P4C800-E Deluxe blown USB-controllers

  • Thread starter Thread starter Kai Bohun
  • Start date Start date
K

Kai Bohun

Can this be true: now the 4th P4C800-E MoBo within 6 Month with blown
USB-Controllers!!! Without warning all former properly funktional hardware
can't be recognized.

I've googled - looks like I'm not alone.

Any help/advise available here, folks?

Kai
 
Yeah - It's happened to many. The cause has been attributed to static charge
blowing components on the board - mainly through the USB ports, but it not a
100% certainty.
What were you doing when it happened?
Were you using the front USB ports?
What kind of case do you have?

All I can tell you for sure is that the ICH5R chip on the board is partly
fried. It may be visibly burned.
There is nothing you can do to make the USB ports work again.
There are precautions you can take to avoid a repeat. ie.. disable the
on-board USB ports and install a PCI USB2 card
It's not the way it should be, but someone (Asus or Intel) screwed up
somewhere.
You can have your board replaced if you ship it back.

Good luck
Dave
 
Dave, thanks to you!

To answer your question: I tried to connect an external Hauppauge WinTV
PVR-USB2 TV-Card.

Inbetween ASUS responded to my emails. They're suggesting the use of another
power supply with more "power" on the +5V wire (with the next MoBo,
hahaha!).

But how much? Mine has 30A DC Output on +5V, so the technical paper says,
but I never messured it. (But how to messure it?!)

Kai
 
Kai Bohun said:
Dave, thanks to you!

To answer your question: I tried to connect an external Hauppauge WinTV
PVR-USB2 TV-Card.

Inbetween ASUS responded to my emails. They're suggesting the use of
another
power supply with more "power" on the +5V wire (with the next MoBo,
hahaha!).

But how much? Mine has 30A DC Output on +5V, so the technical paper says,
but I never messured it. (But how to messure it?!)

Here is a post from PAUL from a previous thread. Lots of good info:

Dave.
********************************************************

Pinout is on page 19:
http://www.formfactors.org/developer/specs/atx/atx2_1.pdf

(pin 1) (pin 11)
+3.3VDC +3.3VDC
+3.3VDC -12VDC
COM COM
+5VDC PS_ON# <--- connect to COM to start PSU.
COM COM <--- Plastic lock latch is
+5VDC COM <--- next to these pins.
COM COM
PWR_OK -5VDC
+5VSB +5VDC
+12VDC +5VDC

You have six voltages to test:

+5VSB - standby supply for sleeping computer, also powers start up
circuit on PSU. If not at full 5 volts, PSU may not work
properly, as supervisor cct in PSU will not be properly powered.
For the PSU to be completely turned off, PS_ON# should float
near the same +5VSB voltage. When PS_ON# is grounded (to COM),
the supply runs.
+3.3VDC - Used by PCI/AGP cards for chip power. May be used for
memory power, via linear regulation.
+5VDC - Used to power Vcore switching regulator on Asus AthlonXP boards.
Video card power, disk drive power.
+12VDC - Used to power Vcore switching regulator on P4/Athlon64 boards.
Video card power, disk drive power.
-5V - Historically, used for ECL or really old tech, three rail
DRAM memories. Maybe some kind of video card RAMDAC.
Typically not used today. Possibly used in linear (op amp)
circuits.
-12V - On Asus boards, can be used for RS-232 converter chip. Also
potentially used for op amps in linear regulators. It is hard
to say which negative rail Asus would use on their op-amp
circuits, as negative swings are not really required. In any
case 0.1 amps of consumption on a motherboard is probably
a reasonable estimate.

A quick test, is to connect PS_ON# to a COM pin. The plastic lock
latch will help guide you to the right cluster of pins. The purpose
of a quick test, is just to check that the PSU fan starts to spin.
That would only prove that the supervisory circuit is capable of
turning on, in a situation where, say, a motherboard refuses to start.

To quantify operation, you need a load on the PSU. One choice
is to apply the minimum load necessary for proper regulation.
Some supplies have minimum current loads specified on the label
on the side of the supply. If you draw that minimum current, then
the output is guaranteed to be within 5% of the proper voltage.
One would hope, under those conditions, it is a lot closer than
that.

