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I have searched for a few hours on the internet and in some circuit design books, but have been thusfar unable to find a circuit that produces absolute value around a number other than zero. Does anyone here know of such a circuit? An example IO would be: At Vin = 10, Vo = 10 At Vin = 9, Vo = 11 At Vin = 11, Vo = 11

I'm hoping to accomplish this with a single absolute value circuit, rather than subtracting 10 beforehand, taking the absolute value, and adding 10 or offsetting the result.

Thanks for any help

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  • \$\begingroup\$ Can you just take the normal absolute value circuit and offset its power supply? \$\endgroup\$ – Hearth Jul 31 '18 at 12:30
  • \$\begingroup\$ @Jake - Every component you add will introduce error. \$\endgroup\$ – Bort Jul 31 '18 at 12:37
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    \$\begingroup\$ What’s the real application? \$\endgroup\$ – PDuarte Jul 31 '18 at 12:49
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In general you can just take an absolute value circuit and substitute a 10V reference for whatever it uses for a zero reference (which may be a power supply or something else). You will have to take care that there is enough power supply headroom, of course.

Here is a modification of a standard circuit:

enter image description here

to do what you want with, an LTspice simulation:

enter image description here

The blue trace is the input, the green is the output. Where the output goes flat at the beginning and end, it is saturating near the positive rail (+20V in this case).

This particular circuit does need a negative supply, however the +10V reference input is high impedance so you may be able to use a voltage divider etc.

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Does anyone here know of such a circuit?

The circuit below can do what you want: -

enter image description here

You can set input voltage \$e_2\$ to be at 10 volts and A3 (an instumentation amplifier) can have its reference pin set to 10 volts (not shown in the circuit above).

\$Z_L\$ can be a resistor and it could equal R. The instrumentation amplifier can have gain set to unity by not fitting the gain resistor hence the equation becomes: -

$$e_O = |e_1 - e_2|+ e_2$$

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  • \$\begingroup\$ @Jake I've altered my answer - I didn't originally see that part. However, you will still need a 10 volt reference voltage. \$\endgroup\$ – Andy aka Jul 31 '18 at 12:40

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