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I derive the Vout . (Vout=23Vref-22Vsensor)

Then, I want to find the value of Vref so that it can remove a 1.8V DC offset from a sensor output. That is, when Vsensor=Vreal + 1.8 and G is a constant gain,Vout=G(Vreal).


Solution: -22*1.8+23V_ref=0


Please explain this equation.

(I already knew the characteristic of op-amp, V+=v- and how to derive the vout in the op-amp)

  • \$\begingroup\$ Please quote or include all source material. Please explain were the 22 and 23 and 1.8 V comes from. You also have a 1.7 V in the question? \$\endgroup\$
    – skvery
    Mar 29, 2020 at 19:08
  • \$\begingroup\$ This circuit will also invert your sensor signal. Is this intended? \$\endgroup\$
    – Jens
    May 22 at 21:42

1 Answer 1


When it comes to op-amps, think "virtual ground". You want to have the same voltage at both terminals.

If you have \$V_{ref}\$ at the \$+\$ terminal, you need \$V_{ref}\$ at the \$-\$ terminal. Knowing that, you can work out the current flowing through \$R_f\$ in terms of \$V_{out}\$ and \$V_{ref}\$. You can also work out the current flowing through \$R_s\$ in terms of \$V_{sensor}\$ and \$V_{ref}\$.

Additionally for an ideal op-amp, no current flows into the \$+\$ or \$-\$.

That means any current flowing through \$R_f\$ must be equal but opposite to the current flowing through \$R_s\$. Knowing that, simply equate your two equations for \$I\$ and rearrange to get an equation for \$V_{ref}\$.

  • \$\begingroup\$ I add the equation of getting Vref from solution of the question but i still don't understand it. Can you explain according to the equation? Thank you very much. \$\endgroup\$
    – L michael
    May 6, 2018 at 10:59

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