Computer Power +ve & -ve requirements of AT & ATX plugs

Okay, I have the pinouts of both the AT & ATX plugs (and others) of a PC
power supplies. My understanding of these is that most power is consumed
on the +ve side of the power supply.

What I am trying to find out is what proportion of power is consumed by
the -ve side in your average wintel compatible PC.

Or to put it another way, if I take a 100Amphr battery and supply the
+12v & +5v power lines, what size AmpHr battery would I need to supply
the -12v & -5v? (ignoring battery discharge considerations).

Yes, I know old laptops are only a few hundred dollars, but it is the
satisfaction of doing this that is the interesting part.

In scrapping what was labelled as a a Osbourne 386SX desktop, it turns
out to have a Cyrix 586GXM-AV motherboard (microformat) with a 266GP
CPU, so I am considering re-casing the mobo for various
luggable/portable datalogging uses. I do not want to do the full battery
to inverter to PC supply route, so the obvious repacement is two 12 volt
batteries and regulators.

TIA
__
ex-bongo for direct.

"Terry Collins" <(E-Mail Removed)> wrote in message
news:42785808$0$94178$(E-Mail Removed)...
> Okay, I have the pinouts of both the AT & ATX plugs (and others) of a PC
> power supplies. My understanding of these is that most power is consumed
> on the +ve side of the power supply.
>
> What I am trying to find out is what proportion of power is consumed by
> the -ve side in your average wintel compatible PC.
>
> Or to put it another way, if I take a 100Amphr battery and supply the +12v
> & +5v power lines, what size AmpHr battery would I need to supply the -12v
> & -5v? (ignoring battery discharge considerations).

Do you have ANY understanding of electronics???????

Current flows from the negative terminal, via the circuitry, back to the
positive terminal. In theory, whatever current leaves the negative terminal
should reach the positive (some slight loss due to heat, etc). The battery
you use to supply the +ve should also supply the -ve rail (in fact, usually
the -ve is common and the +ve goes via the regulator to provide the desired
voltage)..

Perhaps, given your question, you should leave this to someone with some
basic understanding of how it all works?

Q! wrote:

> Current flows from the negative terminal, via the circuitry, back to the
> positive terminal.

Then what is the ground terminal for? {:-)

On Wed, 04 May 2005 15:02:05 +1000, Terry Collins
<(E-Mail Removed)> wrote:

>Okay, I have the pinouts of both the AT & ATX plugs (and others) of a PC
>power supplies. My understanding of these is that most power is consumed
>on the +ve side of the power supply.
>
>What I am trying to find out is what proportion of power is consumed by
>the -ve side in your average wintel compatible PC.
>
>Or to put it another way, if I take a 100Amphr battery and supply the
>+12v & +5v power lines, what size AmpHr battery would I need to supply
>the -12v & -5v? (ignoring battery discharge considerations).
>

in a typical modern PC - the -12 and -5 draw bugger all, and I would
be surprised if they are even needed on modern motherboards. (except
of course for the serial port needing the -12 v to swing from +12 to
-12, if you are using the serial port that is). you would probably
get away with a 1-2 AH battery in proportion to the 100ah battery on
these other rails.



The other exception being if you had an optional device attatched to
the machine that specifically needed one or both of these voltages.
Older motherboards may need these voltages too as in the late 70's
processors like the 8080, 2708 Eproms etc did need these supplies, but
again I doubt that they drew any significant current.

>Yes, I know old laptops are only a few hundred dollars, but it is the
>satisfaction of doing this that is the interesting part.
>
>In scrapping what was labelled as a a Osbourne 386SX desktop, it turns
>out to have a Cyrix 586GXM-AV motherboard (microformat) with a 266GP
>CPU, so I am considering re-casing the mobo for various
>luggable/portable datalogging uses. I do not want to do the full battery
>to inverter to PC supply route, so the obvious repacement is two 12 volt
>batteries and regulators.
>
>TIA
>__
>ex-bongo for direct.

