6 Dec
2002
6 Dec
'02
9:51 p.m.
I'll bet it used a NiCd, which gives ~4.8 volts for
the vast percentage of a charge - which is "kinda" in
tolerance of most +5 logic. Those things were so
cheap, do you think they'd use a step-up DC-DC
converter? Nah.
--- Tony Duell <ard(a)p850ug1.demon.co.uk> wrote:
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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.
Take a look at 'low dropout regulators'. National
Semiconductor make
(made?) them -- LM2940 series I think. They will
work down to about 0.6V
difference between input and output (so for a 5V
regulator, you need at
least 5.6V in). These are similar to the 7805 -- 3
terminals, and you
need to put a couple of decoupling caps near the
chip.
That probably won't let you use all the capacity of
your '6V' battery,
but it should let you use some of it.
Incidentally, I assuem the Zip drive produced 5V
internally from this
battery pack. Any ideas what it used?
-tony