I am using the following circuit for two purposes:

  • As a voltage divider, to reduce DC sensor output to a range suitable for my ADC.
  • As a lowpass filter, to remove high frequency noise from the signal.

CircuitLab Schematic 3gmsz3afv7rm

It works great, but my choice of component values has largely been a matter of trial and error, so I have a few questions:

  1. What is this type of circuit called? Is it a low-pass filter, a voltage divider or both?
  2. What is the equation for \$f_c\$ (cutoff frequency) for this circuit and how is it derived?
  3. What is the output impedance \$Z_{out}\$ of the circuit in terms of \$R_1\$, \$R_2\$ and \$C_1\$?
  4. If the design requires a specific \$f_c\$, \$Z_{out}\$ and DC \$\frac{V_{out}}{V_{in}}\$, is it possible to derive simple equations for \$R_1\$, \$R_2\$ and \$C_1\$ in terms of these?

3 Answers 3

  1. Not that I'm aware of, call it a filtered voltage divider :)

  2. \$fc=\frac{(R_1+R_2)\sqrt{(10^{3/10}-1)}}{2\pi*R_1*R_2*C_1}\approx \frac{(R_1+R_2)}{2\pi*R_1*R_2*C_1} \$

  3. \$R_1//R_2//C_1 \$

  4. There are other considerations to take into account. In your case for instance, the conversion speed (the time it takes to charge the ADC's internal sampling capacitor) which would have an impact on the max series resistance, max Vin voltage and so on.

  • \$\begingroup\$ How is tau \$τ \$ calculated? With \$R_1\$, \$R_2\$ or both? \$\endgroup\$
    – puncher
    Jan 10 at 8:45
  • \$\begingroup\$ I think it's useful to mention that Vin, R1 and R2 can be converted to the Thevenin equivalent circuit of just Vth = Vin * (R1/(R1 + R2)) and Rth = R1 || R2. Then you only have one resistor and you can apply the simple low-pass RC filter equations. \$\endgroup\$
    – gbmhunter
    Feb 16 at 22:42

The answers to 4) are:

\$R_1 = \frac{V_{in} R_{out}}{V_{out}}\$

\$R_2 = \frac{V_{in} R_{out}}{V_{in} - V_{out}}\$

\$C_1 = \frac{1}{2 \pi f_c R_{out}}\$

Where \$R_{out}\$ is the DC output impedance (\$ \omega = 0\$) and \$f_c\$ is the low-pass cutoff frequency.


1) I call this a voltage divider with a filter cap. There's no special name for it as far as I know

2) \$f_c = \frac{1}{2\pi * (R_1//R_2) * C_1}\$

Where \$R_1//R_2\$ is the value you get when \$R_1\$ and \$R_2\$ are in parallel

3) can be answered by learning some basic network theory. This is explained in any textbook about electronics or electronic networks.

4) yes, by learning what is mentioned under 3)


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