Path: utzoo!attcan!uunet!husc6!mailrus!tut.cis.ohio-state.edu!bloom-beacon!mgm.mit.edu!wolfgang
From: wolfgang@mgm.mit.edu (Wolfgang Rupprecht)
Newsgroups: sci.electronics
Subject: Re: 7805 regulator
Message-ID: <5820@bloom-beacon.MIT.EDU>
Date: 18 Jun 88 05:50:17 GMT
References: <592@picuxa.UUCP> <5770@bloom-beacon.MIT.EDU> <1138@stracs.cs.strath.ac.uk>
Sender: daemon@bloom-beacon.MIT.EDU
Reply-To: wolfgang@mgm.mit.edu.UUCP (Wolfgang Rupprecht)
Organization: Freelance Software Consultant, Boston, Ma.
Lines: 46

Me:
>> ~+8v in				    +5 out
>>   o---+----\	    /-------------------+----o
>>       |     \	   /			|
>>       |      v  /			|
>>       |    ---------	  +5v reg.      |
>>       |        |	 +--------+    	|
>>       | 2.7 ohm|	 |        |     |
>>       +--\/\/\-+-------+ in  out+-----+
>>       |        	 |  gnd   |	|
>>       |         	 +---+----+	|
>>      --- 0.1 uf //         |         --- 0.1 uf // 10uf tantalum
>>      --- 10 uf tant        V         ---
>>       | 		    	       	|
>>       | 				|
>>       v 				v

>al@cs.strath.ac.uk (Alan Lorimer) writes:
>Seems a bit involved - I had to think for a while to figure out how this
>was meant to work, I'm still not sure that is correctly solves the load
>sharing problem though, is the load taken by the PNP transistor related
>to the HFE of that transistor, and the 2.7 ohm resistor? Perhaps you
>could explain the reasoning?

The transistor is the OUTPUT buffer.  It is meant to carry the bulk of
the current for the high-current case.  

For low currents the PNP is turned off (by the 2.7 ohm resistor) and
the total output current flows through the regulator.  The PNP will
start conducting as soon as the voltage drop reaches ~0.6v across the
2.7 ohm resistor.  This happens when the output current is
(~0.6v/2.7ohm) or about 220 ma.  (The resistor will be dissipating a
constant 0.13w from this point onward.)

The regulator only supplies the first 220 ma. plus whatever base
current the transistor needs.  Even at a conservative HFE of only 20,
that's only an additional 240ma of base current for 5 Amps. output
current.  This means that at this 5A. point the regulator will only be
carrying roughly 10% of the output current.

If you can keep the input voltage fairly low, let's say 8-9 volts, you
will get a power loss of 3-4 Watts/Amp output current.  For 5A thats
15-20 Watts.  Of this power, less than 1.3-1.8 Watts will be
dissipated by the regulator itself.
Wolfgang Rupprecht	ARPA:  wolfgang@mgm.mit.edu (IP 18.82.0.114)
TEL: (617) 267-4365	UUCP:  mit-eddie!mgm.mit.edu!wolfgang