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Power supply

Discussion in 'Power Electronics' started by Skyking ALexander, Jan 7, 2013.

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  1. Skyking ALexander

    Skyking ALexander

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    May 2, 2011
    I asked a while back about an LED circuit that I needed help with and was directed to an excellent thread that helped a lot. My question now it that I have reworked the circuit but I have an issue with the mAh. The circuit shown below is 13.95V and all of the LEDs in the circuit are 20 mA and all but the red led are 3.2v and the red LEDs are 2v each.

    The issue I am forseeing is that the the power supply as 9 357 button cell batteries will be large also the mAh of the battery is only 170 while the circuit will draw 160 mA. This means that I will only get 63 min out of the 9 batteries.

    My question is, is there a way to extend the mAh of a power supply.

    [​IMG]
     

    Attached Files:

    Last edited by a moderator: Jan 7, 2013
  2. davenn

    davenn Moderator

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    yeah, get better batteries :)

    you are never going to get much current oput of button cells
    why are you using button cells ? is there need for total portability ?

    if not, use a 12V Gelcell say a 7.2A/hr one will give you many hours of operating between charges

    Dave
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You can stretch it slightly by placing those two single LEDs in series.

    You're only dropping 0.36V across those 18 ohm resistors. Running from batteries, this means that (1) the brightness will vary significantly as the battery voltage drops, and (b) the current limiting will be ineffective.

    However limiting those other strings to 3 in series is going to increase your number of strings by 2, and that will increase your total current draw.

    I think you need a bigger battery (too).
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    As some crazy Scottish guy said, "Ye cannae change the laws o' physics, Cap'n".

    You can't increase the mAh capacity of a battery. Some battery types have higher capacity per unit volume, or per kg of weight. They are also more expensive.

    You might benefit slightly from using a step-up switching converter with current regulation, and putting ALL of the LEDs in series. Such a converter would typically have an efficiency of about 80~85% so you would lose 20% in the converter. You need to compare that with the percentage of power lost in your series resistors to see whether it's worthwhile going down that path.

    With a step-up switcher with current regulation, all your LEDs would run at the same current. You would get better regulation and better operation during the "death throes" of the battery as its voltage starts to taper off dramatically.

    Google LED boost converters or look at companies like Linear Technology and Maxim.

    This might improve things somewhat, but for a really significant increase in run time, you NEED a battery with a higher watt-hour capacity.
     
  5. Skyking ALexander

    Skyking ALexander

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    May 2, 2011

    Attached Files:

  6. BobK

    BobK

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    Sorry, but you are not going to get anywhere near 1 hour at 160ma. The battery is rated at 150maH a drain of 0.17ma! I doubt that it will produce 160ma at all.

    Bob
     
  7. Skyking ALexander

    Skyking ALexander

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    May 2, 2011
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Several things. That shows a final voltage of 0.9V, so if you're starting with about 6.2V (from 4 of them), you'd end with 3.6V. This change in voltage may not be sufficient to even illuminate your LEDs (scale for the number of batteries you're using).

    This means you may not be able to fully discharge the cells, so you can't use their entire capacity.

    But more importantly is the internal resistance. From the datasheet, I assume it is low compared to their recommended load (6.81k).

    As you increase the load (reduce its resistance) the effect of the internal resistance becomes greater. More and more power is dissipated in the internal resistance. This will lead to a lower output voltage at a given current, and also to an estimate of the short circuit current (i.e. when the voltage falls to zero).

    In any battery, increasing the current you draw will result in the effective mAh capacity reducing, but even more, it will reduce the amount of power available due to the reduced voltage from the battery.

    If you measure the short circuit current from one of these batteries, you can use this to calculate an approximate value of the internal resistance.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    And Skyking, "sudo make me a sandwich"
     
  10. Skyking ALexander

    Skyking ALexander

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    May 2, 2011
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    echo Thanks >> Skyking

    They would be far better.

    You need to take care in charging these though. Also in discharging them

    Also note that the voltage can vary between 3V and 4.2V.

    Never mix cells with a different state of charge together in a battery.
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    If you're working with a wide voltage range, you may find it worthwhile to use a current controlled driver for the LEDs rather than current limiting resistors.

    Calculate the LED current at minimum and maximum battery voltages using your chosen current limiting resistors. You'll find the difference is quite significant. If you want a reasonably fixed LED current, you can either drop a larger percentage of the battery voltage across the current limiting resistors, or use a current driver.

    A constant current source can be made pretty easily with a few components. Have a look at http://en.wikipedia.org/wiki/Current_source. About half way down there are several diagrams for simple current sources using one or two transistors. Check the power dissipation in the pass transistor at maximum battery voltage though.

    You will need a current source for each string of LEDs. You can use a single base reference voltage driving multiple pass transistors. That way, each string needs only two components (a resistor and a transistor), instead of just one current limiting resistor, so you wouldn't be adding many components.

    I think probably a better solution would be a boost switching supply with output current regulation driving all the LEDs in a single string, as I suggested in post #4 on this thread. This will give a fairly constant efficiency and constant LED current over a fairly wide range of input voltage. There are lots of examples; Google led boost converter current regulated.
     
  13. tedstruk

    tedstruk

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    Jan 7, 2012
    You want flashlight batteries to light lights. Even an LED needs some support.
    figure 1 amp per LED, and pot the volts so they will dim. meter it. Make sure the scope is within tolerance, then use a variable resistor to attain the meter.
     
  14. davenn

    davenn Moderator

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    and how do you figure that, when we know the LED's he's using are only ~ 20mA per LED ??

    Dimming of LED's has also be disciussed in various threads at length on this forum
    PWM is usually the best way

    Please dont post inaccurate info, it only confusess the person posting the questions

    Dave
     
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