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From 12V DC to 5+ V DC How to?

Discussion in 'Electronic Basics' started by [email protected], Aug 24, 2006.

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  1. Guest

    Hi all,

    I should probably know this but I am not sure.
    I have a motorcycle that has a 12V output jack. I have a device that I
    need to connect to it which is rated for 5V(+). I also have another
    device which is a 12V device.
    I need to connect both devices to the Jack but I need to reduce the
    voltage for one of them from 12 to 5.
    I am guessing this is done with a voltage regulator but I haven't used
    one so I was hoping to find out if this is the right solution and if so
    how to do it so its nice and clean. Is there voltage regulators that
    are nice and modular that I can put in between the source and the
    device?
    Do I need to consider anything else?

    Thanks
     
  2. Guest

    I should also point out that the power supply for the device shows the
    following:
    OUTPUT: +5V 1.5A
    Output Power: 8W MAX
     
  3. Use diodes in series. Each will drop the voltage by about 0.6 volts.

    R
     
  4. ehsjr

    ehsjr Guest


    You have to use a voltage regulator, because the input
    voltage will increase while the bike is running. A simple
    series resistor or a group of series diodes won't work.
    Here's a circuit:

    10 W -----
    +12 ---[2.5R]---+---Vin|LM350|Vout---+---> +5V
    | ----- |
    | Adj |
    [.1uF] | |
    | +--[270R]--+
    | | | +
    | [820R] [1uF]
    | | |
    Gnd ------------+---------+----------+

    You will need a good heatsink on the LM350, something
    rated 7 degrees C per watt or better. Digikey HS276-ND
    is rated 2.5 degrees C per watt and costs $1.44 - it
    will keep the LM350 under 60 degrees C, which is great.
    The 2.5 ohm resistor needs to be 10 watts. The 270 and
    820 ohm resistors can be 1/2 watt. The 2.5 ohm resistor
    will dissipate 5.625 watts, and the LM350 will dissipate
    close to 9 watts, worst case.

    For a smaller and non-heat producing solution, you could
    use 445-2435-ND from Digikey. It is a DC-DC converter
    module that accepts 9 to 18 volts in, and produces
    5 volts out at up to 2 amps. But it is more expensive.

    Ed
     
  5. default

    default Guest

    Probably no

    That is a good "cheap and dirty" way to reduce the power. They don't
    cut down the voltage they just switch it on and off rapidly - the
    average power is less and efficiency is high - but it won't work for
    devices that are sensitive to voltage or anything with a large filter
    capacitor - the cap will charge to the peaks and you'd have 12V going
    into the 5V electronics.

    It can be made useful - with an inductor, pulse width modulator
    circuit, and a degree in electronics - and cheaper to design from
    scratch.

    That particular gizmo is a controller for a brushless motor. It is
    designed to supply both a variable voltage and variable frequency to a
    small three phase motor to allow it to run from batteries - very
    efficient way to control a motor speed - but not what you say you
    want.

    If you have a 5 volt motor and want to run it on 12 V just build a
    simple single phase pulse width modulator to lower the average power
    to the motor (a brush type motor doesn't care that it's getting 12
    volt pulses for 40% "ON TIME" versus the 5 volts it is rated at)
     
  6. default

    default Guest

    Are you sure you need 1.5 amps? A lot of simple there terminal
    regulators will fit the bill - and a lot of them only put out 1 amp.
    You can get up to 5 amp three terminal regulators but cost and
    complexity increases.

    Lot of places on a bike would make a good heatsink - the old Triumphs
    used to mount a big 50 watt zener diode to the air intakes for the
    carbs - that was the "regulator" for the electrical system.
     
  7. HKJ

    HKJ Guest

    Your can use this program to design a power supply with a voltage regulator:
    http://www.miscel.dk/MiscEl/miscelPowerSupply.html

    It might be a good idea to add some overvoltage protection before the
    regulator. A power resistor in series with the supply and a 20 volt
    transient suppressor diode to ground will probably be enough.
     
  8. Guest

    okay what about this thing will this work? The only thing I am worried
    about is that I am going from 12 to 5 which is a 7V drop. I am told
    that will generate lots of heat. But I would think this device would
    take care of that

    http://www.dimensionengineering.com/DE-SWADJ.htm
     
  9. ehsjr

    ehsjr Guest

    That's a *nice* looking device. Don't know if one will
    work for you, tho - you might need 2. Your original
    power supply is capable of 1.5 amps - maximum for the
    device at the url is 1 amp. If your current requirement
    is < 1 amp, one should work fine. The url says they can
    be put in parallel, so if you use 2 in parallel you can
    exceed 1.5 amps.

    Heat is not a problem with switching regulators like it is
    with linear regulators, and DE-SWADJ is a switcher. Sounds
    like you would be in good shape with one (or perhaps 2)
    of them.

    Ed
     
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