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_at_p850ug1.demon.co.uk> wrote:
> > 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
>
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Received on Fri Dec 06 2002 - 21:51:00 GMT