Rating of PC power supplies?

[Terry Pinnell]

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Rheilly Phoull]

One day Terry Pinnell got dressed and committed to text

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

--
Terry Pinnell
Hobbyist, West Sussex, UK

G'Day Terry
Pity you didnt check at the output on the PSU, could have shown up any volt
drop in your load leads. The regulation feedback comes from those
connections.
Then again perhaps I'm wrong (as usual


--
Regards ..... Rheilly Phoull

 

[Franc Zabkar]

On Sun, 29 May 2005 07:10:38 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

IME most PC PSUs regulate on the basis of a weighted average of the
+5V and +12V rails. If your +12V rail is unloaded, then the +5V rail
will also not regulate properly. I'd check the capacitors on the +5V
rail (or add an external cap for testing purposes), and I'd measure
the +12V rail in all four cases.


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[Terry Pinnell]

"Rheilly Phoull" <Rheilly@bigpong.com> wrote:

One day Terry Pinnell got dressed and committed to text

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

--
Terry Pinnell
Hobbyist, West Sussex, UK

G'Day Terry
Pity you didnt check at the output on the PSU, could have shown up any volt
drop in your load leads. The regulation feedback comes from those
connections.
Then again perhaps I'm wrong (as usual

Actually I did, Rheilly - just didn't include the correct column in
the posted results! Here's the fuller table, which as you rightly say
gives the proper picture:

Voltage
Current At PSU output At load terminals
------- ----------------- ------------------
5.1 A 4.96 4.77
7.5 A 4.78 4.51
11.0 A 4.60 4.20
13.0 A 4.50 4.02

My 'PSU output' was at the end of one of the red '5V' output wires,
about 30 cm long, the first location accessible without opening the
case.

Wiring from there to my load (and back to the PSU 0V cable, a similar
distance from case) therefore totaled about 0.04 ohms.

So, much better, with 4.5V at 13A. Not sure how to estimate 20A. Do
those numbers look acceptable?

BTW, correct operation only started with a load of something over 1A
or so. Presumably a characteristic of this type of PS?

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Walter Harley]

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> wrote in message
news:ti1j9194ll8canr4i5f46ms5hm6ngs00rs@4ax.com...

BTW, correct operation only started with a load of something over 1A
or so. Presumably a characteristic of this type of PS?

Yes, that's correct. You can Google around for information on how to
successfully use PC PSUs for general purposes.

Also be aware that there is more than one 5V output, at least if we're
talking about a standard ATX supply, e.g.,
http://pinouts.ru/data/atxpower_pinout.shtml. I don't know whether your
unit was rated for 20A at *each* 5V output, or for them all put together.
They should all be coregulated, indeed they might even all just be in
parallel; but it sounds like the issue you're facing has something to do
with IR drop between the sense point and the load, so it makes a difference.

 

[Terry Pinnell]

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> wrote in message
news:ti1j9194ll8canr4i5f46ms5hm6ngs00rs@4ax.com...
BTW, correct operation only started with a load of something over 1A
or so. Presumably a characteristic of this type of PS?

Yes, that's correct. You can Google around for information on how to
successfully use PC PSUs for general purposes.

Also be aware that there is more than one 5V output, at least if we're
talking about a standard ATX supply, e.g.,
http://pinouts.ru/data/atxpower_pinout.shtml. I don't know whether your
unit was rated for 20A at *each* 5V output, or for them all put together.
They should all be coregulated, indeed they might even all just be in
parallel; but it sounds like the issue you're facing has something to do
with IR drop between the sense point and the load, so it makes a difference.

Thanks. Yes, there are 3 '5V' outputs, apparently in parallel. I'll
check whether connecting them makes any significant difference.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Guy Macon]

Terry Pinnell wrote:

I dusted off an old '200 W' ex-PC power supply yesterday and was
playing with it with a view to using it for some motor tests.

No-name PC power supplies are notorious for being junk. Buy a
lab supply, build your own supply, or buy a PC PS from PC Power
and Cooling (they make very good PC power supplies).

 

[Rich Grise]

On Sun, 29 May 2005 18:34:02 +0100, Terry Pinnell wrote:

"Walter Harley" <walterh@cafewalterNOSPAM.com> wrote:

"Terry Pinnell" <terrypinDELETE@THESEdial.pipex.com> wrote in message
news:ti1j9194ll8canr4i5f46ms5hm6ngs00rs@4ax.com...
BTW, correct operation only started with a load of something over 1A or
so. Presumably a characteristic of this type of PS?

