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I have an analog voltage signal which is in the range [+2.1875 V ... +2.8125 V], centered around 2.5 V. I would like to scale and shift this signal so that it is mapped to the range [-3V ... +3V], and now centered around zero volts.

I am happy to use multiple op amps, but am constrained to using ± 12V power supplies. Is this possible?

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  • \$\begingroup\$ Not really a limitation since regulators exist. \$\endgroup\$
    – DKNguyen
    Feb 18, 2022 at 15:21
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    \$\begingroup\$ Yes absolutely. Given a +/-12V power supply you can produce voltages within the -12 to +12V. In this case you'd add a -2.5V DC offset and then amplify by 9.6X. \$\endgroup\$ Feb 18, 2022 at 15:21
  • \$\begingroup\$ Thank you - can you provide a sketch / schematic of the arrangement in an answer? \$\endgroup\$
    – teeeeee
    Feb 18, 2022 at 15:24
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    \$\begingroup\$ Does this answer your question? Change range of a DC signal without changing the offset with op amps The other guy's answer is better though even though mine was marked at best. My approach is just easier to follow but less efficient circuit. This asks to not change the offset but it's all the same anyways since offset fiddling is needed to keep it the same anyways due to amplification/scaling. \$\endgroup\$
    – DKNguyen
    Feb 18, 2022 at 15:24
  • \$\begingroup\$ Possibly. I am looking to scale the signal and center it around zero though. Does this need to be first done with a subtraction of 2.5V, and then a second op amp for the scaling. Or can it be done with one op amp? An sketch with example resistors would be great... Thank you :) \$\endgroup\$
    – teeeeee
    Feb 18, 2022 at 15:29

1 Answer 1

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can it be done with one op amp?

If one op. amp. and one cheap voltage reference would be acceptable, the circuit below includes all the formulas, which were initially based on this TI app. note..

enter image description here

There is an implicit connection between RS, RK, RR and RG: RS must guarantee enough current goes through the TL431 reference, so, if RG is too small, the feedback resistor should be increased. If RR is too large, the current that enters the "ref pin" should be taken into account. Finally, for the same reason, if the current through RK is too large, RS should be decreased.

So, this is not (yet) fully automatic. If there is not enough current entering the K pin of the voltage reference, check the resistor values or increase the current through RS (set to 10 mA in the equations).

As stated in the comment, this circuit topology works for positive gain (max > min in both input and output) and negative shift output midpoint is lower than input midpoint.

These are the results for the selected input parameters:

enter image description here

Regarding the additional questions:

is there a reason you would choose to use a reference chip, rather than just a resistor network?

Accuracy, stability and more immunity to ripple/noise from Vcc. You could use a simple voltage divider from Vcc and buffer it using another op. amp. In fact, in the app. note I linked the solution with the voltage divider plus buffer is the suggested one. It is an option if relying on Vcc is enough for your requirements and you are using an IC with 2 op. amps.

should I be looking to buffer the offset voltage (your Vreg) using a unity gain follower in this situation?

This reference already has an internal amplifier so, there is no need unless you replace it by a simple voltage divider (see above).

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    \$\begingroup\$ @teeeeee Resistor networks aren't references. They are ratiometric to the supply voltage and change with it, and the supply voltage is not a reference. \$\endgroup\$
    – DKNguyen
    Feb 23, 2022 at 14:36
  • \$\begingroup\$ @teeeeee The reference already works in closed loop. You only need to guarantee it receives the minimum required current for regulation (see the paragraph about the resistors and the edit that followed the additional questions). In one sentence: you only need to reduce RS if needed, no buffer required. \$\endgroup\$
    – devnull
    Feb 23, 2022 at 15:12
  • \$\begingroup\$ When I go through the calculations, I see that Vreg needs to be 2.791 V. If I choose RR = 2.7 kΩ as you have done, then I need RK = 314 Ω. In practice, should I do this with a potentiometer? \$\endgroup\$
    – teeeeee
    Feb 24, 2022 at 11:26
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    \$\begingroup\$ If "standard" precision resistors will be used, fine tuning may be required, depending on the accuracy requirements, by the use of 2 trimpots (the multi-turn ones). The one that you mention and another for RG or RF to adjust the gain (which affects the offset). It may be easier to fine tune if instead of only trimpots, a series association with a resistor is used (e. g., for the example you mentioned, a 270 ohm resistor in series with a 100 ohm trimpot, instead of only a 470 ohm trimpot). \$\endgroup\$
    – devnull
    Feb 24, 2022 at 12:23

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