Divider and lowpass combined

Question

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.

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?

Answer 1

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.

Answer 2

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.

Answer 3

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)

Answer 4

I want to answer question 2 "What is the equation for fc (cutoff frequency) for this circuit and how is it derived?", specifically the derivation.

The corner frequency for a basic low-pass filter is \$f_c = \frac{1}{2\pi RC}\$.

We can turn the voltage-divider low-pass into this form using a Thevenin equivalent circuit by combining the elements in the blue dashed square.

By isolating this portion of the circuit, we do the Thevenin process and imagine another path exists between the nodes A and B. Now we have different paths and can use Kirchoff's laws to produce 3 equations that can be condensed into one that gives us the Thevenin equivalent voltage and resistance.

1. \$V_{in} - V_{AB} = I_uR_1\$ (drop across first resistor)
2. \$V_{AB} = I_1R_2\$ (drop across second resistor)
3. \$I_u = I_1 + I_{AB}\$ (Kirchoff's law)

Plugging equations 1 and 2 into 3 we get \$\frac{V_{in} - V_{AB}}{R_1} = \frac{V_{AB}}{R_2} + I_{AB}\$. Re-arranging this and solving for \$V_{AB}\$ as a function of \$V_{in}\$ and \$I_{AB}\$ we get

\$V_{AB} = \underbrace{\frac{R_2}{R_1 + R_2}V_{in}}_{\text{thev eq voltage, } V_T} - \underbrace{\left(R_1||R_2\right)}_{\text{thev eq resistance, } R_T}I_{AB}\$.

Re-drawing this we get the result we were looking for, a basic low-pass with only one resistor whose equivalent resistance is the parallel addition of the voltage divider resistors. Thus the corner frequency is \$f_c = \frac{1}{2\pi\left(R_1||R_2\right)C}\$.