Switching transistor question
- From: default <default@xxxxxxxxxxxxx>
- Date: Sat, 30 Jun 2007 11:51:26 -0400
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?
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