I have a strip of 9 V LEDs. Every one of them has a 200 kΩ (204) resistor in parallel. What is its purpose?

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Power supply and controller PCB enter image description here

Works with 220V AC.

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The two ICs on the left are PWM controllers BP2876D. IC in the middle is an unknown microntroller probably.

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    \$\begingroup\$ maybe not resistors ... please post a picture \$\endgroup\$
    – jsotola
    Dec 29, 2021 at 17:50
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    \$\begingroup\$ @jsotola Added some pictures \$\endgroup\$ Dec 29, 2021 at 18:30
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    \$\begingroup\$ I have used resistors in parallel in cases where there were small leakage currents. The leakage current may be enough to make the LED glow faintly when it is supposed to be off. In my case 100 k was enough to prevent it from glowing. Red LED's especially will glow a little even with just 10 uA or something like that. This may not be the reason in your case, though. \$\endgroup\$
    – user57037
    Dec 31, 2021 at 6:04

3 Answers 3


Quite possibly, the LED string is AC fed and, when AC fed, you have to watch out for the maximum reverse LED voltage not exceeding the limits for each LED. So, the 200 kΩ resistors would tend to balance out the reverse voltages across each LED so that they each share the same fraction of total reverse voltage (rather than one dropping most of it and failing).

Bonus answer: whatever circuit you have it attached to, you need some extra circuitry that will adequately limit the maximum LED forward current.

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    \$\begingroup\$ to add to this, the max reverse LED voltage is -5V, after which depending on the process they can avalanche like zeners in a smaller junction that cannot withstand the same power as the forward direction with higher density and rapid degradation. This is observed easily on a Lissajou V-I plot by the increase in junction capacitance due to a shrinking of the insulation by metalization growth and smaller gap just before shorting out if voltage is increased in reverse then burning open by fusing. \$\endgroup\$ Dec 29, 2021 at 18:09
  • \$\begingroup\$ I measured the voltage on it while it was running from the lamps circuitry and it has 100V DC on it. There are 12 LEDs connected into strips. Unfortunately the circuit died after a while. \$\endgroup\$ Dec 29, 2021 at 18:34
  • \$\begingroup\$ I can upload a picture of the PCB later. Maybe you will be able to figure out if it is DC or AC? \$\endgroup\$ Dec 29, 2021 at 18:44
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    \$\begingroup\$ @AndrisJefimovs the picture gives no indication what the supply is. \$\endgroup\$
    – Andy aka
    Dec 30, 2021 at 15:10
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    \$\begingroup\$ @AndrisJefimovs well, I believe you have your answer now anyway. \$\endgroup\$
    – Andy aka
    Dec 31, 2021 at 10:55

This is an acceptable way to equally share the reverse voltage in a string so that -15V may be shared equally to -5x3 when AC is applied within this range of <= 15Vp Or multiples of this in larger voltages.

How did they select this value?

Red LEDs are often rated at 1uA leakage and White 10uA due to the dielectric with reverse voltage -5V max. Thus this is equivalent to 500Kohms @ -5V. In order to prevent unequal series voltages from over-stressing one component, leaky resistors split the voltage more equally when sufficiently lower than the component leakage.

So by being conservative on peak reverse voltage and thus slightly higher effective leakage resistance, 300 k will dominate the voltage drop. Without specs, not much more can be specified.

The same method is often used for electrolytic caps to share Vdc equally as both the LEDs in reverse and e-caps in forward are in "insulation mode". In the opposite polarity they are both in conduction mode, although unhealthy for e-caps and luminous for LEDs.

A better way is to include reverse diodes in the chip done on only the best components.

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    \$\begingroup\$ Would placing just one single reverse diode in parallel to the entire series of LED's suffice? Or would each LED require its own? \$\endgroup\$
    – ManRow
    Dec 29, 2021 at 18:16
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    \$\begingroup\$ No will not share the much higher voltage equAlly. \$\endgroup\$ Dec 29, 2021 at 19:08
  • \$\begingroup\$ Why is conduction mode unhealthy for e-caps? \$\endgroup\$ Feb 21, 2022 at 20:01
  • \$\begingroup\$ This occurs when a > 10% of the insulation voltage rating is applied in reverse polarity, then electrolysis dissolves the aluminum oxide free the electrons allowing them to conduct easily (low resistance) with thermal insulation creating a rapid thermal runaway condition and explosive condition. It's also unhealthy to breathe in these vapours. \$\endgroup\$ Feb 21, 2022 at 20:11

I’d say it’s to ensure that the LEDs switch off cleanly; without the resistors the LEDs could continue to glow dimly for several seconds after power is switched off because of residual charge in the driver; this can be annoying in a dark room and can be observed even with very tiny currents.

As an aside, the datasheets for many white LEDs warn against applying any reverse voltage.

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    \$\begingroup\$ LED persistence of that lifetime would have to be due to energy stored in the driver circuit. Resistors in series totaling over 1 Megohm would not be an effective method of quickly discharging the capacitance. \$\endgroup\$ Dec 30, 2021 at 5:08
  • \$\begingroup\$ True, so why might the designer choose not to? Perhaps because a single resistor would need to be rated for a relatively high voltage (a weak reason) or perhaps because one or two LEDs might glow briefly after power-down after the others have extinguished (cosmetic but yuk). \$\endgroup\$
    – Frog
    Dec 30, 2021 at 8:54
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    \$\begingroup\$ "choose not to" what, exactly? (1) these circuits are generally designed to minimize cost. (2) any additional functionality such as 'clean' shut-off circuitry must be paid for. (3) component costs are marked up many times to cover design, manufacturing, sales, shipping, etc. overhead. (4) your tastes seem to be strongly against any residual glow, however ask around and find out how many say they would pay another $1 to have the LEDs go entirely dark immediately. (5) build a competing product, then discover that they buy the less-expensive one instead. \$\endgroup\$ Dec 30, 2021 at 21:57
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    \$\begingroup\$ My comments come from internal discussions in the design lab rather than customer feedback. An afterglow for a couple of seconds is ok, even desirable, but a longer period is considered to be a bad thing (by my client anyway). A particular issue with strip lighting as distinct from point lighting is that it’s relatively easy to pick up induced voltages from other equipment that’s nearby. Putting a bleed resistor on each LED is the most reliable way to avoid this. The incremental cost is around 0.1c per resistor, plus cost of placement. \$\endgroup\$
    – Frog
    Dec 31, 2021 at 0:58

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