Motherboard Power Question

[roger]

I am concerned that if my machine is connected to a power source, my
motherboard is receiving power even though it is not doing anything
until I switch my machine on. How does this impact things like
adding/changing RAM? Motherboard connections to hard drive (power,
Sata, IDE)? USB? Do I need to continue to extract the power plug
before messing with any of these? I have been.

Older motherboards had an LED that by being on made it obvious that
the motherboard was under power. New boards don't seem to have that.
So now, any impact is hidden unless you happen to know that power is
being provided.

Just curious

Big Fred

 

[philo]

I am concerned that if my machine is connected to a power source, my
motherboard is receiving power even though it is not doing anything
until I switch my machine on. How does this impact things like
adding/changing RAM? Motherboard connections to hard drive (power,
Sata, IDE)? USB? Do I need to continue to extract the power plug
before messing with any of these? I have been.

Older motherboards had an LED that by being on made it obvious that
the motherboard was under power. New boards don't seem to have that.
So now, any impact is hidden unless you happen to know that power is
being provided.

Just curious

Big Fred

The best thing to do is unplug the machine, then press the power button
to drain the residual charge left on the capacitors.



Of course I never bother to do that and have never had a problem...
but I really advise to err on the side of caution.

 

[VanguardLH]

I am concerned that if my machine is connected to a power source, my
motherboard is receiving power even though it is not doing anything
until I switch my machine on. How does this impact things like
adding/changing RAM? Motherboard connections to hard drive (power,
Sata, IDE)? USB? Do I need to continue to extract the power plug
before messing with any of these? I have been.

Older motherboards had an LED that by being on made it obvious that
the motherboard was under power. New boards don't seem to have that.
So now, any impact is hidden unless you happen to know that power is
being provided.

Just curious

Big Fred

For an ATX style PSU, and as long as it is plugged into a powered
source, it provides 5 V standby to the motherboard. That's because
there is no direct link between the Power switch and the PSU. The Power
switch goes to the motherboard where logic is used to tell the PSU (via
the PS-On green colored wire in the 20+4 PSU connector to the
motherboard) when to power up. For ATX, the motherboard can powered up
the PSU even without using the Power button, like Wake-On events in
BIOS. ATX PSUs and motherboards use "soft power" with a momentary
contact switch where the motherboard dictates when it gets fully powered
up versus the old AT style that had the Power toggle switch hardwired to
the PSU so it was the PSU that decided when to power up.

http://www.pcguide.com/ref/power/sup/funcSoftPower-c.html

Remove the power cord from the PSU end if you want to be most safe.
There may even be a LED on the motherboard to let you know when it has
power so wait until it goes dark within a couple minutes. If there is
no LED, just wait 5 minutes (pressing the Power button as suggested by
philo won't work to drain the caps because the PSU won't power up when
it's not connected to a powered source). Then do your inside work.

 

[Paul]

I am concerned that if my machine is connected to a power source, my
motherboard is receiving power even though it is not doing anything
until I switch my machine on. How does this impact things like
adding/changing RAM? Motherboard connections to hard drive (power,
Sata, IDE)? USB? Do I need to continue to extract the power plug
before messing with any of these? I have been.

Older motherboards had an LED that by being on made it obvious that
the motherboard was under power. New boards don't seem to have that.
So now, any impact is hidden unless you happen to know that power is
being provided.

Just curious

Big Fred

Unplug power supply from the wall.

Wait 30 seconds at least, for the power supply to drain
the capacitor, while it attempts to keep making +5VSB.

Then it's safe to change RAM, change a video card, as
the slots involved could have power in them otherwise.

Once the power supply is unplugged, and you've waited
a bit before starting work, then the slots should be
drained of power. Then you can safely pull or
insert new cards.

*******

For your own amusement, purchase a USB powered reading
lamp. You can use that as a status indicator.

1) For this test, you don't want the OS running. This
is for when the computer OS is shut down, BIOS is
not attempting to POST, that sort of thing. The fans
should not be spinning. That's a good indication you're
ready for this test.

With power supply plugged in, and switched on at
the back, verify the USB reading lamp lights up.
In the "sleeping" state (no ATX fan running),
the only power should be +5VSB. Using the reading
lamp, is a quick way to verify the USB port is
powered by +5VSB. Once we know the USB port uses +5VSB,
then later, we know we have a means to observe +5VSB.

2) Now, switch off the power supply at the back.
It should take around 30 seconds for the USB powered
reading lamp, to be extinguished. Now, it's safe to
work inside the computer.

