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12vdc->9vdc


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In my wheelchair, I have two 12V batteries wired in series.
I would like to wire a few devices directly to these batteries.

The first items are a few transmitters (garage door buttons [9v battery]).

What circuit do I need to put between 12/24V battery and the transmitters?

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I should have mentioned I am not an electrical engineer.  :-[
If the lm7809 will work, how do I use it?
Can someone draw out a circuit with whatever info I may need, plus part specs so I can buy the parts (hopefully available at my local radio shack)?

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Hi Scott

my first thought is that you could take a connection from one of your 12v batteries and plug it in to the garage door button unit.  I suspect that the difference in voltage would not be significant, ie, 12/9*100 or roughly 30% more, it really depends on the design of the unit and whether it is capable of withstanding the overvoltage.  My first task would be to contact the manufacturer if you can and ask that question - on the answer will depend whether you need go to the trouble of making major modifications

Best of Luck

Ed

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thanks for the quick replies, this is a cool forum, too bad most of the topics are over my head.

For the standard garage door buttons, I would try hooking directly up to 12v. But I have a couple specialized transmitters I do not want to risk frying.  ;D

I think it will be worth my effort to do this, plus a learning experiance. Not to mention, I am bored!

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Hi Scott

boredom is a good thing, it's amazing what you can learn

I lean towards always finding the simplest way of making something work, apart from my earlier post, the next way forward would be to look at the next simplest way.

With your specialized transmitters, if you or someone could measure the current drawn by the unit whilst in operation you could possibly use a simple resistor as a solution.  For example, suppose a unit operated at 9v and 100mA and you wish to use 12v.  In this case you need to drop 3v at the same current, this equates to (using Ohms Law) R=V/I, =3/0.1, =30 Ohms and with a power dissipation in the resistor of V*I = 3*0.1, =0.3 watts.  So, a simple resistor would cure the problem in this case

All the Best

Ed

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Guest Alun

I'm sure it won't be a problem in this instance (as the load's so small) but if these batteries are connected in series and more load is put on one thant the other it can cause problems with unbalenced cell voltages which is even worse if they're charged in series or the wheel chair uses regeneritive breaking.

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