Yes, that's correct. You can Google around for information on how to
successfully use PC PSUs for general purposes.

Also be aware that there is more than one 5V output, at least if we're
talking about a standard ATX supply, e.g.,
http://pinouts.ru/data/atxpower_pinout.shtml. I don't know whether your
unit was rated for 20A at *each* 5V output, or for them all put together.
They should all be coregulated, indeed they might even all just be in
parallel; but it sounds like the issue you're facing has something to do
with IR drop between the sense point and the load, so it makes a
difference.


Thanks. Yes, there are 3 '5V' outputs, apparently in parallel. I'll check
whether connecting them makes any significant difference.

I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

 

[Terry Pinnell]

Rich Grise <richgrise@example.net> wrote:

I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Franc Zabkar]

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

See this ATX example for an explanation of how a typical PSU
regulates:

http://www.pavouk.comp.cz/hw/en_atxps.html

The regulation formula is:

5.0V * 3K01/(3K01 + 3K01) = V12 * (3K09||150K||5K6) / (27K +
3K09||150K||5K6)
+ V5 * (3K09||150K||27K) / (5K6 +
3K09||150K||27K)

This reduces to:

2.50 = V12 * 0.0678 + V5 * 0.327

For this PSU, if the +5V rail increases by 50mV, say, then the +12V
rail will fall by 241mV.

If you want to guarantee a stable +5V output, then remove the +12V
sense resistor and recalculate the remaining +5V sense resistors such
that the midpoint of the potential divider is +2.5V. I have modified
several PSUs this way. I have one which I have reprogrammed for +6V,
and another for +13.8V.


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[legg]

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> wrote:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

Only one of the output and return terminals will be sensed. Use all
releventterminals in parallel to remove harness drop from your
measurements.

You can identify the sensed terminals by visual inspection. These will
have two wires crimped into a single power pin.

RL

 

[Pooh Bear]

Terry Pinnell wrote:

Rich Grise <richgrise@example.net> wrote:

I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

Usually spare drive power connectors.

Graham

 

[Pooh Bear]

Terry Pinnell wrote:

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

I'd expect it to range from 5.1 to 4.9 volts approx. They're normally not
too bad. Shouldn't really be more than 200mV out of spec.

Bear in mind that it's a multiple output supply and an absence of load on
the 12V output is likely screwing things up.

As another poster noted, these supplies *have* to sense a combination of
both the 5V and 12 V outputs so no load on the 12V is bound to be an
issue. The 12V is probably high ( due to no load ) and therefore dropping
the 5V low.

The -5V and -12V outputs are low current and just rely on the reflected
voltage. They won't need loading.

Graham

 

[Pooh Bear]

Franc Zabkar wrote:

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

See this ATX example for an explanation of how a typical PSU
regulates:

http://www.pavouk.comp.cz/hw/en_atxps.html

The regulation formula is:

5.0V * 3K01/(3K01 + 3K01) = V12 * (3K09||150K||5K6) / (27K +
3K09||150K||5K6)
+ V5 * (3K09||150K||27K) / (5K6 +
3K09||150K||27K)

This reduces to:

2.50 = V12 * 0.0678 + V5 * 0.327

For this PSU, if the +5V rail increases by 50mV, say, then the +12V
rail will fall by 241mV.

If you want to guarantee a stable +5V output, then remove the +12V
sense resistor and recalculate the remaining +5V sense resistors such
that the midpoint of the potential divider is +2.5V. I have modified
several PSUs this way. I have one which I have reprogrammed for +6V,
and another for +13.8V.

That works fine if you have the schematic to calculate the values. ;-)

I suspect Terry's old PSU is likely an AT type btw.

Graham

 

[Terry Pinnell]

Pooh Bear <rabbitsfriendsandrelations@hotmail.com> wrote:

Terry Pinnell wrote:

I'm not familiar with switched mode power supplies so hope this
doesn't prove to be a dumb question. I dusted off an old '200 W' ex-PC
power supply yesterday and was playing with it with a view to using it
for some motor tests. (For the time being I settled on a set of 4
NiCads instead, but I may pursue it later.) Its spec says that its
'5V' output is rated at 20A. Arbitrary tests gave me these results:

Current Voltage at load terminals
------- -------------------------
5.1 A 4.77
7.5 A 4.51
11.0 A 4.20
13.0 A 4.02

I didn't go further, but even at 13A isn't that voltage rather low? I
wonder what it would drop to at close to 20A? What is regarded as an
'acceptable' voltage from the '5V' output of these types of supply
please?

