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How to get 1.5 volts out of a wall-wart power supply's 3 voltoutput...I am new to this group I belie

Discussion in 'Electronic Basics' started by Daniel, May 28, 2013.

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  1. Daniel

    Daniel Guest

    Hi, I have recently been trying to get 1.5 volts (or anything other than the power supply's rated 3 volts for that matter) out of a wall-wart 3 volt power supply. The wall-wart power supply is dc in output at 3 volts and evenif I connect a 300 ohm resistor in series with the wall-wart power supply and the load (a 1.5 volt clock, old style analog clock)and when i measure the voltage with the voltmeter in place of the load (but remember the resistor is still in series) I keep getting 3 volts dc measured. I even put a 20 ohm resistor in parallel with the 300 ohm resister and still no change in measurement. I bought a older style wall-wart with a 1.5 volt setting and itmeasures at 3 volts with the voltmeter in series with the output leads. I can't get 1.5 volts (or anything other than 3 volts for that matter) no matter what I do and if with the older style wall-wart power supply I measure at the output leads with that power supply set to 12 volts, etc. the measured voltage is very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had tohave been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!
     
  2. Phil Allison

    Phil Allison Guest

    "Daniel"Hi, I have recently been trying to get 1.5 volts (or anything other than the
    power supply's rated 3 volts for that matter) out of a wall-wart 3 volt
    power supply. The wall-wart power supply is dc in output at 3 volts and even
    if I connect a 300 ohm resistor in series with the wall-wart power supply
    and the load (a 1.5 volt clock, old style analog clock)

    ** Does this clock normally use a 1.5V, AA cell ?

    Does the cell last about 1 year - assuming Alkaline ?


    The math shows that dropping 3 volts to 1.5 and allowing a current of .08
    amps requires ...


    ** How did you get the 0.08 amps figure ?

    A 1.5V, AA cell would last about 1 day at that rate.

    FYI:

    Typical battery wall clocks consume about 150 micro amps - or 0.00015
    amps. Two resistors of say 1000ohms wired in series ACROSS the 3V supply
    will divide the voltage in half.

    Add a 47 microfarad, 10 to 25volt, electro cap across the clock and you have
    it done.


    .... Phil
     
  3. Guest

    Bottom posted.

    Hi Phil! Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5 volts. I accidentally lost track while typing and combined two projects, one powering a 1.5 volt AA mp3 player with a wall-wart power supplyand of course the mentioned analog old style 1.5 volt alkaline wall clock.The mp3 took .08 amps and as you mentioned the all clock uses a lot less current than that. I don't follow your math for the solution but I am very happy with what you said so I can complete this project soon - thanks a million! Again, thanks a million Phil! By the way – I previously solved the mp3 issue by powering it with a wall-wart power supply that has a usb power source socket so I am happy with that. Thanks!
     
  4. Phil Allison

    Phil Allison Guest

    <>
    Daniel wrote:

    Hi Phil! Yeah, the clock operates on one AA alkaline battery for about 1
    year at 1.5 volts. I accidentally lost track while typing and combined two
    projects, one powering a 1.5 volt AA mp3 player with a wall-wart power
    supply and of course the mentioned analog old style 1.5 volt alkaline wall
    clock. The mp3 took .08 amps and as you mentioned the all clock uses a lot
    less current than that.

    ** Only about 530 times less.


    I don't follow your math for the solution

    ** It's really simple.

    You need a 1.5V supply that can deliver 150 microamps average. The clock
    actually takes short pulses of about 10 times that current every second,
    hence the 47uF electro to handle the pulse current.

    The two 1000 ohm resistors just split the voltage coming from the wart and
    have low enough resistance so the average load current is not enough to make
    any difference to this. You can connect the clock & electro across either
    resistor, long as the polarity of both is right.



    ..... Phil
     
  5. Phil Allison

    Phil Allison Guest

    <>


    Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
    volts.


    ** Changing to a wall wart will cost much more in electricity than using AA
    cells does.

    PLUS if the AC power ever goes off for a while, the clock will stop and
    thereafter show the wrong time.

    Till you figure that fact out.

    Wot a dumb idea.


    .... Phil
     
  6. Jamie

    Jamie Guest

    s very high (19 volts for the 12 volt setting). Now, I know that without a load certain power sources don't measure the same as with a load but if I measure the voltage (I wish i didn't have to do that as the voltage measured was 3 instead of the 1.5 the clock is rated for, I might have done tiny damage to the clock for all I know), if I measure the voltage across where the battery would normally be it measures at 3 volts! SO I had to have been taxing the clock with it running at 3 volts instead of 1.5! Anyways - can someone shed some light on this? I am at the minimum above average in electronics skills and this problem I have been having has been driving me bonkers. The math shows that dropping 3 volts to 1.5 and allowing a current of .08 amps requires 18.75 ohms and I provided that like said above (300 ohms in parallel with 20 ohms equals 18.75 ohms) and it still measures at 3 volts dc! any help appreciated before I tear my hair out...Thanks, I think!

    The wallwart most likely isn't a regulated type and it most likely has
    a cap in there, which gives you a higher reading.. But putting a load
    on the supply normally brings things down..

    If you start with a 1.5 RMS (AC internally), that equates to ~2Volts
    after rectified and filled wit out load. So lets try something different.

    For what you're trying to do does not require anything very special.

    Using 2 silicon diodes in series will drop your voltage ~ 1.2 volts

    for every Si diode in series there is an ~650 mv drop.

    use diodes that can handle the current rating..

    Jamie
     
  7. Phil Allison

    Phil Allison Guest

    "Tim Wescott"

    ** Huh ???

    Red LEDs have forward voltage fro 1.7 to 2.2 volts - depending on current
    flow.

    The 2 R and 1C solution I posted is the only one certain to work.

    But " work" means increased running cost and far less reliable time
    display.


    .... Phil
     
  8. amdx

    amdx Guest

    Yep, Daniel dodged a bullet.
    His Usenet life could have been ruined!

    Mikek :)
     
  9. Phil Allison

    Phil Allison Guest

    "David Eather"

    ** The voltage is too low at 1.1V to 1.3V to reliably operate a 1.5V clock -
    often they will not run on a NiCd or NiMH cell.

    Three 1 amp diodes and a 330 ohm resistor might be the go.



    .... Phil
     
  10. Guest


    It will???

    Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.

    Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).

    1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.

    $0.27 >> $0.012


    No argument there

    M
     
  11. Guest


    Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD

    One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)

    M
     
  12. Guest

    On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:

    ....

    Yes... looks like if the wall wart wastes more than 0.2 watts, it will cost more than the battery... hmm...



    Yes! I also was struck with the genius of Phil's suggestion. Quite elegant, and needs few parts.

    Michael
     
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