# In Theory, Which Wiring Scheme Should Result in Longer "Battery Run Time"?



## toddalin (Jan 4, 2008)

I want to power a 6 volt regulator and I have two 9 volt battery packs. In theory, would I get a longer run time if I connect the two packs in parallel (9 volts but double the capacity), or in series (18 volts)?

Recognize that the regulator needs ~7.5 volts to operate.

So with two packs in parallel (6 cells per pack), there is twice the capacity, but each cell can only run down to 1.25 volts before failure. But in series, each cell could run down to 0.63 volts (or flat) before failure, but more energy is dissipated (lost) through the regulator.


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## FRedner (Mar 20, 2011)

What is the actual voltage of the battery packs? 

The basic answer is the setup that wastes the least voltage in the regulator... and also as you clearly understand, the discharge voltage you can handle... 

Notice in parallel, you have TWICE the amp hour capability... so a quick guess is put them in parallel... but back to my first question, what is the actual voltage of the packs? 

Your statement of 1.25 volts seems to indicate NOT nicad or nimh ... the only batteries that give you exactly 9 volts and are rechargeable are lead acid chemistry... 

FR


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## Del Tapparo (Jan 4, 2008)

It sounds like you are using NiMh batteries, which ideally, you would like to discharge to 1.0 volts per cell (6.0 volts for the packs in parallel, which is the way to get the most runtime). If you need an overhead of 7.5V to stay in regulation, that means you can only go down to 1.25V / cell, as you stated.

First of all, how close do you need to regulate the 6 volts. It won't really drop too much, even though it gets below the 7.5V.

You can also use a low dropout voltage regulator, like the LF60CV. Max drop out voltage is 1.0V, so you can get down to 7.0V, or 1.17V per cell, and likely lower than that.


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## toddalin (Jan 4, 2008)

"I want to power a 6 volt regulator and I have two 9 volt battery packs. In theory, would I get a longer run time if I connect the two packs in parallel (9 volts but double the capacity), or in series (18 volts)?"

Two packs each with six AA alkaline cells. Runs through a bridge then a 6-volt regulator.

Circuit is designed to take batteries, track power, or lighting voltage, depending on what's on at the time. Circuit draws ~ 200 ma.


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## Greg Elmassian (Jan 3, 2008)

Parallel.... voltage starts at about 1.6, battery expires around 1.4... twice the amp hours... 

Greg


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## TonyWalsham (Jan 2, 2008)

If you are using rechargeable 9 volt "transistor" batteries make sure they are actually 9 volts. Most are actually 7.2 volts. Some are 8.4 volts and usually more expensive. 
Also, the same rules apply about running NiMh cells in parallel. You need isolating diodes to prevent voltage bleed from one cell to the other. 
Even with two in parallel they cannot handle much more than 100 ma. Best to use much larger capacity single cells wired in series. 

What is the 6 volt regulated voltage going to be used for? There could be better ways of doing this.


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## toddalin (Jan 4, 2008)

Posted By TonyWalsham on 05 Apr 2011 03:47 AM 
If you are using rechargeable 9 volt "transistor" batteries make sure they are actually 9 volts. Most are actually 7.2 volts. Some are 8.4 volts and usually more expensive. 
Also, the same rules apply about running NiMh cells in parallel. You need isolating diodes to prevent voltage bleed from one cell to the other. 
Even with two in parallel they cannot handle much more than 100 ma. Best to use much larger capacity single cells wired in series. 

What is the 6 volt regulated voltage going to be used for? There could be better ways of doing this. 

As I said, two packs of six AA alkaline cells.

It is used for the RigiDuo that runs off a 6 volt motor and with all of the circuitry, draws ~200 mA when the motor actually operates. I use a "trailer connector" so I can run off track power, battery power, or my lighting circuit.


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## Dan Pierce (Jan 2, 2008)

Maybe it is time for a DC to DC converter. Depends on ratings for power loss as to which could be better, regulator or DC to DC converter. 

At 18 volts to 5 volts, a DC to DC converter sounds like a better way to go, but I have not checked for the losses.


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## Greg Elmassian (Jan 3, 2008)

The most efficient would be arranging the 12 cells in 6 volts, as parallel as possible... 

Instead of 2 parallel sets of 6 batteries in series and losing through a regulator, put the 12 batteries together as 3 parallel sets of 4 batteries... that is simple, inexpensive and will give you the most run time. 

Remember that when new, they are about 1.65 volts per cell... you can rewire those plastic holders easily. 

Greg


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## toddalin (Jan 4, 2008)

Posted By Greg Elmassian on 08 Apr 2011 08:22 AM 
The most efficient would be arranging the 12 cells in 6 volts, as parallel as possible... 

Instead of 2 parallel sets of 6 batteries in series and losing through a regulator, put the 12 batteries together as 3 parallel sets of 4 batteries... that is simple, inexpensive and will give you the most run time. 

Remember that when new, they are about 1.65 volts per cell... you can rewire those plastic holders easily. 

Greg 

This was also my initial thought also, but I hate to screw up my two good battery packs.

In fact, I was going to use one set of four cells to power the electronics and two sets of four cells to power the motor. That way when the motor wears down it's two packs, it will just slow down instead of freakin' out the 555 chip due to low voltage.

But most of the time this will be track powered so I would still need the regulator. I was thinking of installing a plug between the regulator and electronics/motor that would allow me to just plug in 6 volts while disconnecting the regulator.


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