battery sizing question

[d. smith]

Does anyone have any experience powering small electronics items (low
wattage) using a power inverter and a 12 volt lead-acid battery?

If so, could you give me some input as to how long you think I could
operate the following setup:

I will operate a vcr player that has a 14 watt rating at 120 volts. I am
considering an 18 Amp-hour AGM type battery. The power inverter is rated
for up to 350 watts and the no load current draw is given to be 0.4 Amps.

I need to be able to get about a 5 to 6 hour runtime. I do not plan to do a
test prior to giving it a go the first time.

 

[Jerry G.]

I did the math very quickly, and found that it should last about 6 to 7
hours. This is providing that everything is ideal, and your battery is
in very good condition to meet the amp hours rating that you are
mentioning.

Try it for curiosity, and email us all back if it works.


--

Jerry G.
======

"d. smith" <dsmith@ss.net> wrote in message
news:1098555854.YMUan/hKkVE8ZT5EKu1oqA@teranews...
Does anyone have any experience powering small electronics items (low
wattage) using a power inverter and a 12 volt lead-acid battery?

If so, could you give me some input as to how long you think I could
operate the following setup:

I will operate a vcr player that has a 14 watt rating at 120 volts. I am
considering an 18 Amp-hour AGM type battery. The power inverter is rated
for up to 350 watts and the no load current draw is given to be 0.4
Amps.

I need to be able to get about a 5 to 6 hour runtime. I do not plan to
do a
test prior to giving it a go the first time.

 

[CWatters]

"d. smith" <dsmith@ss.net> wrote in message
news:1098555854.YMUan/hKkVE8ZT5EKu1oqA@teranews...

I need to be able to get about a 5 to 6 hour runtime. I do not plan to do
a
test prior to giving it a go the first time.

You must test the kit for a short period at least. We've had a a lot of
problems running computers/laptops from mains inverters. The cheaper types
can be very noisy. You might get noise on the tape if recording.

 

[John Fields]

On Sat, 23 Oct 2004 22:41:14 GMT, John <look@sig.net> wrote:

Quick worst-case ballpark calculation:

0.4 amps (no load) * 6 hours = 2.4 amp/hours

14 watts output @ 50% conversion efficiency = 28 watts input

28 watts / 12 volts = 2.3 amps

2.3 amps * 6 hours = 13.8 amp/hours

13.8 amp/hours + 2.4 amp/hours = 16.2 amp/hours

---
When the inverter is loaded, I think the no-load current becomes part
of the efficiency spec and doesn't get added in separately.

--
John Fields

 

[John Fields]

On 24 Oct 2004 20:20:10 -0700, "D Smith" <76lolmc02@sneakemail.com>
wrote:

Hi John,

With an ammeter hooked in series, I see the Vector 350 watt inverter
pulling a constant 0.6 Amps at no load. (Packaging says it is 0.4.)

When the vcr gets plugged in, it goes to 1.10 with the vcr power on at
idle and goes to a constant 1.47 amps when I start recording. This is
because it takes more power to record than to do nothing except be on.
At any rate, I have an amplified antenna that brings the above totals
slightly higher to 1.23 and 1.54 amps respectively. The amplified
antenna doesn't introduce much extra load.

It appears I am looking at about a constant 1.54 amp draw off the
battery. I was testing with a car battery as I do not yet have the AGM
Power Sonic battery that this place said they sell. All I know it is an
18 amp-hour rated AGM battery. I would assume that this is C/20, as
that is the standard. Sorry I don't have more data for you on the
battery.

---
Here's what I found on the PowerSonic site:

PERFORMANCE SPECIFICATIONS
Nominal Voltage...12 volts (6 cells in series)

Nominal Capacity
20 hour rate (900mA to 10.50 volts) ... 18.0 A.H.
10 hour rate (1.7A to 10.50 volts) ... 17.0 A.H.
5 hour rate (3.2A to 10.50 volts) ... 16.0 A.H.
1 hour rate (11.1A to 09.00 volts) ... 11.1 A.H.
15 min rate (34.3A to 09.00 volts) ... 8.6 A.H.

Approximate Weight... 12.6 pounds (5.7 kg)

Energy Density (20 hour rate)... 1.54 Watt-hours/cubic inch
(94.0 Watt-hours/l)

Specific Energy (20 hour rate)...16.5 Watt-hours/pound
(36.6 Watt-hours/kg)

Internal Resistance (Fully Charged Battery)... 12 milliohms
(approximately)

Maximum Discharge Current ( <_ 7 Min.)... 54 amperes

Maximum Short-Duration Discharge Current ( <_ 10 Sec.)... 180 amperes

Terminal configurations ... PS-12180 NB: Tin plated brass post with
5mm nut & bolt connectors PS-12180 F: Quick disconnect AMP, INC.
Faston tabs, 0.250" x 0.032"

Shelf Life - % of nominal capacity at 68o F (20o C)
1 Month... 97%
3 Months...91%
6 Months...83%

Operating Temperature Range
Charge... -4oF (-20oC) to 122oF (50oC)
Discharge... -4oF (-20oC) to 140oF (60oC)

Case .. ABS Plastic

Here is my conclusion though so far from testing: I want to try some
lower rated power inverters to see if I can get that no load draw down
from 0.6 amps into something like maybe 0.3 or even 0.4. Vector
Manufacturing sells a 175 watt model and a really small one at like
maybe 120 watts. I am sure the really small one would probably consume
less power and increase my run time and possibly the 175 watt model
might do the same.

