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Hopefully simple question about DC power supplies

Discussion in 'Electronic Basics' started by [email protected], Nov 11, 2008.

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  1. Guest

    Hi,

    I'm hoping that this is a simple question to answer. Although I'm not
    a technical ignoramus, I know very little about power or cabling, so I
    would like some advice with what I'm trying to do!

    What I need is:

    * PORTABLE (i.e. non-mains) 5V DC power into 'Device A'
    and
    * PORTABLE (i.e. non-mains) 12V DC power into 'Device B'

    What I have at my disposal (so far) is:
    * A mains adapter for 'Device A' that seems to output 5V at up-to 2A
    via a small-ish DC power jack (3mm outside diameter, I think)
    * A mains adapter (arriving in the post tomorrow) for 'Device B' that
    should output 12V at up-to 500mA via a suitably-sized DC power jack
    * A 'rechargeable laptop battery pack', that can output 12V DC via a
    number of different jacks, and 5V DC via a USB cable. (details here:
    http://www.maplin.co.uk/Module.aspx?ModuleNo=48490)

    What I originally wanted to do was to power both devices from the
    'rechargeable laptop battery pack'. But the only 5V output from that
    is via a USB cable, which will not fit the device that needs a 5V
    input.

    I cannot use either of the mains adapters, so I need to get DC power
    from the 'rechargeable laptop battery pack' (or some equivalent).

    So what should I do? I have a few ideas, but I'm not sure if they are
    any good:

    (a) Could I somehow construct a USB-to-DC power jack for the 5V power?
    I'm not sure such a thing exists to buy, but how difficult would it be
    to construct?

    (b) Could I split the 12V output to two destinations (1) 'Device B'
    directly; and (2) 'Device A' via some way of converting it to 5V? How
    would I do this?

    I'm sure what I'm trying to do shouldn't be too difficult. But I'm
    just not sure how to do it, and I don't want to damage my devices or
    start a fire or electrocute myself. So any advice would be
    appreciated!

    Thanks,

    David

    P.S. In terms of getting hold of kit to do this, I'm UK-based, so
    ordering from UK companies would work best for me.
     
  2. Tom Biasi

    Tom Biasi Guest

    Lets cut out some of the words:)
    You have a 12 volt battery that you need to power two devices, one with 12
    volts the other with 5 volts.
    Obviously both devices will stay with the battery and you need a means to
    charge the battery.
    It would help if you weren't so cryptic and named the devices and your
    intent.

    Tom
     
  3. Guest

    Hi Tom,

    Thanks for your quick response.

    The problem here is that I can be criticized for both providing too
    much information (hence your need to "cut out some of the words") and
    for providing insufficient information (i.e. being "cryptic")! I had
    tried to find the right balance with describing my problem, but
    obviously I didn't get it quite right. Sorry.

    I'll try to distill it down to what I think the core problem/question
    is.

    I have a need for a 5V DC feed via a small DC jack, and a 12V DC feed
    via a slightly larger DC jack. But what I have (at the moment) is a 5V
    DC feed via a USB cable, an a 12V DC feed via a (hopefully suitable)
    DC jack.

    The problem is how to get the 5V DC feed via a jack.

    Do I:
    (a) Try to convert the USB connector to a DC jack? If so, how?
    (b) Try to split the 12V feed into two, and somehow convert half of it
    to 5V? If so, how?
    (c) Do something else?

    As background info, I am trying to setup surveillance of my garage,
    where stuff seems to be regularly getting pilfered and generally
    messed around with. So, 'Device A' is a small handheld DVR (with 5V
    'charging' input), 'Device B' is a small CCD camera, which takes 12V.

    Thanks,

    David
     
  4. Lord Garth

    Lord Garth Guest

    Hi Tom,

    Thanks for your quick response.

    The problem here is that I can be criticized for both providing too
    much information (hence your need to "cut out some of the words") and
    for providing insufficient information (i.e. being "cryptic")! I had
    tried to find the right balance with describing my problem, but
    obviously I didn't get it quite right. Sorry.

    I'll try to distill it down to what I think the core problem/question
    is.

    I have a need for a 5V DC feed via a small DC jack, and a 12V DC feed
    via a slightly larger DC jack. But what I have (at the moment) is a 5V
    DC feed via a USB cable, an a 12V DC feed via a (hopefully suitable)
    DC jack.

    The problem is how to get the 5V DC feed via a jack.

