# Amplified Active Low Pass Filter with Capacitor in Place of Resistor

I'm an electrical engineering student learning about OP-Amp circuits, and I ran accross this circuit for an amplified low pass filter. I was wondering how the output of the circuit would be altered due to one of the resistors in the voltage divider that normally provides amplification being replaced with a capacitor. What is the output formula of this type of circuit. Is it the same as that of a standard amplified low pass filter? ## 1 Answer

That is more a non-inverting integrator, if you analyze the input, it is a voltage divider, the voltage at the non inverting input is

$$V_p(s)=\frac{\frac{1}{sC_1}}{R_1+\frac{1}{sC_1}}V_1(s)$$

which simplified gives

$$V_p(s)=\frac{1}{1+sR_1C_1}V_1(s) ...(1)$$

Now, R2 and C2 can be analyzed like a regular non-inverting amplifier

$$V_O(s)=\left(1+\frac{1}{sR_2C_2}\right)V_p(s)$$

Rewriting

$$V_O(s)=\left(\frac{1+sR_2C_2}{sR_2C_2}\right)V_p(s) ...(2)$$

If you substitute (1) in (2), you get $$V_0(s)=\left(\frac{1+sR_2C_2}{sR_2C_2}\right)\left(\frac{1}{1+sR_1C_1}\right)V_1(s) ...(3)$$

Now here comes the interesting part, if $$\C_1=C_2=C\$$ and $$\R_1=R_2=R\$$

You get

$$V_O(s)=\frac{1}{sRC}V_i(s)$$

The inverse Laplace transform gives

$$v_o(t)=\frac{1}{RC}\int v_i(t)$$

• Thank you so much! This clarifies things a lot. – Ashtionation Dec 2 '20 at 4:04
• The circuit is one of the well-known non-inverting integrator stages. It is called "Balanced-Time-Constants Integrator" (BTC)". However, it has the same problems as the classical MILLER integrator (missing DC feedback). – LvW Dec 2 '20 at 8:05