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10 watt LED

conntaxman

Jun 17, 2011
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I made a desk lamp out of a 10 watt led ,it is the type that has the 10 very small leds in a small block about 3/4 x 3/4 of an inch. I have it hook up to a 12vdc battery and I have a 3watt 1 ohm resistor on the pos leg. It has been working very good for about 3 months, for about say 3 hours a day, but now it started to FLIKER a little bit and then it stops flikering and stays on. I do think that it got a little bit dimmer then when I first got it.yes at the same voltage.I have a large enough heat sink on it. I bought it on ebay.
any one know why the flikering?
tks
John
 

(*steve*)

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Asking just after Christmas is a good thing.

I have some Christmas lights that contain LEDs in series/parallel. Some of them exhibit a failure mode where some of the LEDs flicker, or have delayed turn on initially. I expect that they are damaged, and since they work after a while, I further expect that temperature is a factor.

I presume those LEDs are in series/parallel. If you could break apart the strings and provide each with a current limiting resistor, you may be better off (well, you may have been before they got damaged).

If you have a variable current source, or a variable voltage power supply, you could try increasing the current (voltage) slowly and see if some LEDs misbehave at low currents.
 

(*steve*)

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No, it's 9 LEDs -- the fact that they don't have separate leads is another issue.

A single LED would have a Vf of 3.4V or thereabouts.

Placing them in series/parallel like that is kinda alright as long as the LEDs are closely matched and in good thermal contact with each other. It's no guarantee though.
 

BobK

Jan 5, 2010
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Did you try measuring the current? You have probably been over driving it and thus reduced it life time. If you read the description, it says you need a constant current driver for this, which is a good idea.

Bob
 

conntaxman

Jun 17, 2011
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Did you try measuring the current? You have probably been over driving it and thus reduced it life time. If you read the description, it says you need a constant current driver for this, which is a good idea.

Bob

Bob, wouldn't the 1 ohm resistor be doing the same as a Driver would do.
Tks
John
 

Resqueline

Jul 31, 2009
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Only provided Vf is exactly 11V and the battery is exactly 12V. The only way to know for sure is to measure all three voltages involved.
If Vf is 10V (which it should be, since it's rated as a 10W LED) then you get 2V across the resistor, resulting in a 2A drive current.
 

conntaxman

Jun 17, 2011
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10watt led

Resq. the forward/ vf said 12vdc. It didn't have a higher or lower voltage, like vf is 10-12.
it just had 12. so then Im pushing it too hard with the 12volts.
could I use a lm317 to lower it to 10.
I have two 20 watt leds also from ebay.
20W Warm White High Power 1600LM LED Lamp Bulb Light
Light color: warm white
Color temperature: 3000 - 3500 K
Voltage: DC 12V
Power: 20W
Lighting angle: 140°
Lumen: 1400 - 1600LM
----------------------
and those are what they give for the spec.
so should I use 10 vdc for that also.
tks
John
 

Resqueline

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Vf is said to be 9~12V. This is a too wide span to calculate a theoretical Rs when using a 12V source.
You can use a resistor but then you'll have to measure the actual Vf for your individual LED's.
An LM317 is impractical to use for these power levels and especially from a battery. It needs at least 3-4V headroom.
And no, you should not regulate the voltage down, or think in terms of voltage when driving LED's. Think current.
Switchmode LED drivers will take care of feeding a constant current regardless of conditions, plus they won't waste power.
 

(*steve*)

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The other problem with a fixed resistor is that Vf falls as Tj rises.

This means that the current tends to increase as the device gets hotter, which increases the current, which makes it hotter...

This is the normal thermal runaway issue that you get when powering a LED from a constant voltage.

However the same thing can happen with a resistor. The issue is, will it run away?

(edit: runaway means increasing heat --> increasing power dissipation. Even without this, the power dissipation can cause the temperature to rise above the limit where the device fails, but that is a thermal failure, not a thermal runaway.)

Let's model the LED as a resistor.

If the LED current is nominally 1A, and the battery voltage is 12V and it has a 1 ohm resistor in series with it, then it must be acting as if it has a resistance of 11 ohms.

