The short idea:
What I want to do is switch ~1/2 amp with no more than .2 volts drop (would be happier with .15) . . . all this on a 3 volt supply, burning no more than about 150 milliwatts, and with no moving parts.
Any other ways to skin this animal?
The application: I'm building a time-lapse camera using an el-cheapo 1.3 megapixel electronic camera.
The camera is happy with about 2.2 volts at ~500 ma when fully turned on - providing the source impedance is low and is designed for a pair of AAA cells.
Plan A was to power the whole thing with a pair of AA batteries. To that end, I got it working beautifully, but had to use a relay to switch the camera on. The relay only sucks down 40 milliamps and is only on for a short period so it is a viable way to do it.
Ideally I would want a semiconductor switching the camera on and off if that's possible, but it isn't looking so easy with only 3 volts or less to work with.
The sequence is to apply power to the camera. Wait 5 seconds or so for it to initialize its processor, toggle the shutter low for a brief period then leave the camera on another 9 seconds so it has time to store the picture in its flash memory, then the camera has power removed and the processor goes to sleep for ten minutes and the cycle repeats.
So far the only semiconductor that will switch the ~ 500 ma the camera needs when on has been a Darlington pair - but the CE voltage is too high for the camera to power up reliably with only a 3 volt supply.
To bias a small NPN transistor on and into saturation with 500 ma in the collector would be great but I haven't found a transistor with enough gain at that current.
Plan B is to use a supply of 4.5 volts - but that is essentially wasting one whole cell just to satisfy the voltage drop across the Darlington.
Is there some way of doing this without the penalty of going to an extra battery cell or driving it with more than 40 milliamps?