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 Post subject: Question about battery volatage
PostPosted: Fri Feb 11, 2011 5:57 pm 
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I've drawn up a schematic and was wondering if it was correct.
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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 6:03 pm 
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http://batteryuniversity.com/learn/article/serial_and_parallel_battery_configurations


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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 7:00 pm 
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Sorry, that didn't really help.


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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 7:33 pm 
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How didn't it!

It sure helped me to under stand how to do a Serial and Parallel connection...


SERIAL increases the VOLTAGE ONLY and NOT THE AMPS.
PARALLEL increases the AMPS ONLY and NOT THE VOLTAGE.


Thanks for the link Muniosi. Appreciated...


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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 8:15 pm 
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You can't simply just split the wires; the power will come from the same spot and will have the same voltage.
If you want to lower voltage, you'll have to use a resistor or a regulator.

And Haunted, it's "series," not serial.

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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 9:20 pm 
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So if I did that, both wires would be the same voltage?


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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 9:29 pm 
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Muniosi wrote:
So if I did that, both wires would be the same voltage?


Yes.

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 Post subject: Re: Question about battery volatage
PostPosted: Fri Feb 11, 2011 9:43 pm 
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hailrazer wrote:
Muniosi wrote:
So if I did that, both wires would be the same voltage?
Yes.
If they have an equal load. So you have to be powering 2 of exactly the same thing for it to be split in half equally (in simplest terms).

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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 9:24 am 
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I'm confused. Would both wires be 4.5 volts or 9 volts?


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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 9:43 am 
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Muniosi wrote:
I'm confused. Would both wires be 4.5 volts or 9 volts?

With the diagram you had, each wire would be 9 volts.

Here's how I would explain it in the simplest terms:
Parallel:
Image
This gets you 9 volts for each thing, but will draw more power from the battery (mAH).

Series:
Image
This gets you 4.5 volts for each thing.

Disclaimer: This is approximate, if the load (resistance) of the two different devices varies it won't be quite this exact, but it'll still be close.

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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 12:00 pm 
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jdmlight wrote:
Disclaimer: This is approximate, if the load (resistance) of the two different devices varies it won't be quite this exact, but it'll still be close.

Actually, you can have two series loads with the same resistance, but if one's just a bunch of resistors and the other involves some sort of inductive load, good luck. That's why I don't really like using simple voltage dividers, they don't always behave the way you'd expect.


If you need exactly 4.5V, something like this would work, if you can get away with using 5V then this is simple plug-and-play, no extra components needed to set the output voltage.

Or if you really want to save on battery power, maybe look into a DC-DC converter (like this), one of those would probably do the job too. (More money, and somewhat larger, but you do get better efficiency.)


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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 12:35 pm 
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jdmlight wrote:
Series Diagram (removed)
Disclaimer: This is approximate, if the load (resistance) of the two different devices varies it won't be quite this exact, but it'll still be close.

That seems very wrong. I know that if both loads are of an equivalent resistance they will have the same voltage drop across them, but they won't be at the same voltage.
Wouldn't the first thing be "fried" because it has 9V?

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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 12:50 pm 
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bassmasta wrote:
jdmlight wrote:
Series Diagram (removed)
Disclaimer: This is approximate, if the load (resistance) of the two different devices varies it won't be quite this exact, but it'll still be close.

That seems very wrong. I know that if both loads are of an equivalent resistance they will have the same voltage drop across them, but they won't be at the same voltage.
Wouldn't the first thing be "fried" because it has 9V?

Nope. Let's simplify, say each load is a resistor. Say you have two identical resistors in series. Each is the same. Each will drop the same voltage, just like parallel identical resistors drop the same amperage. If you have a 10V source, with a 50 ohm, a 30 ohm, and a 20 ohm resistor in series, then the 20 ohm will drop 2V, the 30 ohm will drop 3V, and the 50 ohm will drop 5V. (((single resistor ohms / total series ohms) * voltage) = voltage dropped by single resistor)

Loads in parallel will all see the same voltage, but will split amperage the same way as parallel loads split voltage.

Oh yeah, and for n resistors in parallel? The total resistance of the parallel set as a group when you want to drop the set in series with something else is (1/((1/R1)+(1/R2)+(1/R3)+...+(1/Rn)))

This is all for pure resistive loads, of course. Things get a little weird when you throw in inductive loads of course. That's why you don't see voltage dividers used so much in real-world situations.


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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 1:13 pm 
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What are you wanting to power?

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 Post subject: Re: Question about battery volatage
PostPosted: Sat Feb 12, 2011 1:48 pm 
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vskid wrote:
What are you wanting to power?

I'm powering a Hip Gear screen pad and a weird plug-and-play thingy that runs on 3 AA's.


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