Help Needed: Will this work?

Discussion in 'The Projects Forum' started by viperstd, Apr 3, 2008.

  1. viperstd

    Thread Starter Member

    Apr 2, 2008
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    Hi! My first post. I have spent a while digging around here since I was pretty stuck on how to make my project for a "reasonable" amount of money and I am completely new to electronics.

    I programmed a 16F84A to make patterns and everything worked great up to there! However, I want to take the TTL from the PIC and use it to control some much higher load sources. I kept coming back to SSRs since I knew relays from cars, but they were very expensive and big. This is what I came up (thanks to the search function here!!! :D ):

    1. Take the 5V TTL and put it into a TLP250 optocoupler (Toshiba Photocoupler) ~$1.90

    2. Take the 10-15V DC from my power source (aka a car) and send it to an LM317 (national semiconductor adjustable regulator) ~$1.00

    3. Use the current regulated out from the LM317 to my load (3 luxeon rebels in series) $5.00x3

    I would like to know if this will work. Thank you in advance!

    p.s. sorry about the napkin-esque drawing here and please be gentle, this is my first serious project and I slept through my only two EE classes about 5 years ago :S.

    [​IMG]

    [​IMG]
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    I recognize the LM317 in constant current regulator mode. I've used it myself. One thing that jumps out though, are you really using .7A for those LEDs? That is a magnitude too big, LEDs need around 20ma to turn on entirely. That is 0.02 Amps, .07 would probaby be OK, but I don't think the LEDs in the circuit are long for this world. Figuring they drop about 2.5 Volts each they are probably disappating around 2W in the current setup, over 1 W if they drop 1.5 volts each. That is enough to burn your fingers if you touch them, definately not something LEDs should do. I'll have to look up the TLP250 shown before I know anything about it.

    Those kind of circuits are always fun to play with, aren't they? Nothing wrong with the drawing, it is definately good enough.
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    Hi Bill,
    Welcome to the year 2008.
    Luxeon Rebel LEDs are extremely powerful and bright. If they are heatsinked properly then you can feed them 1A continuously! There are many Chinese copies on E-Bay that blow up.
     
  4. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    Thanks for the quick reply. The 700mA is correct! I think there is even a 1.2A version.

    One major question that arose looking at it is:
    Where the heck am I going to find a 1R74 ohm resistor in .... ..ummm 10W (~.7A*15V)? How do I calculate the resistor (R2) wattage?

    Do such resistors exist? I would like to not run 10 in parallel :S.
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    Ohm's Law simply calculates ohms, current and voltage.
    The Power Law simply calculates power (heat) in a resistor.

    A resistor operated at its max power rating is extremely hot (almost incandescent).

    25W resistors are made in a finned aluminum heatsink case that can be bolted to a larger heatsink.
     
  6. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    ... what I was really asking is,
    "what voltage does R2 see?"
    "where do I find less than 10ohm resistors?"

    V=IR, P=IV=I^2*R... I am comfortable with that. :D Intuitively (probably a mistake with my level of understand here), I would think that since I know my current will be 689mA and I know that each LED will drop ~3.85V:
    V=3.85*3=11.56V -->P=.689A*11.55V = 7.96W.

    Can I buy a 1.74ohm 10W Resistor? where and is it cheap?

    Thanks in advance!!
     
  7. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    It might be clever to integrate the resistor into the PCB design, since it may save a lot on costs in the end. I thought, since you're coupling the led to a heatsink, you can share the heatsink.

    Otherwise, you can choose from a lot of places, caddock is a pretty popular one

    Steve
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Actually, your constant current circuit's power dissipation is going to vary a bit.

    When you first start your vehicle, the alternator may put out up to 14.5v to quickly recharge the battery. You'll need a decent-sized heat sink for both the LM317 and the resistor.

    Allied carries Ohmite 20 series vitreous enamel conformal coated wirewound resistors in various wattages.
    http://alliedelec.com
    You could use a 1 Ohm 10W (296-0076, $1.56) and an 0.75 Ohm 10W (296-0978, $1.90) resistor in series to get pretty close. You could actually reduce the wattage for the 0.75 Ohm resistor, but it might be convenient to have them both the same size. Oh, those prices were from the 2007 catalog.

    Rather than using a constant current source, you might investigate using a PWM circuit. That would considerably reduce your power dissipation problem. Have the PWM circuit intercept/modulate the ON signal between your PIC and the TLP250.
     
