when linking two high voltage batteries in parallel, what certain
things do i have to take in to consideration, to have a stable output
current?  i have two 144 Volt batteries, each made up of 20 strings of
6 1.2 volt batteries linked in series.  i know that theoretically if i
do link them in parallel i should get a constant voltage of 144, and
double the output current.  if this true in the real world, and if not
then what things do i have to take into consideration.

Hello awl,

Firstly we should note that I believe the actual voltage you would be
dealing with in this scenario is 130 Volts. ( You had me going to the
calculator :-)

20 Batteries X 6.5 Volts each = 130 Volts. In series we simply add all
the voltage together to get the total amount. I will just use the 130
Volt reference to answer your question, because I do understand what
you meant.

You are correct however, you would most certainly get 130 Volts total
if you wired these two batteries in parallel.

"When linking two high voltage batteries in parallel, what certain
things do i have to take in to consideration, to have a stable output
current?"

The first and foremost thing to take into consideration is the amount
of current flowing through each leg if you plan on using a parallel
circuit. In parallel, each 'leg' will get the same amount of voltage,
and you can add resistance to each leg to change the desired circuit
current state. A simple example of resistors in parallel can be found
here :

http://ourworld.compuserve.com/homepages/g_knott/elect61.htm

With another simple schematic of how current actually works in a
circuit:

http://ourworld.compuserve.com/homepages/g_knott/elect35.htm


Here are some more key factors to take into consideration:

- Current Draw. What is the amount of current that the load (device
being powered) in question will use? Are you supplying enough current
for the circuit? Ohms law will help you figure out what you will need
to do. Here is the Ohms Law equation :

E = I X R

Where E ( Or sometimes V ) is the voltage, I is the Current, and R is
resistance. I found a website that explains it in far more depth than
I could in this answer:

http://ohmslaw.com/ohmslaw.htm


- Adding a fuse to your circuit. Adding a fuse to your circuit will
protect the power source and the load ( item drawing the current )
from an excessive current flow problem.

- When configured in parallel, batteries will last twice as long
(using the configuration in the link below ) as opposed to the same
circuit idea, wired in series. This is because only half the 'amount
of work' is put on one battery.

Example Circuit
http://ourworld.compuserve.com/homepages/g_knott/elect27.htm.

- We have then created an instance where you would need to be
concerned with the 'rating' of the components you would use in this
circuit. Resistors for example, come in many different power ratings.
For your circuit, you would again use Ohms law to determine your
overall requirements. Here is great reference for resistors :

http://www.nzart.org.nz/nzart/exam/Amateur%20Radio%20Study%20Guide/Course%20Files/RESISTORS%20and%20the%20COLOUR%20CODE.htm

- Power generation. As current comes from your Batteries it flows
through the resistors and creates power consumption. If this project
were for example, a power conserving effort, then you would have to
adjust and tweak all your circuit variables for the best overall
effect.

Example of Power
http://ourworld.compuserve.com/homepages/g_knott/elect63.htm

I hope this touches on your concerns, and I think you will find the
pages I sent you to a great resource as you go. The main page for the
site can be found here:

Electronics and Microcomputing 
http://ourworld.compuserve.com/homepages/g_knott/index1.htm


I found the answer to your question by searching for :

Paralell Batteries stable current
://www.google.com/search?q=parallel+batteries+stable+current

I also pulled on my own experience in the electronics field. As you
can see - most of the concern is based on what you wish to do with the
batteries in question.

Thanks for the great question!
SgtCory

---

Clarification of Answer by sgtcory-ga on 21 Oct 2002 18:53 PDT

Hello again -

The same answer still applies - I simply just realized what a
tremendous effort you had taken in wiring the batteries together :-)
They are indeed 144 volts. I took the reference 6 1.2 as being 6 1/2 (
one half ).

Please ask for any clarification if you need it. I'd be glad to assist
-

SgtCory

---

Clarification of Answer by sgtcory-ga on 21 Oct 2002 19:11 PDT

And further explanation brings me to this : The 20 strings of 6 1.2
volt batteries was the full 'fender bender'. This translated into 20 -
6 1/2 volt batteries is series on my end. Either way the situation is
still the same, as the end concern is using them in a parallel
capacity.

