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Old 29-06-2009, 17:28   #1
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Use of Inverters

Electrical is not my strong suit. I understand the concept of how many watts an inverter puts out and the basic use of limiting actual usage to I've been told about 75% of the rated wattage available. That is of course a rule of thumb. I suspect the inverter has some surge capacity above it's rated value.

However, How do I compute the draw down on my 12V batteries if you wife wants to use a hair dryer for 15 or 20 minutes. I guess if there is such a thing, I'm looking for the magic bullet formula that will tell me that if I use 15 amps at 120V I will need X amount of amps in the batteries to make that available.

Glen
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Old 29-06-2009, 17:49   #2
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I know the practical output of many inverters is a bit lacking.

I had a black and decker one I bought from Wal-Mart for 50 bucks. It was rated at 450 watts continuous. I could power my electric drill no problem, even through a 100 foot extension cord.

However, it couldn't charge my laptop which has a power adapter rated at only around 200 watts.
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Old 29-06-2009, 18:53   #3
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15amps x 120volts=1800 watts

1800/3. 1/3 hour or 20min
600watts/12. =12vamp hrs
50 amp hrs from batteries

Here's my guestimate
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Old 29-06-2009, 19:00   #4
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Originally Posted by drew.ward View Post
I know the practical output of many inverters is a bit lacking.

I had a black and decker one I bought from Wal-Mart for 50 bucks. It was rated at 450 watts continuous. I could power my electric drill no problem, even through a 100 foot extension cord.

However, it couldn't charge my laptop which has a power adapter rated at only around 200 watts.
I have a 500watt inverter and I too am unable to use my laptop which is rated at 65watts. My battery has approximately 200 amphours and is always fully charged by solar panels. Anybody know what gives?
hank
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Old 29-06-2009, 19:02   #5
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Since 12V Dc is appox 10% of 110VAC, the current draw rating of a 110V appliance will be approx 10 times that at 12V. Your 15amp hair dryer will therefore draw 150 amps at 12VDC. Divide by the fraction of an hour of expected usage (15 minutes) equals 25% of 150 amps or approx. 37 amps.

If you have a 100 amphr rated battery, you just used 37% of it. You can do the math to extrapolate to whatever you really have for batteries and assuming you understand the concept of not drawing down batteries beyond 50% capacity, can figure out what size bank you would need just for the hair dryer.

This can get more complex but the about is a simplistic description.
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Old 29-06-2009, 19:02   #6
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Originally Posted by Hankthelank View Post
I have a 500watt inverter and I too am unable to use my laptop which is rated at 65watts. My battery has approximately 200 amphours and is always fully charged by solar panels. Anybody know what gives?
hank
Probably your laptop doesn't like square wave AC power.
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Old 29-06-2009, 19:29   #7
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Agree with S?V Illusion. You need a more sophisticated Sine wave inverter for computers, some electronics, clocks and printers.
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Old 29-06-2009, 19:40   #8
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We must remember a nasty little thing called the Peukart Effect.
You may think you just drew 50 AmpHrs of electricity in 20 minutes. You didn’t if it was:
Battery Size..........Rate of discharge....Hair dryer...Amps drawn/...Actual Used
AmpHrs (20 hrs)............20 hrs................Amps.........20 hr rate......... AmpHrs
400 ............................20 Amps.................150...............7X....................80
100 ..............................5 Amps.................150..............30X......... .........Dead
You were discharging the battery bank at a rate 150 Amps. Rated batteries capacity is the multiple of 20 hrs times the rate of discharge that empties the battery in that time.
If you want to use a hair dryer think about a 1000 AmpHr battery bank.
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Old 29-06-2009, 20:06   #9
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I had so much trouble trying to post that table I missed a lot.
If you are using a modified sine wave inverter for your labtop computer it will ruin the batteries in your labtop or any rechargable batteries. It's because the charger is a transformer less capacitive charging device that just does not like modified sine wave. It overcharges your rechargeable batteries, it cooks them. It also uses a lot more power as you have noticed doing that.
Generally most motors need 3 times the rated wattage (amperage) of the motor, refrigerators need up to 5 times.
This all has to do with the amperage lagging a bit behind the voltage changes.
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Old 29-06-2009, 20:23   #10
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Battery bank size
If you use 100 AmpHrs/day the battery bank should be at least 6 times that or 600 AmpHr rated capacity. That is 6 golf cart batteries (Trojan T105) or 3 8D batteries.
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Old 29-06-2009, 22:57   #11
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most laptops nowadays will charge just fine on a modified sine wave. some of the really cheap invertors will not provide this. same goes for power tool battery chargers.
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Old 30-06-2009, 01:13   #12
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I have cooked a lot of batteries and ruined a lot of equipment over the past ten years using a modified sine wave inverter before I learned why.
It happens with sustained daily use over months. You will not notice much of a battery capacity drop charging your batteries once in a while.
It is very difficult to “filter” or “smooth” out an instantaneous drop in voltage followed by a short period of zero voltage followed by an instantaneous negative voltage. This is complicated by as the voltage of your battery bank drops the ”peak” voltage also drops. To compensate for this voltage drop the time the voltage is on lengthens and the time the voltage is off or at zero shortens. This is hard to deal with.
Inductive loads such as motors run slower, have less power and run noisier. The power is no longer simply the amps times the volts. For a short time there is a lot of voltage but no amps therefore no power. At the other end of the “½ cycle” there is no voltage but the amps are flowing, therefore no power. Power factor is a way of describing this problem, a simplified way of expressing power factor for a modified sine wave inverter is;
Power factor = time when both volts & current are present / time when current is present
Note; current in the denominator can be replaced by voltage in this simple explanation
A power factor of 0.75 means a 1 hp motor now produces only ¾ hp. That is why a skill saw or any power tool lacks power running off an inverter. Fridges and freezers are also affected.
Electronic equipment has difficulty dealing with this issue of full on, nothing and then full on of the opposite voltage. They prefer smooth sinusoidal changes.
Any equipment such as some sewing machines or motors using microprocessors to control the speed may become damaged. With sustained every day use they will eventually fail.
It can cause problems for electronic equipment such as printers, etc,
Over a period of time a modified sine wave inverter may ruin any battery and/or equipment that uses a transformer less capacitive voltage conversion system.

If you are planning on using an inverter on a full time basis for years to power labtops and other electronic equipment it would be prudent to install a pure sine wave inverter .
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Old 01-07-2009, 00:35   #13
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Use of Inverters

Many thanks for the great imput! I've learned a great deal. Which was not hard as this is pretty much an empty vessel when it comes to electrical issues.

Glen
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Old 03-07-2009, 02:05   #14
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Powering MacBook laptop from boat batteries?

Over a period of time a modified sine wave inverter may ruin any battery and/or equipment that uses a transformer less capacitive voltage conversion system.

If you are planning on using an inverter on a full time basis for years to power labtops and other electronic equipment it would be prudent to install a pure sine wave inverter . [/QUOTE]


I need to power a MacBook Pro in my sailing boat.

Is there a way, however, to power the Mac directly from the boat batteries, keeping into account that the boat batteries are around 12.5 V max and the Mac laptop rechargeable battery is (at the moment):

Amperage (mA): -1947 (recharger not connected)
Voltage (mV): 11113

Thank you.

Paolo
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Old 03-07-2009, 05:53   #15
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Years ago I powered a Toshiba computer from a 12 volt battery. The nominal voltage of the battery pack was 12.5 volts. The output of the charger was 15 volts.
It worked just fine. It never seemed to fully charge the batteries in the computer but it worked and did not appear to hurt the labtop’s battery.
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