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.
simulate this circuit – Schematic created using CircuitLab
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4 Answers
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6
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.
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3\$\begingroup\$ @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 :-) \$\endgroup\$– TonyMJan 20, 2018 at 9:43
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1\$\begingroup\$ 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. \$\endgroup\$ Jan 21, 2018 at 3:07
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1\$\begingroup\$ 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? \$\endgroup\$ Jan 21, 2018 at 3:11
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2\$\begingroup\$ @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. \$\endgroup\$– boinkJan 23, 2018 at 19:31
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2\$\begingroup\$ +1. The term I usually come across in the US in dealing with UL is 'Single-Fault Tolerant.' \$\endgroup\$ Jul 30, 2019 at 1:22
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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.
From this Panasonic datasheet.
From this Vishay datasheet.
answered Jan 19, 2018 at 6:29
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\$\begingroup\$ 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 \$\endgroup\$ Jan 19, 2018 at 6:38
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6\$\begingroup\$ 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. \$\endgroup\$ Jan 19, 2018 at 6:47
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1\$\begingroup\$ 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 \$\endgroup\$ Jan 19, 2018 at 6:49
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\$\begingroup\$ 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/… \$\endgroup\$ Jan 19, 2018 at 6:51
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2\$\begingroup\$ @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. \$\endgroup\$ Jan 19, 2018 at 6:56
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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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1\$\begingroup\$ 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) \$\endgroup\$– Doktor JJan 19, 2018 at 20:33
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With the values given, the maximum dissipation in each of the 1 Megohm resistors is 40mW (V^2/R = 200.200/1.10^6).
Therefore there aren't 2 of them for power dissipation reasons.
However a single resistor is unlikely to be rated for 400V. Therefore 2 are used in series. If the input had to tolerate higher voltages still then more, say 3 or 4 would be used.
Note that the size of those resistors is also larger (1210 rather than 1206) which may also be related to the maximum operating voltage too.
Even with leaded components such as a generic 1/4W resistor, the voltage rating, despite the part being larger than SMT is still typically less than 400V, so requiring a similar approach.