Maybe I really do need glasses awl? :-)

SgtCory

---

Request for Answer Clarification by awl-ga on 22 Oct 2002 09:52 PDT

thank you for you help, but i am still a little confused about concern
of power ratings for components, and how they should be tweaked?  In
your example you have a resistor as a load, i'm not sure if you are
saying that i need a resistor, in my circuit, or does the resistor
just represent my actual load, which in this case is an electric
motor.  Lastly in my circuit i have to two battery packs linked in
parallel and now due to your answer i'll probably add a fuse in
between the motor and the packs, what things can i actually tweak to
conserve power as you have stated in your answer?  once again i really
appreciate you taking time to help this college student, thanks again
bye

Andy

---

Clarification of Answer by sgtcory-ga on 22 Oct 2002 10:39 PDT

Hello awl,

Not a problem. That's why we are here - to help answer your question.

"power ratings for components, and how they should be tweaked.."

I was referring to the actual power rating of a resistor for example.
They come in different levels of power dissipation. Depending on your
circuit, you would/could change this variable. As to how it would work
into your overall equation would probaly require some schematics and a
higher level of participation as far as building the circuit is
concerned. I just mentioned this as a hypothetical angle of approach.

In short :

The larger sized resistors allow for greater dissipation.

"In your example you have a resistor as a load, i'm not sure if you
are
saying that i need a resistor, in my circuit..."

I was just giving reference to typical configurations. Without knowing
exactly what you are building that is all I can really do. What type
and size of resistor you will be using is up to your desired end
result. One thing to note in your clarification is that the load is
your motor. My answer is based on the fact that you might of had a
load in mind - i.e. the motor. The resistor would/could be used to get
a desired voltage drop if you needed it. Somewhat easier to understand
info on resistors and voltage drop can be found here:

http://www.eworld.contactbox.co.uk/data/itemsmr/r_theory.htm.


"...what things can i actually tweak to conserve power as you have
stated in your answer?"

I was referring to power dissipation in the example. The formula for
power is :
P = I X E  (think of apple PIE :-)

Now that you have the formula, you would use that in the overall
equation to make adjustments to your variables, I - the current and E
- the voltage.

When I say the word 'tweaked' - I was simply referring to changing
your variables.

I hope that clears up the answer for you.

Thanks again -

SgtCory

Very helpfull and at the same time very willing to help,  thanks a lot

The answer is 'in principle yes'.  You do not double the output current,
of course,
(which still depends on the load). You double the output current capacity.


In real world I would worry about possible overheating.
Since batteries are not identical there will be an inner loop 
current even when there is no load.  It will go to zero soon,
but will generate some heat before it dies off.

You are better off going from AA's to As, Bs, ..  Ds and so on.

Yes - thanks for the helping hand hedgie. You do double the output
current "capacity", and as stated in my answer - one of the main
concerns is current draw. You will only 'use' as much current as the
load requires.

As stated in the original answer -

"Adding a fuse to your circuit will protect the power source".

If properly rated for the circuit in quesition this will also preclude
any thermal problems we may run into.

Thanks again for the clarification -
SgtCory

Yes - you are welcome Sergant.

 In my haste I have perhaps not elucidated that the heating
 I am concerned about is due to the 'inner loop' current.

 The fuse, as normally used, would not do help here. It would
 protect from a 'short' - in the EXTERNAL loop as shown here:

Like this:
  --------------fuse---------
 I      I                    |
 I      I                 LOAD

  ........                   |

 I      I                    |
 -----------------------------
                  external loop
  

    Internal loop is  all on the left,
   Runs from up one battery pile and down  the other,

when actual voltages differ a bit, even if nominal are same.

 Left column may add up 146 V, right to 140 V  (since batteries are not
  identical).   In this case you would have 6V driving the loop
 with just two internal resistances of the two piles, even if load has 
 very high resistance.

   Sorry for addding this bit to your comprehensive answer. 
I was just concerned about possible safety issue here.

hedgie,

Not a problem at all. The more of an answer - the merrier.

I was answering on the facts given :

- "i have two 144 Volt batteries, each made up of 20 strings of
6 1.2 volt batteries linked in series"

This means the battery is already made, and has lost all of its inner
loop current. I didn't want to make the answer more complicated.
However - I am 100% sure that awl will find your addings to be very
valuable information for future use :-)

- "when linking two high voltage batteries in parallel"

Again - awl was referring to a completed circuit, which is where I
started with my answer. Had this of been a question about the battery
itself, the approach would have been much different.

Thanks for all your comments!

SgtCory

hedgie -

That last comment I made still is not very clear. In a nutshell I was
working as if the circuit was already built, and in a quiescent state.
What concerns come then?

Your insight has opened a few new thoughts for me and I sincerely do
appreciate you taking the time to comment.

Thanks

SgtCory