YOU THINK YOU KNOW, BUT WE'LL SEE

[Dave]

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

 

[Byron A Jeff]

In article <319c4f99.0309300855.83c9896@posting.google.com>,
Dave <dmcomm_ads@yahoo.com> wrote:

Ok...

No actually not OK. This post has two problems:

1) You have posted individual messages to multiple newsgroups. You need to
crosspost such messages so that readers of multiple newsgroups will see it
once.

2) I let it go the first time in sci.electronics.design, but your title
absolutely sucks. It gives no substative information, issues a challenge, and
screams at the reader all at the same time, definitely a strikeout.

One always attracts more responses with honey than vinegar. A short to the
point subject in a single newsgroup will do much more than this multiple
copy of a post here.

BAJ

 

[Fred Abse]

On Tue, 30 Sep 2003 17:55:50 +0100, Dave wrote:

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???

Voltage doesn't flow, it pushes.

_Current_ flows, conventionally out of the positive terminal of the
battery, and eventially back into the negative terminal.

Aside from the heated debates there have been on this NG regarding the
pros and cons of conventional current, that's the way it was, is now and
ever shall be.

If it says +5v, you connect that terminal to the positive of whatever 5V
supply you use. Likewise +12V.


--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[Peter Bennett]

On 30 Sep 2003 09:55:50 -0700, dmcomm_ads@yahoo.com (Dave) wrote:

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???

"Conventional" current (which most people talk about) is a flow of
positive charge from the positive terminal of the power source,
through the load, and back to the negative terminal of the power
source, regardless of the type of battery or power supply.

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

NiCad and NiMH are only about 1.2 volts for most of their useful life,
so you should probably use 10 cells.

If you are drawing 3 amps (1.5 amps each for +12 and +5), this battery
pack will only last for a half hour or so. When drawing high currents
from a battery, you won't get the full advertised amp-hours from it.

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.

If you are reducing the 12 volts to 5, you'll be wasting over 10 watts
as heat in the regulator. It would be better to use a DC to DC
converter (available from http://www.digikey.com and elsewhere), as
these are much more efficient.

In any case, you shouldn't use linear regulators in parallel, as they
won't share current equally. The National LM323 is a 3 amp 5 volt
regulator, and would be a better choice if you want to stay with a
linear regulator.

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.

The IN terminal connects to the positive supply, GND (ground) to
negative, and OUT is the +5 volt output.

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

You need some bypass and filter capacitors on the regulator. Look at
National Semiconductor's website for information on their 78xx series
of voltage regulators.

Novice in need.

Thanks for any help,

-Dave

--
Peter Bennett VE7CEI
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Steve]

dmcomm_ads@yahoo.com (Dave) wrote in message news:<319c4f99.0309300855.83c9896@posting.google.com>...

I was always under the impression that voltage flowed from
the negative side of batteries???

Is this a troll?! ahh, I'm bored and don't want to do real work, I'll
take the bait!

forget "voltage flow", volts don't flow, they "push"

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

no they won't. You'll get 9.6V for a very short period of time, then
it will start to droop. Read the print on the side of your batteries.

For the 5V requirement, I will use 2 heatsinked 1A +5V voltage
regulator chips.

You're planning on drawing close to 1 amp @ 5V from 1825mAh
batteries?! Hmmm...

Since I do not understand the circut flow

BINGO! You said it! Use google to find articles on "conventional
current flow" and "electron flow".

Basically, what ACTUALLY happens, and what people THINK happens is
totally opposite. For a real quick and greasy, go here;
http://www.mi.mun.ca/~cchaulk/eltk1100/ivse/ivse.htm

Not sure what device you are using as a regulator, but here's a data
sheet with a typical application circuit diagram, see figure 13;
http://www.farnell.com/datasheets/3251.pdf

 

[John Popelish]

Dave wrote:

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???

The positive sign has nothing in particular to do with which way
charge flows to make up a current. Though electrons flow from
negative to positive, holes, and positive ions (including naked
protons) go the other way. Now, forget all about which way what is
going.

The positive sign indicates that the negative side of the 5 and 12
volt supplies are tied together and that potential is referred to as
common or ground (even if it has no connection to Earth). This common
node is the potential that all signals and supplies are measured with
respect to, so this node is defined as the zero volt potential for all
other measurements in the circuit. So if the negative side of the 5
volt supply is connected to the node that is defined as zero volts,
the positive side must be at a potential of +5 volts with respect ot
the zero volt reference node.

Same for the 12 volt supply. See? Nothing to do with what moves in
what direction.

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.

That rating is probably 1825 ma hours, so if you draw 3000 ma (both
1.5 volt supplies) from it, it will last about
1825 ma hours/3000 ma = .6 hours. Fortunately for you, those supply
ratings are peak values, and the average drain may be quite a bit
less.

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.

Getting those two regulators to equally share current is an added
complication. Better to go with a single 3 amp regulator. Better yet
would be to get a prepackaged 2 amp switching regulator that will
convert 12 volts to 5, instead of waste 7 out of 12 to make 5. It
will cost more, but will make the batteries last a fair amount longer.

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.

'In' means the connection to the supply the 5 volts will be reduced
from. This is the +12 volt node. The +5 comes out the 'out' pin, and
is regulated to be 5 volts more positive than the 'gnd' pin, so you
want ot connect that one to the negative side of the 12 volt supply,
since that is the node defined as zero volts in this system.

