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An LT-spice simulation suggests that the output is always near -15V, that is, saturation. Assuming linearity, I come up with only two linearly independent equations: $$I_{in}=\frac{Vin}{R}$$ $$I_{R3}=\frac{-Vout}{R3}$$

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    \$\begingroup\$ Is the bottom side of R2 at -15 V ? R2 is not in your equations. \$\endgroup\$ Jan 12, 2016 at 13:02
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    \$\begingroup\$ That circuit is a mess. If you saw this somewhere, explain the context, and be sure that you copied it right. Otherwise, if you dreamed this up in some hallucigen-induced stupor, forget about it and move on. Also keep in mind that simulators aren't designed to model parts correctly when used way off spec. In fact, real parts aren't predictable either when used way off spec. \$\endgroup\$ Jan 12, 2016 at 13:03
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    \$\begingroup\$ @OlinLathrop This is an exercise in my analogue electronics course. Obviously, it's just there to test our knowledge on this very subject. \$\endgroup\$ Jan 12, 2016 at 13:05
  • \$\begingroup\$ @FakeMoustache All I can see is that the bottom of R2 is at some voltage we presume to be the positive supply voltage. \$\endgroup\$ Jan 12, 2016 at 13:06
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    \$\begingroup\$ It is still unclear to me what pins 1, 4, 5 and 7 do. For any pins on Opamps other than inputs (marked + and -) and the output, clearly mark what pins these are for. Offset ? Bias setting ? Output enable ? Supply rails ?? \$\endgroup\$ Jan 12, 2016 at 13:25

1 Answer 1


I will make the following assumptions: pin 7 is the positive supply, and is connected to a positive supply V+ (but assumed, not shown), and pin 4 is connected to a negative supply V- (again, assumed but not shown.)

Then the two left-hand resistors provide a 1:1 divider to V+. To bring the - input to ground, current must balance, and $$\frac{V_{IN}}{R} + \frac{V+}{R} + \frac{V_{OUT}}{R} = 0$$ and since the R's drop out$$V_{OUT}= -(V_{OUT} + V+)$$


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