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AC/DC Power Adapters: Input & Output Specifications

  1. Jul 7, 2011 #1
    in regards to power adapters, mainly the brick-type ones that come with laptops and external hard drives, how is the output current greater than the input current?

    for example, i see a laptop power brick with the following:
    input 100-240V~1.7A 50-60Hz
    output 18.5V == 3.5A 65W


    how exactly does this work with a branch current method of analysis if the branch going into the power brick is 1.7A? i see that you can pull current from voltages, but if only 1.7A is going into the network, where are the other carriers coming from to bump up the output to 3.5A?

    i've been trying to think this out but i can't arrive at anything. what i left off on is that 1 ampere = 1 coulomb per second. if the input is 1.7A = 1.7 coulomb per second operating at 50-60Hz, perhaps it actually collects enough current to drive 3.5A?

    i'm a bit lost.


    thanks!
     
  2. jcsd
  3. Jul 7, 2011 #2

    davenn

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    Gold Member

    yes but its 1.7 A at the mains voltage 100 - 240V. But you then drop the voltage and you can increase the current. I dont have the formulae for showing the workings. Some one else is likely to be able to do that.

    Same thing happens around the other way. I have a PSU that can supply for example... 12V at 5A. If I double the voltage, I'm going to 1/2 the current capability
    That is. The PSU has a particular max Wattage rating V x I (current) = Watts if you change V then I is going to change to keep that max Wattage

    Thats a basic way of looking at the problem :) I'm sure others could explain it in more detail :)

    cheers
    Dave
     
  4. Jul 7, 2011 #3

    MATLABdude

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    Welcome to PhysicsForums!

    There's a transformer inside that converts the high voltage AC to lower voltage AC (which is fairly close to the output DC voltage). This lower AC voltage is then converted to the final DC voltage.

    The formula for power, given voltage and current, is:
    [itex]P=I*V[/itex]

    Since power is conserved (more or less, some of the energy is lost as heat during the transforming), you can find the input and output voltages:
    [itex]P=I_{in}*V_{in}=I_{out}*V_{out}[/itex]

    More on this topic at:
    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/transf.html
     
  5. Jul 8, 2011 #4
    Great! thanks a lot!
     
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