  9. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    I don't follow the resistor comment, however, caddock is a good lookout. I didn't know that site. I see they have MV311 0.1-50ohm 10W resistors. Perfect. ;)
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    A 1 ohm resistor + 0.75 ohm resistor in series is 1.75 ohms, and the wattage is split between them. I've had resistors dissipating high wattage melt the solder holding them at considerably lower ratings, so you'll need to be sure they are way off the PCB and lots of lead length for heat disappation. This probably explains the heat sink comment, it would be nice to have a place to dump the waste heat.

    The LM317 has a TO3 case style, which is a lot easier to heat sink than the TO220 shown. Just a thought.

    Learned something today, I hadn't seen that flavor of LEDs before. Impressive.
     
  11. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    I did expect that, but I am not particularly concerned if I am dissipating 7W (.7A*10V {dead battery}) or 10.5W (.7*15V {overzealous voltage regulator on the alternator}).

    Thanks! I will check it out.

    Well, maybe a bit more information is needed. The signal coming from the PIC is already PWMing the signal. The end-project that I am working on is for an emergency signaling device for my father's (a firefighter) personal vehicle. Rather than buy a commercial one, I wanted to make one.

    I am strobing the individual light segments at varying duty cycles (up to ~45Hz strobe). Additionally, flash patterns mean that any given light is not on for more than .4s at a time.

    Or did I completely misunderstand you? Again, I am very new to this and I jumped in the deep end off of the 6m board w/o my floaties.

    Cheers
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    From what I've seen it's been pretty well thought out. I'm looking forward to hearing how it works out.
     
  13. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    Ok... am I calculating the power dissipation in R2 correctly.
    Assuming that I have 0.689 A to the string of LEDs, and each LED sees a 3.85V drop, that leads to (3*3.85)*(0.689) =~ 7W...

    or is the voltage seen by R2 the difference between the supply voltage (10-15V) and the voltage drop on the leds...
    @ worst case: ~15.0V-3*3.85V->3.45V*.689A= 2.38W
    @ (12V-3*385V)*.689 = .31W
    What happens if the auto voltage drops below 11.55V in this case? Does my seen current just drop, or will the LM317 crap out?

    I know that the LEDs are going to eat up some power no matter what... but they are only 1W each. It seems wrong that I am going to have 10W in the resistor and 3W on the series of LEDs. Seeing as I am planning on having 7 segments, that would be a whopping 91W! That seems way out of hand. How can I get it closer to the 21W from just the LEDs?

    Thank you again.
     
  14. Wendy

    Moderator

    Mar 24, 2008
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    The 317 stops regulating at the calculated current, but you still get the maximum current it can provide, which is less than the regulated current. No problem.
     
  15. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    Graceful failure! I like that!
     
  16. Wendy

    Moderator

    Mar 24, 2008
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    It's always nice when parts cooperate in a design.
     
  17. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
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    just noticed that on my LM317 the Vout and ADJ are swapped. It also seems that if I assume that the LEDs are dropping 3.6V per, the Vref is 1.2V and the the LM317 drop is 1.5V, that I am not going to have enough juice to run this at 700mA. (13V-1.5-1.2 = 10.3V remaining. /3leds = 3.4V/led).

    Do I need to go to 2 LEDs per segment? I'm not sure if my logic was sound, but I found that I would end up with ~3.4V per LED as shown, and via the Lux datasheets, that is only about 580mA. I will be losing quite a bit of luminosity at that current. Should I just go for broke and run 2 per segment at 800mA?
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    Add all the voltage drops:
    2V for the LM317, 1.25V for the current-setting resistor, 1.8V for the controller. 7.7V for two LEDs. The total minimum input voltage needed is 12.75V.

    The resistor has only 1.25V across it, not 15V. So it dissipates 0.7A times 1.25V= 0.875W. Use a 1.5W or 2W rersistor.
    The LM317 has 0.7A and a max of 4.25V across it. So it dissipates a max power of 3W. Use a little heatsink.
     
  19. viperstd

    Thread Starter Member

    Apr 2, 2008
    13
    0
    doh... I forgot the drop across the TLP250. Thanks. This is turning into a mess now. I am starting to think that I would be better off with a larger number of traditional LEDs instead of 2 k2s per segment.
     
  20. Pich

    Active Member

    Mar 11, 2008
    119
    4
    The voltage across R2 is constant at 1.2 v (enternal reference of LM317) tthrefore R2 would dissipate W=E*I 1.2*.7=.84W. The rest is dropped by the LM317 and the diodes.
     
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