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Drop from 5v to 3v?

Discussion in 'General Electronics Discussion' started by bonedoc, Apr 17, 2012.

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  1. bonedoc

    bonedoc

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    Dec 21, 2011
    So, I have a 3V 5mW laser. But, my rail is 5v. My board is really cramped. Its a stupid question, but since my laser is a load and has resistance, cant I just put 1 resistor in series with it so that the laser gets 3V and the resistor gets 2V?

    Laser resistance

    V= IR

    R = V/I

    W = VI

    I = W/V

    R = V^2/W = 3^2 / .005 = 1800 Ohms resistance of the laser load

    With the laser connected to the 5V rail and then in series with a 1200 ohm resistor to ground, that divides it up as 2V on the resistor and 3V to the laser. Am I correct. Just to make sure, the current flow would be .001667A through this, and the laser is .005W/3V = .00167A. Is this fine?
     

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  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,813
    2,753
    Nov 17, 2011
    In theory: yes.
    In practice: not a good idea.

    A constant current source is a much better solution. Many laser diodes also have an integrated photodiode which can be used by a laser diode controller to regulate the current such that a very stable light output is ensured.


    Some reading: http://www.teamwavelength.com/info/laserdiodedrivers.php

    Harald
     
  3. bonedoc

    bonedoc

    122
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    Dec 21, 2011
    Thanks for the link! My setup is very simple compared to that. It is a self contained laser module that just needs 3V...king of like the guts of a laser pointer. My 5V is coming from a voltage regulator. So, I think it will be reliable. Is this reasonable?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    You can find 3.3V regulators that will operate from 5V -- three terminal devices. Just get one of these to produce a regulated 3.3V rail.

    It's the simplest solution (there may be cheaper ones, but not by a great deal, and if space is an issue, one of these won't be that much larger than a resistor. If you can, a surface mount to TO-92 case version would be great.

    Something like this.
     
  5. bonedoc

    bonedoc

    122
    0
    Dec 21, 2011
    Ok, so I have 2 questions.

    -Will the resistor method work reasonably?

    -If the laser spec says it needs 3V, will 3.3V not harm it? It seems like there are a lot of 3.3v versions, and a few 3v versions.

    I accidentally did apply a full 5v to it when I first got it and it worked just fine.
     
  6. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    I would say a few diodes in series is just as simple ;)
     
  7. bonedoc

    bonedoc

    122
    0
    Dec 21, 2011
    Hmm...I do have some diodes that are 1.7v forward voltage....but then I was thinking....I have an LED that indicates when it is on. I could just use that in series. Im sure I could jut get some 2V drop leds too.

    This would work voltage wise. However, not sure about current situation. The seller (in china) said the laser is actually 40mA. The max LED I will use is 20mA, so my 2n2222 has no trouble providing this. But....if the led and laser are in series, will a current resistor be needed for the LED sake? If so, ho do I make sure I am getting enough current for the laser too?
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

    11,813
    2,753
    Nov 17, 2011
    Since it is a module chances are that the current limiting is done within the module. So just providing 3 V or 3.3 V will be fine. If you use a series LED to drop the voltage from 5 V to 3 V you will not need an additional resistor but an LED that is rated for at least 40 mA.
    The module will limit the current to 40 mA and since it is a series circuit the same current flows to the external LED.

    If your external LED caanot stand 40 mA, you can add a resistor in parallel. I know that sounds very unusual for operating an LED but it is the right thing here since the module acts as a current source (not voltage source). The resistance should be R=Vled/(Imodule-Iled). SInce module current and LED forward voltage are not well defined, this can be only a coarse approximation, but will be good enough to detract current from the LED.



    Harald
     
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