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I came across this schematic design of a driven-leg circuit, designed in order to enhance the performance of an ECG amplifier. enter image description here

As far as i can see, the solution provided here is a little bit different from the normal approach, where two R5/2 resistances would be used and the common node voltage would be gathered from their common connection. The above presented approach would make sense to me if two buffers were used in order to repeat the signal and then divide it on the two 47K resistors. But why use such a configuration, where the non-inverting input is tied to the output? I have never seen such thing and by simulating this i got a strange result.

Any thoughts? Thanks.

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  • \$\begingroup\$ I believe you'll find that U3A/B are used to produce a square wave drive for U3C. While you don't show the rest of the circuit, I'd suggest that U3 is using a 3V supply. \$\endgroup\$ – Jack Creasey Jul 31 '19 at 17:37
  • \$\begingroup\$ If it helps, here is the full circuit people.ece.cornell.edu/land/courses/ece5030/ECG/hands.pdf . My concern is -what does such a circuit even do? (U3A/U3B) While having a positive feedback from the output, is this supposed to act also as a buffer? Thank you for your reply. \$\endgroup\$ – Steven Jul 31 '19 at 18:46
  • \$\begingroup\$ It's not a buffer, it's a level detector. \$\endgroup\$ – Jack Creasey Jul 31 '19 at 19:10
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The above presented approach would make sense to me if two buffers were used in order to repeat the signal and then divide it on the two 47K resistors. But why use such a configuration, where the non-inverting input is tied to the output?

It's a faulty circuit - U3A and U3B should not have positive feedback - I would say that the inputs on both those op-amps need to be swapped around for this to work.

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  • \$\begingroup\$ Thank you for your answer! I'm a noob when talking electronics, but i have never seen such a configuration- namely U3A/U3B acting as what, buffers? In theory, what should happen if it's wired as it is- wire from IN+ to OUT? Shouldn't it act as a comparator? \$\endgroup\$ – Steven Jul 31 '19 at 19:02
  • \$\begingroup\$ The output will be hard against one supply rail or the other so definitely, it’s not meant to be wired that way. They should act as unity gain buffers. \$\endgroup\$ – Andy aka Jul 31 '19 at 19:09

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