Today I disassembled a power strip (power extension) and found a LED connected directly to AC with a normal resistor (not a power resistor). How is it working as I thought it would be burned in no time? The resistor value is 220k.
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asked Apr 9, 2016 at 9:22
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3\$\begingroup\$ Are you sure it's an LED and not a neon? \$\endgroup\$– Roger RowlandCommented Apr 9, 2016 at 9:30
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11\$\begingroup\$ Maths. It's important. 220kOhm, even directly across 230VAC without LED or Neon lets only slightly over 1mA flow, and dissipates P = v^2 / R = (230*230)/220000 = 0.24W \$\endgroup\$– AsmyldofCommented Apr 9, 2016 at 11:21
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21\$\begingroup\$ You should smash that power strip to bits and put it in the trash. The mains power connections are made with solder joints that are not mechanically secure. If it is ever overloaded, it will fail in spectacular ways. There's no way that this design has passed any sort of safety certification. It just isn't worth the risk. \$\endgroup\$– Dave TweedCommented Apr 9, 2016 at 13:57
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2\$\begingroup\$ @DaveTweed looks like the earth pins aren't connected either. \$\endgroup\$– ratchet freakCommented Apr 9, 2016 at 19:15
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2\$\begingroup\$ @DaveTweed im aware of the poor design and quality of the product thats why i opened the box found this LED, i will not be using this in future \$\endgroup\$– mohammed ashkerCommented Apr 9, 2016 at 19:54
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3 Answers
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Use Ohm's law:
$$ {220\:\mathrm{V} \over 220\:\mathrm{k\Omega} } = 0.001\:\mathrm{A} = 1\:\mathrm{mA} $$
And power is the product of voltage and current, so:
$$ 0.001\:\mathrm{A} \cdot 220\:\mathrm{V} = 0.22\:\mathrm{W} $$
A 1/2 W resistor could connect across 220V just fine without burning up.
The LED is also a diode, and lets current flow only in one direction. So half the time there's no current at all, so the power is actually half this.
And if you are in a 120V country, the power is even less.
That explains why it doesn't immediately burst into flames, at least. However as others have mentioned, this plug strip has a number of other problems which might will result in other dangerous failures.
answered Apr 9, 2016 at 11:30
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1\$\begingroup\$ Re: "without burning up". Provided the resistor is rated for such a voltage drop over it, sqrt(2)*220 V ~ 350 V. \$\endgroup\$ Commented Apr 9, 2016 at 19:28
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The resistor power dissipation is safe as Phil Frost stated. However through my 54 year old eyes it looks like a SFR16 resistor that has a Voltage rating of 200V. I would always put 2 resistors in series when using SFR16. The actual resistor in the photo could be some generic or copy or ripoff so the voltage rating could be bad. If the LED is just a standard one things will be dim. A high efficiency or super bright one will provide a credible indication. What I always did was use a bridge rectifier made from 4 1N4148s. This stopped flicker and increased apparent brightness due to the doubling of average current. The poor LED in your powerbox is getting negative mains pulses albeit current limited across it half the time. Is the LED a special back to back LED or does it have an integrated reverse protection diode across it? If not you have an unreliable product.
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\$\begingroup\$ The peak voltage is nominally 311 V. Shouldn't the rating be more like 400 V? \$\endgroup\$ Commented Apr 9, 2016 at 19:31
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\$\begingroup\$ @PeterMortensen Given the shoddy overall construction of this power strip, I wouldn't be surprised at all if they'd used the wrong kind of resistor too. \$\endgroup\$– user39382Commented Apr 9, 2016 at 23:01
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The problem here is when LED is in reverse voltage mode. Reverse voltage can go beyond LED's specs and burn it.
