Archived from groups: alt.comp.periphs.mainboard.asus (
More info?)
In article <41551f76$0$20248$cc9e4d1f@news-text.dial.pipex.com>, Mike
<cotech.shoes@ntlworld.com> wrote:
> My mobo just passed away, it collapsed halfway through booting, it
> bravely made several more attempts to boot but was getting weaker each
> time but now, alas, it is gone :-( I laid it out on the slab to do a
> postmortem and discovered the cause.
> In the corner between the fixing screw and the ram socket resides
> component Q11, an NEC K2941 81L, which Google tells me is a n-Channel
> MOSFET, which fell off when I touched it. The component isn't charred
> and all else seems OK.
> What are my chances of resurrecting the mobo if I replace this? Can the
> component be easily checked with a meter? Does anyone know of a supplier
> in the UK that stocks a K2941? Is there an equivalent that's more
> readily available?
> All help gratefully received.
>
> Mike (spamcheck - remove my footwear to reply)
This web page shows the chip controlling the VIO circuit in the
upper right hand corner of the board, is the HIP6019
http://tipperlinne.com/p2b-dsvio.htm
If you go to
http://www.intersil.com/data/fn/fn4587.pdf and
download the datasheet for the HIP6019, then go to page 15, you
can see a schematic for a typical motherboard circuit.
As near as I can tell by looking at the picture of the motherboard,
in the P2B-DS user manual, there are two voltage regulator chips.
There is a smaller one next to the ATX 20 pin connector. The other
one, a HIP6019, is between the two slots. The HIP6019 has two switching
regulators. The main one uses two MOSFETs and drives Vcore on one of the
processor slots - possibly the slot to the right of the regulator chip.
The other switching regulator can be used to control the VIO regulator.
If you look at page 15, and look to the left hand side of the chip,
there are components Q3, CR2, L2, C19-23. These are the main power
handling components for VIO. Q3 might correspond to the 2SK2941
used on the P2B-DS. The reason CR2 is used, is a switching regulator
doesn't need to use two MOSFETs - two MOSFETs improves efficiency,
but in fact a diode can be used in place of the lower MOSFET. That
is what CR2 is doing in that circuit - it means the HIP6019
doesn't need another gate drive circuit, which would take extra pins
to drive and monitor.
In terms of fault modes, if CR2 fails, it will likely fail shorted.
This will cause a lot of current to flow through Q3. So, before
soldering Q3 back into place, you would want to use your ohmmeter
across the equiv of CR2, to see if it is shorted. If it is, you'll
have to find a replacement for the MBR2535 (possibly onsemi.com
makes those - the MBR number is former Motorola, and onsemi got the
diode business from them). The equivalent of C19-23 are the four
capacitors next to the floppy connector. Examine those for bulging
or leaking and replace them before soldering the MOSFET back, if
they show signs of having failed. If they go ohmic, they can draw
a large current through the toroid and burn the MOSFET.
So, those are the things to look for. Check the equivalents of
Q3, CR2, L2, C19-23 for signs of failure, and if they are all
visually intact, and the node where Q3, CR2, and L2 join isn't
shorted to ground, then it is probably safe to power it back up
after the MOSFET is reinstalled.
To check the MOSFET, you would need two supplies, one for Vgs and
one for Vds. Page 6 of this Berkeley lab, shows how a student
might measure a MOSFET characteristic.
http://www-inst.eecs.berkeley.edu/~ee43/f03/labs/LAB7.PDF
I would use a 9V transistor radio battery and a
couple of 1K ohm resistors (power rating doesn't matter). I
would run two test cases. One is with the gate at +9V. The
other is with the gate at 0V (ground). In one case, using a
voltmeter, you'll find the full nine volts between the +9V
terminal and the D (drain) terminal. In the other case, there
will be roughly zero volts or close to it, between +9V and D,
and that is the case when the MOSFET is switched off and is
drawing no current through the drain load resistor. As long
as the transistor exhibits two states of operation (i.e. the
test cases do different things), the transistor is ready to
be reused. (Note: This test circuit is applicable to full
swing MOSFETs, and may not be the best for a MOSFET with
"logic" gate inputs, in which case a 5V supply should be used
on the gate test input. AFAIK, the HIP6019 uses full swing
MOSFETs. With "logic" gate MOSFETs, it pays to check the
gate breakdown voltage of the device, before selecting a
gate test voltage to use.)
+9V <--------Measure +9V
| with |
Resistor ---voltmeter Resistor
| / |
|--+ D <--/ |--+ D
+9V ---Resistor---G--| +--Resistor---G--|
|--+ S | |--+ S
| | |
GND GND GND
What will be the hardest part of doing the above test circuit ?
Figuring out the pinout of the MOSFET
You may be able to
deduce this by looking at the layout of the motherboard where the
MOSFET is connected. Compare the layout to the schematic in the
HIP6019 datasheet, and that may give you a hint as to which pin
is which. Compare what you see to my best guess, which is:
_______
G---| | |
D---| |D|<--- Metal tab joined to center pin
S---| | |
-------- Device with metal tab down,
plastic body facing up, device
resting flat on your table.
I'm sorry I cannot help with datasheets for devices - I've tried
looking, but transistor companies change hands so often, that
old datasheets are impossible to find. Even a datasheet I found
on web.archive.org for a different device a while back, is no
longer available. So much for archives :-(
HTH,
Paul
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