I'd expect it to range from 5.1 to 4.9 volts approx. They're normally not
too bad. Shouldn't really be more than 200mV out of spec.

Bear in mind that it's a multiple output supply and an absence of load on
the 12V output is likely screwing things up.

As another poster noted, these supplies *have* to sense a combination of
both the 5V and 12 V outputs so no load on the 12V is bound to be an
issue. The 12V is probably high ( due to no load ) and therefore dropping
the 5V low.

The -5V and -12V outputs are low current and just rely on the reflected
voltage. They won't need loading.

Graham

Many thanks, and for the other follow-ups. I'll load the 12V output
and re-test.

Note that my later post prompted by Rheilly gave the more relevant
readings:

Current V (At PSU output)
------- -----------------
5.1 A 4.96
7.5 A 4.78
11.0 A 4.60
13.0 A 4.50

I'd be more than happy with even 4.8V at 15-20A. Crude motor tests
apart, that would be handy if I ever build any more stuff from my
ancient stocks of TTL. (Don Lancaster's 'TTL Cookbook' recommends +/-
250mV.)

Must say I'm impressed at the performance of these things. I expect
there are downsides in hf noise, but they're so light and compact in
comparison to a conventional transformer-based equivalent. I made a
heavy duty 20A car battery charger years ago, still in use. OK, it's
12 not 5V, but I have to wheel that monster out on an ex-shopping
trolley <g>.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Franc Zabkar]

On Mon, 30 May 2005 02:09:08 +0100, Pooh Bear
<rabbitsfriendsandrelations@hotmail.com> put finger to keyboard and
composed:

Franc Zabkar wrote:

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

See this ATX example for an explanation of how a typical PSU
regulates:

http://www.pavouk.comp.cz/hw/en_atxps.html

The regulation formula is:

5.0V * 3K01/(3K01 + 3K01) = V12 * (3K09||150K||5K6) / (27K +
3K09||150K||5K6)
+ V5 * (3K09||150K||27K) / (5K6 +
3K09||150K||27K)

This reduces to:

2.50 = V12 * 0.0678 + V5 * 0.327

For this PSU, if the +5V rail increases by 50mV, say, then the +12V
rail will fall by 241mV.

If you want to guarantee a stable +5V output, then remove the +12V
sense resistor and recalculate the remaining +5V sense resistors such
that the midpoint of the potential divider is +2.5V. I have modified
several PSUs this way. I have one which I have reprogrammed for +6V,
and another for +13.8V.

That works fine if you have the schematic to calculate the values. ;-)

Not necessarily. The most common PWM controller IC is the TL494 or its
equivalent, KA2500B/C. All you need to do is to confirm the voltages
at the inputs to the IC's error amp. One input will usually be 2.5V (=
internal 5.0V reference divided by 2), the other input will be derived
from the +12V rail and/or the +5V rail via a resistive potential
divider. All you need to do is to recompute the resistor values in the
feedback network so that the target voltage, eg 13.8V, produces 2.5V
at the input to the error amp.

I suspect Terry's old PSU is likely an AT type btw.

The two that I modified were both AT types. I had the schematics for
neither. Only very minor reverse engineering was required.

Graham

- Franc Zabkar
--
Please remove one 's' from my address when replying by email.

 

[Terry Pinnell]

Franc Zabkar <fzabkar@optussnet.com.au> wrote:

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

See this ATX example for an explanation of how a typical PSU
regulates:

http://www.pavouk.comp.cz/hw/en_atxps.html

The regulation formula is:

5.0V * 3K01/(3K01 + 3K01) = V12 * (3K09||150K||5K6) / (27K +
3K09||150K||5K6)
+ V5 * (3K09||150K||27K) / (5K6 +
3K09||150K||27K)

This reduces to:

2.50 = V12 * 0.0678 + V5 * 0.327

For this PSU, if the +5V rail increases by 50mV, say, then the +12V
rail will fall by 241mV.

If you want to guarantee a stable +5V output, then remove the +12V
sense resistor and recalculate the remaining +5V sense resistors such
that the midpoint of the potential divider is +2.5V. I have modified
several PSUs this way. I have one which I have reprogrammed for +6V,
and another for +13.8V.


