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Quick Regulator Question

Discussion in 'Electronic Basics' started by Captain Blammo, Oct 26, 2005.

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  1. I'm going to run a camcorder viewfinder and a digital camera off a ~7v
    camcorder battery. They will be needing 5v and 3.3v respectively. I have
    settled upon these two regulators for the task:

    UA78M33CKC
    LM2940CT-5.0

    Does this seem like a wise choice? I don't know much of what's out there,
    and I'd like to squeeze as much battery life out as possible whilst avoiding
    SMT components if I can.

    Thanks for any advice!

    CB
     
  2. Before selecting a regulator, you need to know what current their
    loads will consume.

    The combination of "as much battery life out as possible" and your
    choice of linear regulators is a contradiction. Linear regulators
    waste all the extra voltage as heat. If battery life is important,
    you might look into prepackaged switching regulators, if you can find
    ones that can take the 7 volt input. They convert energy at one
    voltage to almost the same energy at another voltage.

    They are also a lot more expensive than your linear regulators.
    Here is a page from the Digikey catalog that lists some candidates:
    http://dkc3.digikey.com/PDF/T053/1532.pdf

    For example, near the top, find the PT6305A 3.3 volt 3 amp regulator
    rated for 4.5 to 9 volts in.

    Finding a 5 volt integrated switcher that works over that range of
    input voltage will be harder.
     
  3. The combination of "as much battery life out as possible" and your
    I actually have no idea as to just how much is wasted by linear regulators.
    If I run 5v at 1A from a 7v, 2Ah battery, roughly how many Ah will be wasted
    as heat? Is there a formula? I realise that it will vary with choice of
    regulator, but a ballpark idea would be nice.

    Is switching regualtor wastage negligible? Those things aren't so cheap! I
    guess I should have said that I want an optimal balance of cost and battery
    life :)

    CB
     
  4. 5 amperes at 1 amp is 5 watts. But if that same current comes from a
    7 volt battery, that consumes 7 watts. That would drain a 2 amp hour
    battery in 2AH/1A= 2 hours. If you Have a switching regulator with
    90% efficiency, the load on the battery would be 5.5 watts. %.5 watts
    from 7 volts needs a current of 5.5W/7V= 0.786 A. That would drain a
    2 amp hour battery in 2AH/.786A= 2.55 hours.

    The 3.3 volt case is more dramatic. If you need to supply 2 amps to
    the 3.3 volt load, that represents 6.6 watts. A linear regulator will
    take that 2 amps directly from the battery for a life of 2AH/2A= 1 hour.

    A 90 percent efficient switching regulator would take 6.6W/.9= 7.33 W
    from the battery which requires 7.33W/7V= 1.05 A, for a life of
    2AH/1.05= 1.9 hours.

    So a switcher for the 3.3 volt output may be more important for
    battery life if the load current approaches or exceeds that needed by
    the 5 volt output.
    Perfection is never cost effective.
    A good compromise might be a switcher for the 3.3 volt output and a
    low drop out regulator for the 5 volt output.

    Have you come up with load current estimates, yet?
     
  5. I don't think the camera is likely to use much more than 0.3A on a bad day
    (that's a total guess), since it won't have the LCD on or be taking
    pictures, just feeding video out, and I know the camcorder viewfinder uses
    around 1A.

    If I understood you correctly, that means that I'll be looking at 1.3A of
    load with linear regulators, giving me 2Ah/1.3A=1.538 hours of life. I'm
    assuming that plonking the regulators in parallel with the battery will make
    the loads additive, but my electronics is still pretty sketchy :)

    In any case, I can't see the loads being any higher than what I quoted, so I
    should get at least 1.5h of life. According to the calculations below
    (please correct me if I'm wrong), I'm only looking at an optimal battery
    life of 2.134h with 90% efficient regulators; only a 36 minute difference.
    All things considered, I think a fully linear solution will do my wallet and
    my project much justice. Thanks very much for the help!

    [email protected]=0.9W
    0.9W =90%eff=> 1.000W
    1W/7v=0.143A

    [email protected]=5W
    5W =90%eff=> 5.555W
    5.555/7v=0.794A

    0.794A+0.143A=0.937A
    2Ah/0.937A=2.134h of life


    ==
    CB
     
  6. The 5 volt and 3.3 volt load currents add together with linear
    regulators to find the battery current load.
    That looks right to me.
     
  7. Jasen Betts

    Jasen Betts Guest

    you'll get 2Ah out of te battery, but with a switching regulator you
    could get more than that out of the regulator.

    switching regulators use a transfoomer like effect to convert voltage and
    current...
    Last time I looked (several years ago) it was about 80% efficiency, (but
    that has probably improved), so for converting 7V to 5 probably not
    worth-while, but for the 7V to 3V it'd halve the power consumption.

    Bye.
    Jasen
     
  8. ^^^^^^^ ^^^

    Really? I think you have a typo in there somewhere, John. ;-)
     
  9. Could be.
     
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