To measure the output, you will need a voltmeter. Home units are
typically not that good, and my $100 CDN meter is good to about
1.5% to 2% accuracy, depending on range used. That means there is
room for doubt in the readings, to those precentages.

Drawing the minimum supply current ensures the outputs will be in
spec. A second test would be a load test, using a representative
load. Using a load like this, proves the power supply will be
able to run a real motherboard - the load circuit takes the risk
out of having to connect a real motherboard, to prove the power
supply works.

Here are some representative loads measured on systems here:

A7N8X (3200+) P4C800-E (2.8GHz)
3 DIMMs 4 DIMMs
+3.3V 5.5A 13.75A
+5.0V 16 + 5.5A 0.6 + 5.5A
+12.0V 0.5 + 0.9A 6.4 + 0.9A

The added currents shown above (5.5A and 0.9A) are for a 9800pro
AGP video card while gaming. The 6.4A for the P4, is 6 amps for the
processor and 0.4A for the processor fan.

To build a load tester, you will need a cable harness (snip up
a 20 to 24 pin adapter cable - even a 24 pin connector can be
used for the project, as there is only one way to plug a 20 pin
power supply into such a connector, and have all the pins
connected). You will also need resistors.

To work out a load resistance:

V/I = R_load = 3.3V/5.5A = 0.6 ohms

V*I = power_dissipation = 3.3V*5.5A = 18.15 watts
Can also be calculated as V*V/R, if only V and R are known.

More than one resistor can be used to build a load. Two 0.3 ohm
resistors put in series, would give a 0.6 ohm load. Two 1.2 ohm
resistors put in parallel, would give a 0.6 ohm load. The
power dissipated in each case, is now split between the two
resistors, so two smaller resistors can be used to make a larger
resistor. In the third example below, a square matrix of resistors
of equal value, can be used. In that case, each resistor only
needs to handle 1/9th of the total power to be dissipated.

----- 0.3 ------ 0.3 ------ = 0.6 ohms

---+--- 1.2 ---+--- = 0.6 ohms = R1*R2/(R1+R2)
| |
+--- 1.2 ---+

----+--- 0.6 ---+--- 0.6 ---+--- 0.6 ---+----- = 0.6 ohms
| |
+--- 0.6 ---+--- 0.6 ---+--- 0.6 ---+
| |
+--- 0.6 ---+--- 0.6 ---+--- 0.6 ---+

(Hint - the parallel configuration may be easier to construct
- a ladder constructed of stiff copper house wiring
could be used to mount resistors in parallel.)

There are many styles of power resistors. In the two examples
below, one is designed for free-standing operation. The other
one is intended for use with a heat sink (but no thermal data
is given?). In the load examples in the table above, the 5V
at 21.5 amp load would be the hardest one to cool, at 108 watts.
For some resistors I've been looking at, they appear to be
intended to dissipate the rated power without forced air, but
if you run a fan over them, you might not burn yourself on
them.

There are some candidate resistors here. To reach our target
resistances using these products, in some cases, multiple
resistors are needed (in parallel) to get the resistance low
enough. In other cases, multiple resistors are needed to get
enough power handling capability. As the Digikey catalog
doesn't have all possible resistor values, sometimes a higher
power resistor must be substituted to get a necessary resistor
value. (I.e. The selection process is a mess. I did all the
math necessary, to build my load box, while standing in front
of the power resistor display rack at my local electronics
store. That is because you never know what resistors will be
in stock...)

FVT series: The FVT are fixed value resistors. Vitreous enamel
looks to be a high temperature material, so you can burn the
hell out of these.

http://dkc3.digikey.com/PDF/T052/1075.pdf

The TMC ones (second item from top of page) have a casing suitable
for use with a heatsink. Now, there simply isn't enough data
here, to design with these and understand what you are doing.
(No thermal resistance data, for example.) How many of these
could you bolt to one heatsink ?

http://dkc3.digikey.com/PDF/T052/1076.pdf
==> http://www.heiresistors.com/aluminum.htm

A7N8X (3200+) P4C800-E (2.8GHz)
3 DIMMs 4 DIMMs
+3.3V 5.5A => 0.6ohm 18W 13.75A => 0.24ohm 45W
+5.0V 21.5A => 0.23ohm 108W 6.1A => 0.82ohm 31W
+12.0V 1.4A => 8.6ohm 17W 7.3A => 1.64ohm 88W
totals 143W 164W

now select some resistor combinations...

