I'm building a front end for a data logging project where I need to measure the voltage of a solar PV string. The max voltage I want to design for is 600V.

My question is: when choosing resistors R1 and R2 (see schematic below), do I need both of them to be rated for 600V? In other words, do I design for the scenario where the connection between R2 and GND is broken, thus exposing R2 to 600V (since no current flow through R2 means that there is no voltage dropped by R1).

The problem is that the only through-hole 10k resistors I can find that meet the voltage requirements are only accurate to within +/-5%, which makes them pretty useless for use in an ADC voltage divider. See search results on element14.


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ You said no current flows through R2 if it loses the ground connection- So what's the voltage across R2 then? \$\endgroup\$ – John D Sep 16 '16 at 18:52

I don't see the need for R2 to be rated at 600V. If you lose connection to GND, the circuit will be open and both resistors will have 0V across them.

As a side note, if you find it dificult to source high voltage resistors, you can always use several lower voltage resistors in series. It's a common practice in multimeter design.

enter image description here

  • \$\begingroup\$ Are you saying that when a resistor max voltage rating is 600 V that it is designed to drop a maximum of 600 V across itself when admitting current? Hmm that makes sense. For some reason I was thinking that a 600V rated resistor meant that it could only be exposed to a max of 600V at one of it's terminals (i.e. it would start short-circuiting if exposed to more than 600V). \$\endgroup\$ – macdonaldtomw Sep 16 '16 at 19:11
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    \$\begingroup\$ @macdonaldtomw I'm saying that without current, there is no voltage. V = I * R. If I = 0, then V is also 0. \$\endgroup\$ – Armandas Sep 16 '16 at 19:14
  • \$\begingroup\$ Think of it this way: If the ground connection at the "bottom" of R2 is lost, then that is like a very very high resistance has appeared in R2's place. The V drop along each R in a series circuit is proportional to the R. So very very close to all of the applied voltage is dropped across that very very high R, leaving almost 0 to be dropped across the actual resistors. \$\endgroup\$ – Jamie Hanrahan Sep 17 '16 at 0:09
  • \$\begingroup\$ @macdonaldtomw "600V at one of its terminals" - there is no such thing as "600V at one terminal", there is only "600V difference between one terminal and the other terminal" or "600V difference between one terminal and the down-pointing triangle in the circuit diagram". \$\endgroup\$ – immibis Sep 17 '16 at 1:34

R2 only has to be rated for low voltage. Even if the ground were to be broken or the resistor not populated, the current would be limited to a few hundred uA.

The MCU input protection network would clamp that to a few volts and, provided that didn't cause the power supply voltage to rise too much, nothing bad would happen. You can add a zener to ground (12V, say) or a low leakage diode to the supply if you are concerned.

Of course if R1 got shorted, many bad things would happen, but R2 might be of relatively little concern- the MCU would be fried, for starters. You need enough creepage distance and clearances around R1 to prevent problems and you have to ensure it can't get wet or whatever. The problem is not so much R2 but the damage and possible safety issues that could result if 600VDC showed up at the low voltage side.


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