    Do I:
    (a) Try to convert the USB connector to a DC jack? If so, how?
    (b) Try to split the 12V feed into two, and somehow convert half of it
    to 5V? If so, how?
    (c) Do something else?

    As background info, I am trying to setup surveillance of my garage,
    where stuff seems to be regularly getting pilfered and generally
    messed around with. So, 'Device A' is a small handheld DVR (with 5V
    'charging' input), 'Device B' is a small CCD camera, which takes 12V.

    Thanks,

    David


    Hi David,

    Many CCD cameras can operate from an unregulated supply. My experience
    is with the small cameras used to observe store fronts. Do check your
    camera
    for its power requirements as an AC input to a DC jack would prove
    disastrous.

    The cameras we use do not specify a polarity to the connection point on the
    rear
    which means the regulator circuitry is inside the camera itself. A typical
    installation
    is with a cable we call 'siamese' because it consists of a coax for the
    video and a
    two conductor 16 gauge stranded cables connected to the insulation of the
    coax.

    Most of the power supplies are full wave bridge with a few filter
    capacitors. Some
    of the cameras have been connected in reverse with respect to the others
    (say the
    red wire was placed on the left rather than the right). The camera operated
    perfectly
    this way.

    You do not mention the current requirements of your equipment but that is
    important
    as well as the voltage (and polarity if your camera requires DC). A
    voltmeter will
    show the voltage drooping if your current demand is beyond the capability of
    the
    supply.

    It is possible to supply power over the same coax as the video signal and
    there are
    adapters made to do this...a simple capacitor will block the DC but pass the
    video.

    You can also opt for an IP camera that supports PoE or Power Over Ethernet.
    This
    will allow the camera to be powered over the same CAT5 as the video plus you
    can
    view the image on any browser and most of these cameras come with a software
    DVR that supports motion detection. Further, they can email you when motion
    is
    detected or in some cases, text your mobile phone. A Trendnet TV-201 is
    such a
    camera. A good price is from www.provantage.com
     
  5. ehsjr

    ehsjr Guest

    We don't know how much current either your 5 volt or your
    12 volt device draws, nor the capacity of your laptop
    battery, nor how much current it can provide via the
    usb cable.

    That said, the general hookup is this:

    Battery
    -----
    | +|---+---------------------> +12V to 12V device
    | | |
    | 12v | | -------------
    | | +---| 5V regulator|---> +5V to 5V device
    | | -------------
    | -|-------------------------> - to both devices
    -----

    The 5V regulator could be a "linear regulator" which
    wastes power but is easy and cheap to get parts for
    and build, or a "DC-DC converter" which is harder to
    build/get parts for and more expensive. YMMV

    Not being mains connected is going to defeat you, in any
    event. The setup will work only as long as the battery
    does not need to be recharged. You will need to charge the
    thing on a regular basis, but without knowing how much
    current will be drawn and what the battery capacity is,
    we can't say how often that will be. If the adapter
    for the 5V device (which you said puts out up to 2A) is
    any indication, you probably won't get 1 day from the
    setup before you need to recharge. That guess is as
    exact as the spec you gave, so don't make any wagers
    based on it!

    Ed
     
  6. Guest

    Thanks Tim,

    I am inclined to follow your advice and go for option (a). Of course,
    there are still the questions (from other posters) about power
    consumption, etc, which I'll try and provide some info on.

    David
     
  7. Phil Allison

    Phil Allison Guest

    <>


    I am inclined to follow your advice and go for option (a). Of course,
    there are still the questions (from other posters) about power
    consumption, etc, which I'll try and provide some info on.


    ** If you use the same battery to power both devices - then you will wind
    up with a common negative for both DC supplies. Each device will then work
    but they may not tolerate being interconnected via the video leads without
    malfunction or damage.

    AC adaptors ALWAYS have floating outputs and many equipment makes RELY on
    this being the case.



    ..... Phil
     
  8. Tom Biasi

    Tom Biasi Guest

    Hi Tom,

    Thanks for your quick response.

    The problem here is that I can be criticized for both providing too
    much information (hence your need to "cut out some of the words") and
    for providing insufficient information (i.e. being "cryptic")! I had
    tried to find the right balance with describing my problem, but
    obviously I didn't get it quite right. Sorry.

    I'll try to distill it down to what I think the core problem/question
    is.


    Notice I had a :) after "cut out some of the words".
    I was just trying to boil down you situation. Usually don't try to give a
    scientific dissertation of what you want just say what you want to do.
    Others have supplied info so I need not say more.
    Best Regards,
    Tom


    ..
     
