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

From: Dwight K. Elvey <dwightk.elvey_at_amd.com>
Date: Fri Dec 6 19:29:00 2002

>From: ard_at_p850ug1.demon.co.uk
>
>> 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
>
>

Hi
 He didn't say what kind of batteries he was using
to provide 6V. Different batteries have different
discharge voltages. Even though a lead-acid battery
is fully discharged at 5V. Using it until 5.5V and
then recharging is a good idea for longer life.
 Even if the regulator drops below 5V on the output,
most circuits will work down to 4.8V someplace
or lower. A low dropout regulator, as you suggest,
might still be the best option.
Dwight
Received on Fri Dec 06 2002 - 19:29:00 GMT