tegwin
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| posted on 29/3/08 at 04:55 PM |
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LEDs resistors and voltage drop
Im having one of those blonde moments...
Can someone see what im doing wrong?
I have 1 white LED- Forward voltage 4.8v 700mA
The power supply is 12V....
By my calculations, the resistor I need in series with the LED is 12R at aproximatly 6W
I dont have a 12R 6W resistor...
So I have used two 33R 7W resistors in parallel....that should give me 16R....
The LED lights up with a nice bright light, but the resistors get REALLY REALLY hot...After about 60 seconds of running they are too hot to
touch...
Obviously not ideal..
I can appreciate that I need to loose some of the voltage somewhere,...But is there an easier/cooler way of running the above LED on a 12V supply? I
dont really want to start a fire
If I found a single resistor rated at 12R 6W would that be cooler?
[Edited on 29/3/08 by tegwin]
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blakep82
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| posted on 29/3/08 at 05:04 PM |
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i'm wondering if 2 resistors in parallel might reduce the heat on each one... or 2 in series... can't remember how it works.
what about 12v Leds?
________________________
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don't write OT on a new thread title, you're creating the topic, everything you write is very much ON topic!
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tegwin
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| posted on 29/3/08 at 05:09 PM |
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Unfortunately I cant find a decent source of ultrabright white LEDs....
Unless I want an 18W 12V monster, which is a wee bit to bright for my application!
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RazMan
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| posted on 29/3/08 at 05:31 PM |
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Are you sure you are putting the resistor in series with the LED? To generate that amount of heat you must be passing a lot of current.
ps If you put two 7W resistors in parrallel then you double the wattage - but if they are getting hot that means you must be passing over 1 amp!
Cheers,
Raz
When thinking outside the box doesn't work any more, it's time to build a new box
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tegwin
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| posted on 29/3/08 at 05:35 PM |
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Raz....Yes...deffinately have the resistors in series with the LED....the resistor is on the +ve wire
The LED is deffinately only rated at 700ma...so im not sure how im managing to draw more than 1A...
Its for a sidelight in a car tintop...is it possible that the standard bulb the other side of the car is somehow drawing current though the same
cable...I wouldnt have thought so, but....
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RazMan
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| posted on 29/3/08 at 05:40 PM |
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Is it a standard type of wedge bulb or something special?
I have tried all sorts of LED 'bulbs' which have internal resistors so no faffing involved. I have probably got some spares lying around
somewhere if it helps.
I've used LEDs for every light on my car except for headamps which are HID, so the current consumption is about as low as you can go. If you put
dissipating resistors in the indicators you just burn the same amount of power but with the heat generated by the resistors instead of bulbs.....
which can be a fire risk.
The best type is the Luxeon - far far brighter than anything else on the market.
[Edited on 29-3-08 by RazMan]
Cheers,
Raz
When thinking outside the box doesn't work any more, it's time to build a new box
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tegwin
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| posted on 29/3/08 at 05:48 PM |
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Its a wedge bulb...
I cut the top off the bulb and fitted the LED inside the bulb...
Slotted it back into the car....then fitted a resistor on the +ve supply to the light....
I dont get why its not behaving atall!
What brigthness of LEDs do you have for sidelights?
I have successfully converted my rear light clusters to LED...Sidelight, indicator, brakes... Just got to do the front now...oh..and replace the
damn HID bulbs
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Confused but excited.
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| posted on 29/3/08 at 05:49 PM |
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A 16.5R resistor (2x33R in parallel) with 12 volts across it will pass 727mA, or just under 3/4 of an amp.
This will produce just under 9 (8.72)watts of heat. So they will run warm.
You can get power resistors with an integral heatsink.
In addition you will get 0.7 volts drop across the LED.HTH.
Tell them about the bent treacle edges!
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RazMan
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| posted on 29/3/08 at 05:54 PM |
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Check out this link for starters for a good reliable source - all
those external resistors are a fire risk imo!
I used
these as side lights and they are actually brighter than a filament bulb but consume a fraction of the
current.
these ones are a wedge type which might suit you.
The main point to watch is the angle of view - try and use the reflector more than the lense if you see what I mean.
[Edited on 29-3-08 by RazMan]
Cheers,
Raz
When thinking outside the box doesn't work any more, it's time to build a new box
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redscamp
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| posted on 29/3/08 at 05:56 PM |
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when doing led calcs rember your battery
is 12 volts but it jumps to 13.8 volts
with the alternator on.
14 volts is a better starting point.
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BenB
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| posted on 29/3/08 at 06:00 PM |
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Are they definately 33R resistors???
Two 33R 7W resistors in parallel gives you 16.5R 14W equivalent and the resistors will be burning off just over 5W in your set-up.
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BenB
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| posted on 29/3/08 at 06:35 PM |
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quote:
Originally posted by Confused but excited.
A 16.5R resistor (2x33R in parallel) with 12 volts across it will pass 727mA, or just under 3/4 of an amp.
This will produce just under 9 (8.72)watts of heat. So they will run warm.
You can get power resistors with an integral heatsink.
In addition you will get 0.7 volts drop across the LED.HTH.
But there isn't 12v across the resistor. Surely the resistor will be dropping 12-4.8=7.2v to get the 4.8v forward voltage. The wattage of the
resistor needs to be 7.2V*0.727A which is 5.234W
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redscamp
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| posted on 29/3/08 at 06:36 PM |
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you say 700mA max.
leds have an absolute max rating, but a
lower normal running current.
are you sure you need to max it out?
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minitici
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| posted on 29/3/08 at 07:39 PM |
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Here is a handy LED resistor calculator
Calculator
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segan2b
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| posted on 29/3/08 at 08:42 PM |
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For power LED's you really need a proper regulator, it is not a good idea to use low value resistors with a large voltage drop. Try something
like
THIS
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matt_claydon
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| posted on 29/3/08 at 10:42 PM |
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There is nothing wrong with your resistors getting very hot - that is what they are supposed to do to dissipate the power. Think how much heat a 7w
bulb puts out - the resistor will put out more heat than that since it has to dissipate the same power but none of it is emitted as light (that
doesn't mean it will get as hot as a filament, but it will give out as much heat energy).
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tegwin
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| posted on 30/3/08 at 01:24 AM |
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What a muppet......Of course I could easily use a 5V voltage regulator..
Infact, I have a bag of the buggers sat on my desk...
Why didnt I think of that!
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RazMan
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| posted on 30/3/08 at 03:55 AM |
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Exactly - why dissipate it when you don't need it in the first place? 
Cheers,
Raz
When thinking outside the box doesn't work any more, it's time to build a new box
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Macbeast
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| posted on 30/3/08 at 06:17 AM |
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A voltage regulator is of course just a form of resistor ( unless you're using some kind of sophisticated switching regulator ). So you will
still be dissipating 14 - 5 V = 9V@700 ma = 6.3 W and it will get hot. The advantage of a voltage regulator is perhaps that it has a metal tab you can
use as mounting and heatsink
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MikeRJ
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| posted on 30/3/08 at 11:37 AM |
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The only way to drive LED's with low power dissipation is to use a switching regulator. As said a standard linear regulator will dissipate
almost exactly the same as a resistor (a tiny bit more in practice as the regulator itself uses some current).
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Macbeast
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| posted on 30/3/08 at 04:24 PM |
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Good God !!
I've just noticed I'm a posting freak !!
Maybe I should get on with building ?
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