6 Dec
2002
6 Dec
'02
4:31 p.m.
I think there are some forms of the 7805 that only need .6V (one diode drop) of
overhead so you should be able to run them on as little as 5.6V and still get 5V output.
Also you may look inside your battery pack and find that it uses a higher voltage
internally and then regulates it down to 6VDC. If so you can take power off ahead of the
intenal regulator.
Joe
At 12:44 PM 12/6/02 -0800, you wrote:
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
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