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Will this high-current regulator work?


Wubs
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Hi, everyone!

I'm a novice hobbyist, not a student, so please don't laugh too loudly when reading this and please be kind and don't use big words in your replies!  ;)

I want to use my two old 14.4VDC @ 4Amp batteries from two old drills as a supplemental power source for my laptop when I go camping. I have two battery packs and two charges, so this is convenient and helps keep things out of the landfill.

After searching Google for project ideas, I decided to use an LM317 because I have them and I am on a tight budget.

Laboring through various manufacturers datasheets on the LM317 to boost up current output, I have come up with this design. I'm sure I've messed up the calculations somewhere.

About the circuit:

  • Transistor Q1 and resistor R1 form the short-circuit protection portion of the schematic.
  • I don't know what the value of R1 should be, since the diagram I pulled it from only listed the value of Q1 and said the resistor causes it to conduct when the load drops (of course, I can't find the website where I pulled this from, so I can't give the exact explanation).
  • I'm leaving the circuit as adjustable, so that with the use of my DMM, I can adjust R4 so that I get exactly 19vdc.
  • The output amperage is critical, my laptop's PSU calls for 19vdc @ 4.74amps.


My questions/problems:

I'd like to keep Q1 & R1 in the circuit, but if no one can help me determine the value of R1, then I can live without the added protection of Q1 & R1.

Are my other values correct?

(I realize that an LM338 will give me 5A out, but my local supplier doesn't stock them and I do have all the above components in my parts collection).

Thanks in advance for your assistance!

post-18151-14279144099971_thumb.png

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Certainly it doesn't work because the 2N3055 is not a PNP transistor.


Don't I feel foolish! I got the specs for the transistor from the Motorola datasheet for the MJ2955 which also contains the 2N3055 NPN compliment, but otherwise the specs are the same.

Thanks for pointing out this typo. I wonder what would have happened if I built it and powered it up using the NPN transistor?  Probably my neighbors would wonder why the fire engines were on their way to my house!  ;D

Thanks for your input!
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It's not possible to have a linear regulator with an input current of 4A and the output current is 4.75A. With a linear regulator, the output current is always equal or lower than the input current.

You need to use a switched mode power supply if you want to reduce the voltage but boost the current. Obviously it's not a perpetual motion machine and the power out is still less than the power in. Fortunately, in this case converting 28.8V to 19V will draw less than 4A.

Note that the batteries are probably rated in Ampere hours, not amps - they can supply 4A for one hour, 2A for 2 hours or 8A for half an hour. This  is a a bit of a simplification because the capacity of most batteries decreases with current draw is increased; you only get 4Ah at very low currents.

I would still recommend using a switch mode power supply anyway because you need the batteries to last as long as possible.

With 28.8V and [email protected] out the input current will be 3.45A, assuming 90% efficiency.  Obviously the laptop won't continuously draw 4.75A and the battery capacity will be <4Ah so it's difficult to estimate how long the batteries will last between charges; it will depend on how you use your laptop.

The laptop probably w

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