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I am making a regular old voltage divider. The Req is 602k ohms which results from 6 or so resistors. But I was thinking that using 1M resistor or something on that order would be more cheap. I don't have a problem decreasing the current but decreasing the number of resistors would save cost.

The reason I didn't do this is because I don't see people do that very often with input voltages greater than 10. Maybe I am wrong, but I thought that this may be because the tolerance on many smaller resistors together makes a more precise voltage division as opposed to using bigger (and less) resistors with the same tolerances. I could do the math for that and get a better idea but I feel there would probably be another factor I am missing so I might as well just ask on here. For my application, the resistors have a Vin max of 100V. enter image description here

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    \$\begingroup\$ where does this requirement of 602k come from? \$\endgroup\$
    – user16222
    Jul 8, 2021 at 21:42
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    \$\begingroup\$ @Ahmed Please ask a specific question. \$\endgroup\$
    – Voltage Spike ♦
    Jul 8, 2021 at 21:50
  • \$\begingroup\$ So you want 30.1k/602k ratio, or 1:20? How precise the divider must be, i.e. what tolerance you need? \$\endgroup\$
    – Justme
    Jul 8, 2021 at 22:54
  • \$\begingroup\$ basically i was wondering if i can get away with something like one 570k resistor in the top part and one 30k resistor in the bottom. at 60V input, the output should be close to 3V +/- 1%. the output is being compared with a 3V Vref if that helps. \$\endgroup\$
    – Ahmed Anwer
    Jul 9, 2021 at 17:48

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I can see the 602k is used as a 20:1 Vac divider. perhaps from 230V rms or 350Vp.

Since the current is shared the voltage breakdown of 100V is marginal using half of the resistance considering transient spikes of 1kV are common and the risk of burnout is high.

Normally you would use at least 10x the nominal pp voltage and with protection and probably parts rated for 250V each.

But here R25 (324k) will be overstressed the most out of the 602k string.

It makes no sense to use small series R values when the tolerance of the largest exceeds the others. 1% of 324k = 3.24k is greater than R27=2.05k.

That would require calibration testing and choosing custom values. A better solution is a custom laser trimmed part with required breakdown voltage rating and tolerance for needed accuracy.

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  • \$\begingroup\$ oh thanks for pointing that out. i kind of chose the resistors randomly but it does not make sense to use resistors that are the tolerance values of a larger resistor. the 324k will have the most stress but does that mean it just needs to have the highest power rating to withstand the stress? \$\endgroup\$
    – Ahmed Anwer
    Jul 9, 2021 at 17:43
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    \$\begingroup\$ It would be very expensive to have a 1kV rated 2W resistor. So it is better to choose a voltage withstanding twice your peak voltage and more than twice the power dissipate so only 50 % temp rise to 100’C above ambient. Thus if using 100V 1/8W parts you need about 8 x 100k parts and with a shunt to scale down to your range with bias on AC. \$\endgroup\$
    – Tony Stewart EE75
    Jul 9, 2021 at 18:18
  • \$\begingroup\$ thank you that helps! \$\endgroup\$
    – Ahmed Anwer
    Jul 11, 2021 at 0:40
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Placing resistors in series does increase the resistance by the sum of the resistors. The tolerance also increases with each resistor making it less accurate. You can hand sort and get the starting resistance very close. Resistors have a maximum voltage rating and sometimes they use several because of the voltage. You are looking at semiconductor probably digital circuits. Check analog circuits then download some vacuum tube circuits and you will find a lot of resistors having more than 100V applied. Resistors also have a noise specification they are definitely not all the same. Am I correct in assuming this is a school problem?

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    \$\begingroup\$ How does the tolerance increase and makes it less accurate? If you take a hundred 1kohm resistors with +/- 1% tolerance, the worst case is you end up with 100kohm resistance with +/- 1% tolerance. \$\endgroup\$
    – Justme
    Jul 8, 2021 at 22:20
  • \$\begingroup\$ You are correct! I was thinking of something else. Thanks for the catch! \$\endgroup\$
    – Gil
    Jul 9, 2021 at 0:27
  • \$\begingroup\$ Very good overview of tolerance of combination of resistors here: electronics.stackexchange.com/questions/228345/… \$\endgroup\$
    – crj11
    Jul 9, 2021 at 0:35
  • \$\begingroup\$ not a school problem but a circuit for high voltage indication on a car. \$\endgroup\$
    – Ahmed Anwer
    Jul 9, 2021 at 17:39

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