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_at_yahoo.com>
To: <cctalk_at_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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Received on Fri Dec 06 2002 - 14:53:00 GMT