# Find Vref of the level shifter when a offset is removed 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).

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Solution: -22*1.8+23V_ref=0

V_ref=1.7V

Please explain this equation.

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

• 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? Mar 29, 2020 at 19:08
• This circuit will also invert your sensor signal. Is this intended?
– Jens
May 22 at 21:42

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 $V_{ref}$.