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7354Re: NSLU2 using 7805s and a 12V battery

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  • efilvasers
    May 5, 2008
      First off, I'd like to thank you all again for your help.
      I should have done a bit more research before jumping head first into
      this project.
      Next time around, I'm definitely going with a switching regulator.
      My quick solution is this:
      A 2N3055, bolted to an old CPU heatsink (which still had some thermal
      compound on it), with the base connected to a 7805 (with a small heat
      sink), collector to the battery, and emitter to the NSLU2 (and other
      Also, the ground pin of the 7805 is connected to the real ground
      through a 1A diode; when I measured the emitter voltage without this,
      it read about 4.4 volts, went up to about 5.2 after adding it (runs
      about 4.9 with the NSLU2 on).
      The 7805 heatsink doesn't even warm up, while the 2N3055 heatsink
      warms up after a couple of minutes, but went back down to near room
      temperature when I placed an extra 80mm fan next to it.
      I'm using the same solution for the 9 volt regulator (both sharing the
      same fan).

      --- In nslu2-general@yahoogroups.com, Philip Pemberton <ygroups@...>
      > efil vasers wrote:
      > > So far, I've tried a few different regulators (rated 1 or 1.5
      amps), and
      > > when I turn the NSLU2 on, after about 30 seconds, the regulators
      are too hot
      > > to touch and soon shut themselves down.
      > OK, this is going to be a long post, mainly because I don't like
      saying that
      > something is going wrong, I also like saying exactly *why* it's
      going wrong
      > and how I came to the conclusion. If you've seen Big Bang Theory, think
      > 'Sheldon with slightly better social skills' :)
      > Two things that could cause this:
      > You're trying to drop too much power over the 7805s.
      > ----------------------------------------------------
      > 7805s are linear regulators (and horribly inefficient ones at that,
      > they're cheap and pretty much the 'jellybean standard' so everyone
      uses them).
      > That means that any voltage drop has to be dissipated in the form of
      > So let's do the math:
      > Your input voltage is 13.8V (assuming a fully charged SLA battery).
      > using 7805s, so your output is 5V. That means a voltage drop of 8.8V
      by virtue of:
      > 13.8V - 5.0V = 8.8V
      > You're also dumping a peak of 2A over the the regulators (though ~1A
      > the USB ports are rated for 500mA each for a total of 1A). That's
      17.6 Watts:
      > 8.8V * 2A = 17.6W
      > Or with the USBs 'open' (nothing connected), we assume 1A because:
      > 2A - (500mA * 2) = 1A
      > Thus:
      > 8.8V * 1A = 8.8W
      > So your regulators have to dissipate between 8.8 and 17.6 Watts of
      energy as
      > heat. And probably more than that because of internal
      inefficiencies. That's a
      > lot of heat!
      > Your heatsink isn't big enough
      > ------------------------------
      > You need to get rid of 17.6W of heat, and keep the temperature of
      the chip die
      > itself below the thermal cutoff point (70 Celsius if I remember
      correctly). I
      > honestly can't remember the calculations for heatsinks off the top
      of my head,
      > but you can probably find them in 'The Art of Electronics' by
      Horowitz and
      > Hill. Or on the internet.
      > OK, so your possible solutions would be:
      > - A switching regulator. These are pretty much turn-key these
      days, but you
      > have to be careful with them. Good PCB layout is essential,
      heatsinking less
      > so. Also, because you're not wasting as much energy as heat, your
      battery will
      > last longer.
      > - A bigger heatsink. Probably something about the size of a PC
      > heatsink, with a fan, and drilled to take your regulator. You'll
      need some
      > thermal grease and an insulator pad too.
      > - A pre-regulator. Add another regulator in front of the 7805 to
      > pre-regulate the 12V down to 8V, then feed the 8V to the 7805. Still
      > inefficient, but you're distributing the load over two separate
      chips, which
      > reduces the heat output per chip.
      > > I connected the regulators in parallel with each other to provide
      up to 3
      > > amps (more than the NSLU2 power supply), so I figured that I
      wouldn't need
      > > heatsinks for them. Am I wrong, or is my NSLU2 just behaving
      > > How many people out there have tried this?
      > You shouldn't parallel 7805s. One chip will end up with a lower
      voltage than
      > the others (due to internal tolerances), and that'll take most of
      the load.
      > Result being that it ends up getting roasted, then when it
      eventually blows
      > you get a cascading failure of all the other 7805s. Lots of fun,
      > burned plastic, melted silicon and smoke that really isn't good for
      > health or the chips'.
      > You can add an external pass transistor though -- though it's been
      ages since
      > I've done it. I do recall it involves a 10-ohm resistor and a beefy PNP
      > transistor though. Cost is likely to be more than a switcher, though.
      > Thanks,
      > --
      > Phil. | (\_/) This is Bunny. Copy and
      paste Bunny
      > ygroups@... | (='.'=) into your signature to help him gain
      > http://www.philpem.me.uk/ | (")_(") world domination.
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