So by simple calculation, it must be dissipating 11W (I'm using simple figures, so ignore the excess dissipation)

Lets say that it gets heated by this and the internal resistance falls to 10.5 ohms

The current rises to 1.043A, and the power dissipation rises to 11.4W.

So now it's going to get even hotter...

Let's instead, assume that the series resistor is 13 ohms and the current is 1A and the LED appears to have a resistance of 11 ohms. (Which means our power supply must be 24V).

The LED sees the same power, so its resistance drops, lets say to 10.5 ohms again. This time the current rises to 1.021 A, giving a power dissipation of 10.95W -- The power falls! So the LED will cool slightly.

In the first case, the LED current can still run away, in the second it reaches an equilibrium quickly (actually it will reach equilibrium in the first case too, but at a current dramatically higher than the nominal current.

In these 2 cases, we have assumed a voltage source with zero internal resistance and a series resistor. This is exactly the same as a real (imperfect) voltage source with an internal resistance.

It's pretty clear that increased internal resistance is a good thing in this case because it stabalises the current.

Wouldn't it be great if we could have an infinite internal resistance? Impossible?

Enter the current source...

A perfect current source is an infinite voltage in series with an infinite resistance. Because the load resistance is always low compared with the impedance of the supply, the current does not change as the load resistance changes.

In practice a constant current source employs neither infinite voltages not infinite resistances (and with it infinite power dissipation), but employs some circuit with feedback to ensure a relatively constant current over some range of load resistances.

So we have the LED with an assumed resistance of 11 ohms connected to a 1A constant current source. The current is clearly 1A, and the power dissipated is 11W, as it heats up and the resistance falls to 10.5 ohms, the current remains the same and the power drops to 10.5 watts, which is a lot lower. The device can still get hotter, and can still exceed Tj(max) but since the current does not increase, and the power dissipated does not increase, failure will be from excess temperature caused by the nominal current, not by thermal runaway caused by ever increasing current.

If you use resistors, you can't exclude the possibility of thermal runaway, but you can ensure that device power falls as current increases. To do that, at least half the supply voltage must be dropped across the series resistor. Clearly for high currents (i.e. high power LEDs) this is extremely wasteful.

Things are actually a little worse than I've suggested above. LEDs are not resistors. Their resistance does not fall as their temperature rises, their Vf does. Vf is also dependant on current, so in the three cases above, if the second case reflected what actually happened, the first would be worse (more current, higher power), and the last would be better (lower Vf even lower power).

If you've read my tome on driving LEDs, you'll note that I recommend constant current sources for high power LEDs. The above is an explanation that I've not put in there which justifies that recommendation.
 
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(*steve*)

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Looks like the right tool for the job.

Just make sure that you have adequate heatsinking.

The low voltage overhead is a plus in this case.
 

conntaxman

Jun 17, 2011
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10watt led

steve, you seem very smart on this stuff, me NOPE. well anyway i just ordered some led Drivers from ebay, im going to try them out and try to see what parts and cir they use.
tks to all the guy/girls on here for all the answere.
John
I use to do Plumbing,Heating.airC. remodeling. Not building cir.
 

(*steve*)

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You should see me try to do plumbing... :D
 

BobK

Jan 5, 2010
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In electronics we do a smoke test, in plumbing its the water test. In my last plumbing effort, everything worked fine except the one joint I forgot to solder, so I failed the water test miserably.

Bob
 

jackorocko

Apr 4, 2010
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I always found it much easier to dig up a busted water line and patch it back together then trying to analysis an electronic circuit. to each his own I guess.
 

tedstruk

Jan 7, 2012
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leds

:D
you shouldn't have targeted the resistor.
The LEDS that were originally made, were ISA protocol of 1a per bulb so you would need 10 amps of power to light 10 bulbs.
They make lots of different amps and uses of LEDS.
depending on the LED, your nice array might last forever if you used a standard 1v line, and 1 amp per bulb. Resistors are used to reduce current, not amps. I tested an LED once, it burned out in about 6 minutes on a straight 1.5v DC current. I assumed that the bulbs were designed for millivolts instead of straight volts...probably:)
 

Resqueline

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tedstruk; you're making absolutely no sense... As an engineers joke it's good though.. :D
 

(*steve*)

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tedstruk, your post is almost indistinguishable from nonsense.

edit: beaten to it.
 
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