# Switch between power sources

R

#### RR

Jan 1, 1970
0
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

T

#### Tom Biasi

Jan 1, 1970
0
RR said:
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
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.

R

#### RR

Jan 1, 1970
0
Tom Biasi said:
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?

"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.
What you are looking for is called an automatic power transfer switch.

Thanks. Obvious name now that you've told me.

thanks again,
RR

D

#### Dan Hollands

Jan 1, 1970
0
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
[email protected]
www.QuickScoreRace.com

B

#### Bill Bowden

Jan 1, 1970
0
"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.

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

R

#### RR

Jan 1, 1970
0
Bill Bowden said:
I would suspect the cost of replacing the batteries will be greater
than the cost of the power saved.

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%.
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.

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 R #### RR Jan 1, 1970 0 Dan Hollands said: 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. Yes, thanks for that. I'd get an electrician to hook it up, so that's their job. cheers, RR B #### Bill Bowden Jan 1, 1970 0 It costs$10000 for the special meter etc. and
they only let you feedback 2.4kw

$10,000 seems high. Somebody is making a lot of money selling sinewave inverters. I suspect it can be done for$1000 or less.
my computer alone cost nearly $500 per year run. Really? My computer runs 24 hours a day and costs$7.50 a month.

-Bill

R

#### RR

Jan 1, 1970
0
Bill Bowden said:
$10,000 seems high. Somebody is making a lot of money selling sinewave inverters. I suspect it can be done for$1000 or less.

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!
Really? My computer runs 24 hours a day and costs $7.50 a month. 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

J

#### Jasen Betts

Jan 1, 1970
0
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).

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

Jasen

R

#### RR

Jan 1, 1970
0
Jasen Betts said:
which input current? measuring the AC input will give a false impression.

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

J

#### Jasen Betts

Jan 1, 1970
0
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?

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

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

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

B

#### Bill Bowden

Jan 1, 1970
0
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?

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

-Bill

R

#### RR

Jan 1, 1970
0
Jasen Betts said:
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 )

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

R

#### RR

Jan 1, 1970
0
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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