  9. Guest

    I've tried to dig out some power figures...

    * DVR operation current: 650mA. Although I've seen conflicting
    information here.
    * camera operation current: 90mA
    * battery capacity: 4700mAh (80Wh energy, apparently)

    Does this mean I'll get somewhere in the region of 6 hours out of it?
    It's a long time since I studied all this stuff at school!
     
  10. Jasen Betts

    Jasen Betts Guest

    very easy to build, take a USB cable and chop one end off and fit a
    suitable plug. google for the pinout (the outermost contacts are the
    power ones)

    2A could be an issue. USB is only rated for 1/4 of that (500mA), check
    what the documentation on the 'rechargeable laptop battery pack' says,
    you may have to ask maplin to look it up for you.
    you would use a regulator.

    you could start with circuit on the LM7805 datasheet (which may need
    a large heatsink and will waste more energy than it ptovides to the
    5V device) or buy a prefabricated 12v to 5V converter. something like
    this one perhaps.
    http://cgi.ebay.com/XM-Car-Power-Ch...ite_Radio_Accessories?_trksid=p3286.m20.l1116
    (I just typed 12V-5V into the e-bay search box and hit "go")
     
  11. ehsjr

    ehsjr Guest

    Well, no. 12 volts at 4700 mAh would be equal to 56.4Wh,
    not 80 Wh. (To figure watts, multiply amps times volts)

    Can't figure the current taken from the battery by the
    650 mA or the 90 mA without knowing which runs on 12 volts,
    and which runs on 5 volts, because the 5V regulator will
    waste some power. There's nasty little things to consider,
    like the discharge curve for the battery and the regulator
    efficiency.

    For example, say the 5V regulator is ~70% efficient, and it
    is powering the 650 mA device. That means it is actually
    drawing about 929 mA to provide 650 mA to the load. Add to that
    the 90 mA used by the 12 volt load, and you're using 1019 mA
    every hour. So in very general terms, it would last a little
    over 4 hours. But it can get a lot worse. Battery capacity
    ratings usually are for 20 hour discharge. If that is true of
    your battery, then the 4700 mAh rating is based on a current
    draw of about 235 mA - and you're drawing over 4 times that
    in this example. Generally that means you'll get a lot less
    than 4700 mAh.

    Tha above is not to give you any real numbers - it is just
    a general outline. To get better numbers, you need the
    detailed battery specs and more info about which device
    uses which voltage and current.

    In any event, whatever the detailed specs, you'll need to
    charge the battery at least 4 times daily based on the
    specs you've posted so far. Since it is not to be mains
    connected, you'll need at least 2 batteries - one to power
    the equipment while the other is charging. And you'll
    need 4 to 6 battery changes each day, minimum, for 24 hour
    coverage. That is impractical. You need to get mains
    connected. Even if you use equipment that draws _far_
    less current, you still have a problem. Say you can get
    equipment that draws a total of 100 mA. You'll need to
    charge the battery every 2 days. Guaranteed you won't
    keep that up for long. The only possible option I see,
    other than mains connection, is equipment with very low
    current draw, powered from a battery that is connected
    to a solar charger. I have no idea if that is practical
    for you. You might consider using a much bigger deep
    cycle battery, but you would still need to charge it
    periodically. It will be heavy and awkward to move and
    take a long time to charge, so solar power seems like
    a better option.

    Ed
     
  12. ehsjr

    ehsjr Guest

    The discussion was of a regulator, not a DC-DC converter.
    Your description is of a DC-DC converter, which you
    incorrectly call a regulator. They are not the same
    thing.

    Ed
     
  13. ehsjr

    ehsjr Guest

    Would you call a 10.769 ohm (approximate) resistor in series
    with the 12V battery and the load a DC-DC converter? Are
    you calling a 7805 a DC-DC converter? Both seem to fit
    your idea of a DC-DC converter, providing the load 5 v
    at 650 mA.

    Both a resistor and a linear regulator convert some of the
    electrical energy to heat energy. That is not a DC to DC
    conversion. The electrical energy that is not converted to
    heat is passed to the load.
    No, because it is incorrect. I totally ignored the
    input - output voltage difference and looked only
    at current. Stupid mistake.
    Yes. :)

    The 7805 _is_ a regulator, and that is the term that should be used.
    Shouting is not needed.

    Ed
     
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