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
Best Answer
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.