A7N8X (3200+) P4C800-E (2.8GHz)
3 DIMMs 4 DIMMs
+3.3V 5.5A => 0.6ohm 18W 13.75A => 0.24ohm 45W
(3) FVT25-2.0-ND in (4) FVT50-1.0-ND in
parallel $9.93 parallel $19.16

+5.0V 21.5A => 0.23ohm 108W 6.1A => 0.82ohm 31W
(4) FVT50-1.0-ND in (1) FVT50-1.0-ND parallel with
parallel $19.16 (1) FVT50-5.0-ND $9.58

+12.0V 1.4A => 8.6ohm 17W 7.3A => 1.64ohm 88W
(1) FVT25-10-ND $3.28 (1) FVT100-2.0-ND parallel with
(1) FVT25-10-ND $11.49

-5V (1) FVT25-25-ND $3.28 (1) FVT25-25-ND $3.28
will draw 0.2 amps will draw 0.2 amps

-12V (1) FVT50-50-ND $4.50 (1) FVT50-50-ND $4.50
will draw 0.24 amps will draw 0.24 amps

+5VSB (1) FVT25-10-ND $3.28 (1) FVT25-10-ND $3.28
will draw 0.5 amps will draw 0.5 amps

Total $43 for resistors $51 for resistors
+ 80mm fan for cooling 80mm fan for cooling

Wiring the A7N8X example:

+3.3 ----+----+----+
| | |
2ohm 2ohm 2ohm
| | |
COM ----+----+----+

+5.0 ----+----+----+----+
| | | |
1ohm 1ohm 1ohm 1ohm
| | | |
COM ----+----+----+----+

+12.0 ----+-----------+
| | (red)
10ohm 80mm fan
| | (black)
COM ----+-----------+

-5V ----+
|
25ohm
|
COM ----+

-12V ----+
|
50ohm
|
COM ----+

+5VSB ----+ PS_ON+ ----+
| |
10ohm (toggle switch)
| |
COM ----+ COM ----+

Note: This is just to give you some ideas for how to construct
your own PSU load box. If you can find cheaper resistors, you
will be able to use more of them, and perhaps have them run a
bit cooler. The FVT100-2.0-ND resistor on the right hand example
is going to get pretty hot. When using the 80mm fan, it might
be a good idea to build a "wind tunnel" with sheet metal, to
shape the airflow around the resistors.

You will also want some place to probe with your volt meter.
So, if building vertical ladders with copper wire, screwed down
to a wooden base, leave enough excess wire on the ends of the
ladder, so you can probe with the voltmeter.

When testing the PSU, run it with the representative load for
a couple of hours, and see if all voltages are still within
5% of the stated values.

There is no point testing the "full power" rating of the
supply, because many commonly available power supplies will
disappear in a cloud of smoke if you do that. The idea of a
load test, is to present a representative load, similar to
the motherboard you will be using.

The power supply test products that have LED indicators on
them, could be using a window comparitor circuit. For example,
if one analog comparator checks for 11.4V, and another one checks
for 12.6V, using some logic gates, those comparators can be
used to check that the voltage is within the "window" of
11.4V <= V_out <= 12.6V. Under those conditions, the LED comes
on. If the voltage is outside the window, the LED goes off.
You can get the same information by simply measuring the output
with a voltmeter.

P.S. I'm not responsible if you burn yourself on the resistors!
They can get really hot (resistors rated to 350C degrees), so
don't be poking them with your fingers. As long as there is a
decent airflow with the cooling fan, that should prevent them
from getting all the way to 350C. Make sure there is some
separation between the resistor and any combustable material.
Using bare copper wire to hook up the resistors will avoid
the embarrassment of having the wire insulation start to burn.
And, above all, don't run this gadget unattended - turn it off
and unplug it, if you leave the room. Think of having to explain
to the insurance agent about power supply testers :-)

Is it worth $100 for test gear (tester+voltmeter), to test a
$30 power supply ? Only you can answer that. I built my tester,
just to make sure I don't kill any expensive motherboards. My
tester draws less current than the examples above.
 
Back
Top