9V with 4Ah issue

[Security]

Ok, let's get smart here...I'm leading you somewhere...

Answer these questions...in order...

1) What is the NOMINAL voltage of a standard AA NiMH cell (not the peak voltage, not the depleted voltage, not the voltage you read from the meter right now, but the NOMINAL voltage according to any NiMH, or NiCAD datasheet)?

2) How many cells do you have in your pack?

3) Multiply the number of cells in your pack by the NOMINAL voltage of each NiMH cell in that pack. What is that value?

You are asking the wrong questions. There is space for 6 batteries which amounts to 7.2Volts as you have already figured out. Adding two batteries is not possible here without creating a new adapter. And I am talking about creating in an injection molding sense.

If you got that opportunity you are lucky.

So, replacing the 78xx series regulator by a LT 1083 ( 7.5 A )could be an idea ... instead of adding cells.

Yes. As the one mentioned requires additional components to adjust the voltage a new circuitry is needed. This is what I desire to avoid.

[skimask]

You are asking the wrong questions.

I don't have to ask ANY questions. I'm just asking the questions so you can figure out what you're problem with the circuit is.

There is space for 6 batteries which amounts to 7.2Volts as you have already figured out. Adding two batteries is not possible here without creating a new adapter. And I am talking about creating in an injection molding sense.

If you've got a mold already, looks like you should've prototyped the circuit a bit better before creating a mold.
At any rate, whatever you decide to go with, if you look at the regulator's datasheet, you'll find a spec called the 'dropout' voltage or maybe (on the datasheet I'm looking at) 'Minimum Input voltage required to maintain line regulation'. In this case, it's 7.2v. You're battery pack is 6 cells, 7.2v NOMINAL. Put a decent load on those batteries, and you WON'T have 7.2v, could easily drop to ~6v even with a C/10 load (check the batteries spec sheet). And at 6v, you WON'T have 5v at the output, if you have an output at all.
If you change to a low dropout regulator (i.e. not a 7805 type, maybe an LT1129-5.0 type), you may get better performance from it (i.e. lower dropout voltage differential), thereby enabling your circuit to run at a lower input voltage.
Otherwise, looks like it's time for a redesign.

If you got that opportunity you are lucky.

Which opportunity is that?

[Acetronics2]

Hi, Selcurity

Adding two batteries is not possible here without creating a new adapter.

reaching one hour autonomy nor ...


For once, could you tell us what exists as a regulation ... you want to make work YOUR thing without changing anything ...

and it doesn't work yet !!! ... really funny.




Heuuuu, by the way ... LT 1083 EXISTS in fixed 5v output ...

Damned, ... its possible !

Alain

[skimask]

Heuuuu, by the way ... LT 1083 EXISTS in fixed 5v output ...

I wonder how far the battery voltage will drop under a 4A load...and if that battery voltage will be enough for even the LT1083 to still regulate properly...

[Acetronics2]

Hi, Ski

1.3v @ 7.5A ... that's Ok !

Note the MIC 2975x ... might only drop .3 to .5 v ( from my souvenirs )

But now ... 6 AA "hicap" NiMh cells should be around 0.45 Ohms internal resistance ... (0.223 - 0.3 measured for 4 CELL !!! )

a bit more for the wiring and awful connections ...

YOU WIN !!!

But 6 Sub C ... hé,hé ...

Alain

[skimask]

But now ... 6 AA "hicap" NiMh cells should be around 1.5 Ohms internal resistance ...
a bit more for the wiring and awful connections ...

You mean like this?
http://www.picbasic.co.uk/forum/show...2&postcount=32

YOU WIN !!!

What do I win?

But 6 Sub C ... hé,hé ...

Apparently that would require some sort of redesign...
Sub-C just might handle the load

[Acetronics2]

You mean Ted's work, I suppose ...

for sure !

Alain