Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)

From: J.C.Wren <jcwren_at_jcwren.com>
Date: Fri Dec 6 16:03:00 2002

        You can use a switcher chip, but that's little bit of a pain to build by
hand, since the majority are surface mount, and you need a good ground plane
(not really hard, just not as easy as a 7805 and a bridge rectifier).
Basically, the problem is that a diode drop is .6 volts, so you need a
minimum of 5.6 volts of input, and that's with a low drop-out regulator.
That doesn't leave much room for your battery to die. A normal 7805 really
wants 7V minimum to regulate reliably.

        And depending on the current load, 15V on a 7805 is a Bad Idea, because
it's got to dump that 10V drop as heat. That's 10 watts at 1 amp, which is
a damn lotta heat in a TO-220 case. Ideally, a 7805 should be run at 7V or
8V to keep heat to a minimum.

        If you do want to build a switcher, just about everybody has a switcher
chip that will work about .2V to .4V above the regulated output voltage.
Maxim is the first place I'd look. Get some double sided copper clad board
for a ground place, use an Xacto or a Dremel tool to route traces. Some of
the companies also offer free or very low cost eval boards. For instance,
National Semi has a switcher eval that takes 4V to 30V and produces 3.3V out
at 1 amp. They're $8. It has two caps on it that cost me $2 each, plus
having boards turned, the switcher itself, and a few other incidental parts.
It's cheaper for me to buy them than it is produce my own (I use them a lot
in prototyping projects).

        When trying to get 5V from 6V, especially on batteries, linear regulators
aren't really the way to go. You'll dump a lot of the battery's energy as
heat, which is wasteful. Same for a Zener type regulator. You could use a
a couple diodes in series, but your output voltage will sink as your battery
voltage drops, since the drop across the diode is the same.


-----Original Message-----
From: cctalk-admin_at_classiccmp.org [mailto:cctalk-admin_at_classiccmp.org]On
Behalf Of Ethan Dicks
Sent: Friday, December 06, 2002 15:45
To: cctalk_at_classiccmp.org
Subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first
good find!!!)

> ...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.


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 - 16:03:00 GMT

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