"Michael A. Terrell" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> "Q!" wrote:
>>
>> Do you have ANY understanding of electronics???????
>>
>> Current flows from the negative terminal, via the circuitry, back to the
>> positive terminal. In theory, whatever current leaves the negative
>> terminal
>> should reach the positive (some slight loss due to heat, etc). The
>> battery
>> you use to supply the +ve should also supply the -ve rail (in fact,
>> usually
>> the -ve is common and the +ve goes via the regulator to provide the
>> desired
>> voltage)..
>>
>> Perhaps, given your question, you should leave this to someone with some
>> basic understanding of how it all works?
>
>
> Do you have any understanding of multiple output power supplies with
> a single common return? AT and ATX supplies have four or more separate
> output voltages produced from different windings on the switching
> transformer.
>
> A common AT/ATX power supply has:
> + 12 Volt output
> + 5 Volt output
> - 12 Volt output
> - 5 Volt output
>
> ATX power supplies also have a small, separate 5 volt supply for
> standby mode. Each of these is connected to the common rail, which is
> grounded to the power supply case and computer chassis. Some supplies
> have a 3.3 volt supply for the CPU as well.
>
> Learn what you're talking about before you criticize the skills of
> others.

And, pray tell, what has that got to do with the fact that the current flows
from -ve to +ve? All you are stating here is that there is more than just a
5V and 12V supply rail, with all connected back to the -ve common ( "single
common return")..

That changes nothing about my reply, or the original question of "what
capacity battery to connect to the -ve rail if (x) value is connected to the
+ve rail".

Bloody yanks...

"Terry Collins" <(E-Mail Removed)> wrote in message
news:4278922d$0$94186$(E-Mail Removed)...
> Q! wrote:
>
>> Current flows from the negative terminal, via the circuitry, back to the
>> positive terminal.
>
> Then what is the ground terminal for? {:-)

Ground, which is NOT the same as the negative or common, although sometimes
the negative is connected to the chassis of the equipment to provide a
'common' connection. Ground is used on the AC side, not the DC side, and
strangely enough connects to the GROUND! (i.e. bloody big copper rod driven
into the ground outside with hefty green wire running to supply the
electrical ground connection). In case of a fault it provides a path for
the AC to ground, which helps the residual current device type circuit
breakers to trip.. (they sense that the current in the active and neutral
are equal - if some current is leaked to ground then the current in the
active and neutral will be different and the breaker will trip,
disconnecting the supply. If there is no path to ground then that doesn't
happen!)

Yes folks, that IS a very, very, very basic, simplified (and from the
professional angle, ridiculous) explanation so don't start - but given the
obvious level of OP knowledge (or lack thereof) I'm not entering into a full
on theory lesson with pretty drawings and connect the dot type
lessons........... If someone else wishes to do so, please go ahead!!

Q! wrote:

>
> And, pray tell, what has that got to do with the fact that the current flows
> from -ve to +ve?

Sigh, as they say "you can lead a horse to water but you can not make
him drink".

2nd try. If I have two 12v batteries in series and I connect a GROUND to
the common lead, then I have +12V and -12V with respect to the common
ground.

KLR wrote:
> in a typical modern PC - the -12 and -5 draw bugger all, and I would
> be surprised if they are even needed on modern motherboards. (except
> of course for the serial port needing the -12 v to swing from +12 to
> -12, if you are using the serial port that is). you would probably
> get away with a 1-2 AH battery in proportion to the 100ah battery on
> these other rails.

Thanks. It was what I suspected, but was looking for someone with
practical knowledge to give some feedback.

I will probably end up having the second battery as I am planning to use
the serial inputs for sensors and keyboard inputs. It could also be used
to run lighting, etc anyway.

Q! wrote:

> Yes folks, that IS a very, very, very basic, simplified (and from the
> professional angle, ridiculous) explanation so don't start - but given the
> obvious level of OP knowledge (or lack thereof)

You're obviously an engineer.
[aps to REAL engineers].

Michael A. Terrell wrote:

> I would use a DC to DC converter to supply the negative voltages at
> the very least. Some ICs use multiple supply voltages and can be
> damaged by a missing supply. Two batteries will not discharge at the
> same rate. There are 12 DC in PC power supplies on the market, but a
> small inverter and a standard power supply is a cheap and easy way to
> go. Its cheaper than designing the separate switch regulators, and
> regulates the 12 volt supplies as well.

I will keep your points in mind if it becomes something permanent.