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Is this possible?

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gcewing:

--- Quote from: rstofer on December 04, 2021, 08:21:49 pm ---I had to get a little more realistic about the resistors.

--- End quote ---
10k ohms seems realistic enough to me. Did you read it as 10 ohms?

fourfathom:
I did it in my head using 1 Ohm (and 2 Ohm) and an ideal opamp.  As long as you can add, divide by three, and multiply by two, it all works out (if you mind the polarities).

741:
By superposition, mentally short out one of the signal voltages (thus keeping only the series resistance).

For each signal, the 'bottom' resistor is effectively 0.5R.

Therefore, we can sum the two separate currents, which combine in the "0.5R" effective bottom resistor.

I1 = 1/(1 + 0.5),  I2 = 2/(1 + 0.5), I_SUM = (1+2)/(1.5) = 2

V+ = I.R = 2*0.5 = 1V

Now, The top part is simply a non-inverting amplifier, gain = 1 + (2/1) = 3


Finally, Vo = gain * V+ = 3*1 = 3V

fourfathom:

--- Quote from: 741 on December 05, 2021, 04:35:39 pm ---Finally, Vo = gain * V+ = 3*1 = 3V
--- End quote ---

I see that you also missed the polarity in that "1V" source.  I did too.

rstofer:

--- Quote from: gcewing on December 05, 2021, 05:57:35 am ---
--- Quote from: rstofer on December 04, 2021, 08:21:49 pm ---I had to get a little more realistic about the resistors.

--- End quote ---
10k ohms seems realistic enough to me. Did you read it as 10 ohms?

--- End quote ---

When I solved just the input voltage, I used 1 Ohm.  That's fine because the divider is based on ratios, not values.  1 Ohm, 1k Ohms, same result as long as they're all the same.

When I added the op amp, LTspice didn't think much of 2 Ohms and 1 Ohm (or at least it seemed that way at the time) so I multiplied everything up to 1k and 2k.

The values aren't important other than as a load on the op amp.  It's the ratios that matter.

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