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i am working on CAN_bus capacitance sniffer meaning take the 2 data line and without cutting the wire to recrate the signal. and i came across this configuration. i can’t undusted it, can anyone help me? enter image description here

enter image description here

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  • \$\begingroup\$ Welcome to EE.SE. That schematic is wrong and incomplete at best. With capacitors it could be a gyrator (active inductor), but normally the inputs are not so directly connected. \$\endgroup\$ – Sparky256 May 1 '18 at 7:58
  • \$\begingroup\$ Any news on where the circuit connects in the bigger picture of things? \$\endgroup\$ – Andy aka May 1 '18 at 7:58
  • \$\begingroup\$ hi thank you happy to be here i have the full schematic i send only the part i didn’t understand how is the 2 inputs are connected is there any current flows trow R43? \$\endgroup\$ – Daniel Brosh May 1 '18 at 8:11
  • \$\begingroup\$ added the full schematic on the original post \$\endgroup\$ – Daniel Brosh May 1 '18 at 8:15
  • \$\begingroup\$ Which parts can you understand? \$\endgroup\$ – Andy aka May 1 '18 at 9:25
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The first circuit, as drawn is a comparator. You can draw it like this, where the ground symbol refers to the mid-rail 1.65V pseudo ground created by U6A:

schematic

simulate this circuit – Schematic created using CircuitLab

There may be something wrong with your schematic.

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  • \$\begingroup\$ thank you all, i think that indeed something is not right with my schematics (it is done using reverse engineering of a working circuit) so if i have 5mv of a square wave input and i want to offset it by 1.6V all using one opamp what is the best way? \$\endgroup\$ – Daniel Brosh May 2 '18 at 5:28
  • \$\begingroup\$ AC couple it to a reference voltage (1.6V) and amplify from there. That's similar to what you've drawn but there should be another resistor from R26/inverting input to the 1.6V "ground". \$\endgroup\$ – Spehro Pefhany May 2 '18 at 5:58
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Answer to the first circuit diagram:

The shown circuit is a unity-gain amplifier.

  • What is the advantage of this circuit - if compared with the commonly known simple unity gainamp (opamp with 100% feedback)?

  • Answer: Bandwidth and stability margin !

In this configuration, you are allowed to use a broadband opamp which must not be compensated for unity-gain stability. Hence, it has a larger bandwidth. Stability is not a problem because there is not 100% feedback; the feedback factor is k=R43/(R43+R26) and can be made so small that the loop gain is in the stable region (The influence of R90 can be neglected due to its large value).

The trick of the circuit is that there is no direct dependency between closed-loop gain and loop gain (which determines stability proprties, like phase margin etc.).

Analysis (Calculation):

Vin=Vp; Vn=Vp (opamp ideal);

Therefore: Vn=Vin >>> no current through R43 and no current through R26.

Result: Vout=Vn=Vin >>> Vout/Vin=1

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