I want to use my thinkpad battery(11.1v) hooked up to a LM7805 to charge a USB Device.

What are the downsides to such a setup?

  • 1
    \$\begingroup\$ Other than pissing away half the power in the battery? \$\endgroup\$ Apr 8, 2014 at 11:08
  • \$\begingroup\$ What would be a more efficient way of stepping down the voltage? \$\endgroup\$
    – Tejas Kale
    Apr 8, 2014 at 11:11
  • 4
    \$\begingroup\$ Using a switching regulator. \$\endgroup\$ Apr 8, 2014 at 11:11
  • \$\begingroup\$ Or a DC-DC converter. See Digikey/Mouser/Jameco \$\endgroup\$
    – gwideman
    Apr 8, 2014 at 11:12
  • \$\begingroup\$ What charging current to the USB device? \$\endgroup\$
    – Andy aka
    Apr 8, 2014 at 12:42

2 Answers 2


The LM7805 is a linear dropout regulator. These are classic, robust devices that do what they do pretty well. But their intended function is to modify small voltages a little bit to maintain relatively constant supplies for things that do not use much power, like a lot of modern low voltage logic and regulation ICs. Even the supplies that do use these regulators have things in place to kind of ball-park the voltage before the 7805 rounds off the top.

All voltage dropped from one end of a 7805 to the other is blown away as waste heat across some resistance. Its not a big deal for a small voltage drop for a bias supply, but you would literally be venting about half of your input power for no reason. A fairly common chopper buck circuit or half bridge switching supply could do this with efficiencies in excess of 90%. You could even make a reasonable one out of a 555 timer, a transistor, some resistors, three capacitors and a zener diode, if you just feel like you have to play with something.

  • \$\begingroup\$ The 7805 is a linear regulator, but it is defintiely NOT a low-dropout regulator. The 78xx series needs a minimum of about 3V of overhead (\$V_{in} - V_{out}\$). \$\endgroup\$
    – Dave Tweed
    Apr 8, 2014 at 15:22
  • \$\begingroup\$ Of course you are very correct. Hadn't slept in a while. Just what I was thinking, I'm not sure. Mixing up my components. \$\endgroup\$
    – user39962
    Apr 8, 2014 at 16:38

Disadvantages of 7805 in this application as already noted by others: Wasting a lot of heat, draining battery twice as fast as you might otherwise.

DC-DC converter might be a good off-the-shelf solution for you:

500mA $2.84 http://www.digikey.com/product-detail/en/R-78E5.0-0.5/945-1648-5-ND/2834904

1 Amp $7.49 http://www.digikey.com/product-detail/en/R-785.0-1.0/945-1038-ND/2256218

Hope that helps.


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