Powering ½W white LED's

Discussion in 'The Projects Forum' started by jj_alukkas, Nov 21, 2011.

  1. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
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    I used to build 1W LED lights with current regulators, but recently happened to try some ½W white 5mm leds. I have a couple of them and want to install them at different locations. However it wouldnt be economical if I use a linear current regulator like LM317 as it would cost me 2 times what the LED itself costs. I have a +5V DC source everywhere, so that would be the supply. So my question is, is it okay to drive a ½W LED with a resistor? I tried a bench test with a 12ohm ½W resistor on +5V. It lights bright, runs on 4.6V and 68mA current. Is this normal? AFAIK, white ½W leds run at 3.7-4.5V and at 120-150mA. The 68mA current is fine, but what about voltage? The overall LED power seems at 0.313W for this 0.5W LED which is also ok. I would only be using this for small durations and not a permenantly on setup. ANy help here please?

    Edit: Running on a linear current regulator of LM317 and a 12ohm resistor and 12v source, I get 4.3V@ 106mA on the led which is perfect, but cant it even be close with a resistor?
     
    Last edited: Nov 21, 2011
  2. DickCappels

    Moderator

    Aug 21, 2008
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    "...is it okay to drive a ½W LED with a resistor?"
    Sure, its done all the time!

    "...runs on 4.6V and 68mA current. Is this normal?"
    It depends on the specs for the LED. The voltage across the LED at the specified current should not exceed the maximum voltage on the spec. sheet, otherwise either 1) you have a measurement error, or 2) you have a defective LED. Since you obtained the acceptable voltage/current on a later measurement with an LM317, then most likely, you made a measurement error when operating with a resistor.
     
  3. John P

    AAC Fanatic!

    Oct 14, 2008
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    It seems to me as if your 12 ohm resistor was too big. Calculating from the 12V circuit, you should use a 7 ohm resistor. Then you'd get .7V drop across the resistor with .1A flowing, and the LED should work the same as it did with the regulator. Power dissipation would be .07W in the resistor (so a 1/4W package should be fine) and .43W in the LED.

    However, with such a small drop across the resistor, you'd be vulnerable to small changes in the power supply voltage causing large changes in current flow. If that's a possibility, you might need to reconsider.
     
  4. iONic

    AAC Fanatic!

    Nov 16, 2007
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    How stable is the 5V supply? Is it a battery or a regulated 5V supply. If the supply does not fluctuate and you are O.K. with the brightness with the 12 ohm resistor then your fine.
    I wouldn't worry too much about the voltage as long as the current is at or under the spec and you have a resistor in series you'll be O.K. The LED will drop the voltage it needs to obtain the current you have allowed it.
     
  5. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    I always run ordinary 3mm led strings with resistors, but this 1/2W thingy doesnt seem to run at the current and Voltage I calculate it for!!

    There is my problem. My dealer gives me LED's which has no part number or name. He even doesnt know those things! But for the ordinaries as well as 1W's I didnt ever need a spec sheet, as they all worked after a bit trial and error on the bench. And there isn't much chance for a measurement error as the multimeter though i use is not of high precision, it shows values of accuracy +/-0.01V from a known source. And I have bought 4 of them for this testing purpose and all act the same. The problem comes as I cant get the voltage in range as I control the current with the resistor. When I run the same led with 2 12ohms in series, it runs on 3.6V@42mA !! Now the voltage is in range, but current is 1/3rd of what rated for a 1/2w led! and thats what I can't seem to get!
     
  6. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    Actually on reducing the resistor, the voltage is shooting up way over 4.5V at the led. I just tried with 7.4ohms in series with the led, it runs on 4.95V@93mA. I just cant seem to control the voltage when I regulate the current as Im a bit scared of running them on 4.9V! Supply is stable, I have a regulated 5V all the time.
     
  7. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    The supply uses a 12V battery and a 7805 regulator. Since its an inverter battery, I can load it pretty well and the 1A regulator is the only bottleneck if I need to, so I expect it to be stable. I am ok with all the things you said, current, resistor in series, brightness, but not the voltage which is 4.7V which is a bit too high, isnt it? Also what scares me more is the current is going up slowly by 0.1 mA, though voltage is stable. In my exp during the past bench tests I performed with ordinary white led's, if the current for an led is not stable, the resistor is not ok. Could you help me with this doubt?
     
  8. John P

    AAC Fanatic!