That will give you some idea, whether things are working
as expected. Then, you can use the observation in (2),
to verify it's drained. But only if you used step (1),
to actually verify the USB port is powered by +5VSB.
Once you've proved (1), then you can use (2) thereafter,
for that machine. The reading lamp has become the equivalent
of the Asus motherboard green status LED.

http://www.amazon.com/Gino-Housing-Flexible-Gooseneck-Reading/dp/B00ARBTMGA

Paul

 

[philo]

Remove the power cord from the PSU end if you want to be most safe.
There may even be a LED on the motherboard to let you know when it has
power so wait until it goes dark within a couple minutes. If there is
no LED, just wait 5 minutes (pressing the Power button as suggested by
philo won't work to drain the caps because the PSU won't power up when
it's not connected to a powered source). Then do your inside work.

Yes it does, even though I said I don't bother to do so...I actually do
quite often. As soon as I push the button, I can see the on-board LED's
go out. Waiting five minutes may or may not drain the caps.

 

[VanguardLH]

Yes it does, even though I said I don't bother to do so...I actually do
quite often. As soon as I push the button, I can see the on-board LED's
go out. Waiting five minutes may or may not drain the caps.

Whether you push the *momentary* Power button switch or not, the PSU
will continue providing its 5-volt standby power. So the PSU will drain
when you pull its power cord no matter what you do with the Power
button.

Perhaps I wasn't clear. Pushing the Power button will not power up the
PSU when it's not connected to a powered source. Pushing the Power
button will not make its caps drain faster, especially since it is just
a momentary contact. The PSU continuing to provide 5 VSB to the mobo
after it loses its power source is what drains its caps.

 

[Paul]

Perhaps I wasn't clear. Pushing the Power button will not power up the
PSU when it's not connected to a powered source. Pushing the Power
button will not make its caps drain faster, especially since it is just
a momentary contact. The PSU continuing to provide 5 VSB to the mobo
after it loses its power source is what drains its caps.

Actually, it will. Trust me

You can do the circuit analysis yourself, right here.

http://www.pavouk.org/hw/en_atxps.html

The power supply is divided into two halves.

The switcher in the lower left hand corner, makes +5VSB.

The switcher in the upper center, makes 3.3V/5V/12V... (main rails).

These circuits share the energy stored in C5 and C6.

The *momentary* power switch, is monitored by a chip
on the motherboard. The monitoring can be done, for as long
as +5VSB is available. In other words, the motherboard
can convert a "pulse" on the power switch, to a solid level
on PS_ON#, for as long as +5VSB is available. It'll release, if
+5VSB goes away.

*******

So let's say, the computer was not running the OS, the OS is
shut down. The fan stopped spinning. No voltage on 3.3V/5V/12V.

With your multimeter, you'll see +5VSB is still running.
If your computer is "Sleeping", that's what keeps the RAM powered.

At the same time (checking the Pavouk schematic), capacitor C5 and
C6 are *fully* charged. The bridge rectifier keeps them charged
And using 1/2CV**2, they hold a pretty large number of joules.
Dangerous stuff. That's how the +5VSB circuit can run for
30 seconds after you pull the power plug. C5 and C6 make
it possible.

Now, as Philo says, we pull the power cord. The lower left
switcher is still running. For the next 30 seconds (or so),
+5VSB is available. The motherboard chip can convert a
*momentary* press of the front power button, into a
steady ON level on the PS_ON# signal to the PSU. The
PSU then engages the main switcher (upper center section).
The switcher starts to run, and proceeds to *quickly*
drain the 300V or so on C5 and C6. Eventually, the main
switcher shuts off, when the capacitor voltage drops too low.
The +5VSB might run for a couple seconds longer. Eventually,
both switchers give up. All switchers have some minimum voltage,
below which, they won't run.

At this point, you can pull or insert RAM, pull or insert PCI cards.
because both 3.3V/5V/12V... and +5VSB, are dead.

A residual charge is still left on C5 and C6. Resistors R2 and
R3 are bleeder resistors. After a few R*C time constants,
C5 and C6 terminals will be safe to touch. Only an idiot
would touch them of course, without further attempts to
discharge them with a standalone bleeder (one where you've
verified the resistor didn't blow while you were using it).
Never trust R2 and R3 to be functional. Always assume they
burned out.

*******

*Never* stick a screwdriver across the terminals of C5 and C6.
Always use a bleeder resistor. An appropriately sized bleeder
resistor. (You must keep several sizes at your desk, and
reach for the right one.) The reason for this warning, is
the noise it makes. I used to use a screwdriver when working
on tube equipment, but no longer do that. It's dumb. I wouldn't
dare use a screwdriver on C5 or C6, or, on the capacitor
inside a microwave oven. My hearing is bad enough as it is.

