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voltage regulator HELP!!

Discussion in 'General Electronics Discussion' started by drcarnine, Mar 11, 2012.

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  1. drcarnine

    drcarnine

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    Mar 11, 2012
    I am needing to put an external battery pack on a security light. I want to use a 4.5 AH 6 volt battery and connect it to the 3.6 volt internal battery pack. I am trying to use an
    LM317T . By putting a 3.6 ohm resistor as R1 and an 18.6 as R2 I am getting 3.8 volts out. I need to get about 500 milliamps when the LED light comes on.This setup gets really hot. I am using 1/4 watt resistors. If I increased the resistors to 1/2 watt, would that make it run cooler?
    Is there a better small inexpensive circuit that I could get the 500 MA with this 6 volt battery connected to the 3.6 volt battery?
    Thank you in advance. I am running blind here.
    Dennis
     
  2. davenn

    davenn Moderator

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    Sep 5, 2009
    hi Dennis
    welcome to the forums :)

    What is getting hot ? the resistors or the voltage regulator or all 3 ?

    you could try a DC-DC Buck converter instead, they are generally more efficient and therefore wont waste so much power as heat

    Dave
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    It's going to be remarkably inefficient to do it that way. In addition, you need to prevent current from flowing from one battery to the other.

    I would tend to recommend that you replace the internal battery with a larger battery rather than try to use two battery packs simultaneously.

    Does the security light have some charging mechanism (e.g. solar) or are the batteries charged externally? In either case, you would need a new charger to charge the new battery pack due to its higher voltage.

    I would recommend that you get a switchmode LED driver and use that rather than a linear regulator. But you would need to (presumably) power other parts of the internal circuitry and for that you may need a constant voltage that a LED driver would not provide. The device itself may already have some form of switchmode LED driver and mat be tolerant of a higher input voltage, but there is no guarantee that it would be tolerant of a 6V battery.

    What you're trying to do can be done in a simple manner, but the performance is likely to be very poor. To do it properly would require a lot more information about the circuit involved.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Oh, the main reason your resistors are getting too hot are that they may be in the right ratio (which I also doubt), but their values are way too low.

    The resistor connected between the ADJ pin and ground should normally be around 120 ohms and the resistor connected between the ADJ pin and the output as then adjusted to set the voltage. In your case it should be about 220 ohms. This will give an output voltage of about 3.5 volts.
     
  5. drcarnine

    drcarnine

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    Mar 11, 2012
    voltage regulator

    Thank you for answering my request. I have been selling these lights for a few months now. They are used for controlling wild hogs in Texas. They have a small 3-AAA internal battery pack. There is no room for any other kind of battery. The light only cycles about 30 times on the original batteries. My customers need more cycles so I am trying to get the 6 volt external battery pack to work. I have the battery box and 6 volt solar panels. The reason I want to use both batteries is that a portable light can be setup using the 3.6 volt solar panel or the light can be used semi-permanent by using the 6 volt battery and its solar panel. I had a 220 R1 and a 1.3 K for R2. output was about 4 volts but only allowed 2 ma of current to pass. That is why I went to the smaller resistors. With the setup I have now that is getting hot. when the light comes on the draw is about 200 ma--less if the3.6 volt batteries are fully charged. I see one of the persons answering me talked about a DC-DC Buck converter but I do not know what that is or how expensive they are. I am trying to keep the cost down as much as possible. With the resistors mentioned would the current flow be enough or would it restricted as was my original trial?
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    If your regulator won't allow sufficient current then you've wired it up wrong (most likely)

    Show us a circuit diagram of how you wired up the regulator.

    Are you saying you have an external battery and solar panel and you want to wire that in place of the internal AAA batteries?
     
  7. drcarnine

    drcarnine

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    Mar 11, 2012
    Thank you for the help, I found the problem in my interperation of the schematic as you said
     
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