- Franc Zabkar

Nice post, thanks. As you see from my other reply. I'm off to the
shed/workshop to try it soon! Will probably take the easier approach
of trial and error loading, until I get a satisfactory minimum, and
will report back on results.

When I do get around to opening the case of such a unit, that
schematic and notes will be invaluable.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[SioL]

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message news:429A66F2.E1D7AC6@hotmail.com...

I'd expect it to range from 5.1 to 4.9 volts approx. They're normally not
too bad. Shouldn't really be more than 200mV out of spec.

Bear in mind that it's a multiple output supply and an absence of load on
the 12V output is likely screwing things up.

I have experimented with these a bit, albeit an older variety (AT type) of todays
power supplies. What I found is that 5V is usually pretty accurate, while
12V is somewhat flimsy. This particular PS was only sensing the 5V
output, relying the others would be more/less close to the design ratio;
it can be off quite a bit for +12V and varies a lot with load.

The only output that had to be loaded was the sensing output, that was 5V.
My reason to play with these was conversion into 10-15V variable voltage PSU,
which involved cutting sensing track from 5V and running it to 12V output
via adjustable voltage divider. Loading was now only required for 12V output
and fan did a good job of that (required some LC filtering to make it quiet electrically).

Made for a nice encased dirt-cheap hobby power supply with current protection. Some
of these PSU's had chromed enclosured, looked fairly good.

Hopefully someone will benefit from this.

Take care with high voltage, can be painfull/lethal!!

--
Siol
------------------------------------------------
Rather than a heartless beep
Or a rude error message,
See these simple words: "File not found."

 

[Terry Pinnell]

Franc Zabkar <fzabkar@optussnet.com.au> wrote:

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

As per my reply to SioL, I did not have to load the 12V output, only
the 5V.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

Franc Zabkar <fzabkar@optussnet.com.au> wrote:

On Mon, 30 May 2005 02:09:08 +0100, Pooh Bear
rabbitsfriendsandrelations@hotmail.com> put finger to keyboard and
composed:

Franc Zabkar wrote:

On Sun, 29 May 2005 21:49:17 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> put finger to keyboard and
composed:

Rich Grise <richgrise@example.net> wrote:


I'd parallel all of the black wires as well; forgive me if I'm being
redundant here.

Cheers!
Rich

No harm in trying it, so I will! But without any expectation that it
will affect any measurements. I'm pretty sure I recall when this PSU
was inside my old PC that several of the output 'sets' were floating
loose in their plugs, presumably awaiting service for additional hw.

You *must* load the +12V rail with a dummy load, otherwise you will
*never* achieve proper regulation on the +5V rail.

See this ATX example for an explanation of how a typical PSU
regulates:

http://www.pavouk.comp.cz/hw/en_atxps.html

The regulation formula is:

5.0V * 3K01/(3K01 + 3K01) = V12 * (3K09||150K||5K6) / (27K +
3K09||150K||5K6)
+ V5 * (3K09||150K||27K) / (5K6 +
3K09||150K||27K)

This reduces to:

2.50 = V12 * 0.0678 + V5 * 0.327

For this PSU, if the +5V rail increases by 50mV, say, then the +12V
rail will fall by 241mV.

If you want to guarantee a stable +5V output, then remove the +12V
sense resistor and recalculate the remaining +5V sense resistors such
that the midpoint of the potential divider is +2.5V. I have modified
several PSUs this way. I have one which I have reprogrammed for +6V,
and another for +13.8V.

That works fine if you have the schematic to calculate the values. ;-)

Not necessarily. The most common PWM controller IC is the TL494 or its
equivalent, KA2500B/C. All you need to do is to confirm the voltages
at the inputs to the IC's error amp. One input will usually be 2.5V (=
internal 5.0V reference divided by 2), the other input will be derived
from the +12V rail and/or the +5V rail via a resistive potential
divider. All you need to do is to recompute the resistor values in the
feedback network so that the target voltage, eg 13.8V, produces 2.5V
at the input to the error amp.

I suspect Terry's old PSU is likely an AT type btw.

The two that I modified were both AT types. I had the schematics for
neither. Only very minor reverse engineering was required.

Graham


- Franc Zabkar

Mine has no manufacturer's name on the case, only 'Model PTP-2006'. It
is 200W and has +5V @20A (red), +12V @ 8A (yellow); -5V @ 300mA
(white); -12V @ 300mA (blue); 0V (black); and an orange wire marked
'PG', the purpose of which I'm unsure about.

--
Terry Pinnell
Hobbyist, West Sussex, UK