From: Kevin McMurtrie
Subject: Re: 10A Power supply
User-Agent: MT-NewsWatcher/3.2 (PPC Mac OS X)
Date: Sat, 07 Sep 2002 06:04:20 GMT
NNTP-Posting-Date: Fri, 06 Sep 2002 23:04:20 PDT
In article <email@example.com>,
"Josh Hillman" wrote:
>*** post for FREE via your newsreader at post.newsfeed.com ***
>I'm looking for a power supply design around LM317 and 2N3055. I found
>something on the web that says:
>"I don't know what circuit you are working from, but a crude high-power
>linear regulator can be made by taking the output from the LM317 and
>connecting it to the base of a high power NPN (like a 2n5686 or 2n3055),
>with the collector connected to the input of the LM317, and the emitter used
>as the output. The regulation is not great, as the output will drop by the
>base-emitter voltage of the transistor (0.7 v typically) under load, and the
>power dissipated by the transistor and the LM317 can be large, depending on
>your input voltage, i.e. watts dissipated = ((Vin - Vout) * Current). You
>may need a large heatsink and a fan. The DC current gain of the transistor
>times the LM317 output current will determine the output current capability.
>Is that a workable solution? Has someone seen a similar better design?
Put a PNP transistor on the _input_ to the regulator and bypass the
transistor E-B with a 1 Ohm resistor. Now when the regulator draws more
than 600mA, the transistor will turn on to pass some current around it.
Perfect regulation will be maintained. This trick is in the application
notes of most of these regulators.
PNP E: V+, 1 Ohm pin 1
PNP B: 1 Ohm pin 2, regulator input
PNP C: regulator output
10 Amps is a lot for this type of regulator. It regulates by wasting
excess voltage as heat. See the formula you quoted. It will be easier
to build (or buy) a switching regulator if you have a voltage drop of
more than a few volts.