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using battery to power device not intended to be

Discussion in 'Electronic Basics' started by Patrick Chu, Feb 6, 2004.

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  1. Patrick Chu

    Patrick Chu Guest

    How, if at all, can I use batteries to power a device that was meant
    to be powered by USB (universal serial bus)?

    Here are the specs:

    Power Consumption: 0.4 Watts
    Current: 75 mA +/- 5 mA
    Input Voltage: 5V DC

    Instead of having to have it plugged into a USB port, I'd like to be
    able to use batteries to power it free-standing. What kind of
    batteries and other electrical hardware would I need, if this is even

  2. XandMS

    XandMS Guest

    try to use 9V batteries, and use 78L05 to convert 9V to 5V. It's a simply
  3. Patrick Chu

    Patrick Chu Guest

    Actually, let me try, using my Google-assisted layman's knowledge, to
    answer my own question, and hopefully an expert can verify or correct

    If I take a 9-volt battery and some kind of voltage regulator, I could
    drop it down to 5 volts.

    Then, if ohms = volts/amps, and the required current is something like
    75 mA, or 0.075 A, then I would need a 5/0.075 ~= 65 ohm resistor
    somewhere in the mix?

    Using another formula to double-check, ohms = (volts^2)/watts, the
    ohms comes out to 25/0.4 = 63, which is around mid-60s like the
    previous calculation said.

    So to recap, I need a 9V battery, a +5V voltage regulator, and a ~65
    ohm resistor? Can all be found at Radio Shack?

  4. Yes. A linear regulator wastes the extra 4 volts.
    Actually, that is the effective resistance of a 5 volt load that draws
    ..075 amp. The regulator that wastes the extra 4 volts would have an
    effective resistance of 4/.075~53 ohms. But if the load current
    varies a bit, or the battery voltage decreases, the regulator will
    have to have some other resistance to waste only the excess voltage.
    Fortunately, regulators are an active resistance that varies as
    A 5 volt regulator and a couple capacitors are all you need. It can
    handle a load current up to 100 ma. Try a 78L05. See the data sheet
    for the typical capacitors recommended.
    available from Digikey for $0.40
  5. CFoley1064

    CFoley1064 Guest

    From: (Patrick Chu)
    Thanks for a good problem description. 75 mA is quite a lot to be drawing from
    a 9V "transistor" battery. It won't last very long. If you just need a quick
    fix, you might want to try one of the powered USB hubs -- that would be your
    quickest fix, and wouldn't be expensive in comparison with batteries,
    perfboard, IC regulator, caps, switch, enclosure, &c. That might be your best
    solution if you've got an outlet handy.

    If not, or if you really want to use batteries, you could try 6 "C" or "D"
    batteries in series to give you the 9V to start with. You could then use the
    information given by other respondents about the 78L05 to get a solution that
    will last more than a day or two.

    Good luck
  6. You might want to look into a switching regulator instead. It will
    save a lot of battery life. It's a slightly more complicated circuit
    and I don't have any part numbers off hand, but look into it - your
    battery life could easily increase by 2/3.

  7. vic

    vic Guest

    Well, if you use a 9V battery, the exceeding 4V will be converted to
    heat by the regulator and you will lose this energy. You'd better use 4
    1.5V batteries and a low drop regulator or a simple diode (1N4004) wich
    has a drop voltage of 0.6V, so this is an easy way to get 5.4V or 4.8V.
    (your circuit should work OK with this kind of supply voltage.

    Also you do not need to insert a current limiting resistor. In fact, if
    you do so, supposing the circuit draws 75mA, the supply would be :
    5 - .075 * 65 = ~0V which is not what you want.

    The circuit will draw itself the current it needs.

    As others have suggested you may add 100uF in parallel with 100nF to
    filter out parasites and transients.

    Note that if you use a 9V battery the lifespan will be about :
    300mAh / 75mA = 4 hours (rough estimation for an alcaline battery)

  8. George

    George Guest

    Try an LM2574N-5.0 or something like it. It's more expensive, but more
    efficient. $1.86 at digikey with about 80% efficiency. You can get 90%
    efficiency if you are willing to pay $3.10 for an LM2674N-5.0.
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