I found some 12V LEDs that output 800 lumen and consume about 1A. I am looking at creating a floodlight. Would putting 10 LEDs in series work with a direct plug into an AC circuit? could it be that simple?

  • \$\begingroup\$ At a minimum, you'll need a 120 ohm resistor able to handle the wattage: mouser.com/Passive-Components/Resistors/_/… \$\endgroup\$ – Wayfaring Stranger Feb 13 '14 at 3:22
  • \$\begingroup\$ @WayfaringStranger: Since the OP said these are 12 volt LEDs, I'd assume that they include a suitalbe current limiter - but he should provide a link to a datasheet so we can verify that. \$\endgroup\$ – Peter Bennett Feb 13 '14 at 3:29
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    \$\begingroup\$ Highly unlikely that your LEDs would work directly from AC, as LEDs and LED assemblies normally are intended to work from DC. Can you provide a link to a datasheet or description of your LEDs? We would need more information on them before making any definite comments. \$\endgroup\$ – Peter Bennett Feb 13 '14 at 3:32
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    \$\begingroup\$ If this is a typical Chinese 10W LED assembly with ~12V \$V_F\$, and you're thinking of plugging them into the wall-NO!, it will die instantly. The peak line voltage is almost 170 volts and it will destroy the LEDs, which need either a series power resistor, or (preferably) current-limited driving electronics. \$\endgroup\$ – Spehro Pefhany Feb 13 '14 at 4:02

I'm assuming you just want to series 10 LEDs across 120VAC.

Sure, it can work as you've described. Many cheap night lights are just a LED or two with a current limiting resistor directly on the AC circuit.

This makes a half wave rectifier, with each LED dropping 12V. But its not a very good way to do it and your likely to damage the LEDs eventually. This is because the 12V forward voltage is not fixed. The LED may drop less or more, and will not be current limited to only 1A. Also, the 120VAC is not fixed, I've seen it as low as 113VAC and as high as 128VAC. Also, this is RMS voltage, the peak voltage will be 1.414 times that, so peak can be as high as 128VAC x 1.414 = 180V peak. You also have to consider the reverse breakdown voltage of those 12V LEDs. Many LEDs have low reverse breakdown voltages, and might require a diode across the LED in the opposite direction and a current limiting resistor to protect the LED.

So, although it can work it is not as simple just wiring the LEDs in series without thinking about the consequences of no current limiting and the other half cycle of the AC in reverse.

It's always much better to power a LED string with current limiting.

So, if you still want to continue this way, you need to design for the peak voltages and assume each LED will drop it's minimum forward voltage drop. So let's assume the AC is at 128VAC, 180V peak, and let's say that the minimum forward voltage of each LED is 11.5V. So x10 = 115V across the LED string. You will need to drop the remaining 65V across a current limit resistor. Using ohms law, R = V/I and thus R=65/1 = 65 ohms. So a 65 ohm resistor in series will limit the current to 1A, allowing each LED to drop 11.5V and the AC line voltage to go as high as 128VAC (180V peak). But the current will go down if the LEDs drop more voltage or the line voltage is lower, and the current will go up if they drop less or the line voltage is higher. nominally, the line voltage is 120VAC (170V peak)

Finally, it's much, much better to use a proper constant current voltage supply to power a LED series string like this. I don't condone this method of direct connection to the AC power line, but it's often done for a cheap night light.

EDIT: I added a schematic to show you what I meant by a reverse diode connection to protect the LEDs from reverse current. The R2-D2 path is designed to carry about 350 uA of reverse current. This path should be lower impedance than the other diode string path, to ensure that all the reverse current follows this path back to the source instead of trying to tunnel through the LEDs via the reverse leakage current of the LEDs themselves. I think about 350uA should be enough, I don't think the LED will leak that much reverse current (LEDs are light emitting devices and thus the datasheets usually do not specify reverse leakage currents for LEDs), and besides, the D1 in that path will definitely not leak that much reverse current, it's rated at 50uA peak reverse current.


simulate this circuit – Schematic created using CircuitLab

[ pretend those two back-to-back zeners is a MOV. Circuit Lab doesn't have a real MOV part to put there ]

Wayfaring mentioned in the comments you can put a full bridge and a fat capacitor too. You can also put a high-valued capacitor on this half wave rectifier too, just after D1 , connected to the other side of the AC source. At 1 Amp current in your LED string it means the capacitor would need to be above 470uF @ 250V rated, to be of any real value, and that can take up a lot of space you might not have. You can still reduce strobing somewhat with a lower valued cap, say 100uF at 250V, but it will be less effective.

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    \$\begingroup\$ You're making a 120Hz strobe light this way, the LEDs light only for a brief moment every half cycle. \$\endgroup\$ – jippie Feb 13 '14 at 6:22
  • \$\begingroup\$ yes, good point.. with 10 LEDS like this it will only have enough voltage at the peak of every cycle to briefly light the LED string. It really needs to be a constant current supply \$\endgroup\$ – Brian Onn Feb 13 '14 at 10:11
  • \$\begingroup\$ A full diode bridge and a fat capacitor helps a lot with the strobing; without going to a full regulated DC supply. Still some people are more sensitive to strobing than others. \$\endgroup\$ – Wayfaring Stranger Feb 13 '14 at 14:42
  • \$\begingroup\$ These LEDs are the $1 variety from amazon. Thanks for the in depth answer.@jippie your eyes can only see at 30 or so hertz so you would not see a strobe. Since the turn on voltage is 9V I may just have to try putting 12 in series... \$\endgroup\$ – banno Feb 14 '14 at 12:59
  • \$\begingroup\$ @banno I added a schematic to show you what I meant by the reverse diode.. I added the schematic because I think this is an important requirement to protect the LEDs \$\endgroup\$ – Brian Onn Feb 14 '14 at 19:33

You mention: "with a direct plug into an AC circuit". This is kind of vague, it would lead some one to think you have built a power supply circuit that supplies AC voltage. However in context, I will guess you are referring to the main power plug. Assuming this is the case, then please, don't do that.

I personally would just go with a simple method. So I don't fuss with the dangers of messing with main power, I would just buy a wall wart that outputs a nice DC voltage.

You could (and probably should) use some sort of voltage regulator setup after this, either an LDO or switching regulator, with the appropriate discreet components that go along with them.

Finally, I would personally choose to wire the LEDs in parallel and not series with a current limiting resistor so you don't fry them.

Best of luck.


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