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I have a project that is powered by a 12v power supply. I need as bright of an LED as possible and decided to use this 9v LED. According to the datasheet, its max current is 1A, which is what I would like to get it to. In order to do this, I am using this 3 Ohms resistor

According to my calculations, (12v-9v)/1A = 3 Ohms. This should get my LED to the correct current.

To test the current, I placed a multimeter in series with the 12v power supply, LED, and resistor. However, I am only reading 0.639A. I tried several power supplies and meters and it's all the same.

What am I doing wrong? Is my test setup providing too much resistance, thus decreasing the current?

enter image description here

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    \$\begingroup\$ And what voltage did you measure accross the LED? \$\endgroup\$ – Marko Buršič Oct 2 '18 at 15:32
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    \$\begingroup\$ The datasheet shows a typical forward voltage of 9V at 400mA, not 1A. \$\endgroup\$ – Finbarr Oct 2 '18 at 15:34
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    \$\begingroup\$ First, if you're dropping 3V across a 3 ohm resistor, you're dissipating 9Watts!! That's a 3W resistor. \$\endgroup\$ – Scott Seidman Oct 3 '18 at 16:57
  • \$\begingroup\$ @Finbarr , The graph indicates 9V 350mA at 55'C which is more realistic than 25'C. Daniel, try my design instead. using std. power NFET and any old NPN with jumper wire calibrated for 560 mOhm. 1/2W or correct part. in 1W \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 3 '18 at 17:06
  • \$\begingroup\$ @ScottSeidman I adjusted the resistor and am dissipating 2.1v, with around 1A current. Does't that make it about 2 watts? \$\endgroup\$ – Daniel Frenkel Oct 10 '18 at 16:53
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According to the chart on page 17 of the datasheet, the terminal voltage of a nominal-9V LED rises to between 10.25 and 10.5 V (depending on temperature) at 1 A. You need to size your resistor accordingly.

But it would be far better to use an active current regulator to feed this kind of LED. Then, the current wouldn't depend on temperature. Or on your actual source voltage, for that matter.

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    \$\begingroup\$ Thank you for this clear explanation and guidance. I had looked at that chart in the past and simply did not understand it, but I do now. Thanks again \$\endgroup\$ – Daniel Frenkel Oct 2 '18 at 16:04
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    \$\begingroup\$ @DanielFrenkel Just to add something else to consider to put more emphasis on a current regulator of some kind. The temperature curves showing voltage drop vs operating current should be assumed to be "typical" and not cast in concrete for every device you get. You may have to contact Cree to get a \$2\sigma\$ boundary range at your desired current. Assume that over temperature and specific device you buy, the range is from 10.1 V to 10.6 V (being optimistic), a \$1.5\:\Omega\$ resistor might allow anywhere from .9 A to 1.3 A. They make poor regulators with such low voltage overhead. \$\endgroup\$ – jonk Oct 2 '18 at 18:32
  • \$\begingroup\$ Dave you should educate the importance of thermal design and 50% tolerances on Ri of LEDs ( and all diodes) -1 but otherwise your simplistic advice is ok but oversimplified. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 3 '18 at 17:00
  • \$\begingroup\$ Dave, do you recommend I use the following to replace the resistor? link ... I've never adjusted the output on a voltage regulator like this, so will probably create a new post to see if I am doing it correctly. \$\endgroup\$ – Daniel Frenkel Oct 10 '18 at 19:44
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From the data sheet: enter image description here

Look at these characteristics. Measure the temperature as well. For example if the case temperature (that means right under the diode) has a temperature of 85˚C, then you should get a forward voltage 10.25V at 1000 mA. That would need a resistor of approx. 1.7 ohm.

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  • \$\begingroup\$ Can you provide an attribution for where you got this graph? How is it related to the OP's LED? \$\endgroup\$ – Elliot Alderson Oct 2 '18 at 15:42
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    \$\begingroup\$ @ElliotAlderson from the datasheet that OP linked. \$\endgroup\$ – Marko Buršič Oct 2 '18 at 15:43
  • \$\begingroup\$ At 105 ˚C there is no 10.25V or 1A rating. It's dead. I assume you meant the 85 ˚C rating. \$\endgroup\$ – Passerby Oct 2 '18 at 22:02
  • \$\begingroup\$ The attribution should be part of the answer. Thanks to @Passerby for taking care of that. \$\endgroup\$ – Elliot Alderson Oct 2 '18 at 22:51
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It could well be stray resistance. If you are only getting 640mA, then the resistance you are seeing is (12-9)/0.64 = 4.69 Ohms. That is an extra 1.7 Ohms. Your resistor is 3 Ohms. It looks like it is a 5% tolerance, which means it could be as high as 3.15.

