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I should probably know this but... LED power question — Parallax Forums

I should probably know this but... LED power question

GICU812GICU812 Posts: 289
edited 2008-08-27 07:52 in General Discussion
Okay, Ive got a laser diode module, AKA, a hacked up laser pointer. It originally took 2 x 1.5 v batteries, or 3v. My whole setup runs on 5v, without knowing the mA draw, or value of whatever resistance is allready in there, how can I run this from 5v? Im okay with guestamating, its not an expensive module, and I would error on the side of caution, since I dont really need the 1 mile range or whatever, just a few hundred feet. So loosing a little output wont hurt me.

Advice?

Thanks

·

Comments

  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-08-24 13:48
    3V_LD.jpg
  • GICU812GICU812 Posts: 289
    edited 2008-08-24 20:53
    Hmm.. I was hoping for something simpler.

    Actually, looking at it, it appears to have a 360 Ohm resistor. So its 3v with a 360ohm resistor... So that means the diode is using 83ma?

    The diode has a resistance of 120k. Maybe someone can explain how I can determine the specs on this.

    Does that mean total resistance is 480 = 63ma?

    Post Edited (GICU812) : 8/24/2008 9:05:07 PM GMT
  • GICU812GICU812 Posts: 289
    edited 2008-08-24 21:15
    Then again, I guess it doesnt get much simpler than that schematic above, I mean I have to have a transistor to turn the the on and off anyway, and I can just use a pair of resistors as a voltage divider.

    Still, Id like to know if its possible to calculate the resistor needed to drive it directly from the specs I have found. One of the "in the future" things.




    Post Edited (GICU812) : 8/24/2008 9:56:40 PM GMT
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-08-24 21:52
    I didn't want you to spoil your pointer and I anticipated your need to switch it via an output pin (that's what it's all about, afterall.)

    "Id like to know if its possible to calculate the resistor needed to drive it directly from the specs I have found."

    If it's simply a matter of a "diode" and a resistor, then I'd go about it this way:

    whatsmyline.jpg

    ·Post Edit -- R = (VR1 + 2V) / mA1


    Post Edited (PJ Allen) : 8/24/2008 9:59:43 PM GMT
  • metron9metron9 Posts: 1,100
    edited 2008-08-25 05:27
    PJ, Now I am confused. True I don't play with transistors much but could you explain, I thought you usually tie the Emmiter to ground and put the load to the collector so I went back to my What is a microcontroller book, (I still have it) on page 263 and 264.

    I guess i just learned a few things about emitter followers. Reading the chapter in Art of Electronics right now. Cool how they can go negative voltage as well to a point.


    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think Inside the box first and if that doesn't work..
    Re-arrange what's inside the box then...
    Think outside the BOX!

    Post Edited (metron9) : 8/25/2008 5:56:07 AM GMT
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-08-25 12:24
    I've been extolling the wonders of the utility of the common-collector amplifier ("emitter follower") for nearly as long as I've been on the Parallax Forum, a lone voice crying in the wilderness.·

    In case you missed it, the "emitter follower" puts the heft in a voltage regulator.

    You couldn't do better than to whump up that circuit on your breadboard and test it out.· You'd do better, experiment-wise, to add a resistor (470&#937[noparse];)[/noparse]·in series between the pot and the input voltage, so that you don't go full-tilt (0&#937[noparse];)[/noparse].·

    [noparse][[/noparse]Dwg attached.]
    EF_exp.JPG
    401 x 247 - 10K
  • metron9metron9 Posts: 1,100
    edited 2008-08-25 14:14
    Should come in handy in a circuit I am playing with Resistor/PWM DAC http://www.k9spud.com/traxmod/pwmdac.php

    I will post my sin wave and audio clip of it when it's ready.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think Inside the box first and if that doesn't work..
    Re-arrange what's inside the box then...
    Think outside the BOX!
  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2008-08-25 18:00
    The transistor is in a 'current limiting' mode. That is the beauty of the circuit.

    P. J. correct me if I am wrong.
    From what I understand, the LED needs a minimum voltage to turn on - like a normal light.
    But, it also needs to have the amount of current limited or it just burns up.

    With a conventional light, all you have to worry about is the correct voltage. The current takes care of itself. No so with the LED.

