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Switch between power sources

Discussion in 'Electronic Basics' started by RR, Dec 1, 2005.

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  1. RR

    RR Guest

    Hi,

    I'm looking for a device that has two inputs and one output. It passes one
    of the inputs to the output.

    It defaults to one input on power-up, and then switches to the other input
    when a signal is applied.

    What's it for?

    I want to run a battery bank through an inverter to supply one or more house
    circuits. When the batteries are getting low, I want to switch
    (automatically) to mains power.

    So, the device takes my inverter as one input and mains power as another
    input and selects between them. Ideally it would take a voltage measurement
    from the batteries and make the switch when this voltage goes below a
    certain point.

    This is so simple, I can't believe I haven't found such a device yet.

    I guess it's probably because I don't know what it would be called, so I
    don't know what to search for.

    tia,
    RR
     
  2. Tom Biasi

    Tom Biasi Guest

    Usually people want just the opposite. They run on mains and when mains fail
    the switch is to battery and inverter.
    Why would you want to run a battery down and then switch to mains?
    What you are looking for is called an automatic power transfer switch.
     
  3. RR

    RR Guest

    "Free" power. The batteries will be charged by renewable resources (solar,
    wind) and I want to reduce my electricity bill by running some low current
    circuits (e.g. lighting and computers) from this.

    However, when the batteries are exhausted, I don't want to be left in the
    dark.

    As an example, we currently spend $400-$500 per year just for the computer
    equipment. It's hard to know what the lights contribute, but we have some
    expensive-to-run halogens in the main living areas. I've replaced some
    incadescents with low-power fluros, but I may as well hook as much as
    possible up to this if I'm going to go to the trouble of doing it.

    Our house draws 1kw on average (about 24 kwh per day) and then another 15kwh
    per day for the off-peak hot water service.
    Thanks. Obvious name now that you've told me. :)

    thanks again,
    RR
     
  4. Dan Hollands

    Dan Hollands Guest

    You should talk to your local power company about this to make sure that
    whatever you used is an approved device/method.

    Power companies are very concerned about the possibility that during a loss
    of mains power that the invertor will put power out on the power lines when
    the lineman believe the line is dead.

    Dan

    --
    Dan Hollands
    1120 S Creek Dr
    Webster NY 14580
    585-872-2606

    www.QuickScoreRace.com
     
  5. Bill Bowden

    Bill Bowden Guest

    "Free" power. The batteries will be charged by renewable resources (solar,
    I would suspect the cost of replacing the batteries will be greater
    than the cost of the power saved. A more practical approach may be to
    put the extra power on the grid so your wattmeter runs backwards and
    you get credit for the power delivered.

    -Bill
     
  6. RR

    RR Guest

    Maybe. I still have to do the precise numbers on this. A well treated
    battery can last 5-10 years or more and my computer alone cost nearly $500
    per year run. $500 will buy me about 400Ah worth of batteries, which should
    last at least 5 years. Roughly speaking, I should be ahead by at least 50%.
    Not worth it where I live. It costs $10000 for the special meter etc. and
    they only let you feedback 2.4kw. You then have the capital cost to
    generate that much electricity and, for solar, you only get it around 8
    hours per day.

    If I could actually feedback 2.4kw for 8 hours per day, it would take 4
    years to break even on the meter cost. Feeding back a smaller amount
    (because I'm using some) will take up to 20years to break even. Electricity
    costs 13.7cents per kwh where I live.

    cheers,
    Russell
     
  7. RR

    RR Guest

    Yes, thanks for that. I'd get an electrician to hook it up, so that's their
    job. :)

    cheers,
    RR
     
  8. Bill Bowden

    Bill Bowden Guest

    It costs $10000 for the special meter etc. and
    $10,000 seems high. Somebody is making a lot of money selling sinewave
    inverters. I suspect it can be done for $1000 or less.
    Really? My computer runs 24 hours a day and costs $7.50 a month.

    -Bill
     
  9. RR

    RR Guest

    That's not the inverters. That's *just* the meter and the right to feedback
    into the grid.

    The inverters, and batteries, and solar panels are *on top* of all that!
    I meant to say "computers". I've measured the input current to our two
    UPSes and we burn 400W 24 hours per day (a bit less at night when the LCD
    screens are switched off).

    We pay 13.7 cents per kwh. Works out at $480 per year.

    cheers,
    RR
     
  10. Jasen Betts

    Jasen Betts Guest

    which input current? measuring the AC input will give a false impression.

    Jasen
     
  11. RR

    RR Guest

    Really? If I measure the AC input to the UPS (ie. from the mains power),
    that's surely how much power I'm paying for?

    BTW, I used a clamp-style multimeter to measure the current.

    regards,
    RR
     
  12. Jasen Betts

    Jasen Betts Guest

    no they don't charge (bill you) by current they charge by energy.

    and because AC is varying in strength and direction you can't just take an
    RMS current measurement and multiply that by the RMS voltage because in
    cases where the current doesn't track the voltage that will give you the
    wrong answer

    /\ t=start
    |
    energy is | current * voltage
    |
    \/ end


    Google for (or try the wikipedia) for "power factor" for a better explanation.

    If that UPS uas a rectifier-capacitor input stage the power factor could
    down around around 10% (this does not mean 10% efficiency it means the
    ratio between Volts x Amps and Watts )
    if your multimeter has a watts range (using both contact probes and the clamp)
    use it

    Otherwise unplug the power and see what draw there is on the batteries
    (measure mean currrent if you can - not RMS) - I think most DC ranges
    measure mean current. multiply that by the measured battery voltage
    and you'll get an aproximation of what the computer is really using.

    Bye.
    Jasen
     
  13. Bill Bowden

    Bill Bowden Guest

    Really? If I measure the AC input to the UPS (ie. from the mains power),

    Not necessarily because you need to factor in the phase angle between
    the voltage and current. For example, a large capacitor of say 100uF
    will draw 4.5 amps from the 60 cycle line, but there is no energy used
    since the capacitor stores the energy and then releases it back into
    the line. The phase angle in this case is almost 90 degrees meaning the
    current will reach a maximum when the voltage is at minimum( zero). But
    the AC ampmeter will read a continuous 4.5 amps while the wattmeter
    will read zero.

    -Bill
     
  14. RR

    RR Guest

    Ahhh.... a glimmer of understanding is emerging in my old grey matter.

    The UPS manual says it has a power factor of 0.7. But it lists this in the
    OUTPUT section and says nothing in the INPUT section. Maybe a misprint. My
    limited knowledge tells me that specifying a power factor on OUTPUT is
    meaningless. No?

    However, the latest brochure for this model (mine is several years old) says
    it has an input power factor of > 0.95
    (http://upsonic.com.au/pdf/propower.pdf).

    So, if it has PF 0.95, I am burning around 380W, and if it has a PF 0.7,
    then I'm still burning around 280W. Sounds right?

    thanks,
    RR
     
  15. RR

    RR Guest

    Oh, and this UPS has a PFC circuit, which seems to be "Power Factor
    Correction" and seems to be part of the rectifier.

    More confused....

    tia,
    RR
     
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