My question intersects with "What does this op-amp circuit do? (part of an ECG)", but I need some more detailed clarification on the opamp integrator of this right leg drive circuit as found in the LT1168-Datasheet, especially what calculation is behind C1, R1, R2 and the frequency response of that integrator…
If I understand correctly the integrator acts like a low pass filter because of the capacitor C1 in the feedback path that alters the output gain with increasing input frequency. I reckon "POLE AT 1kHz" is meant as the point where the integrator starts to attenuate the output signal with increasing input frequencies (and so, amplifying only the wanted common mode signal (50Hz/Europe)). Can you please explain how the "1kHz" calculation come about? I am confused by two resistors in the feedback path. What does R2 do?
Best Answer
An ecg is an instrumentation amplifier. The right leg drive amplifies the common mode noise (mains hum) and injects it into the patient body, so that it can be rejected by the instrumentation amplifier more easily.
I'm not sure if the 30K resistors play any role in setting the gain. If the midpoint of the 30K is considered ground, then the 12K has to be outside the feedback loop (otherwise it does not make sense). The attenuation did not change much from using a single 15K to ground (and the 12k inside the feedback loop). I checked this by doing a bode plot in LTspice. Attenuation is about 38dB till 10Hz and decreases to about 3dB at 1.5KHz (6dB at 1KHz for the 15K). Based on this it may be better to have the 12k inside the feedback loop, but it does not seem to make a big difference.