Circuit Theory/Active Filters/Example90

Start with writing a node equation:

Active filter ... find transfer function

$i_{1}-i_{2}-i_{3}-i_{0}=0$

${\frac {V_{in}-V_{node}}{R_{1}}}-V_{node}sC_{1}-{\frac {V_{node}-V_{o}}{R_{3}}}-{\frac {V_{node}}{R_{2}}}=0$

But the last expression for i_o could have also been written using C₂. And we need to eliminate V_node. So our second equation is associated with the feedback resistor:

${\frac {V_{node}}{R_{2}}}-(0-V_{o}sC_{2})=0$

Now these two equations can be solved for V_o/V_in:

solve([(vin-vnode)/r1 - vnode*s*c1 - (vnode-vout)/r3 - vnode/r2, vnode/r2 - (0-vout*s*c2)],[vout,vin])

The transfer function is going to be the ratio of V_out/V_in so:

vout := -vnode/(c2*r2*s)
vin := (vnode*(r1 + c2*r1*r2*s + c2*r1*r3*s + c2*r2*r3*s + c1*c2*r1*r2*r3*s^2))/(c2*r2*r3*s);
vout/vin
-r3/(r1 + c2*r1*r2*s + c2*r1*r3*s + c2*r2*r3*s + c1*c2*r1*r2*r3*s^2)

Yields this:

$-{\frac {R_{3}}{R_{1}+(C_{2}R_{1}R_{2}+C_{2}R_{1}R_{3}+C_{2}R_{2}R_{3})s+C_{1}C_{2}R_{1}R_{2}R_{3}s^{2}}}$

Which looks like a low pass filter (inverted because of the op amp).