Re: -5V on IIgs

[et472@FreeNet.Carleton.CA (Michael Black)]

Don Bruder (dakidd@sonic.net) writes:
> In article <1173043981.038947.50150@t69g2000cwt.googlegroups.com>,
>  "bieling@terra.es" <bieling@terra.es> wrote:
> 
>> On Mar 4, 8:23 pm, Don Bruder <dak...@sonic.net> wrote:
>> >
>> > The "quick and dirty" method:
>> > Get a 7905 regulator (-5 volts regulated output - 78xx and 79xx series
>> > regulators are good to just under an amp)
>> >
>> > Wire the input of the regulator to ground. Wire the ground of the
>> > regulator to the +12V line. (hook it up to the supply "backwards") Now
>> > use the output to feed the -5V line.
>> >
>> > Presto!
>> >
>> > You can do the same, using a 7805 regulator wired normally - ground to
>> > ground, input to +12V, output to +5V line - for your +5 volt supply.
>> >
>> 
>> Excuse me Don, you want him to fry his gs or what ?
>> What you say would give +7v, not -5v !
>> 
>> Jorge.
> 
> You've managed to misunderstand, methinks... Or missed something while 
> reading... Or SOMETHING! :) 
> 
But you'll never get -5v out of the scheme.  At best you are creating an
artificial ground so there will be something -5v in reference to that
artificial ground, but you've done nothing to compensate for the artificial
ground.

You are treating the existing ground as a minus terminal, and making your
+12v line the "ground" line.  YOu don't even need a negative voltage
regulator to do that, you can just come up with a +5v supply and reverse
the leads.  

The problem is, you no longer have a common "ground".  The things running
off the +12 and +5v lines expect the current ground to be ground.  But the
things running off the -5v line expects the +12v line to be ground.  But those
expectations are not isolated, so you have everything expecting the current
ground to be ground.  So at best, you have +7v out of that 7905 (the
artificial "-5v" in reference to the existing ground is actually 
+12 -5v=+7volts).

Your scheme might sort of work if you were doing something like a voltage
divider across the +12v line, so across one resistor you get 5v and across
the other resistor what remains, ie 7v, and then you make the center point
of the divider your actual ground.  Of course, that doesn't really work, since
you need to start with a higher voltage supply if you want +12 and -5v.  And
it also gets messy, since the resistors have to handle the current, and
as the different voltage lines draw different current, the voltages may
vary.

  Michael