of the filter. I found this similar question, but the answer doesn't give any explanation.

P.S - I would also appreciate if anyone can give suggestions on what type of capacitor to use, since I read electrolytic capacitors cannot be used with AC voltage.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Simply calculate the cutoff frequency for both cases, with and without the resistor. What do you find ? It's only an RC circuit, this is quite basic. \$\endgroup\$ – Bimpelrekkie Mar 17 '16 at 15:46
  • \$\begingroup\$ That's the problem. I don't know how to calculate the cut-off frequency when both R1 and R2 are connected. \$\endgroup\$ – Hassaan Mar 17 '16 at 15:49
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    \$\begingroup\$ @M.Hassaan - Then you need to learn about Thevenin equivalent networks. This will let you learn about your two resistors. Then you should Google on "high pass filter", and learn about that. Now go study. \$\endgroup\$ – WhatRoughBeast Mar 17 '16 at 16:05
  • \$\begingroup\$ Isn't this a part of study? I know about Thevinin equivalent circuits and high pass filters. Please elaborate on how to use those concepts here. \$\endgroup\$ – Hassaan Mar 17 '16 at 16:13

The "trick" you lack may be using the superposition principle to "ignore" (temporarily) the DC source and consider it to be 0 V while studying only the AC source. Using the superposition principle, you can show that V_out consists of a DC component equal to approx. 1V plus an AC component which depends on Vin and the high-pass filter:

Without \$R_1\$ and the DC source, the cutoff frequency is given by : $$ \omega_c = \frac{1}{2\pi R_2 C_1} = 0.15 Hz $$ (This is a typical RC circuit)

Adding \$R_1\$ and the DC source, the circuit changes a bit, since you are adding a resistor parallel to \$R_2\$, the cutoff frequency is now $$ \omega_c = \frac{1}{2\pi (R_2//R_1) C_1} = 0.238 Hz $$ Since \$R_1//R_2\$ is lower than \$R_2\$, you increased the cutoff frequency, but it's still two decades lower than your expected 50 Hz so it shouldn't be an issue.


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