Battery Life

Discussion in 'General Electronics Chat' started by jerseyguy1996, Jun 23, 2011.

  1. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    If I have a 4.5 Amp Hour SLA battery and I have a circuit that is drawing one milliamp does that imply that this battery would last 4500 hours attached to this circuit?
     
  2. iONic

    AAC Fanatic!

    Nov 16, 2007
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    Ideally, I believe so. This of course would not take into consideration the batteries self-discharge characteristics which would involve time & temperature.
     
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  3. #12

    Expert

    Nov 30, 2010
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    4500 hours is more than 6 months...187 days. Low quality batteries can lose a percent a day to self discharge mechanisms. This is significant if you're trying to do something real. If you're just checking your understanding...never mind.
     
    Last edited: Jun 23, 2011
  4. Audioguru

    New Member

    Dec 20, 2007
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    I have a lithium-ion battery from my daughter's first cell phone. It is about 10 years old and it still is fully charged. I was going to use it to power an electric RC model airplane until I tried lighter-weight Li-Po batteries.

    Throw away the antique lead-acid motorcycle battery.
     
  5. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    It needs to last 7 days while intermittently powering a top off pump for my aquarium. I want to include a battery voltage monitor in my design so that if the voltage across the battery falls below 11 volts or so the system will shut itself off completely and then not turn on again. I was going to run the output of a TC54 voltage monitor to an NPN transistor that would supply current to both the top off circuit as well as back to the voltage monitor. The thought was that if the voltage monitor output goes low due to a low voltage across the battery that it would shut the transistor off which would kill power to the TC54 and the rest of the circuit preventing it from coming back on again. The goal is to avoid a situation where the battery is getting low and the start up current of the pump motor causes the voltage to drop which causes the TC54 to shut it off but then because the load is removed the voltage rises and the TC54 powers the circuit up again and you will get an endless oscillation which will burn up the pump. Once it is off I want it to stay off. The 1 ma figure came from the base current of the NPN. I really don't know if that number is correct but the datasheet for the NPN seems to use that number a lot as far as operating the transistor in its saturation region. I figure that is a constant current drain that will be happening 24 hours a day seven days a week until the battery is dead.
     
  6. #12

    Expert

    Nov 30, 2010
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    Ahhhhh! My head hurts! Can you slow that down a bit?

    Seriously, a schematic would work wonders in this case. You have so many ideas going by that I can't make a drawing by reading the words.

    ps, 11 volts is a disaster for a lead acid battery. Better check your specs at batteryuniversity.com
     
    Last edited: Jun 23, 2011
  7. iONic

    AAC Fanatic!

    Nov 16, 2007
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    It now depends on the current demands of the top-off pump, but a week does not sound unreasonable. If including circuitry to shut the system down, the 11V you are selecting is way to low. Your battery will not last very long. Generally, I wouldn't let it go below 11.7V, ideally I wouldn't let it fall below 12.4V before charging it. When the battery falls below 12.4V it will start to build up sulfate on the electrodes and shorten its lifespan.
     
  8. #12

    Expert

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    Yeah, what he said.
     
  9. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    I am working on a schematic for the whole circuit which I will post. Thanks for the reply!
     
  10. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    Top off pump is a peristaltic pump that I will expect to run for about 30 seconds each hour. Maybe even less than that. Under no load it draws .32 amps and under full load (me grabbing the shaft and holding it as tight as I can...not stalled though) it draws up to about an amp. I will change the divider values to make the low voltage 12.4 if you think that is more acceptable.
     
  11. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    Well I put together a schematic of what I am trying to do. The TC54 is a voltage monitor. The output is high whenever the voltage on the input is above 2.7 volts. The divider is supposed to present a voltage higher than 2.7 volts to the TC54 as long as the battery voltage is above 12.15 volts. I want the circuit to shut down completely and stay off once the TC54 trips to avoid any on and off oscillations caused by the start up current of the motor when the battery starts to get low. My thought is that I can press SW1 momentarily to turn on the circuit. Please if anyone has some time can you look this over and tell me if it is an appropriate implementation of what I am trying to do?
     
  12. #12

    Expert

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    As soon as the TC54 draws current to activate the 2n3904, the current will have to come from R6 and that will stop everything. Your logic is wrong.

    You need to think in terms of, "when the TC54 draws current and fouls its own supply, that will insure the 'off' condition persists".
     
  13. jerseyguy1996

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    Feb 2, 2008
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    I'm confused because when the 2n3904 is "on", isn't the circuit made up of R6, R7, and Q3 identical to Figure 4.1 on page 6 of the attached datasheet?
     
    • TC54.pdf
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  14. #12

    Expert

    Nov 30, 2010
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    You have properly copied fig 4.1
    and when the output of the TC goes high to send current to the 2n transistor, where does it get that current?
    How much current will go through R5?
    and what will be the voltage at the junction of R6-R7?
    and what will the TC chip do when it sees that voltage on its input?

    ps, you need to use the VC version to get a "high" output,
    just in case you labeled that chip wrongly.
    consider fig. 4.2 for "on until battery goes low".
     
    Last edited: Jun 24, 2011
  15. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    So the hamster in my head is struggling a bit on his wheel. I am thinking that the problem is when the output goes high it is basically equivalent to sticking a resistance (R5 + any resistance in Q3) in parallel with R7 which is going to throw the ratio that I have established in my divider off. Am I close?

    Oh and the chip is the VC version....specifically mouser part # 579-TC54VC2702EZB
     
  16. #12

    Expert

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    Yes. The current to drive the 2n transistor has to come from R6, and there is very little of that available. The voltage that the TC senses on its input (Junction R6-R7) will immediately switch it off, as in "low battery" condition.

    You need to rearrange the switching logic, possibly by using the mosfet inverter of figure 4.2

    That might require rearranging something else!
    I'm sure you can do it, but it's not a 5 minute job.
     
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  17. jerseyguy1996

    Thread Starter Active Member

    Feb 2, 2008
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    Back to the drawing board then. I really appreciate all of your help with this!!!
     
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