    Post script
    Thanks for the circuits P.J. as I have 3 laser pointers that I have been thinking of hacking into and using for robot navigation. Your schematic allows me to fine tune their output.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    It's sunny and warm here. It is always sunny and warm here.... (unless a typhoon blows through).

    Tropically, G. Herzog [noparse][[/noparse] 黃鶴 ] in Taiwan
  • StarManStarMan Posts: 306
    edited 2008-08-25 19:13
    Kramer,

    My experience in with LEDs is limited to lighting applications, not laser or indicator LEDs.· And my background is mechanical engineering so it has been a struggle.· Here is what I've found.

    An LED will begin conducting at a certain minimum forward voltage but after that there is a pretty steep curve of voltage vs current, meaning a small increase in voltage will produce a relatively large increase in current.· A series resistor will flatten out the curve somewhat.·

    LEDs typically have a safe current limit where the heat·produced in the diode junction is is safely drawn away.· Most of the low power LEDs have a safe current spec at 20 mA and the forward voltage at that current is usually published in the spec.· The heat produced in an LED is usually drawn away from the junction through the cathode.· The life of the LED is dependent on how well the heat is dissipated.·

    In my designs, I·typically try to limit the current to 17 or 18 mA and connect the cathode directly to·a ground plane of my pcb.· For the low power LEDs, current is limited with a regulated voltage supply and the correct resistor (calculated with PJs equation above).

    High power LEDs (Luxeon et al) have different construction and therefore different requirements.· Finned heat sinks and constant current sources are typically required.

    Chris I.
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2008-08-25 23:05
    The "emitter follower" is a current amplifier; its voltage gain is (always) < 1, but it's called 'unity' anyway.

    So far as an LED goes, it's a current-driven device, and that current must be controlled.· Once it's conducting, the voltage developed across it will stay fairly constant as current increases till it goes "pop" [noparse]/noparse]I[sub]LED (max)[/sub.·
  • GICU812GICU812 Posts: 289
    edited 2008-08-26 00:20
    I'll have to try that once I get a new laser.... I needed to tack a little piece of metal on my mount, thought it was far enough away to be okay, and I quenched it immediately but it must have melted the lens a little or something because the beam is very distorted now. Oh well.

    As for the circuit, I tried a LM317 voltage regulator, but my setting resistors got stupid hot for some reason. I dont know why, but the diagram on the back of the package, as always, is more of a guess, they never tell you which side of the device is facing you GRR!

    Im trying to calculate the values for a voltage divider in place of the pot in the above schematic. Does R1 :300 and R2 1k sound about right?
  • metron9metron9 Posts: 1,100
    edited 2008-08-26 05:28
    Use google to find a datasheet for those, they have better diagrams usually.

    Also you can use a cheap LED for your initial experimenting. You can find out what the forward voltage is by slowly increasing voltage to the laser till it starts to light up a tiny bit, (point it at a white piece of paper!) You must have a voltage meter I would think so you can measure the current if it has a current mode setting or measure the voltage across a resistor in the circuit to calculate the current.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Think Inside the box first and if that doesn't work..
    Re-arrange what's inside the box then...
    Think outside the BOX!
  • GICU812GICU812 Posts: 289
    edited 2008-08-26 06:36
    I can try that still with this laser LED, the optics are just distorted.
  • GICU812GICU812 Posts: 289
    edited 2008-08-27 07:52
    Well I picked up 2 more today, found a couple of things at $2. One was a laser\bright LED\pen, the other just a laser and bright LED.

    The pen broke when I took it apart, the laser diode came out of the end and fell apart.

    The other one, I disected neatly, removed the LED and switches and trimmed the circuit board. Soildered on the positive and all was well. Then I soldered on the negative, and something went wrong, damanged the Laser, now its dim. The negative lead was pretty far from the laser, but I must have overheated it or something, I tested with the led just touching there before, and it worked fine.

    Its probably okay, both these were diffrent from the first one, the optics were almost exactly like a LED housing, just spaced away from the juntion. Anyway, when I shone it on the house across the street, it appeared the dot was 6" or bigger, so I think it would have diffused too much in my setup, shining on the fenceposts.

    I ordered 5 more of the first kind I had, $13 shipped for all of them, not bad.

    These new ones are going to be 4.5 volt though, so I'll have to adjust power to compensate.
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