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In my project I'd like to check the 20 contacts of a particular relay type. A simulator power every contact with 12 VDC (max) and this goes to LMxxx comparator logic that drive some BJT transitors connected to LED bi-color for visual check. The same voltage need to be scaled down gradually (based on the contact resistance) to 5 VDC (max) for Attiny85 ADC, because I would like to show the Ohm contact value on a LCD screen.

How to get a precise voltage translation?

Here a schematic example of the circuit (SW1 is like relay contact):

circuit

Example:

  • 12V -> 5V
  • 10V -> 4.1(6)V
  • 8v -> 3.(3)V
  • 6V -> 2.5 V
  • 4V -> 1.(6)V

and so on.

Thanks in advance.

EDIT 1

Many thanks for your help and time.

The voltage divider solution would be good but the resistance values, obtained by calc., are different to commercial ones and hard to achieve (with precision). We can combine them but there is still a slightest inaccuracy.

divider

Searching on web I found this solution with an op-amp with Ao set to 0.41(6). op_amp_0416 Resistance values are a little closer to commercial ones but there is still a slightest measurement inaccuracy.

Since I'm only an electronic apprentice what do you think about it? There could be other solutions you might suggest me please?

Thanks again in advance.

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  • \$\begingroup\$ What's wrong with a resistor divider? \$\endgroup\$
    – Justme
    Commented Jul 25, 2023 at 8:49

1 Answer 1

Reset to default
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So, split the 500Ω into 292Ω and 208Ω in series and take the voltage from the midpoint.

The voltage will be 5/12 of the voltage at the lower terminal of SW1.

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    \$\begingroup\$ Just to note with this solution you might have better luck sourcing 1% (E96) resistors in the real world which would be 280Ω and 200Ω instead \$\endgroup\$
    – Mukira
    Commented Jul 25, 2023 at 10:05
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    \$\begingroup\$ @Mukira Or add a trimpot for calibration. Or use multiple E96 1% or 0.1% resistors. Accuracy is unspecified by OP, but probably it will be potentially off by a couple percent without adjustment because of the 5V rail tolerance (which is likely used as the ADC reference). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 25, 2023 at 10:07
  • \$\begingroup\$ Thanks @SpehroPefhany for the reply. I hadn't thought to use the divider like that. However I wonder if there could be other solutions because the resistors are not precise and some values got by calc. aren't available, also to have some resolution in the divider I have to raise the current value and this implies bigger resistors (1W or 2W). If you have any new thoughts, please let me know. \$\endgroup\$
    – philfs
    Commented Jul 27, 2023 at 7:56

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