Power rating of resistor for LED ring

Thread Starter

PFFT

Joined Dec 2, 2022
2
I've been messing around with an angel eye style LED ring - a ring of around 60-odd bright LEDs on an aluminium substrate.

I don't have a datasheet for it, but on the listing it stated:

  • Power: 11W
  • Input voltage:12V DC
  • Allowable maximum current: 1100mA

When I hook it up to my small bench-top power supply set at 12VDC (constant voltage) it draws around 1A and before too long gets pretty hot. It's a lot brighter than I need it to be when I run it like this.

After a bit of trial and error I found that a 3Ω resistor would result in the LED ring drawing 0.35A from the PS (still set at a constant 12VDC) and the voltage measured across the LED ring would drop to around 11V - it was still bright enough for my needs and ran cool enough to the touch.

The resistor I used in my little test has quite a low power rating, I'm sure (perhaps 1/4W) - it does get hot, but you can still touch it without any discomfort.

I'd like to include a 3Ω resistor in with this simple circuit, but I'm at a loss as to how to calculate the power rating I need. I've looked up how tos, and I'm aware of this excellent calculator but the fact that this is an LED ring made up of 60 or so warm white LEDs and not having any info with regard to it's Vf throws everything off - ie. in the example above a 24Ω resistor should drop the 1A down to 0.35A when used with a white LED with a Vf of 3.6 - but with this LED ring it turned out it had to be a 3Ω resistor.

I'm an electronics dunce - this stuff blows my mind - any pointers on how I should proceed? Maybe the answer has been staring me in the face with the stated power of the LED ring being 11W? I dunno, I just can't get the numbers working.

Many thanks!
 

dendad

Joined Feb 20, 2016
4,251
given...
I = .35A
R = 3 Ohms

P = I x I x R
P = .35 x .35 x 3
P = .3675W
So, use a 1Watt resistor and it will be well within the rating.
 

Audioguru again

Joined Oct 21, 2019
5,418
3 ohms x 0.35A= 1.05V. The power in the resistor is 3 ohms x 0.35A squared= 0.37W. A 1/2W resistor will be very hot so use a 1W resistor. The power in the group of LEDs is 12V - 1.05V= 10.95V x 0.35A= 3.83W. Each LED produces 3.83/60= 0.64W. The current in each LED is 0.35A/60= 5.8mA which is not much.
 

Audioguru again

Joined Oct 21, 2019
5,418
Did you measure the forward voltage of each LED and not use the ones that are not identical?
When LEDs are in parallel the one with the lowest Vf uses more current than each of the others and burns out first. Then the next lowest voltage burns out then the next then the next faster and faster.
 

WBahn

Joined Mar 31, 2012
27,893
The power being dissipated in your resistor can be calculated three different ways (all of which are equivalent, sans measurement tolerances):

Voltage across a resistance: P = V²/R = (12 V - 11 V)² / 3 Ω = 333 mW
Current through a resistance: P = I²·R = (0.35 A)² · 3 Ω = 368 mW
Current through a voltage drop: P = V·I = (12 V - 11 V) · 0.35 A = 350 mW

Note that if the resistor is actually 2.86 Ω, then all of these would agree. This would mean that the resistor was slightly less than 5% below its nominal value, which is quite possible unless it's a precision resistor (and, even if it is, you are abusing it with that much power, so it might well be outside it's normal spec range).

Using a 0.5 W resistor would be about the lowest you want to do. This would give you a margin of about 50%. A common rule-of-thumb, however, is to use a margin of 2x, so at least 700 mW, which would take you to a 1 W resistor. If you can get your hands on one of those easily enough, that's the route to go.

A couple of other options: If you have several of those 1/4 W, 3 Ω resistors handy, you can make a 1 W resistor by making two pairs of parallel resistors and then putting two of those in series (four resistors total).

Another option would be to drop your voltage with a diode, but getting one (or a couple in series) to drop right about 1 V at a current of 350 mA is probably more hassle than it's worth.
 

SamR

Joined Mar 19, 2019
4,492
Or do the math on a parallel resistor circuit of 3Ω total using whatever resistors you have that will work out to their lower wattage rating per resistor! Or even 3 1/4W 1Ω resistors in series will get you to a lesser wattage per resistor than a single resistor with the full voltage drop across it.
 
Last edited:

WBahn

Joined Mar 31, 2012
27,893
3 ohms x 0.35A= 1.05V. The power in the resistor is 3 ohms x 0.35A squared= 0.37W. A 1/2W resistor will be very hot so use a 1W resistor. The power in the group of LEDs is 12V - 1.05V= 10.95V x 0.35A= 3.83W. Each LED produces 3.83/60= 0.64W. The current in each LED is 0.35A/60= 5.8mA which is not much.
There are some contradictory assumptions being made here.

First, by assuming that the 3.83 W is ONLY dissipated in the LEDs, the assertion is that there is no current limiting being done. Also, it would be 64 mW, not 0.64 W -- probably a typo.

Second, by assuming that the current is evenly divided across all 60 LEDs, the assertion is that they are all in parallel. But that means that all 60 have the full 10.95 V across them (given the no-current-limiting assumption above).

I'm going to guess that the LEDs are in strings of three (might be just two or perhaps four, depending on the LEDs in use) and that there is a small current-limiting resistor per string.
 

sghioto

Joined Dec 31, 2017
3,707
3 ohms x 0.35A= 1.05V. The power in the resistor is 3 ohms x 0.35A squared= 0.37W.
Where did you get that formula from?
The voltage is 3 ohms x .35A = 1.05V
Power equals current x voltage
.35 X 1.05 = .3675W
dendad had it correct in post #2, end of story.
 
Last edited:

WBahn

Joined Mar 31, 2012
27,893
Where did you get that formula from?
The voltage is 3 ohms / .35A = 1.05V
Power equals current x voltage
.35 X 1.05 = .3675W
dendad had it correct in post #2, end of story.
And what is 0.3765 W rounded to two sig figs (which is not only more than adequate for this purpose, but all that is justified)?

As for where the formula he used comes from (i.e., power in a resistor is the resistance multiplied by the square of the current through it), start from the same place you did:

Power equals current x voltage

What is the voltage across a resistor with a particular current flowing through it?

voltage equals current x resistance

So

Power equals current x voltage which equals current x current x resistance

This equals resistance x (current squared).
 

WBahn

Joined Mar 31, 2012
27,893
I know the numbers work out the same. I don't think I've ever seen using current squared to figure power.
It's extremely common. For a resistor:

P = V·I = V²/R = I²·R

This is also the basis for many measures based on the premise that power is proportional to either the square of the current or the square of the voltage, which is extremely useful in many, many situations.
 

Thread Starter

PFFT

Joined Dec 2, 2022
2
Wow,

Thank you all for such a thorough and complete set of answers. 1W it is, then!

Really appreciate everyone's input here - it's sinking in, but no doubt I'll be back again with more noob questions before too long!

Thanks again!
 

Audioguru again

Joined Oct 21, 2019
5,418
On ebay, a car mod shop in the UK lists a 12V ring of surface-mounted cold-white LEDs that are rated to be "road-legal" and headlights are mentioned.
https://www.ebay.co.uk/itm/120mm-LE...hite-super-bright-12v-Black-PCB-/303795879404

They have about 39 LEDs and thirteen 270 ohms series resistors for each string of 3 series LEDs.
At 13.8V I calculate-guessed a current of 183mA.

In Canada, tampering/modifying a vehicles road-lighting is illegal but many cars have weird lighting. They also have blackened glass, no muffler and are lowered. They stink so they probably have the pollution controls removed.
 
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