# YOU THINK YOU MIGHT KNOW? HMPH, PROVE IT!

Discussion in 'Electronic Design' started by Dave, Sep 30, 2003.

1. ### DaveGuest

Ok...

Have a portable hard drive kit. It's power requirements are DC +5V &
+12V both at 1.5A

First, I don't understand the + for the voltage... As this is battery
powered, I was always under the impression that voltage flowed from
the negative side of batteries???

Need to make this kit more portable for certain reasons.

Therefore, I am in the design stages of a battery power supply.

Will use 8 C 1825mA rechargable batteries that will of course provide
12V.

For the 5V requirement, I will use 2 heatsinked 1A +5V voltage
regulator chips. The 5A regulator is too big for the box and circut
board I want to use.

12V requirement will go strait thru.

Now there are 3 pins on these regulators, the IN, GND, OUT obviously.

Since I do not understand the circut flow (again I understood it flows
from the negative side of consumer batteries, AA, C, etc.) I do not
know what exactly the flow is going to the IN side of the regulator.

Is it the negative from the battery pack? Positive? I know for
instance in a car battery, they flow originates from the Positive
terminal. But for little AA C, D's is this the case?

Are there any other components I should include in this setup?

Novice in need.

Thanks for any help,

-Dave

2. ### Jim ThompsonGuest

You must be a physicist ?

Engineers define the direction of current flow based on the movement
of a positive test charge.
Study the data sheets *carefully*!

...Jim Thompson

3. ### John FieldsGuest

---
There are a few you should get rid of: The 5V regulators and the heat
sinks.

assuming your drive will work on +4.5 instead of +5V, here's all you
need to do:

| | | | | | | |
+--||---||---||-+-||---||---||---||---||--->+12OUT
| | | | | | | | | |
| +-------------------------->+4.5OUT
|
+------------------------------------------>GND (0V)

If it won't, then do this:

| | | | | | | |
+--||---||---||---||-+-||---||---||---||--->+12OUT
| | | | | | | | | |
| |
| +----[1N5400]--------->+5.3OUT
|
+------------------------------------------>GND (0V)

The only caveat I can see doing it this way is that with both loads on
at the same time you're going to be taking 3A from the leftmost cells
and only(!) 1.5A from the rightmost cells, which will deplete the
leftmost cells more quickly than the others.

OTOH, if you do it using linear regulators the way you've described,
you'll be drawing 3A from the entire stack, so everything will be
depleted equally, but there'll be a huge wast of energy to get your 5V
that way.

Probably the best way to do it would be to run a 12V to 5V switcher to
get the 5V.

4. ### Byron A JeffGuest

-
-Have a portable hard drive kit. It's power requirements are DC +5V &
-+12V both at 1.5A
-
-First, I don't understand the + for the voltage... As this is battery
-powered, I was always under the impression that voltage flowed from
-the negative side of batteries???

Nomenclature. Nothing more. It's a waste of time to try to apply real physics
to the issue.

-Need to make this kit more portable for certain reasons.
-
-Therefore, I am in the design stages of a battery power supply.
-
-Will use 8 C 1825mA rechargable batteries that will of course provide
-12V.

Actually it'll provide 12V or less. One issue with most batteries is that
the voltage drops as they discharge.

Also what type of rechargable batteries? NiCad is usually about 1.3V per cell
so 8 will really only give about 10.5V

-
-For the 5V requirement, I will use 2 heatsinked 1A +5V voltage
-regulator chips. The 5A regulator is too big for the box and circut
-board I want to use.

Bad idea all the way around. Linear regulators work by turning power into
heat, therefore wasting it. It's like pouring 60 percent of your battery
charge down the drain.

You need a switching regulator which has efficiencies in the 85-90 percent
range as opposed to the 40 percent you're looking at with the linear regulator.

You'll probably end up needing a bigger box. But you'll be able to ditch the
heat sink most likely.

Something like a National Semi LM2576 simple switcher should do the trick.
\$6.15 USD at Digikey.com. Add one inductor, a fast recovery diode, and a
couple of caps and you're done.

-
-12V requirement will go strait thru.

Probably not a good idea because of the voltage drop off. If you had the
space it would almost be better to organize your cells to 9V using 12 cells
in 2 6 cell stacks that are in parallel, then boosting the 9V to 12V using a
LM2577, and bucking down the 9V cell stack to 5V using the LM2576. You'll get
more power and it'll last longer because the regulators will produce the
correct output voltages even as the voltage of the stack drops.

-
-Now there are 3 pins on these regulators, the IN, GND, OUT obviously.
-
-Since I do not understand the circut flow (again I understood it flows
-from the negative side of consumer batteries, AA, C, etc.) I do not
-know what exactly the flow is going to the IN side of the regulator.

The positive side of the battery stack. The side labelled '+' on the battery.
Don't worry about the physics. This is engineering.

-
-Is it the negative from the battery pack?

No.
Of course.
-instance in a car battery, they flow originates from the Positive
-terminal. But for little AA C, D's is this the case?

The positive of a car battery and one of a flashlight cell are the same for
this particular purpose. '+' goes to IN, '-' goes to GND.

