Electronic – Notch filter ECG

This circuit doesn't work for me either and I've bread borded it. It appears that the Q is so high it the notch depth amplitude disappears. Try reducing the feedback it worked for me! Put a 10K pot between your op-amp output and ground, buffer it with another op-amp (non inverting) and use the output of this opamp as your feedback point. the 10K pot will give your a variable Q.

Besst regards Ray

Assuming an ideal op-amp, you have \$v_1=v_2\$ (as pointed out in a comment by LvW). Also note that your second equation is wrong because there is a current into or out of the output of the op-amp, so you can't just add up the impedances of \$R_2\$ and \$C_2\$.

Introducing another unknown voltage \$v_x\$ at the output of the op-amp, you can write down three equations:

$$(v_i-v_1)sC_1-v_1/R_3=0\\ (v_i-v_1)/R_1+(v_x-v_1)/R_2=0\\ (v_i-v_o)/R_4-(v_o-v_x)sC_2=0$$

where \$v_i\$ and \$v_o\$ are the input and output voltages, respectively, and \$v_1\$ is the voltage at both inputs of the op-amp. These equations can be solved for the transfer function \$H(s)\$, i.e., for the ratio \$v_o/v_i\$. With \$C=C_1=C_2\$ and \$R=R_3=R_4\$ and \$R_1=R_2\$ you get

$$H(s)=R\frac{s^2+\frac{1}{R^2C^2}}{s^2+s\frac{2}{RC}+\frac{1}{R^2C^2}}\tag{1}$$

From the numerator of \$(1)\$ you immediately get for the notch frequency

$$\omega_0=\frac{1}{RC}\tag{2}$$

Note that as long as you choose \$R_1=R_2\$, the actual value of these resistors doesn't show up in the transfer function.