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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\$ – Bimpelrekkie Jan 12 '16 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\$ – Olin Lathrop Jan 12 '16 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\$ – Emir Šemšić Jan 12 '16 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\$ – Emir Šemšić Jan 12 '16 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\$ – Bimpelrekkie Jan 12 '16 at 13:25
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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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