Author Topic: Help Needed.  (Read 1696 times)

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Re: Help Needed.
« Reply #8 on: July 07, 2019, 20:50:24 pm »
What if I used 13 9 volt batteries and got 117 volts to produces 600 watts.
Wouldn't that be under 400 Milliamps per cell?

Stevie, whats your thoughts?

P=U*I 
600= 117 * I
I= 5 amps.  Not 400milliamps

So, your 9 volt batteries are sucked empty pretty quickly

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Re: Help Needed.
« Reply #9 on: July 07, 2019, 20:55:50 pm »
What if I used 13 9 volt batteries and got 117 volts to produces 600 watts.
Wouldn't that be under 400 Milliamps per cell?

Stevie, whats your thoughts?

P=U*I 
600= 117 * I
I= 5 amps.  Not 400milliamps

So, your 9 volt batteries are sucked empty pretty quickly
Tek is right about voltage drop. This happens when yr battery cannot deliver the proper amps.
A 9 volt battery delivers about 250-300 mAh
to be able to provide 5 amps, you must use a couple of them in parallel. 20 strings of 13 batterys in series.
That works.


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Re: Help Needed.
« Reply #10 on: July 07, 2019, 21:03:56 pm »
how exactly is a solar panels built anyhow?
I have 5 panels, each is 37.5 volts and 8.5 amps.

If I connect 2 of the panels in parallel I get double the current production and the voltage remains the same.
But, If I connect 2 of my panels in series, the current production remains the same but the voltage doubles.

If I were to calculate the 600 watts directly from the utility pole, I would say I use 30 Milliamps at 20,000 volts, then onto the 240 pole transformer that consumes  2.5 amps, but my Window unit is 120 volts so I must have 5 amps.

I thought about powering the unit with a 12v  battery at one time since everyone says thats the logical way to do it, but then I decided that 50 amps from that battery at such low voltage was not as appealing as the 120v DC using a H Bridge at 5 amps. But then I realized that If I could just somehow use 240 volts at 2.5 amps I would be more than willing to donate a amp or 2 for the heat loss of stepping it back down to 120, since the unit does operate on 120 volts.


I installed a new unit the other day and the rotor locked up, the locked roter amps was 108 amps at 240 volts. I realized that a locked moter was consuming over 25,920 watts!!! But I dont think anyone hardly noticed since I was only using 1296 Milliamps from the 20kvolt power line wire!  :)

I still dont fully understand how my solar panel is built, Is it like one very long strip that produces 37 volts? and also, how you reckon it produces a whopping 9 amps of current at that 37 volts?

Anyone gave any thoughts to those 52 batteries in the question I ask above yet? Again, Your opinions are welcome, and No answer here has to be perfect. Get the Ideal in your head so I can be cleared up on this, then I will move on to the next step.

http://www.ti.com/tool/TMDSSOLARPEXPKIT

Thats a circuit I think is junk, But the Ideal is Legit, it has way to much computerized junk on it.

A solar panel is build out of small cells who are connected in serie.
I know as i have 40 panels here, pumping 90V dc into my MPPT which charges up my 20kw lit ion batterys
I use actually two mppt's.
I went for a 14S battery setup.

Tell me a bit more what your ideas are...What do you have and what goal have you set?

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Re: Help Needed.
« Reply #11 on: July 12, 2019, 03:56:12 am »
Divide Ah by 10 to be a safe health of the used battery. So if you have 3000Mah for 1.2 volts it would be this sum you would not want to go over per hour.

3000mah or 3 amps/10 wouldnt want to use more than 300Mah constant flow over an hour period for 10 hours.
.350/10 This 9 volt battery can provide 35Ma for ten hours. Its understood this battery can not produce 350 in one hour, it would destroy it. But it can produce 35ma for 10 hours.

So, if I have 100 3000Mah 1.2 volt batteries. Again, I would then have 120 volts!
Consuming at a rate of 600 whats, thats 120 volts 5 amps!

If one did use 100 of these batteries to produce this, how much current would be consumed from each individual cell?
The Load Wants from each battery 50Ma. Because .50Ma X 100 Batteries is 5 amps.

The battery only wants to supply 300 Ma per hour for 10 hours, Since it is a 3000Mah battery.
But we're not asking each cell for 300 Ma in this situation, We are asking for 50Ma.

Please give this a second opinion.

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Re: Help Needed.
« Reply #12 on: July 12, 2019, 08:18:03 am »
In my opinion, if you put a hundred batteries in parallel, the volts will be the same as a single cell, but the current will add up.  Put the batteries in series, and the volts will add, but the current would only be what an individual cell can provide. 

You can't get 600 Watts from 100 batteries unless you can get 6 Watts from a single battery.  That's 4000 mA, continuous, with 1.5 Volt batteries.  Until the voltage starts going down.
« Last Edit: July 12, 2019, 08:33:18 am by tektrical »

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Re: Help Needed.
« Reply #13 on: July 12, 2019, 09:45:50 am »
100 battery,s in serie give the maximum amps of what 1 cell can give.

Batterys in parallel give the totall amps of all of them..


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Re: Help Needed.
« Reply #14 on: July 12, 2019, 16:32:21 pm »
Steve is right.
To achieve what you are thinking of, you need 100x100 batteries.
That's a lot!

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Re: Help Needed.
« Reply #15 on: July 13, 2019, 08:05:51 am »
Thank you so much for solving this confusing iss for me! That's why I ask here, because 4 eyes and ears is better than 2 right?

A little twist here. My car battery is 600 cold cranking amps. That's a whopping 7200 watts of a jolt!

Let's say I want to take the 12 volts and move it's decimal point to the right 3 times. "12000" volts. And the 600 amps decimal point to the left 3 times, ".6".


Who here thinks the same power can create the same torq in horse power with the same wattage?

Move the decimal 4 times.
120,000 .06

Who here thinks a 120k moter that draws 60 milliamps can start a car?

If so, wouldn't the wire be smaller? Would this be possible without changing the size of the starter desined for 12 volts?

Comments and opinions please.
Thanks guys! And that is a ton of batteries in the last few post, but was very correct.