With this tactic, I think the 18 amp hour battery could work.
Especially knowing that I am only pulling 1.54 amps, which is less than
what we thought it might be. Also, if I power down for commercials
either fully or simply halt the recording process this might help save
precious amps but I am not sure how much it would help because cycling
the power introduces 3 second or so current demand spikes that go up to
around 2 amps momentarily and then go back down to the normal readings
I gave.

---
From the data above, at C/10 you can get 17AH, so assuming you could
also get 17AH with a 1.54A load (safe assumption, BTW) the battery
would last for at least 17AH/1.54A ~ 11 hours even if you never turned
anything off.

--
John Fields

 

[d. smith]

I would strongly suggest testing the setup before going "live".

John

Thanks John,

Yeah I'll probably test it and see how it does once I get everything.

I am using a Vector Manufacturing (www.vectormfg.com) 350 watt inverter
that they sell at wal mart for $45. The users manual doesn't list anything
concerning efficiency, although the packaging says "Maximum efficiency is
Approximately 90%". Perhaps I should call and ask them. Do you think I
would be better off using a smaller inverter? For example, the 175 watt
version or even the tiny little ones? I want good heat capacity since I am
running 5+ hours continuous even with a small load (14 watt vcr).

Don't have specs on the battery at this time.

Thanks for your help.

 

[John Fields]

On Sat, 23 Oct 2004 14:24:41 -0400, "d. smith" <dsmith@ss.net> wrote:

Does anyone have any experience powering small electronics items (low
wattage) using a power inverter and a 12 volt lead-acid battery?

If so, could you give me some input as to how long you think I could
operate the following setup:

I will operate a vcr player that has a 14 watt rating at 120 volts. I am
considering an 18 Amp-hour AGM type battery. The power inverter is rated
for up to 350 watts and the no load current draw is given to be 0.4 Amps.

I need to be able to get about a 5 to 6 hour runtime. I do not plan to do a
test prior to giving it a go the first time.

---

On the primary side, if the inverter is running at 50% efficiency and
you've got a 14 watt load on the secondary, the inverter will need 28
watts in. Since


P = IE

then

P 28W
I = --- = ----- = 2.33A
E 12V


With an 18AH battery, if you suck 2.33A out of it, it should last for


C 18AH
T = --- = ------- ~ 7.73H
A 2.33A


depending on the rate of discharge.

That is, if it's rated for 18AH at C/10, (1.8A *10H) then since you'll
be taking 2.33A from it you won't get 7.73 hours out of it, you'll get
something less. You can tell how much less by getting the
manufacturer's data sheet for the battery. If I had to guess, I'd
think you could get 6 hours out of it. To find out for sure you'll
need to get the battery data sheet and the data sheet for the
inverter. Specifically, what you'll be looking for is the inverter's
efficiency with a light load like 14 watts on it.

Get that data and come back with it and we can give you a better
estimate.

--
John Fields

 

[John]

On Sat, 23 Oct 2004 14:24:41 -0400, "d. smith" <dsmith@ss.net> wrote:

Does anyone have any experience powering small electronics items (low
wattage) using a power inverter and a 12 volt lead-acid battery?

If so, could you give me some input as to how long you think I could
operate the following setup:

I will operate a vcr player that has a 14 watt rating at 120 volts. I am
considering an 18 Amp-hour AGM type battery. The power inverter is rated
for up to 350 watts and the no load current draw is given to be 0.4 Amps.

I need to be able to get about a 5 to 6 hour runtime. I do not plan to do a
test prior to giving it a go the first time.

Quick worst-case ballpark calculation:

0.4 amps (no load) * 6 hours = 2.4 amp/hours

14 watts output @ 50% conversion efficiency = 28 watts input

28 watts / 12 volts = 2.3 amps

2.3 amps * 6 hours = 13.8 amp/hours

13.8 amp/hours + 2.4 amp/hours = 16.2 amp/hours

Batteries are usually rated based on a 20 hour discharge rate - for an
18 amp/hour battery that would be a load of 0.9 amps. You need to
check the manufacturers spec sheet for the exact figures for this
particular battery.
With the estimated worst-case load of 2.7 amps (3 times the
specification discharge rate), the battery will NOTdeliver its rated
output. You might get 3 to 4 hours of use.

Since conversion efficiency is typically higher than 50%, the actual
load current could be less than 2 amps, which might give you 5+ hours.

I would strongly suggest testing the setup before going "live".

John

 

[d. smith]

To find out for sure you'll
need to get the battery data sheet and the data sheet for the
inverter. Specifically, what you'll be looking for is the inverter's
efficiency with a light load like 14 watts on it.

