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I saw this circuit while going through some circuits and have been trying to solve it since. I get confused at multiple points while solving for the current.

  1. Since there is a buffer for feedback, does that mean that there is no current going from the Vout+ to this branch?
  2. Going by a technical datasheet , as seen here, I tried solving for Vout+ and Vout-. However the datasheet suggests that Beta1 or Beta2 is 0 when there is no feedback. When resistance is 0, doesn't mean that the feedback is essentially shorted and not open?
  3. I have solved for V (which I guess was pretty easy) but I don't know how to proceed further. V = Vin- (1 + R2/R1) - R2/R1 Vin+ PS: NOT A HOMEWORK ASSIGNMENT Can someone help?

Btw, this was not a homework assignment. I am just really bad at analyzing analog circuits :|

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  • \$\begingroup\$ If the buffer is assumed to be an ideal op-amp, then there would indeed be zero current consumed by both inputs. \$\endgroup\$ – Roger Rowland Apr 11 '15 at 8:17
  • \$\begingroup\$ Great. That's actually where the problem starts for me. Ideally, assuming no current is being fed back to the input, how can I solve the circuit? \$\endgroup\$ – Mathews_M_J Apr 11 '15 at 14:30
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Hint 1: the buffer has zero input current, therefore R3/RL/R3 form a simple voltage divider across Vo+ and Vo-. So you can express V in terms of Vo+

Hint 2: I assume that Vocm is connected to reference ground as it doesn't feature in the circuit. Therefore the differential output voltage, Vod = [(Vo+) - (Vo-)] = 2Vo+

Hint 3: using the input/output relationship for the differential op-amp, Vod = A Vid, where A>>1, you can derive an equation for V in terms of Vi-, Vi+, R1 and R2

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