Switch between power sources

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.

Reply to
Tom Biasi
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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

Reply to
RR

fail

"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

Reply to
RR

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

Reply to
Bill Bowden

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
Reply to
Dan Hollands

$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

Reply to
Bill Bowden

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

Reply to
RR

loss

when

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

cheers, RR

Reply to
RR

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

Reply to
RR

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

Jasen

Reply to
Jasen Betts

LCD

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

Reply to
RR

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

Reply to
Jasen Betts

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

Reply to
Bill Bowden

explanation.

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

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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

Reply to
RR

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

Reply to
RR

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