In the schematic/picture below, what is the purpose of 2 serial resistors for voltage divider? Temperature, thermal runaway, stocks, prices, or something else?

Thank you.

schematic

simulate this circuit – Schematic created using CircuitLab

up vote 31 down vote accepted

It's typically done to meet reliability requirements for safety.

When operating from a hazardous high voltage, a circuit needs to have Single Point Of Failure (SPOF) protection to meet safety approvals such as CE. Specifically, a hazardous voltage is usually that above 50 Vac or 120 VDC, but the requirement is stated in the standards that the equipment must be approved to. It certainly applies to your 400 VDC here.

Designing for SPOF means that the effect of a failure of a single component will have on the circuit needs to be considered, for every component. For SPOF, 'failure' means that the component fails short-circuit or open-circuit. Components don't all fail this way in real life but this is how it is considered in SPOF. The circuit must not cause further hazards, such as fire, harm to people or over-rating of other components, when a single component has failed in this way.

Considering SPOF here, a single series resistor from 400 V could fail short-circuit and deliver 400 V across the 1 K resistor and the output. So two series resistors are used instead, for SPOF-level protection. If one fails short-circuit, the other must still be working since we are considering a single point of failure.

Each surviving resistor must be rated to the handle the full voltage and power it would then have to deal with. So here, you would need 1 M resistors rated for 400 V plus the tolerance of your supply plus a safety margin (500 V or higher?). And the power rating needs to be for the highest 400 V supply voltage across a single 1 M resistor and the 1 K, with derating. So lets look at 160 mW dissipation and use at least a 320 mW resistor e.g. 1/2 W.

Next, if the 1 K fails open-circuit, the 400 V through 2 M source impedance will be delivered to your output. So that needs to be considered also. You could use a second parallel resistor and make both 2 K. A failure of any of the four resistors you've now got will affect the potential divider output voltage, so that must be allowed for. If it's just detecting the presence of 400 V, suitable resistor values would let the output drive an NPN transistor or voltage comparator that'd work from any of the three output voltages caused by the three possible dividers (2M:1K normally, 1M:1K, 2M:2K). If you're trying to measure the 400 V, you could add a second and third identical divider circuit and put them through a majority voting circuit to identify the correct voltage (two of the three voltages nearly the same). There are different ways of doing these things but they all affect cost, peformance, space etc., so they need careful consideration.

This may not be the original reason that your circuit here has two series resistors, I don't know the application or its requirements. But its a reason why it should.

Designing for reliability, safety and EMC are often forgotten in circuit designs over pure function. It is a very good design approach to consider these requirements in the very conception of a circuit, not try to add them later.

  • Maybe it's because I've never worked with high voltage circuits (much less created one) that I find this really counterintuitive, but am I understanding your answer correctly, that even though designing circuits this way decreases MTBF, it's done so that when a failure does occur, it isn't as dangerous in terms of fire or arcing hazards? – user3052786 Jan 20 at 4:34
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    @user3052786, interesting comments but it's a different goal. MTBF is examining the chances of a failure: the reliability of the function. This is examining the consequences of a failure: the reliability of safety, not of function. Have a read also on dual and triple redundancy for components/systems, equally worth learning about. Although far fewer general and domestic applications, you'll feel better about flying :-) – TonyM Jan 20 at 9:43
  • Right, I do see that there are different priorities, I meant to ask if this is intentionally sacrificing the reliability of function to mitigate the consequences of a failure. Because it seems to me that using two resistors as in this case would increase the likelihood of failure (as in impaired function), because it is another component that can fail. – user3052786 Jan 21 at 3:07
  • But even if it does fail, and it fails short-circuit (which I've never seen, but as I've said, I haven't been exposed to many high-power designs), the suddenly nonexistent load doesn't trigger a potentially catastrophic chain of failures in the components downstream, because the other resistor has enough headroom for heat dissipation to handle the higher current? – user3052786 Jan 21 at 3:11
  • @TonyM: the SPOF protection sometimes also goes under the tag "fail-safe" design. That means, in the cause of any failure, the design falls into a non-hazardous state. Another typical example for fail safe design is regarding logic level definitions for alarm signals: i.e. active low signals to detect a cut wire as as alarm or active high signals to detect a short circuit as alarm. – boink Jan 23 at 19:31

Most resistors, especially SMD (even larger 1210 ones) aren't rated for 400V.

So one of the possibilities is that they used 2 in series to divide the voltage requirement.

While higher-rated resistors exist, there are other factors to consider such as cost, availability, the extra time that takes to source them, an extra component to put in the pick and place machine, etc. (i.e. most PCBA houses will have 1M standard resistors, but not likely the high voltage ones). So all things considered it can be cheaper to just use 2 standard ones. It also gives more flexibility in case the high voltage ones go out of stock, etc.

Also consider that even if you have 1210 resistors that can tolerate 400V, PCB creep tolerances might require distances larger than the resistor itself, so you need either a larger resistor or more than one.

enter image description here

From this Panasonic datasheet.

enter image description here

From this Vishay datasheet.

  • Have you looked at these datasheets? You can 700V for 1206 package and 1000V for 1210 package? vishay.com/docs/49876/_tnpve3_vmn-pt0447-1504.pdf vishay.com/docs/28881/tnpve3.pdf – Ugur Baki Jan 19 at 6:38
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    Those are quite specific resistors, I never said they don't exist. I mention "most" in my answer specifically due to that. Most of the times its easier to use 2 or 3 of "normally" available resistors than to source a high voltage ones (then you have PCB creep and distance tolerances, independent of device tolerance). Anyway, since there is not much more information about the circuit/context, other than the fact that its a voltage divider/RC filter, its quite hard to speculate beyond that. – Wesley Lee Jan 19 at 6:47
  • My point is, most engineers or fab houses will have 1M resistors with standard ratings in stock. If you want the high voltage rating ones you'll have to make a specific order, new reel to mount on the pick and place machine, etc etc – Wesley Lee Jan 19 at 6:49
  • Thank you for your attention. Please check these links. First one is 1Mohm and second one is 2Mohm . Same price and same package. Is it possible to use a 2Mohm for this application under same conditions? digikey.com/product-detail/en/vishay-dale/TNPV12101M00BEEN/… digikey.com/product-detail/en/vishay-dale/TNPV12102M00BEEN/… – Ugur Baki Jan 19 at 6:51
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    @UgurBaki - another factor to consider is PCB clearance/creepage distances (they're a bit different). The 1206/1210 packages seem to have 2mm between the solder pads, which is a bit too close for 400VDC and can be unsafe. – Wesley Lee Jan 19 at 6:56

If you have a resistor R of power rating W, with voltage V across it, the power dissipated on R would be \$ V^2/R \$. If it is more than the power rating of the resistor, you have to think of another option. Quick idea would be to use two resistors R/2 of similar power rating W, in series. The total power dissipation remains the same, but individually they dissipate only half the power \$ (V/2)^2/(R/2) = V^2/2R \$

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    This is the simplest (and probably correct) answer. 2x1M @ 1/3W resistors can safely dissipate up to 2/3W, which at up to 400V is 1.6mA, vs 1/3W per resistor for 0.8mA max @ 400V. Even if the load is < 0.8mA, they may be planning for peak currents (inrush, surge, etc) – Doktor J Jan 19 at 20:33

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