-
-Are there any other components I should include in this setup?

Described above. You're not going to be happy with the performance of your
current setup when you see that the batteries discharge long before you expect
them to.

BAJ

5. ### Ben PopeGuest

Voltage doesn't flow... current does.
Well of course, since 1.2*8 = 12.
That'll probably get nice and toasty then.
Hopefully the equipment doesn't need a regulated supply then.
Think of it as voltage source, ground and regulated output then. So connect
"in" to the battery as that is the source or input to the reg, and the out
is the regulated supply to be connected ot the hard drive.
Right, bear with me... Conventional current flow is in the direction of
positive charge carriers. Unfortunately charge carriers (electrons) hold
negative charge. It's sometimes useful to think of the gaps between
electrons as holes, which are positive (as all things are relative) charge
carriers. Conventional current flow is from postive to negative, although
electrons actually go the other way.
An obvious problem that I see is that for a start 8 rechargeable cells will
unlikely produce 12V. 10 will, but the voltage drop after a time may affect
your setup. I think with NiMH batteries the voltage doesn't drop with
charge until the end... this will mean that it's difficult to spot when the
batteries are running low until already there. Fortunately you can top up
charge them with more success then NiCd.

Be sure to check the maximum electrical characteristics of the drive if you
plan to run unregulated... 10% is usually acceptable on these things but
always good to be sure.

Make a calc on how long you expect the device to last on the batteries you
use. And half that time for a more realistic figure.

Ben

Or a technician. Or have we engineers finally taught most
technicians the "correct" direction of current flow?
Hey, I like the humble feel of the subject line.

7. ### MichaelGuest

If you dont want to actually make anything, rip a power supply out of
an old PC, run it from a 12-240V inverter, and run the inverter of a
big gel cell or a motorbike battery.

Obviously not the optimum solutiion, but requires the least skills in
electronics.

8. ### Active8Guest

[snip]
no doubt. catchy. gets one's attention. once you start reading the post,
you forget the bait. must be a salesman.

mike

9. ### DaveGuest

Thanks for the help!

Based on your suggestions, I'm thinking I'll use 2 9.6v RC car packs @
1600maH

Do all think "Typical Application" diagrams are good to build from for
my app???
for the buck down LM2576 http://www.national.com/ds/LM/LM2576.pdf
and boost LM2577 http://www.national.com/ds/LM/LM1577.pdf ??????

The potential problem this chump novice sees in the LM2577 goes to
800ma and I need up to 1.5A (that what provided AC power supply is
capable of).
What are the options here?

Dave

-----
(Byron A Jeff) replyed in message

Bad idea all the way around. Linear regulators work by turning power
into
heat, therefore wasting it. It's like pouring 60 percent of your
battery
charge down the drain.

You need a switching regulator which has efficiencies in the 85-90
percent
range as opposed to the 40 percent you're looking at with the linear
regulator.

You'll probably end up needing a bigger box. But you'll be able to
ditch the
heat sink most likely.

Something like a National Semi LM2576 simple switcher should do the
trick.
\$6.15 USD at Digikey.com. Add one inductor, a fast recovery diode, and
a
couple of caps and you're done.

-12V requirement will go strait thru.

Probably not a good idea because of the voltage drop off. If you had
the
space it would almost be better to organize your cells to 9V using 12
cells
in 2 6 cell stacks that are in parallel, then boosting the 9V to 12V
using a
LM2577, and bucking down the 9V cell stack to 5V using the LM2576.
You'll get
more power and it'll last longer because the regulators will produce
the
correct output voltages even as the voltage of the stack drops.

10. ### Paul BGuest

Electron current flow is from negative to positive of your battery.
That's what happens in 'real life.' There's an old convention,
however, that current flowed the other way around. Nowadays we call
that - appropriately enough - "conventional current flow" and it's an
important distinction to keep in mind. Voltage, OTOH, doesn't 'flow'
at all as such.

11. ### Byron A JeffGuest

-Thanks for the help!
-
-Based on your suggestions, I'm thinking I'll use 2 9.6v RC car packs @
-1600maH

-
-Do all think "Typical Application" diagrams are good to build from for
-my app???

Yes. Also note that the intro web page for both parts has a calculator that

-for the buck down LM2576 http://www.national.com/ds/LM/LM2576.pdf
-and boost LM2577 http://www.national.com/ds/LM/LM1577.pdf ??????
-
-The potential problem this chump novice sees in the LM2577 goes to
-800ma and I need up to 1.5A (that what provided AC power supply is
-capable of).
-What are the options here?

Where do you see that? The NPN switch current limit on page 3 of the datasheet
indicates a minimum of 3A. So where did the 800ma come from?

BAJ

13. ### DanGuest

I've done this before, trouble is the drive I used cut out at 11.3V if I
remember correctly. I was planning a DC-DC converter but then got hold of a
5V laptop drive that was good down to 4.2V. It seems to be fine at 6V so
regulation was never really needed with 5 * 1.2 2000mAh NiMh AA cells.

Dan.

Winston Churchill