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I want to build a voltage reference based on LM4040 (A series 0.1%), the output voltage is switchable with a selector switch:

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming that the op amp have input offset of ~100 uV and resistors are 1% but hand picked to match the values;

  • How can I calculate the output error?
  • Which parameters affects the accuracy (only resistors error,op amp input offset voltage and temperature)?
  • Is there any other downside to this configuration other than lowering the accuracy?
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  • \$\begingroup\$ So you specs are select 1 of 3 inputs with n=1,2,3 x 2.048 V with 0.1% resistors and 100uV Vio Op Amp?? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 at 19:24
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Yes the specs are those that I mentioned in the question, though it may vary a bit when building it... \$\endgroup\$ – ElectronSurf Apr 24 at 19:26
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    \$\begingroup\$ Look at the temperature coefficients of those resistors (in their datasheet). Higher precision 0.1% resistors aren't that expensive and may have lower tempco. \$\endgroup\$ – Brian Drummond Apr 24 at 19:26
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    \$\begingroup\$ Why not? Always decide what you will accept and expect ( write a design spec) not just trial and error \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 at 19:31
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    \$\begingroup\$ Note that as an alternative to buying precision resistors or trying to bin for a match there's also the option of buying TCR matched pairs. The important thing is the ratio, not the absolute tolerance. An example: digikey.ca/product-detail/en/vishay-beyschlag/… \$\endgroup\$ – Peter Apr 24 at 19:48
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enter image description here Simulation

Accuracy is only limited by stack up of identical resistor tolerance errors which can be matched to 0.1% 50 ppm/'C such as cheap resistor arrays in 10k or 100k range.

2 DIP switches select the feedback ratio to multiply Vref by 1,2,3,4.

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  • \$\begingroup\$ Thanks, how can I get the 1024 V from that configuration? again with a voltage divider? \$\endgroup\$ – ElectronSurf Apr 24 at 19:56
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    \$\begingroup\$ corrected ...Rev A \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 at 20:01
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    \$\begingroup\$ edit your question as Rev B. Change my design to rotary switch selecting open circuit, then resistor values R/1, R/2, R/3. then Vout=1.024 x N = N=1,2,3 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 at 20:16
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    \$\begingroup\$ Like this? it's great, thanks again. \$\endgroup\$ – ElectronSurf Apr 24 at 20:30
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    \$\begingroup\$ rotary switch here tinyurl.com/y994yjjx just click on it to advance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 24 at 20:35
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You can calculate the worst-case output error by adding up all the sources of error and assuming their direction adds.

Initial resistor error, temperature, time drift of resistors (also moisture load life), offset voltage of op-amps, temperature drift of op-amp Vos, op-amp gain vs. loading. Reference error and temperature coefficient. Noise of op-amps, resistors, power supply and reference may also be considered. Voltage coefficient of op-amps and change of reference voltage with supply change. You can buy matched resistor networks so the temperature coefficients and values are matched to each other but they're a bit expensive.

Downside? Power consumption, possibly more noise. Hand picking 1% resistors is not a very good solution compared to buying precision resistors in the first place. If you absolutely must, use matched resistors from the same lot. The calculation won't show the improvement however.

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  • \$\begingroup\$ There's A LOT of parameters to consider! based on your experience do you think building a voltage reference like that is practical? any suggestions for a better cheap solution? \$\endgroup\$ – ElectronSurf Apr 24 at 19:48
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    \$\begingroup\$ For 0.5% accuracy in an office environment I think it’s no problem, just buy good parts and do a quick calculation. \$\endgroup\$ – Spehro Pefhany Apr 24 at 19:49

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