When the capacitor in a microwave oven at work arced over,
inside the chassis, and I was standing next to it, I could
not hear for ten minutes! That's how loud it was. Strangely,
the microwave oven was unharmed. (The reason the microwave
arced over, is my co-workers ran a ton of bags of "buttered
popcorn" through that sucker, until the PCB was coated with
an aerosol water vapor, oil, and salt. Even with a conformal
coating and other forms of insulation, sooner or later *kaboom*.
Scared the crap out of me! This is an indication of how much
energy is stored in there. A lot. It uses a 5000V oil-filled
capacitor. Something it's worth working out the 1/2*C*V**2
equation for. V_squared term is 25,000,000. The voltage makes
a big difference to the joules. Even if the cap was 1uF,
that would be 12.5 joules. Nasty. Always work out the joules,
to understand what the shock wave will be like The
capacitor in the ATX supply, makes up for the lower V_Squared
terms (90,000 or so), by using a larger cap (470 microfarads).

Paul

 

[philo]

Whether you push the *momentary* Power button switch or not, the PSU
will continue providing its 5-volt standby power. So the PSU will drain
when you pull its power cord no matter what you do with the Power
button.

Perhaps I wasn't clear. Pushing the Power button will not power up the
PSU when it's not connected to a powered source. Pushing the Power
button will not make its caps drain faster, especially since it is just
a momentary contact. The PSU continuing to provide 5 VSB to the mobo
after it loses its power source is what drains its caps.

Nope, you are guessing wrong... I have machines on my bench on almost a
daily basis and do this all the time...believe me.

 

[VanguardLH]

Actually, it will. Trust me

You can do the circuit analysis yourself, right here.

http://www.pavouk.org/hw/en_atxps.html

The power supply is divided into two halves.

The switcher in the lower left hand corner, makes +5VSB.

The switcher in the upper center, makes 3.3V/5V/12V... (main rails).

These circuits share the energy stored in C5 and C6.

The *momentary* power switch, is monitored by a chip
on the motherboard. The monitoring can be done, for as long
as +5VSB is available. In other words, the motherboard
can convert a "pulse" on the power switch, to a solid level
on PS_ON#, for as long as +5VSB is available. It'll release, if
+5VSB goes away.

*******

So let's say, the computer was not running the OS, the OS is
shut down. The fan stopped spinning. No voltage on 3.3V/5V/12V.

With your multimeter, you'll see +5VSB is still running.
If your computer is "Sleeping", that's what keeps the RAM powered.

At the same time (checking the Pavouk schematic), capacitor C5 and
C6 are *fully* charged. The bridge rectifier keeps them charged
And using 1/2CV**2, they hold a pretty large number of joules.
Dangerous stuff. That's how the +5VSB circuit can run for
30 seconds after you pull the power plug. C5 and C6 make
it possible.

Now, as Philo says, we pull the power cord. The lower left
switcher is still running. For the next 30 seconds (or so),
+5VSB is available. The motherboard chip can convert a
*momentary* press of the front power button, into a
steady ON level on the PS_ON# signal to the PSU. The
PSU then engages the main switcher (upper center section).
The switcher starts to run, and proceeds to *quickly*
drain the 300V or so on C5 and C6. Eventually, the main
switcher shuts off, when the capacitor voltage drops too low.
The +5VSB might run for a couple seconds longer. Eventually,
both switchers give up. All switchers have some minimum voltage,
below which, they won't run.

At this point, you can pull or insert RAM, pull or insert PCI cards.
because both 3.3V/5V/12V... and +5VSB, are dead.

A residual charge is still left on C5 and C6. Resistors R2 and
R3 are bleeder resistors. After a few R*C time constants,
C5 and C6 terminals will be safe to touch. Only an idiot
would touch them of course, without further attempts to
discharge them with a standalone bleeder (one where you've
verified the resistor didn't blow while you were using it).
Never trust R2 and R3 to be functional. Always assume they
burned out.

*******

*Never* stick a screwdriver across the terminals of C5 and C6.
Always use a bleeder resistor. An appropriately sized bleeder
resistor. (You must keep several sizes at your desk, and
reach for the right one.) The reason for this warning, is
the noise it makes. I used to use a screwdriver when working
on tube equipment, but no longer do that. It's dumb. I wouldn't
dare use a screwdriver on C5 or C6, or, on the capacitor
inside a microwave oven. My hearing is bad enough as it is.

