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Thread: Voltage/resistance calculations

  1. #1
    Does he need a reason? Funkstar's Avatar
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    Voltage/resistance calculations

    I used to do this all the time back in school and uni, but thats so long ago, i really have no idea any more

    I need to drop 5v to 3v, what size of resistor do i need?

    or better yet, can anyone point me to a site that has all these calculations/formula listed.

    thanks ever so much

  2. #2
    lazy student nvening's Avatar
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    r=v/i

    v=Potential Difference(volts)
    i=Current(amps)
    r=Resistance(ohms)
    (\__/)
    (='.'=)
    (")_(")

  3. #3
    Does he need a reason? Funkstar's Avatar
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    so i need to know the current of the device i want to power?

    bugger

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    What "device" is it?
    LED perhaps?

  5. #5
    Does he need a reason? Funkstar's Avatar
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    nope. it's a Trinloc USB DAB reciever. It's going to be used on my HTPC and i see no point in using the power brick that came with it when i already have a molex breakout panel on the back for powering external drives.

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    Senior Member chrestomanci's Avatar
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    Quote Originally Posted by Funkstar View Post
    so i need to know the current of the device i want to power?
    Either that or you can make sure that the current flowing through your device is small compared with the bypass current.

    Consider this circit diagram:
    Code:
     +5v --+------+
           |      |
           |      |
          3 Ohm  Load
           |      |
           |      |
         A +------+
           |
          2 Ohm
           |
      0v --+
    If there is no current flowing through the load, (the resistance is infinty) then the total resistance between the power and ground will be 5 ohm. From Ohm's law we can calculate that the current will be 1 amp, and the voltage at point A and C will be +2 volts.

    If we then add a low current device accross the load (with a resistance of hudreads of ohms), then it will see a voltage difference of 3 volts. But in order to provide it with a current of a few mili-amps at 3v, you have wasted a whole amp of current across the load.

    If the device you want to power is draws a lot of current, then you have two choices. If you know it's resistance you can put another resistor in series with it. The problem with that is that you will waste 2/5 of your power as heat, and it will only work if the resistance is constant.

    Alternatively if you need to save energy, and the resitance of the load will vary a lot, you can build a power regulator circit that will chop the 5v power into a series of pulses that average out to 3v, posibly with a capacitor to smooth out those pulse back to a smooth current. This is basicaly how the PSU in you PC works.

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    How about powering it with a couple of brand new fully volted disposable AA batteries (rechargables are only 1.2 volts) in a battery holder (cheap at Maplins) and measure the current drawn (don't forget to pick up a cheap multimeter when you're getting your battery holder at Maplins, if you need one).
    To save going home and then coming back to Maplins when you know what resistor you need, create a security alert by doing the testing in the pub/coffee shop nearest to Maplins!

  8. #8
    Does he need a reason? Funkstar's Avatar
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    Thanks snedger. I think i'll leave this just now, got enough other things to worry about with this system

  9. #9
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    No problem!
    I envy your dilemma.
    A friend asked me to build her a PC for a £500 budget and I told her to go to Dell!
    Then unexpectedly she upped the budget to £1000!
    Can't get started because unbelievably I can't even get happy in choosing a case for her.

  10. #10
    Senior Member chrestomanci's Avatar
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    Your PSU will have a 3.3V rail, perhaps that would be close enough?

    I think floppy drives get a 3.3 volt line, so perhaps you could connect to that?

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