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?

Regulators need a small capacitor from in to gnd and out to gnd, to
make sure they don't oscillate. A pair 0.1 uf 50 volt mylar should
work. they should be connected very close to the regulator.

Novice in need.

Thanks for any help,

-Dave

--
John Popelish

 

[R. Steve Walz]

Dave wrote:

Ok...

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

Okay.

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???
-----------------------

Then you were mis-taught, or taught by the military in some antique
war. They simplified it for rural rubes and screwed it up for a couple
generations of people who had to re-learn it the other way.

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.
-----------------------

You cannot use two of those in parallel, it won't work!

12V requirement will go strait thru.
----------------------------

You'll find you need that regulated as well.

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.
----------------

That's totally wrong, it flows from the positive around through the
circuit back to the negative. The Input of a regulator "wants" the
positive. The ground is the negative.

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?
---------------------

Yes, you probably will not be able to make this work without upgrading
your electronics skills, because you don't understand what's happening.

Novice in need.

Thanks for any help,

-Dave
--------------------

You're going to need a much larger portable supply than C cells.

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

 

[Terry]

Dave wrote:

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

If this isn't a troll!!!! There is some confusion in the original
posting IMHO.
Voltage doesn't 'flow'. Volts are the units used to measure the
pressure or potential difference across a circuit or a device.
The devices (circuits) in my car are designed for 12 volts.
The devicses in my house are designed for 120 volts. (And some
for 240 volts, but that's another story).
OK so far?
Current flows are measured in amps (or subdivisions or multiples
of amps such as milliamps/kiloamps).
It is/was a convention that current flows from positive to
negative. Despite the fact that negatively charged electrons may
move through a device by being attracted to the positive side or
pole of the device.
That doesn't matter; the 'convention' is that Current flows from
positive to negative.
Another comment. Storage cells of different types have different
voltages and different capacities to provide current. Six cells
of a car battery at 2.0 volts each is a 12 volt battery.
NiCad/NiMh cells in a battery of eight will give you about 8 x
1.2 = 9.6 volts.
Then there is Ampere/hour capacity.
If you discharge cells at a heavy rate you will get less capacity
from them than if you take a very small current for many hours. A
heavy drain you may get only a third or less of the ampere hours
stated on the label; the specification of capacity will include a
discharge rate; for example, "Capacity, 'Y' amps for 'X' hours."
Capacity x times y, but only at THAT rate.
Thus you may only be able to start a car (100 amps say) five or
six times from a fully charged car battery; whereas my friend
employed a used car battery to run a low drain radio (a few
milliamps) at a remote wilderness cabin all summer without ever
recharging it.
Have fun, do some more calculations and be clear about some of
the basics involved. Terry.

 

[Kevin Aylward]

Terry wrote:

Dave wrote:


Thanks for any help,

-Dave

If this isn't a troll!!!! There is some confusion in the original
posting IMHO.
Voltage doesn't 'flow'. Volts are the units used to measure the
pressure or potential difference across a circuit or a device.

Consider a pulse on a transmission line. One can certainly argue that
there is a potential difference that is moving along the line, hence a
"flow of voltage"

However, I do agree, that in the context of the original post, that the
statement made no sense.

The devices (circuits) in my car are designed for 12 volts.
The devicses in my house are designed for 120 volts. (And some
for 240 volts, but that's another story).
OK so far?
Current flows are measured in amps (or subdivisions or multiples
of amps such as milliamps/kiloamps).

Current is a flow of charge, that is, current already includes the
concept of movement. A "flow of charge flow", don't make for good
English, mate. Your phrase would be better stated as:

"Charge flow is measured in amps."

or

"Current is measured in amps."


Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Terry]

Kevin Aylward wrote:

However, I do agree, that in the context of the original post, that the
statement made no sense.

Terry had previously written.

Current flows are measured in amps (or subdivisions or multiples
of amps such as milliamps/kiloamps).

Kevin commented that;

Current is a flow of charge, that is, current already includes the
concept of movement. A "flow of charge flow", don't make for good
English, mate. Your phrase would be better stated as:

"Charge flow is measured in amps."

or

"Current is measured in amps."

Kevin: Thanks for the comment. I agree the word 'flow' is
redundant after the word current. Although I doubt if, in
practical day to day parlance (outside of a laboratory or
something), anyone uses the expression "flow of charge"?
Your expression "Current is measured in amps etc." IS a better
way of saying it.
As a result of your comment I now notice, that I do, redundantly,
tend to use the word 'flow' in such situations as "How much
current is flowing in that conductor etc.".
My regards. Terry.

PS. The subject headline "You think you know ..... "is also a bit
puzzling. I couldn't figure out if someone was challenging the
collective knowledge of the group, perhaps based on a
misapprehension of potential difference and current (flow!); or
perhaps it should be read as "One thinks one knows something, but
help me out here ...?". Just a thought anyway!
Funny how often it comes back to basic Ohms Law, I guess each one
of us learns those 'basics' in a slightly different manner.
Cheers to all and thanks for the discussion.
PPS. I also just found out that I didn't know how to spell "r e d
u n d a n t". Never too old to learn!