You have the resistance in the multimeter leads, you have resistance in those white wires, and you also have resistance in your terminal blocks, where the wires connect to the breadboard, and the breadboard itself also has resistance.

The next thing to do is measure the actual voltage drop across the LED. That will have some tolerance as well, so you may find it is not exactly as the datasheet says. This will also have some significance.

Check all of this and recalculate.

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    \$\begingroup\$ The bread board probably ain't helping. \$\endgroup\$ – Passerby Oct 2 '18 at 18:31
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Page 17 of the datasheet shows the IV curve for the 9V version. For 1 amp, you are looking at a Vf of 10.4 volts or so, depending on the temperature of the led.

Note that 1 amp is the ABSOLUTE maximum, and without proper cooling of the led, you are almost guaranteeing a dead led, or multiple internal failures.

Note, if your getting 640 mA with 3 ohm resistor, you may want to measure both the voltage and resistor in use. Your source voltage may be higher than 12V, and the resistor may be on the lower range of its tolerance. 640 mA is the If at 9.6V. Actual measurement of values should always be taken over theoretical values, especially when small variations lead to large changes.

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You have to actually READ THE DATASHEET:

Depending on temperature, the LED will drop about 10.2 to 10.5 V with 1 A thru it. Conversely, with 9 V across it, the current will be about 300 to 450 mA.

The datasheet is really quite clear. I can't see how anyone would think this LED would drop 9 V at 1 A.

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  • \$\begingroup\$ Thank you. My understanding of LEDs has been very limited. After the responses here, I now understand things much better. Thank you for this answer \$\endgroup\$ – Daniel Frenkel Oct 2 '18 at 15:55
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    \$\begingroup\$ @Olin: ... other than it says 9 V right on the intersection of the red curve and the 1000 mA grid line. \$\endgroup\$ – Transistor Oct 2 '18 at 15:56
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    \$\begingroup\$ @Transistor: This snippet from the datasheet is a little out of context, as snippets necessarily are. The "9 V" to the right of the graph identifies the 9 V version of this LED. The same datasheet covers 18 V, 36 V, and other versions of the basic model. That convention is used throughout the datasheet, and what it means is pretty clear when looking at the datasheet as a whole. Note that the Y axis is clearly labeled in volts on the left side, as you would expect. \$\endgroup\$ – Olin Lathrop Oct 2 '18 at 16:00
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    \$\begingroup\$ I know that and you know that. The novice OP could easily be confused by it. \$\endgroup\$ – Transistor Oct 2 '18 at 16:31
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    \$\begingroup\$ 9V refers to the nominal voltage at 350mA at 55'C=Tc only. from which you can estimate Ri. on your components as they may vary +/-25 to 50% as his did, much lower. So Don't rely on the chart alone. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 2 '18 at 22:53
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Short answer:

Use 1.3 Ohm instead of 3 and verify temp and current. It should not instantly burn your finger. You must measure LED voltage rise with current to optimize this R limit value. Or see my design.

What you need to learn:

  • All diodes above nominal current have a **constant differential resistance, but those may be +/-50% tolerance

Ri = ΔVf/ΔIf {min:max} [Ω] ( also applies to LED's, transistors, Zener's(Zzt), MOV's)

Here is a simple adjustable driver using an NFET RdsOn=0.5 Ohm Vgs=4V 2W heatsink warm only 0 to 1 A adj range.

  • Fixed Rs may be adjusted for tuning range to match LED.

  • the VI curves are collected with thermal constant T heatsink 25'C and others.

  • Diodes have a threshold voltage that drops with rising temperature

What the datasheet tells me:

The graph on p17 measures a slope of 4.25 Ohms @ Tc rising to 85’C.

When you added 3 ohms is added in series your current in theory, reduces to
3V/(3Ω + 4.25Ω) = 0.41A Since you measured 0.64A this means ...

Req was (12-9V=3V@ 0.35A) so 3V/0.64A=4.7 = (3Ω+1.7Ω) so your sample had 40% Ri of the graph's slope of Ri = ΔVf/ΔIf ( due to wide tolerances on Ri)

  • now recomputing series R , you need 3Ω-1.7Ω= 1.3 Ohm or so.

Keep in mind if your heatsink compound or slow fan is inadequate, and Tc rises, then the current also rises from the Shockley Effect, due to table worst case Vf@Tc computed as -1.6V/+60’C case rise.

There are many ways to regulate an LED power, linear or SMPS. CC or variable CC. enter image description here

Here using the BJT current Vbe= 560mV so I used Re=560 mOhm which could be made from AWG28 magnet wire with <1W dissipation kept cool by an engineered solution. For best performance use a CPU 1W muffin fan with a Resistor drop for reduced speed with silver paste on LED array.

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