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Small/cheap/easy 3V power supply

Discussion in 'Electronic Design' started by Ignoramus20785, Jan 31, 2007.

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  1. I have a chop saw with a laser beam that shows where the saw would
    cut. Which is nice. What is not nice is that this laser is powered by
    two separate AAA batteries instead of by 110V. It does not work as
    well when it is extremely cold, uses batteries, and is outright stupid
    (but I am sure that the manufacturer saved $1 to make some extra

    So, I would like to find some really small 3V power supply (I am sure
    that it would need to supply some milliamps for this laser), and stick
    it in the battery compartment. I am hoping that I can just find some
    four legged microchip that would do that, and would solder its 3V
    outputs to battery connectors.

    Would anyone suggest a particular supply? 120VAC --> 3.2 VDC?

  2. I found a 3.7V DC wall wart. I will see what is its actual open
    circuit voltage, etc, maybe I can rig up some simple voltage divider
    to reduce voltage a little bit.

  3. John B

    John B Guest

    A 1N4001 will drop .7V for you.
  4. This is VERY smart. I love your idea. I have about 100 of them, or
    similar diodes, if not more. Now it looks like a one hour project
    (time mostly spent making sure that the wire from wall wart is
    securely fastened to the chop saw).

    Thank you!!!!!!!!!!!

  5. Cool idea :) But 5% mains voltage change will be 5 V.....
    I'd personally get a small 9switching) AC/DC adapter (very small ones exist,
    and either combine the cables, or mount it on the tool.
  6. Well, if you have some URL for something, let me know. For now, I
    think that dropping .7 volt (or whatever I need, I will see what is
    OCV on this supply), will do what I want. My mains voltage is pretty
    stable at 123-125 volts. Probably would drop somewhat when the saw is
    actually cutting, but by that time I should not care about the laser

  7. Eh, one came with my Nokia cellphone I think it is 3.7 V.

    Better to use a series capacitor,

    ----||------------ a diode k-------- +4.3V --- diode ---- 3.6 -- diode --- 2.9 V
    |k |
    mains 5V zener ===
    |a ---
    | |
    ------------------------------------- 0V

    Can you calculate C1? For 60Hz and the current you expect?
  8. There are many problems with using a capacitor to drop line voltage, as
    discussed in the thread on the PIC16F84. It might be worthwhile to replace
    the original AAA batteries with rechargeable ones, and connect a charger
    full time. This will stabilize the voltage. However, NiCd and NiMH would
    provide only 2.4 to 2.6 volts, which may not be enough for the laser. A
    Lithium battery with a charger might be better.

  9. Paul, I will check with a scope what kind of waveform I get out of
    this power supply, that will shed some light on what I can do with it.

  10. Don't you think that is overkill? I have several applications that simply use
    a cap...
    I bought some remote conrolled light recently, it too has (I open everything up)
    the cap method.
    As do my PIR alarms... etc..
    Just calculate the current, use 1/jwc to calculate the cap, that is it.
    Make sure you pass AC and not DC, caps do not work on DC :-0
    I have used this method since the eighties in last century ;-)
  11. I assume he meant dropping the 3.7V to 3V.
  12. Chris Jones

    Chris Jones Guest

    He was not proposing dropping the mains voltage with diodes, he was
    proposing using a nominally 3.7V output from a Wall-Wart thing and dropping
    0.7V with a diode. That sounds much more sensible.

    I don't know how sensitive the laser module is to the supply voltage. If it
    is not critical (as one would hope since it is battery powered) then the
    above solution should be OK, otherwise it would be better to use a 3V LDO
    regulator, but these are not all that common/cheap compared to a LM317.
    The problem with the LM317 is that it needs about 6V input to put out 3V
    reliably. If he has a 6V wall-wart and a LM317 then that would be ideal.

    Whatever solution is chosen, it might be good to put plenty of edcoupling
    capacitance right across the battery contacts of the laser module, as I
    have read that they are easily damaged by spikes, static etc.

  13. Jim Yanik

    Jim Yanik Guest

    Use Energizer E2 AAA lithium batteries,they work in low temps.
    They also have a 10 yr shelf life.
    Or you could use a single CR123 lithium cell; rated at 3V,1300 maH
  14. I have used it as well, but I had a problem when I added an 18 ohm series
    resistor to limit line surges. Even with normal current of about 60 mA (60
    mW), a 1/2 W resistor showed burn marks and melted a nearby IC socket. The
    capacitor appears as essentially a short circuit to high frequencies and
    high dV/dT spikes, as may be common when used with a power tool. The zener
    might be damaged with no current limiting. An 18 ohm resistor might allow
    100 amps at 1800 volts, but that is better than the ESR of the capacitor,
    which is probably 0.5 ohms or less. A series inductor may be even better,
    except for the possibility of resonance.

  15. Must have been confuesd by 'have a hundred'.
  16. OK I see your point.
    maybe here in the Netherlands the mains is much cleaner (I scoped it many times,
    no huge short peaks here).
    In the remote light switch I bought they have (looking) they use 100 Ohm
    2 W (wirwound??) and 330 nF (at 230V 50Hz).

    So 330 nF = an impedance of 1 / (6.28 x 50 x 33.10^-8) = 100 000 000 / 10363 =
    the current would then be 230 / 9650 = about 24 mA.
    25 mA in 100 Ohm series is about 6 mW, so no heat in normal use.

    I have always used a series R, and also a series fuse (this thing has what looks
    like a fusable PC track.....
    A 1kV short spike will see 100 Ohm, and power will be dissipated in the series R.
    I forgot to look for the zener, but it looks like all 200mW type stuff in there.
    it switches a relay in fact.

    I have also used this system by using a series R, cap, bridge, zener to 12 V,
    and then mains isolation with a simple self oscillating one transistor
    ferrite core converter 1:1, with only a few turns primary and secundary wound
    with normal wire :)
  17. neon


    Oct 21, 2006
    The whole thing is ridiculous. 3.7 or 3v is the same for a diode it will just get better lumination that is all. These diodes require min voltage and min current to operate. A .7v increase will not make it to blow up. besides you want your 3 v add the proper resistor to I*R drop to get the brightness that is required. The manufacture use battery which make sense to me maybe another 1.5 v will blow it up not because of current but because exceed power dissapation. [but will not last long] use any wall transformer that you find and limit the current I*R to whatever it uses when working properly with 3v. end of story.
    Last edited: Feb 1, 2007
  18. Yes, that is my plan.
    I think that it ought not be too sensible, being battery powered.

  19. ehsjr

    ehsjr Guest

    Determine the current I your laser uses.
    Then do this:

    ------- -----
    | Wall +|---Vin|LM317|Vout---+
    | Wart | ----- |
    | 6V DC | Adj [R] R = 1.25/I
    | | | |
    | | +----------+---[Laser]---+
    | | |
    | -|----------------------------------+

    You can measure the current your laser draws
    from the batteries. The LM317 will limit the
    current to that value in the above configuration.
    The input supply needs to be a few volts higher
    than the 3V you need, thus a 6V wall wart. However,
    you can use 6 volts or higher, so if you have a
    12V wall wart on hand, you can use that. It makes
    the LM317 hotter - it will dissipate (Vin-Vout)*I
    watts. But even the small LM317 TO39 package will
    handle over 1/2 watt, so you should be ok, depending
    on the current your laser needs and the voltage
    of the wall wart.

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