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I use Wheatstone bridge as a dual voltage divider to feed both inverting and non-inverting inputs of differential op-amp: enter image description here Theoretically voltage can be calculated using voltage divider expression: Va = V1*(R2/(R2+R1) = 7.5 mV. Following the same reasoning Vb = 6.0 mV. When I use LTSpice to calculate these voltages for the bridge alone without op-amp, everything is fine. However, when I connect bridge to op-amp circuit, Va = 6.5 mV and Vb = 4.8 mV. If LTSpice doesn't give erroneous results, how should I calculate voltages Va and Vb?

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  • \$\begingroup\$ The voltages at Wheatstone bridge are correct when you do not load the points Va and Vb. You then add the opamp + circuit. Are points Va and Vb still unloaded or does the opamp + circuit draw some current? \$\endgroup\$ – Bimpelrekkie Jun 19 '19 at 8:22
  • \$\begingroup\$ R1,2,3,4 are unbalanced and too high relative to R5,6,7,8 and thus affects the gain of + and -ve signals with different gains. The DIff AMP impedance must use an INA (instr. Amp) for these values to be insensitive to source impedance. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 19 '19 at 8:23
  • \$\begingroup\$ @Bimpelrekkie, it is an exercise from a textbook. Theoretically op-amp input current is 0. \$\endgroup\$ – tenghiz Jun 19 '19 at 8:31
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    \$\begingroup\$ it is an exercise from a textbook That doesn't mean anything to me, what if the textbook shows a circuit with unexpected behavior to teach you how that affects the circuit's behavior. Never assume that a circuit "must work" because it is in a book. Theoretically op-amp input current is 0. I was not assuming that there was current flowing into the opamp's inputs. How about the current through R5, R6, R7 and R8 ? Are they zero? Why / why not ? Look at the circuit and THINK what happens when R5, R6, R7, R8 + opamp are there and when the're not. \$\endgroup\$ – Bimpelrekkie Jun 19 '19 at 8:35
  • \$\begingroup\$ @Bimpelrekkie, I(R8): -4.83844e-008 A, I(R7): -4.83921e-008 A, I(R6): -1.31965e-007 A, I(R5): -1.31973e-007 A. These currents definitely are not 0, because I have a current from voltage source which flows through bridge, R5 and R6 to output terminal of op-amp. \$\endgroup\$ – tenghiz Jun 19 '19 at 8:44
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Many thanks to @Andy aka for the hint.

Also, it would be useful to draw a common ground for the whole circuit. enter image description here In this case it would be obvious that R4 is in parallel with (R7 + R8). The Thévenin resistance for this branch is 37.5 kOhm and Vb = 4.8 Ohm. Then the voltage between R7 and R8 is 3.84 V. According to ideal op-amp model, the same volatge is between R5 and R7. enter image description here Now we can calculate Va: (Vs - Va)/20k = Va/30k + (Va - 3.84)/20k. ==> Va = 6.5 V.

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