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Overheating Regulator and micropuck. 3.7V and 27V Vin, help please

Discussion in 'General Electronics Discussion' started by JPU, Jul 22, 2013.

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  1. JPU


    May 19, 2012
    Hi All

    I have a simple circuit. An LM7815AC and a LM7805C Regulator. They are connected to a micro puck 2009 in a Buck/boost driver configuration as shown in the pic. I have also included links to the regulators and the Cree.

    Please can someone take a look at this for me as the LM7815AC gets very hot and when testing the micro puck burnt out and now fails. I have a bench supply and drove the voltage from 26V to 29V. The mAh shown on the supply remained <400mAh during this period.

    1) I know this is going straight over my head due to my lack of training in electronics but can someone explain what's going wrong please.

    2) The end result is that I want to be able to connect the headlight to a 27V battery. The headlight is normally powered by a 3.7V battery and the micro puck works well at this Volatage. I would like to create a small module so that if the battery goes flat I can then connect the headlight up to the larger battery as an emergency power source. The V in, will have to be dropped to <5V due to the micro puck and other electronics in the headlight (ie 08M2 Picaxe) so as to emulate the voltage from the 3.7V battery. Can you please suggest a way I can achieve this.

    Thanks for looking and I appreciate your help in advance.

    Attached Files:

    Last edited: Jul 22, 2013
  2. GonzoEngineer


    Dec 2, 2011
    The main problem is that linear regulators drop the voltage by dissipating the extra voltage as heat.

    Do some quick calculations, and you will see that at .4 Amps, the 7815 must dissipate 12V @ .4A. That is 5 Watts! The LM7805 loss is another 4 Watts!

    So, you have 9 Watts of lost power, all being turned into heat, essentially.

    You need a pretty big heatsink to dissipate that much heat!

    The smart way would be to use a Switching Power Supply, also known as a "DC to DC converter". They are much more efficient.

    You can check the Farnell website and probably find something cheap.
  3. JPU


    May 19, 2012
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010

    The max input voltage is lower than your input voltage. Also you require a number of other components (at a minimum an inductor, a diode, and a capacitor).

    Find a component (or module) that has a rated input voltage up to about 35V
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