    Oct 14, 2008
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    Something has to be wrong here. At 93mA current you should see a drop of .65V across the resistor. If that's true, then your 5V supply must be at least 0.5V higher than it should be (4.95 + 0.65). Is it possible that the resistor is 0.7 ohms instead of 7.0 ohms? That would make the numbers come out approximately right.

    Anyway, forget the voltage across the LED. The current-voltage relationship in the LED isn't totally predictable, but all you need to do is select a resistor that gives you the current you want. Just concentrate on getting the current right.
     
  9. T.Jackson

    New Member

    Nov 22, 2011
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    This is all OHMS law that you're talking about here. It is algebra.

    Transposing V, I & R.

    Whereas V is your supply voltage, I is your current through the LED and R is the value of the dropping resistor.

    VL is the voltage drop of the LED.

    V = I x R
    I = V / R
    R = V / I

    Therefore ...

    R = (V - VL) / I

    Your power dissipated across R would be ...

    P = (V - VL) x I
    = (n) watts

    or ...

    (V - VL)² / R

    Sometimes refered to as E rather than V
     
    Last edited: Nov 22, 2011
  10. T.Jackson

    New Member

    Nov 22, 2011
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    Am I right or am I rusty?

    I did score A grades once upon a time.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Sure, you did fine.

    We usually use the standard abbreviation for the forward voltage of a PN junction semiconductor aka diode, which is Vf, and I usually explain what Vf means.

    We also usually use the standard "E" instead of "V".

    I'll frequently spell things out a bit more, as many of our members are entry-level hobbyists - not exactly Barney-style, but detailed enough so they can understand why it works.
     
  12. T.Jackson

    New Member

    Nov 22, 2011
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    Vf rings bells.

    I had one teacher who was textbook and used standard exam papers and another who used to write his own and have questions in the test a long the lines of "draw a potato" I thought he was joking and I skiped the question and had marks taken off.

    The guy could draw the internals of some chips on the board off the top of his head for what that says. Most people thought that he was cRaZy.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Have a look at some datasheets to help bring your memory back. Optocouplers would be a good place to start. You'll see numbers like:
    typ Vf = 1.22 @ 10mA; max Vf = 1.3v
    and then they'll give maximum continuous current, and maximum current at a very low duty cycle.

    He understood the value of humor. :)
    People often feel very stressed when taking a test, they are "under the gun" so to speak.
    Throwing some humor in there does a couple of very useful things; it relieves stress, and stimulates peoples' memory and creativity.

    The line between genius and insanity is often quite narrow.
     
  14. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    Yes, you were right. The 7805 was a little bit past 5V. I replaced it and then used a 10 ohm resistor, now it runs on 3.98V@90mA which is pretty good! Thanks for that! Most of the 7805's give exact 5V, but not all, shud have checked again, my bad, ANyway thanks a lot!!
     
  15. T.Jackson

    New Member

    Nov 22, 2011
    328
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    LOL!

    I know a few of them.
     
  16. T.Jackson

    New Member

    Nov 22, 2011
    328
    14
    So you think that an IQ of 135 = a gifted person whilst an IQ of 139 = someone insane?
     
  17. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    @SgtWookie
    According to your experience, what would be the best way to find the Vf for an LED?? would running in constant current mode for the rated current and then measuring Vf do??
     
  18. John P

    AAC Fanatic!

    Oct 14, 2008
    1,632
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    I'll say it again--if your circuit depends on the exact forward voltage of an LED, you haven't got a good design. You can measure it for your own knowledge, and to make sure that your power supply can drive it, but you shouldn't depend on it being predictable.
     
  19. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Using a constant current circuit is the easiest way to determine the Vf at a given current and temperature.

    Note that the Vf will decrease with an increase in temperature; a negative temperature coefficient. If you don't have much of a voltage drop across your current limiting resistor, and the voltage supply has a low impedance, the drop in Vf over temp could lead to thermal runaway and destruction of the LED, or at least a significantly reduced lifespan.

    The more variable your power supply is, the more voltage you would need to drop across the current limiting resistor.

    Active regulation will give far more consistent brightness over a range of input voltages, but linear regulation can get to be quite inefficient.

    The most efficient type of current regulation is a switching supply. There are dedicated ICs that can make this type of regulator fairly easy to implement, requiring just a few parts. It's more expensive and complex to set up in the beginning, but the increased efficiency saves power via high efficiency over the lifetime of the circuit.
     
  20. T.Jackson

    New Member

    Nov 22, 2011
    328
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    Yeah all of these semiconductors are never the same unless from the same batch. Beta gain for transistors is often all over the place. Led brightness varies, sometimes by many mcd. Totally unpredictable.
     
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