I have a Vector Manufacturing 350 watt Max SST power inverter
(http://www.vectormfg.com). I can't find any data on their website or in
the users manual about efficiency at light loads. The only thing it says
are that Maximum Efficiency is "Approximately 90%". When does maximum
efficiency occur when using a power inverter? With small loads like mine or
larger loads? You can find this inverter at your neighborhood walmart.


As far as the battery is concerned, I will have to get back to you on that.
I haven't discussed this issue with the seller of the battery.

I did wonder whether it would be better to go with a smaller rated power
inverter, because even the small 80 watt ones would have more capacity than
the vcr needs. Does anyone know if they would save me alot of power useage
or perhaps their effiency is lower than the bigger ones, like the 350 watt
version. There is a 175 watt Vector inverter but product information claims
it has 0.4 amp no load current draw just like its big brother, the 350 watt
version. And the tech support guy said they both use about the same energy.
My reasoning for going with the bigger inverter is I figured to get a 5+
hour continuous runtime, heat could become an issue and you want something
that has enough heat absorbing capacity to make heat not be a problem.

Thanks for your responses. They were helpful.

Get that data and come back with it and we can give you a better
estimate.

 

[D Smith]

Hi John,

With an ammeter hooked in series, I see the Vector 350 watt inverter
pulling a constant 0.6 Amps at no load. (Packaging says it is 0.4.)

When the vcr gets plugged in, it goes to 1.10 with the vcr power on at
idle and goes to a constant 1.47 amps when I start recording. This is
because it takes more power to record than to do nothing except be on.
At any rate, I have an amplified antenna that brings the above totals
slightly higher to 1.23 and 1.54 amps respectively. The amplified
antenna doesn't introduce much extra load.

It appears I am looking at about a constant 1.54 amp draw off the
battery. I was testing with a car battery as I do not yet have the AGM
Power Sonic battery that this place said they sell. All I know it is an
18 amp-hour rated AGM battery. I would assume that this is C/20, as
that is the standard. Sorry I don't have more data for you on the
battery.

Here is my conclusion though so far from testing: I want to try some
lower rated power inverters to see if I can get that no load draw down
from 0.6 amps into something like maybe 0.3 or even 0.4. Vector
Manufacturing sells a 175 watt model and a really small one at like
maybe 120 watts. I am sure the really small one would probably consume
less power and increase my run time and possibly the 175 watt model
might do the same.

With this tactic, I think the 18 amp hour battery could work.
Especially knowing that I am only pulling 1.54 amps, which is less than
what we thought it might be. Also, if I power down for commercials
either fully or simply halt the recording process this might help save
precious amps but I am not sure how much it would help because cycling
the power introduces 3 second or so current demand spikes that go up to
around 2 amps momentarily and then go back down to the normal readings
I gave.

Thanks.

 

[D Smith]

John,

The only specs I see for the Vector (www.vectormfg.com) 350 watt power
inverter is on the packaging it does say it has a max efficiency of
"Approximately 90%".

My testing shows a current draw off the battery of 1.54 amps. At that
rate, that's what I need to know is how long the 20 Amp hour battery
will run. I will discuss this with the people that sell this battery.
Hopefully they will understand the issues you have brought up here with
respect to C/10, C/20, etc.

I wish I could move up to get some extra capacity but weight is kind of
a critical thing. I can't go much above this particular battery.

 

[D Smith]

I did do some small testing and found that the recording quality was
very acceptable. There is very little extra noise that I noticed that
was introduced by the power inverter. Sometimes if the cables are too
close to each other, you can get a little noise but I guess my power
inverter did a fairly good job with noise.

 

[Jamie]

depends on the inverter.
90% isn't to bad.
all it means is it takes 10 Amps form your
12 volts to supply a 9 amp device.
consequent current maybe what your looking
for or may they might even call is
idling or standby current.


d. smith wrote:

To find out for sure you'll
need to get the battery data sheet and the data sheet for the
inverter. Specifically, what you'll be looking for is the inverter's
efficiency with a light load like 14 watts on it.



I have a Vector Manufacturing 350 watt Max SST power inverter
(http://www.vectormfg.com). I can't find any data on their website or in
the users manual about efficiency at light loads. The only thing it says
are that Maximum Efficiency is "Approximately 90%". When does maximum
efficiency occur when using a power inverter? With small loads like mine or
larger loads? You can find this inverter at your neighborhood walmart.


As far as the battery is concerned, I will have to get back to you on that.
I haven't discussed this issue with the seller of the battery.

I did wonder whether it would be better to go with a smaller rated power
inverter, because even the small 80 watt ones would have more capacity than
the vcr needs. Does anyone know if they would save me alot of power useage
or perhaps their effiency is lower than the bigger ones, like the 350 watt
version. There is a 175 watt Vector inverter but product information claims
it has 0.4 amp no load current draw just like its big brother, the 350 watt
version. And the tech support guy said they both use about the same energy.
My reasoning for going with the bigger inverter is I figured to get a 5+
hour continuous runtime, heat could become an issue and you want something
that has enough heat absorbing capacity to make heat not be a problem.

Thanks for your responses. They were helpful.


Get that data and come back with it and we can give you a better
estimate.