>   For 1A load, and a darlington ( let say h21e = 100 ) will be 10mA
> on command. It's not very huge...

First, that's only the on current into the base of the darlington.  The
resistor from the +V supply to this base needs to provide this 10mA with
as little voltage drop as possible, because any voltage there appears
directly accross the pass transistor.  That will therefore cause a much
larger current when the base is pulled to near 0V to shut off darlington.

Second, the darlington will have significant voltage drop even if the base
were tied directly to the supply.  1.5V drop x 1A = 1.5 watts.  That's
enough to require a heat sink with most packages.

Let's say you spend 500mV accross the base resistor to get the 10mA base
current.  500mV / 10mA = 50 ohms.  15V / 50 ohms = 300mA to pull the base
to 0V to turn off the supply.  15V * 300mA = 4.5 watts just to keep the
supply off!  At the same time the drop accross the darlington when on is
2V (1.5 for the two B-E junctions plus 500mV for the base resistor).  It
will dissipate 2W when on and drop the 15V supply to a 13V supply.

I like Roman's single PNP for the quick and dirty method.  Use a small NPN
to drive the PNP base for the next level, or a P channel FET if voltage
drop and efficiency are very important.


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