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I designed the following circuit so that the output voltage is Vout = 0.625V. But after actually building the circuit and testing I get Vout = 0.95V.

enter image description here

So obviously I was wrong but I can't manage to get the computation right, can you put me on the good track? How to get this Vout = 0.95V?

OP284 Datasheet

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    \$\begingroup\$ Is the op-amp rail-to-rail on the output? \$\endgroup\$
    – winny
    Commented Apr 5, 2017 at 15:19
  • \$\begingroup\$ Add link to datasheet to your question. And then read it. \$\endgroup\$ Commented Apr 5, 2017 at 15:25
  • \$\begingroup\$ I made a typo, it's OP284 and not OP280. I edited and added the datasheet. \$\endgroup\$
    – Kevin P
    Commented Apr 5, 2017 at 15:32

1 Answer 1

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I expect an output of about 1.01 V for your circuit.

Taking KCL at the op-amp's inverting input, you have

$$I({\rm R3}) = I({\rm R2})$$

where both currents are taken as flowing from right to left. Assuming the inverting input is driven to be equal to the non-inverting input by negative feedback and applying Ohm's law, we have

$$\frac{V_o - 1.65}{390}=\frac{1.65-3.3}{1000}$$

where \$V_o\$ is the voltage at the op-amp's output terminal.

Solving this, \$V_o=1.0065\ {\rm V}\$.

Possibly your 3.3 V source has enough equivalent output resistance to disturb this equation and give the 0.95 V you measured. But in any case, that is much closer to the expected value than 0.625 V is.

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  • \$\begingroup\$ Solving your equation I find 2.3V and not 1.0065V. Could you elaborate? \$\endgroup\$
    – Kevin P
    Commented Apr 5, 2017 at 15:58
  • \$\begingroup\$ Try the math again, I get \$V_o=(1.65-3.3)\frac{390}{1000}+1.65 = 1.0065\$. \$\endgroup\$
    – The Photon
    Commented Apr 5, 2017 at 16:00
  • \$\begingroup\$ Tested and approved. It works fine! \$\endgroup\$
    – Kevin P
    Commented Apr 5, 2017 at 20:28

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