Juan said:
From a cold start, my Gateway desktop boots to the Gateway (BIOS) splash
screen and offers <DEL> to enter Setup and F12 to enter Boot Menu.
It then sits there for a minute or more before I get "Starting Windows".
Someone suggested a low CMOS battery. Could he be right?
Wouldn't there be other symptoms if the battery were low?
The battery is about 2 y.o. and is constantly being charged.
CR2032 coin cells, are not charged by the computer.
If you unplug a computer, and provide no alternative
sources of power, the CR2032 lasts approximately three years.
(By calculation, it's slightly less than three years, but
I like nice round numbers.)
If the computer has +5VSB available, that takes the place
of the CR2032. If you leave the computer plugged in all the
time, and leave the switch ON at the back, then the available
+5VSB means the CR2032 won't get used at all. With no load,
the CR2032 lasts around ten years.
On a laptop, the main battery pack may be diode ORed in
a similar way. If the battery pack is present, it could
effectively be running the CMOS clock, when the laptop is
not plugged in. In a laptop, the CR2032 would only be
needed, when the user unplugs the main battery pack, and
no adapter is connected.
You cannot charge a CR2032. The manufacturer limits charging
currents to 1uA (micro amp), and the only reason the level is
that high, is to allow motherboard makers to use Schottky diodes
for an ORing diode implementation. At elevated temperature,
some Schottky diodes leak that much current, and so the battery
rating takes that kind of thing into account. I don't think it
was an accident that particular number was chosen. The battery
manufacturer would choose to set that number to zero, but then
it would make using the battery very difficult.
There are some battery types, charging the battery could
burst it. And for such batteries, the manufacturer may state
what the most charging you could apply to it, without
pressure building up in it.
The two diodes in the ORing diode circuit, picks the "source with the
highest voltage". If the computer power supply is delivering
+5VSB, the voltage at that node is higher on the computer side
than the battery side. And then no current flows out of the battery.
If the computer is completely powered off and put in storage,
the ORing diode then switches to drawing power from the CR2032.
A brand new CR2032 can be close to flat. There were some
batches of Asus motherboards, that shipped with weak batteries.
There have also been instances of motherboards, with much higher
than normal loading characteristics. A normal motherboard is
rated at around 10uA of loading, to run the CMOS 32KHz clock
circuit, as well as power the CMOS RAM. But there have been
motherboards, where leakage in the circuit was near the milliamp
level, and that flattened batteries in no time at all. To put
the 10uA computer drain in perspective, a typical digital watch
draws around 2uA to run a 32KHz time piece. Digital watch
batteries are smaller, but the current drain is smaller
as well.
*******
A dead battery, can stop some computers from starting.
Other than that, there should be no "slowdown" caused by
the battery. The CR2032 battery doesn't run the hard drive
motor or anything
*******
If you want to see the circuit around the CR2032, there is an
example here. See PDF page 18, lower left hand corner.
http://www.intel.com/design/chipsets/designex/BXDPDG10.PDF
Diode D3 and diode D2 are the ORing diodes. They select the
source with the higher source of voltage. The battery has
a 1K ohm series resistor, to limit current flow. Since the
battery never provides current while the computer is operating,
there is no need for a "stiff low impedance" power source.
When the CMOS clock is being relied upon, the current
drain is minimal. And the C15 (0.1 microfarad) capacitor
provides hold-up to transient current flow (when the ripple
counter transitions). When the computer is running,
diode D2 provides current flow, and diode D2 has
no current limiter in it. It is a "stiff" source of power.
D2 runs the CMOS, when the computer is in a running state.
The jumper JP17 in the sample schematic, provides a "safe"
implementation of the Clear CMOS function. You can also
set up JP17, so that it shorts the power source. And if you
then used a "bad" implementation of JP17, then the diode D2
gets burned by the level of current flow. This is why I always
give a warning to "unplug the computer when clearing the CMOS".
That is to provide some measure of protection against
a badly designed JP17 wiring. In that Intel example,
you don't have to unplug the computer, to use JP17.
And no matter how many times companies like Intel
show the other companies how to do it, they still
build the "bad" ones.
On that schematic page, you can also see the 32.768KHz
quartz crystal, and the two loading caps. That's what
keeps time while the computer is unplugged. The 32KHz
signal is divided by a ripple counter, to make a once
a second pulse. Counting those pulses is where the
time-of-day comes from.
Paul