When the capacitor in a microwave oven at work arced over,
inside the chassis, and I was standing next to it, I could
not hear for ten minutes! That's how loud it was. Strangely,
the microwave oven was unharmed. (The reason the microwave
arced over, is my co-workers ran a ton of bags of "buttered
popcorn" through that sucker, until the PCB was coated with
an aerosol water vapor, oil, and salt. Even with a conformal
coating and other forms of insulation, sooner or later *kaboom*.
Scared the crap out of me! This is an indication of how much
energy is stored in there. A lot. It uses a 5000V oil-filled
capacitor. Something it's worth working out the 1/2*C*V**2
equation for. V_squared term is 25,000,000. The voltage makes
a big difference to the joules. Even if the cap was 1uF,
that would be 12.5 joules. Nasty. Always work out the joules,
to understand what the shock wave will be like The
capacitor in the ATX supply, makes up for the lower V_Squared
terms (90,000 or so), by using a larger cap (470 microfarads).

Paul

The power cord gets pulled from the PSU. The PSU has no input power.
There is nothing on the input side of the T2, T3, and T6 transformers.
A charge on C5 and C6 isn't going to get through the transformers. We
were talking about when it was safe to work inside the computer, like
for adding or removing memory, drives, etc, and not working inside the
PSU. While the caps drain and maybe there is a slow reduction in field
strength on the primary coil, I would think that a slowly collapsing
primary field while the caps drain would not induce much usable current
on the secondary side.

As to when you could start working inside the computer after yanking the
power cord to the PSU, how long before the PSU's outputs (+5V, +12V) go
dead? Are you saying an ATX power supply continues supplying power on
its regulated outputs after it has been turned off or lost its power
source? While the ATX specification states upon input power
interruption the PSU should maintain output regulation for a minimum of
17 ms at the maximum rated load. I don't recall the ATX spec says how
long the outputs have some voltage after the PSU is turned off or after
its power source is removed. How is a remaining charge on C5 and C6
keeping the outputs up? How long the regulated outputs remain up looks
to depend on the caps across those outputs. That is, the input-side
caps don't effect the discharge of the output-side caps.

There is a fall off in voltage in the outputs dependent on the RC
circuit design but that's for the outputs. Not only do you have the
resistors across the output caps inside the PSU but there is also the
device load on those outputs which makes they drop pretty fast. When
the PSU is turned off or loses power, I don't see how the input-side
caps are going to keep the outputs going, especially across those
transformers. The input caps are to regulate the input side to smooth
out the input power or accommodate very short outages (like 30 ms?).

It's the +5VSB output that's of concern as to how long it's alive and
also producing its regulated output (above 4.5V) after the PSU's power
source has been removed. It's late now but my eyes see only 2 loads on
the C34 cap across the +5VSB output: the R23 4K resistor and the load by
the motherboard's circuit(s) to which the +5VSB output is connected.
With PS-On floating (probably pulled up by resistor to the mobo's +5V
line) and after pulling the power cord from the PSU, it would be the
mobo's logic draining the C34 output cap. By pressing the Power button
and pulling PS-ON to ground, the R23 4K resistor would get added across
the C34 output cap. So does the addition of the 4K resistor across the
app really affect the voltage drop-off time (but severely shortening
it)?

I don't know what the mobo circuit looks like to know what is its load
on the +5VSB output to know if the draw across the 4K resistor is a
significant added draw. At the start and with +5V immediately
available, the draw across the 4K resistor is only 1.25 mA. What is the
draw on the +5VSB output by the mobo's power-on circuit and "other
peripherals"? 5VSB is not only used for the mobo's power-on circuit but
can also be used by USB ports. A mobo might use th +5V rail for
internal USB ports but uses +5VSB to power the backpanel USB ports;
else, how could a USB-attached keyboard or mouse waken the computer (the
BIOS setting along with a jumper for "power by mouse/keyboard")? The
ATX specs recommends the +5VSB output should handle up to 2 A, or 2000
mA, so 1.25 mA seems a pretty small addition to the potential load.

So besides keeping the Power button pressed after yanking the power cord
to shorten how long before the +5VSB output drops (but doesn't affect
when the other outputs drop), it seems you could also slide around the
mouse and press keys on the keyboard if their USB ports are using +5VSB.
Yet I think the time difference would be small doing any of those.

