6 Dec
2002
6 Dec
'02
2:53 p.m.
You could just put a silicon diode that is forward biased in series with the
6 volts. It will drop close to 0.7 volts. Make sure it has the current
handling capacity that you need. In other words, do not use a signal diode.
At 0.5 amps the diode would dissipate 0.35 watts. Regards - Mike
Mike B. Feher, N4FS
89 Arnold Blvd.
Howell NJ, 07731
(732) 901-9193
----- Original Message -----
From: "Ethan Dicks" <erd_6502(a)yahoo.com>
To: <cctalk(a)classiccmp.org>
Sent: Friday, December 06, 2002 3:44 PM
Subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good
find!!!)
[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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