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does a 9V battery have a current limit?

Discussion in 'Power Electronics' started by dustin02rsx, Feb 8, 2012.

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  1. dustin02rsx

    dustin02rsx

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    0
    May 18, 2011
    I am doing a project with a high power led and i am using a 9V battery as the supply. The problem is, i designed the circuit to have almost 700 mA (670mA) to be exact and im only reading 300-330 mA...

    I am using a 10W 9.1 ohm resistor which i measured to be 9.6 ohms.

    i measured the battery and found it to be 9.68V.

    all the components are brand new.


    E= 9.68V
    R= 9.6 Ohms

    9.68V - 3.25Vf = 6.43V

    I= E/R

    6.43/9.6 = 670 mA....

    why am i only measuring 330mA?

    this is the led i have:
    www.luxeonstar.com/Green-Lambertian-Rebel-125-lm-p/lxml-pm01-0070.htm
     
  2. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    i just read somewhere that without a heat sink that a high power led will be limited to around 100 mA..... Could this be my problem? It was a little warm but definitely not hot.
     
  3. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    700mA sounds like a lot of current for a little 9V battery. How hot is the battery getting?

    Batteries have a internal resistance that will limit their output current. This internal resistance gets worse as the battery discharges.
     
  4. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    battery wasnt hot at all but i was only getting between 300 and 330 mA....

    i just read online that some high power leds currents are limited if there is no heat sink. also i just looked on the led part spec on my led and it says...

    " The thermal pad is electrically isolated from the anode and cathode contact pads."

    im not entirely sure what that means, but im pretty sure i need a heatsink
     
  5. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    A heat sink will not increase the current. What that means is if you run it without a heat sink at it's max current it will blow up.


    Also, did you measure the battery voltage open circuit or under load?
     
  6. KJ6EAD

    KJ6EAD

    1,114
    157
    Aug 13, 2011
    If you'll take a look at this, I think you can answer your own question.


    http://www.powerstream.com/9V-Alkaline-tests.htm


    If you recalculated your current limiting resistor for a 6V supply voltage, you might have decent output for a few minutes.
     
  7. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    If you don't want your LED going 'poof the magic dragon' you won't try it till you get a heat sink though.
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,384
    2,771
    Jan 21, 2010
    In some cases it will actually reduce it (which is a good thing because it means that it was running too hot). It reduces the current (in some cases) because at lower temperature the forward voltage is greater.

    You may like to read this.

    Limiting current to a LED using the impedance of the battery is not a typically wise thing to do. Most importantly, different battery types will lead to different (and perhaps MUCH different) current. Connecting a 9V NiCad (typically only 7.2V) battery might make the current exceed several amps as they have a significantly lower internal resistance.
     
  9. dustin02rsx

    dustin02rsx

    36
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    May 18, 2011
    my led says its rated for up to 700mA without a heatsink... it says you can push more current carefully with a heatsink.

    however i read somewhere that some leds have a limited current without a heatsink.

    i did calculate the voltage drop and to get close to 700mA was the 9.1 ohm resistor...

    now that you (steve) mention it, a 7.2Vf on my battery would put it closer to what im getting but (7.2V-3.2Vled)/9.1ohm = 434mA which is still 100mA off so idk whats wrong...

    its a basic ohms law circuit, i dont see what im doing wrong :(


    and i measured the battery not in the circuit.... i see what you're saying
     
    Last edited: Feb 9, 2012
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    OK, I missed the fact that you are using a resistor.

    Measure the battery voltage when the load is connected. You'll find that it is lower (dramatically lower) than off-load.
     
  11. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    just measured the battery under load and it drops as low as 7.19V.... Still calculates to 432mA

    im using a 10W 9.1 ohm resistor
     
  12. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    anyone have a suggestion for a battery to supply 700mA?

    after reading that link, that one battery started to mechanically fail @1A and i dont want to risk that.
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    OK, you tell us the resistor is actually 9.6 ohms. You tell us the battery voltage falls to 7.19V at 330mA

    That tells me, 3.17V is dropped across the resistor, leaving a calculated 4.02V across the LED.

    That tells me the actual Vf is closer to 4V than 3.25.

    However, your big problems are:

    1) The 9V battery isn't coping well at that current (internal resistance is around 7.5 ohms).
    2) More than half of your power is being dissipated in resistive losses.

    You would benefit from a lower voltage, higher current set of batteries, and a constant current driver for your LED.

    You would benefit even more from a current limited switchmode power supply. If you remove the diodes from that one it will operate below 7V input. For 700mA output current, the input current will be around 350mA
     
  14. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    How come no one ever listens to me? Did I not say to measure the battery voltage with no load? Did I not say the battery has an internal resistance? I will just stop wasting my time I suppose.
     
  15. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    ok thank you for your advice, im going to make a constant current source.... still not sure what battery i should choose though.

    it was your post that i saw i measured the battery out of the circuit and that helped me. i just didnt feel like multi quoting. thank you for your help :)
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    You need one that can provide the current you require.

    I think you were pointed to a datasheet for a battery earlier in this thread. hey generally give curves that show lifetime/capacity vs current. It is pretty clear from these that each battery type has a practical maximum current you can draw from it if you expect it to have a reasonable life.

    Battery datasheets are not always easy to come by, but you can generally assume that the curves from one manufacturer will be similar to those from another manufacturer for similar battery technologies.
     
  17. GreenGiant

    GreenGiant

    830
    6
    Feb 9, 2012
    to answer the question of the original post, there really is no limit to the current that can be drawn from the battery, just how long that current can be drawn.

    A typical 9V battery can provide 600mAh (milliamp hours) which means it will provide 600 milliamps for 1 hour before its dead

    you draw more you deplete the battery faster (1.2amps it will be dead in about half an hour)

    I think you have a voltage and current issue, a 9 volt battery is just not enough to put out what you want

    You may want to use one of the big 6 volt batteries and change the resistor that you have, those can sometimes provide up to 6Ah
     
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,384
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    Jan 21, 2010
    The max current you can draw from a battery is indeed limited by its internal resistance.

    As you increase the current the available capacity also falls.

    So id a particular battery is nominally 1000mAhr for a 50mA load over 20 hours, it may not actually be capable of 1000mA for an hour. You might only get 30 minutes. But at lower currents (say 25mA) you may get slightly more than the predicted 40 hours.

    It's not as simple as it seems.
     
  19. dustin02rsx

    dustin02rsx

    36
    0
    May 18, 2011
    im ditching the 9V battery for sure.... i havent decided what battery i want to use but i just hooked up my led with no resistor (it was a scary moment but i have more) and it only got to 530mA?!?

    The drop across the led was 2.7V
    the battery now measures 8.3V and 5.6 under load.

    (5.6V-2.7V)/530mA= 5.47 ohm internal resistance??? maybe? thats the only thing i can think of

    on a side note, once i get the battery situation figured out i found this current source in a data sheet of a part i have laying around. so ill be using this one.
    [​IMG]
     
  20. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,384
    2,771
    Jan 21, 2010
    That is not a constant current regulator.

    See here.
     
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