 

[Paul]

The power cord gets pulled from the PSU. The PSU has no input power.
There is nothing on the input side of the T2, T3, and T6 transformers.
A charge on C5 and C6 isn't going to get through the transformers. We
were talking about when it was safe to work inside the computer, like
for adding or removing memory, drives, etc, and not working inside the
PSU. While the caps drain and maybe there is a slow reduction in field
strength on the primary coil, I would think that a slowly collapsing
primary field while the caps drain would not induce much usable current
on the secondary side.

As to when you could start working inside the computer after yanking the
power cord to the PSU, how long before the PSU's outputs (+5V, +12V) go
dead? Are you saying an ATX power supply continues supplying power on
its regulated outputs after it has been turned off or lost its power
source? While the ATX specification states upon input power
interruption the PSU should maintain output regulation for a minimum of
17 ms at the maximum rated load. I don't recall the ATX spec says how
long the outputs have some voltage after the PSU is turned off or after
its power source is removed. How is a remaining charge on C5 and C6
keeping the outputs up? How long the regulated outputs remain up looks
to depend on the caps across those outputs. That is, the input-side
caps don't effect the discharge of the output-side caps.

There is a fall off in voltage in the outputs dependent on the RC
circuit design but that's for the outputs. Not only do you have the
resistors across the output caps inside the PSU but there is also the
device load on those outputs which makes they drop pretty fast. When
the PSU is turned off or loses power, I don't see how the input-side
caps are going to keep the outputs going, especially across those
transformers. The input caps are to regulate the input side to smooth
out the input power or accommodate very short outages (like 30 ms?).

It's the +5VSB output that's of concern as to how long it's alive and
also producing its regulated output (above 4.5V) after the PSU's power
source has been removed. It's late now but my eyes see only 2 loads on
the C34 cap across the +5VSB output: the R23 4K resistor and the load by
the motherboard's circuit(s) to which the +5VSB output is connected.
With PS-On floating (probably pulled up by resistor to the mobo's +5V
line) and after pulling the power cord from the PSU, it would be the
mobo's logic draining the C34 output cap. By pressing the Power button
and pulling PS-ON to ground, the R23 4K resistor would get added across
the C34 output cap. So does the addition of the 4K resistor across the
app really affect the voltage drop-off time (but severely shortening
it)?

I don't know what the mobo circuit looks like to know what is its load
on the +5VSB output to know if the draw across the 4K resistor is a
significant added draw. At the start and with +5V immediately
available, the draw across the 4K resistor is only 1.25 mA. What is the
draw on the +5VSB output by the mobo's power-on circuit and "other
peripherals"? 5VSB is not only used for the mobo's power-on circuit but
can also be used by USB ports. A mobo might use th +5V rail for
internal USB ports but uses +5VSB to power the backpanel USB ports;
else, how could a USB-attached keyboard or mouse waken the computer (the
BIOS setting along with a jumper for "power by mouse/keyboard")? The
ATX specs recommends the +5VSB output should handle up to 2 A, or 2000
mA, so 1.25 mA seems a pretty small addition to the potential load.

So besides keeping the Power button pressed after yanking the power cord
to shorten how long before the +5VSB output drops (but doesn't affect
when the other outputs drop), it seems you could also slide around the
mouse and press keys on the keyboard if their USB ports are using +5VSB.
Yet I think the time difference would be small doing any of those.

The two switchers have different operating conditions:

1) C5 and C6 at 300V. Need cord plugged in and switch on back
to ON to charge them.
2) +5VSB switcher runs immediately, if voltage on C5 and C6 is
sufficient.
3) 3.3/5/12V main switcher only runs, if C5 and C6 are sufficient,
and PS_ON# is asserted. To assert PS_ON#, the motherboard drives it.
The motherboard needs +5VSB as well as momentary power press on
the front, to assert PS_ON#. And that is what Philo is doing.

Pressing the power button, accelerates the rate of drain,
as now, both switchers are rinning.

Without assistance, the +5VSB will drain it in 30 seconds.
The Asus green LED makes that easy to monitor, as to when
it is drained. Other motherboards, you don't get a green
monitor LED.

If you press the power button, causing the main switcher
to come on, it'll drain C5 and C6 in no time. Once the
voltage is low enough, that both switchers are off,
then it is safe to work inside the PC. You can't work
inside the ATX supply itself (take off cover), unless
you know about the location and danger of C5 and C6,
and have taken precautions with an external bleeder
resistor. I have to mention that, in case someone
pulls an ATX supply, removes the protective casing
and starts handling the PCB. While most of the
time, it isn't dangerous, you still have to assume
it is dangerous. For your own safety. If the internal
bleeders fail, there could still be 300VDC present
in there.

Paul