# LED Power Supply

I have a Super Bright LED listed for 2.1-2.3 VDC forward voltage, 80ma forward current, and wonder what transformer and resistor would be appropriate to power it, please?

80mA is a lot of forward current. Whichever power supply you use has to supply this or more.
Resistor is calculated so:
Supply voltage V (say 5V) minus LED voltage drop V (say 2.2V) divided by current Amps (say 0.08A)
5 - 2.2 = 2.8
The resistor to suit these parameters would be 35Î©.
Just fill in the numbers for any other scenario.
Tip. If you have access to a current limited power supply connect the LED and adjust current for required brightness and use that figure. You may probably find the 80mA is MAXIMUM forward current and you may not need anything like this.
Cheers Bob

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Hi David,

Thanks for making a post!

The correct resistor will depend on the voltage of your power supply, and the voltage of your power supply depends on your resistor. Itâd be best to select a power supply (any thatâs capable of more than 80mA @ 2.3V - which is any and every power supply just about) will do.

Once youâve settled on a power supply youâll be able to calculate the size of resistor you need.

David
Add on to my last post.
Another method of ascertaining required current if a current limited supply is not available. Assuming a voltage variable supply is available.
Use a series resistor of say 100Î©.
Monitor current with a suitable meter. Non invasive or Hall effect based meter or Digital multimeter. If the DMM, USE THE 10A RANGE. DO NOT use the mA ranges. At these low voltages he voltage burden of the meter will be too high and the 10A range should give you mA resolution. Do not exceed 80mA.

When the required current and power supply is established just plug into above formula to get resistor value.
Cheers Bob

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Hi Oliver, thanks so much for the advice! I wonder if âCoreâ offers an appropriate transformer for this LED? It sounds like a 35 Ohm resistor would be best.

SKU: COM- 08862.

Hereâs the LED I have ordered. Actually re-ordered as I blew the first one âŠ!

David.
âTransformerâ would indicate (to me anyway) an AC supply. Led supply must be DC. LEDs donât take too kindly to reverse voltage. Simply putting a diode in series to rectify does not work too well either. The available DC current of a half wave rectifier is about 28% (I think off the top of my head) of the transformer rated AC current. I do know this figure is 62% for a full wave bridge.
Maybe if your supply was AC this may be the reason the LED âblewâ.
Cheers Bob

PS: Just looked at the specs. It seems that 80mA is not max but recommended is 65mA - 75mA. It has also got a reverse voltage protection diode shunt connected. I donât know how much faith you can put into this Chinese document as it says reverse current is 10Â”A. How this can be with a foreword biased diode (with voltage in the opposite direction) across the LED is beyond me. My comments re AC source voltage still stands.
Cheers Bob

PPS: The protection diode may be a Zener

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Hi Bob, thanks for the info! Is there any particular DC unit that you could recommend? I guess it would then require wiring of the resistor into the circuit âŠ

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David.
Exactly what are you intending to do. If lighting this LED is in conjunction with something else maybe you should consider the whole picture.

This post indicates so far that all you are trying to achieve is lighting this LED with 80mA.
If this is the case then any 5V DC supply which is rated say 1A or better with a 35Î© series resistor will do the job.

One thing. 35Î© is not a preferred value so will not be easily obtainable. This value could be made up with multiple resistors but is not very critical as the LED volts can vary from 2.1 to 2.3.

I would use 39Î© which is a preferred value and at 2.2V LED volts would only reduce the current to â 72mA. I doubt that even the keenest eye would notice any brightness difference. Going the other way with preferred values at 33Î© I think would be too small and increase the current above recommended values.
Cheers Bob

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Hi Bob, all Iâm wanting to do is to light up the Super Bright Red LED. I visited a webpage about train sets which had a very useful link to a âResistor Calculatorâ. With a 6V âtransformerâ, 2.2VDC LED with 80ma, it looks like the R-1 39 ohm resistor would be best. I guess that I can get one at Jaycar. Should it be connected to the positive side of the circuit? Thank~you so much!

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Hey David,

Spot on about which side the resistor will be on. I would add a factor of safety and pick a slightly higher rated resistance (the nearest rated would be a very good option - donât go lower than the calculated value, youâll burn out the LED).

For some more reading Iâd also take a look at Sparkfuns tutorial. Let us know if there is anything else we can help with!

Let us know if there is anything else we can help with!

Liam.

Hi David
The usual convention is to connect to the positive (anode, A) side but as it is in series it does not matter. Circumstances sometimes mean that it is connected to negative (cathode, K) but you have no need to worry about this
I donât know how familiar you are with LEDs but just in case:
The Pos or Anode leg is the longer of the 2.
In case you cut the legs to the same length the neg or cathode will (or should) have a small flat on the edge of the LED body.
If all this fails use the diode check on a DMM. Pos is red lead. Reverse connection will be OL, may be flashing. Forward connection will be the forward voltage drop of the LED and may even light it slightly.
Heed Liams word on power rating. Up is better than down although in this case 0.5W will be OK. It may get quite hot so donât âfinger testâ it.
Cheers Bob
PS: What you term as a âtransformerâ must be a DC supply. Use of âtransformerâ to me only clouds the issue. âTransformersâ can only function with AC.

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