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I have a voltage divider, and the output goes to OP amp.

This is from my schoolwork.

-1 to +1 input | [R1] 1000 ohms | ---> output to op amp | [R2] 1000 ohms | 2v

How do I calculate output that goes to op amp? Regular voltage divider formula is Vout=Vin×(R2÷(R1+R2)) But lets say V1 is -1 and other end is 2v, what do i take for Vin? 3 volts?

Edit: Found the whole exercise enter image description here

V1 is 4v V2 is from -1v to 1v And where the ground symbol is, there should be 2v. What do I take for V2 in formula whem V2 is -1 or +1?

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  • \$\begingroup\$ Could you add a schematic? Your ASCII art is a bit difficult to understand. There's a button up on top of the editor pane that allows you to add a schematic. \$\endgroup\$ – uint128_t Jan 24 '16 at 17:14
  • \$\begingroup\$ It would be more clear what you're asking if you included a schematic. But I think what you are looking for is \$V_o=V_- + (V_+-V_-)\frac{R_2}{R_1+R_2}\$ which is the formula for the output from a voltage divider connected between voltages \$V_-\$ and \$V_+\$. \$\endgroup\$ – The Photon Jan 24 '16 at 17:16
  • \$\begingroup\$ There is no R2 in your schematic, so it's not clear what's the connection from the schematic to your formulas. \$\endgroup\$ – The Photon Jan 24 '16 at 18:42
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It would be more clear what you're asking if you included a schematic.

But I think what you are looking for is \$V_o=V_- + (V_+-V_-)\frac{R_2}{R_1+R_2}\$ which is the formula for the output from a voltage divider connected between voltages \$V_-\$ and \$V_+\$.

You can derive this formula yourself using the principle of superposition.

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