Clear BIOS on Asus &Z170K

short the battery terminals while it is removed.

does capslock LED respond to capslock key?

Thanks Andy :-)

Found the clear RTC RAM pins, so took battery out, put a drive jumper across the pins (good job I kept them, I knew they'd come in useful one day!), took jumper off, put battery back, and it booted first time.

By then I'd moved everything to the Z170P and got that working so I'll leave it for now. I need spare machines, if I treat them like the Meccano set I never had I'm bound to break things now and then!

Andy Burns snipped-for-privacy@andyburns.uk wrote in news:lu4rgmFb19U1 @mid.individual.net:

I'd be surprised if that worked as there's usually a series diode to prevent reverse charging of the battery under fault conditions (or that's the way it certainly used to be done).

Note that Intel reference schematics are not known for their "realism" and real circuits have slightly different details. For example, the Port Angeles on this particular schematic is pure imagination and not real (not a proper SuperIO). Still, you can use this little circuit, to have a discussion about the Clear CMOS jumper and what good (if any), shorting the battery terminal would have.

[Picture]

When running on battery, the switching transients from the ripple counter in the Motorola RTC as small. The 1.0uF capacitor is there for this purpose. The 1K ohm resistor at the battery, is there to limit the discharge rate from the battery, to at most

3mA. But it also ruins the output impedance of the battery in a sense, and the 1uF capacitor restores the electrical performance. When the RTC is receiving write operations, the circuit is powered at that point, from something "derived" from +5VSB and the impedance of that source is a lot lower than this 1K resistor thing. When the circuit is quiet and not much current is drawn, the source impedance does not need to be all that low.

If you short the CMOS jumper while the system is running, the Intel schematic has us believe there is a 4.7K resistor to avoid a disastrous short. Real schematics don't have the 4.7K resistor. The jumper instead is right to ground. A large current flows through the top half of the BAT54 dual diode, burning it so you can't read the legend on top. The BAT54 is a Schottky with low forward voltage drop. The path between "2" and "3" burns if you left the power on while doing this.

I recommend pulling out the PC plug while clearing CMOS, just to ensure your BAT54 is not cooked.

There is an alternate way to clear CMOS, on the more modern chipsets. Intel put an actual RESET signal on the PCH for this. It's probably on the 3.0V powered "pure CMOS" logic section with no ESD protection diodes. This could have relatively low drain characteristics for a logic input. Yet, real circuit board designers *still* don't use this. They've looked at the characteristics and decided the "old way is best", even if one of the fault modes is a burned BAT54.

Anyway, the audience here is good at circuit analysis, and don't need my help to figure out the pathology. a lot of boards use that circuit, but not all of them.

Another nice aspect of this circuit, is on a number of occasions (maybe a half dozen times), the instructions in the manual were WRONG for this thing :-) You would open the box, and a sheet of paper would float out. This is the Addendum to replace the page with the Clear CMOS instructions. Don't lose that sheet :-) The instructions help you avoid the pathology.

A digital watch, draws 2uA of current while running. The RTC on the motherboard, draws 10uA. Both circuits use a 32768 Hz quartz crystal. The 10uA load, means the CMOS CR2032 last for a bit less than three years, if you store a PC in the junk room with no AC power. The fresh battery voltage is a bit higher than

3.0V. The battery is good down to 2.3V and continues to run the RTC at 2.3V. The BAT54 has 0.3V of drop at low current. The PCH stops working for the RTC function, at 2.0V. But the battery stops working at 2.3V. Some motherboards refuse to start when the battery hits 0V, and this is possibly because VBAT is tied to some pin on the SuperIO. And if that pin is logic 0, that has something to do with stopping the board. Not every board stops at 0V on CR2032, but a few do this. A lot of boards will run, and they set the BIOS controls to "default" values.

Paul

Paul snipped-for-privacy@needed.invalid wrote in news:vlqe3k$3sb0a$ snipped-for-privacy@dont-email.me:

TLDR

Can you please state succinctly the point you are trying to make.

I think he its trying to say 'taking the battery out and shorting its terminals may not work (because there is a diode in there) and may cause damage (because there might not be a resistor or diode in there)'.

There are usually some shortable header pins or in extremis, simply removing the battery and letting the data decay from the CMOS should sort it all out.

Of course these days the data is in flash, and wont decay ever. Only the RTC will stop without a battery

Yes, there is, at minimum, a two pin header. A three pin header is used when the company desires a parking location for the provided jumper plug.

There is also a version with a switch on the back of the PC, but the circuit behind that one is different.

Shorting the two pin header, is for clearing the CMOS. On "real" designs, the 4.7K resistor shown in the Intel demo schematic is not there, and you are shorting the 1uF cap directly to ground. The CR2032 is not damaged, as there is the 1K ohm limiting resistor.

However, if you short the header while the +5VSB is available (and it is available even when fans are not spinning!), you can draw an excess of current through the top diode in the dual diode package. (Not seen on that schematic sheet, there is some mechanism for regulating +5VSB down to a voltage for usage on the top leg of the ORing diode.) This burns the diode to the point, you cannot even read the part number off the SOT23 package.

I helped someone with this once. Their dual diode burned. I mentioned a part number they could use. They used a couple 1N914 discrete diodes and soldered that on as a replacement instead. Those have a higher forward Vdrop, shortening the battery replacement interval, but the solution worked for the guy and he was happy to be operational again.

The proper instructions in the manual, say to remove all power, to avoid an "incident". If the power plug is pulled, and you short the two pin CLR_CMOS header or use a screwdriver tip to short the two pads labeled CLR_CMOS, then that discharges the node on the PCH side and drops the voltage on the CMOS well, erasing RTC register contents and CMOS RAM contents (256 locations in PCH).

Most of the time, we expect this "legacy circuit" implementation to be used, with its chance of damage. The Intel logic signal added, the motherboard designers shy away from that. It's possible the machines with the CMOS button on the back, those use the logic signal. Intel no longer makes those demo schematics available without an NDA, so it is not possible to comment on the latest recommended circuit.

The provision of the 4.7K resistor, would not pull the node down enough to guarantee a reset. It would pull down to 80% of 3V, which is 2.4V and still high enough for the CMOS contents to be perfectly happy. (Guaranteed retention at 2.0V and over the temperature range.) Real circuits use zero ohms and pull the node down hard to ensure it is reset. Which is basically what users are attempting to do via poking at battery contacts and so on (which is not on the "correct side" of the problem). But by using zero ohms, there is a burning risk, which is why careful users unplug the power cord first.

Paul

The Natural Philosopher snipped-for-privacy@invalid.invalid wrote in news:vlqqt2$3tscn$6 @dont-email.me:

So pretty much repeating the content of my own contribution but with the /uv (unnecessarily verbose) switch set.

. . . and oh look another ramble has appeared <plonk>

good riddance

-- Of what good are dead warriors? … Warriors are those who desire battle more than peace. Those who seek battle despite peace. Those who thump their spears on the ground and talk of honor. Those who leap high the battle dance and dream of glory … The good of dead warriors, Mother, is that they are dead. Sheri S Tepper: The Awakeners.