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I have this circuit:

enter image description here

It's a basic op-amp that V_out will be like:

$$V_{out} = -V_{in} \frac{R_2}{R_{in}}$$

but I can't find V_{in} and R_{in}.

Simulating, I know that, if R1=R2=R3, then V_out = V_1 + V_2, but what happens when R1 is not the same than R2 or R3?

Thanks!

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Your circuit is an example of a summing amplifier.

You might find this resource useful for understanding more about them, and armed with the proper term, Google will be a good friend :)

Note that your \$V_{in}\$ will always be 0, unless you've driven the op amp into saturation somehow. Since the non-inverting (+) input is grounded, the inverting input becomes a so-called "virtual ground". So your equation is not correct for this configuration.

Your experimental results are in the right direction. I think you'll find that the proportion of \$R_3\$ to \$R_1\$ and \$R_2\$ determines a scaling factor for the output.

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The Vin in your diagram does not correspond with the Vin in the formula: -

enter image description here

Vin (as shown in your diagram) is the virtual earth (or summing point) of the op-amp and it acquires a voltage that is exactly the same as the voltage on the non-inverting input theoretically. So Vin in your diagram is 0V - it has to be 0V in a theoretical op-amp. In practical op-amps it's still a very small number, maybe a few millivolts different to 0V (0V on the non-inverting input).

Regards your formula, this circuit would better apply: -

enter image description here

In the above read R\$_F\$ as R2 in the formula. It isn't a summing amp but it's straightforward to convert the formula you have to include an extra input: -

enter image description here

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  • \$\begingroup\$ And what happens if R1 is not the same as R2? \$\endgroup\$ – Unnamed Sep 25 '15 at 13:50
  • \$\begingroup\$ If R1 is twice R2 then the output will still be a sum but the voltage (call it A) that inputs via R1 would have to be twice as big to make it a true sum of A+B. \$\endgroup\$ – Andy aka Sep 25 '15 at 13:53
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1) In your schematic, eliminate V2 and R1, just for now. Then $$V_{out} = -V_{in} \frac{R_3}{R_2}$$ In other words, V1 is the input, and the resistor connected to it is Rin.

2) Before you try running that, I'd advise changing V3 and V4 to something lower than 10k volts. Like a thousand times less. Try 15.

3) With R3 at 10k, and R2 at 1k, and V1 at 5 volts, your nominal output will be 50 volts. This is not exactly feasible with 15 volt supplies. And more than about 22 volts will kill a 741.

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Since V1 and V2 are both 5 volts, you can replace them with a single 5 volt supply, and then R1 and R2 will be connected in parallel, so their total resistance will be:

$$ Rt =\frac{R1 \times R2}{R1+R2} \text{ ohms} $$

So your drawing, with R1 and R2 reduced to a single resistor, the opamp supplies reduced from plus and minus 10000 volts to plus and minus 10 volts, and Vin located properly, should look something like this:

enter image description here

Next, an opamp's job is to do whatever it has to to force both input terminals to be at the same voltage and, in this case, since the + (non-inverting) input is hard-wired to ground, the output will swing to whatever voltage is needed to make the - terminal go to zero volts.

That voltage will be:

$$ Vout = -Vin \times\frac{R2}{R1} = -5V \times \frac{10k\Omega}{500\Omega} = -100 \text{ volts}$$

The problem is that since the supplies are at plus and minus 10 volts, the opamp's output will go as negative as it can, so it'll rail somewhere short of - 10 volts.

The solutions are to either get a high voltage opamp and high voltage supplies, to decrease Vin to the point where Vout is at a reasonable voltage, or to change the ratio of R2 to R1 so that with Vin at 5 volts Vout will be at a reasonable voltage.

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