# Output voltage of this filter

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.

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

• Is the op-amp rail-to-rail on the output? Commented Apr 5, 2017 at 15:19
• I made a typo, it's OP284 and not OP280. I edited and added the datasheet. Commented Apr 5, 2017 at 15:32

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.

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