9V with 4Ah issue


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  1. #1
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    Question 9V with 4Ah issue

    I am using the attached adapter. It should empower a circuitry with 4A for one hour (4Ah).

    As a reference I have a 9V block battery, rechargeable, 200 mAh.

    Using the 9V bb it works well for a very limited period of time due to it's low capacity. Using the adapter it seems as if not enough current can be delivered, as if there was a bottleneck.

    So the pic does not start it's program using the adapter and a led(+5V -> led -> resistor -> ground) is flashing and shows a very dimmed light. This led is on like normal using a PSU or the 9V block.

    What is wrong with the adapter, how to fix it?
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    Quote Originally Posted by Security View Post
    I am using the attached adapter. It should empower a circuitry with 4A for one hour (4Ah).
    As a reference I have a 9V block battery, rechargeable, 200 mAh.
    Using the 9V bb it works well for a very limited period of time due to it's low capacity. Using the adapter it seems as if not enough current can be delivered, as if there was a bottleneck.
    So the pic does not start it's program using the adapter and a led(+5V -> led -> resistor -> ground) is flashing and shows a very dimmed light. This led is on like normal using a PSU or the 9V block.
    What is wrong with the adapter, how to fix it?
    6 AA NiMN (or NiCad's) do NOT make a solid 9v...Only does about 1.2v per cell, 7.2v nominal...without a load.
    Put a load on it, and it'll drop a bit more.
    Try to run that low-ish voltage thru a 7805 type regulator and you'll be running up against the minimum voltage needed for a 7805 type regulator to regulate properly.

    In other words, add 2 more cells and you should be ok.

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    So you are talking about using three parallel 9V blocks instead of one adapter?

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    Quote Originally Posted by Security View Post
    So you are talking about using three parallel 9V blocks instead of one adapter?

    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?

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    Hi, Folks

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


    but ... I must raise AA cells (even the best : 2800 mah ) DO NOT reach the requested capacity ...

    reminds me someone's projects ... but who's ???

    Alain
    Last edited by Acetronics2; - 23rd September 2008 at 08:51.
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    Quote Originally Posted by Acetronics View Post
    reminds me someone's projects ... but who's ???
    Alain
    Are you thinking who I'm thinking?

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    Quote Originally Posted by skimask View Post
    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.
    Last edited by Security; - 24th September 2008 at 00:18.

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    Quote Originally Posted by Security View Post
    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?

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