On 6/4/2019 1:10 AM, Paul wrote:
Charlie wrote:
On 6/3/2019 3:28 PM, VanguardLH wrote:
T. Ment wrote:

On Mon, 03 Jun 2019 17:00:42 +0100 (BST), rp wrote:

Replace the power supply.

It's a $20 special. I don't care about perfection.

That is way too cheap for a decent power supply.Â* When I do a build,
most of the PSUs that I end up looking at are $80 at a minimum but
usually spend a lot more.Â* However, I don't know what you got for
capacity (VA or watts) for your PSU, or its efficiency (less efficient =
more heat = more thermal stress = less capacity to internal components
since the A/C outlet isn't changing its voltage).

Figure on losing 5% capacity each year with a decent PSU.Â* That's why I
buy PSUs that are much higher in capacity than the load they will
initially need to handle.Â* For a 400W load, I'd get a 750W PSU because
in 8 years (what I like to build for longevity) the PSU will still be
more than enough at 66% of its initial capacity.Â* The 5% is a rough
estimate of capacity reduction, but PSUs do wane over time.Â* Cheapies
die off much quicker.Â* The PSU supplies the life blood to the computer.
If you do the build, you are the hematologist for the computer.

How old is your $20 PSU?

Was it a pre-built computer or did you do your own build?Â* The PSUs that
come in pre-builts are often minimal for the standard configuration by
the model (with the same PSU used within a family of models).Â* They
don't overbuild since they're maximizing their profit margin.

The 5v disk connector may well come from another regulator. At
the very least measure the 5v at the motherboard multiconector
from the psu.

I won't buy another power supply if this one gets me by. My question
is,
what tolerance is safe.

4.75V is the minimum.Â* Less voltage means supplying more current to
handle the same load, and more current puts more thermal stress on the
supply components.

I'm curious as to why you say less voltage means supplying more
current. Â*Doesn't that violate Ohm's law?

I = V / R

If V = 5.00 Volts and the load (resistance) R = say 10 Ohms then the
current I = 0.500 Amps.

If V = 4.75 Volts and the same load then I = 0.475 Amps.

Or is there some kind of perverse circuitry in computer power supplies
that make them work differently?

Charlie

In the engineering profession, as time passes and you become
"seasoned", you begin to realize there are certain topics
you simply "don't wade into". There are things you don't
"shoot from the hip".

In logic design, the trap there, is jotting down a logic
table on a table napkin, and selecting an AND/NAND/OR/NOR/XOR
gate which you think corresponds to the table. 9 times out of 10,
you'll get it wrong when you do that. Stick to your regular
symbolic methods! So when someone tries to trick you, by
making a logic table on a napkin, instead say "that's nice"
and walk away :-) It's not that you don't like a challenge.
It's that this challenge is intended to trick you into
making an unnecessary mistake.

The same happens with power calcs and statements about power.

And by you writing up your statement like that, you sucked
me into your vortex and tried to trick me! Good work :-)

LOL

This is the third time I've had to write this answer.
Again, good work! You tricked me.

At least I'm good for something. ;-)

Seriously, my intent was not to trick you or VanguardLH. I truly wanted
to learn.


*******

The OPs electrical load consists of two parts. This is one
of the boards, where the four +5V wires on the 20 pin connector,
carry the power used by the processor. The VCore circuit, an
SMPS, converts the power from 5V, to say, 1.5V.

Â*Â* +5V --------+------------------+ 5V @ 13AÂ*Â*Â*Â*Â*Â* +--------+ 1.5V @ 43.3A
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â* |
Â*Â*Â*Â*Â*Â*Â*Â*Â* (regular ccts,Â*Â*Â*Â*Â*Â*Â*Â*Â* +--- VcoreÂ* -----+Â*Â*Â*Â*Â*Â* Athlon
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* Ohms law isÂ*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* SMPSÂ*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* CPU (65W)
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* close...)Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* +--- (Constant --+Â*Â*Â*Â*Â*Â*Â* |
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â* power)Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â* |
Â*Â* GND --------+------------------+Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* +--------+

On an SMPS, the SMPS runs at constant power. It's a closed
loop feedback system. It wants to run my mythical 1.5V processor
at a constant 1.5V. And since I'm running 7ZIP compressor right
now, my CPU is railed and drawing 43.3A.

Ah! The perverse circuit I was alluding to (only it isn't in the
computer power supply but downstream of it).

OK, now assume the rail voltage on the inlet side, drops to 4.75V.
The SMPS still needs to make 65W for my processor. Using closed
loop feedback, the inlet current *increases*. (And this is why
some SMPS are equipped with UVLO or Undervoltage Lockout, so
that if the voltage drops too low, something won't burn on the
inlet side.)

Â*Â*Â* 65W
Â* ------ = 13.68A
Â*Â* 4.75V

+4.75V --------+------------------+ 4.75V @ 13.68A +--------+ 1.5V @ 43.3A
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â* |
Â*Â*Â*Â*Â*Â*Â*Â*Â* (regular ccts,Â*Â*Â*Â*Â*Â*Â*Â*Â* +--- VcoreÂ* -----+Â*Â*Â*Â*Â* Athlon
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* Ohms law isÂ*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* SMPSÂ*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* CPU (65W)
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* close...)Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* +--- (Constant --+Â*Â*Â*Â*Â*Â*Â* |
Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* |Â*Â*Â*Â* power)Â*Â*Â*Â* |Â*Â*Â*Â*Â*Â*Â* |
Â*Â* GND --------+------------------+Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â*Â* +--------+

So now we notice there's an item in the circuit, which
doesn't follow Ohms law. And at close to 14 amps, it's
a *major* part of the circuit in this case. The Ohms
law part of the circuit, is well less than 14 amps.

That makes sense. I was unaware there was the need for a 'constant
power' supply (or even that there was such a thing) to the CPU.

And I was this close to pushing the button on the other
two versions of this post, and falling into the same
trap I was going to warn you about. Again, good work :-)

The above isn't a complete workup of the circuit. It's
intended to illustrate the principle of "traps" in engineering.
I've seen a couple of people nearly ruined by stuff like
this. One guy, filled a lab with smoke one day and forced
an evaluation... because he was so sure he was right...
And that's the only time ever, that anyone did something
stupid enough to cause an evacuation.

If you have a device in hand, it's much better to just
*measure* it, than dreamily attempt to analyze it with a half
dozen table napkins.

I agree and would much rather measure something, but I didn't have a
clue how to go about it in this case.
I 'have' lowered my Vcore voltage in the past and that resulted in a
lower CPU temperature. In my mind I equated the lower temperature to
lower power consumed.


Â*Â* Paul (dammit!)

Thank you for your detailed explanation and sorry for wasting table
napkins. :-)

Charlie