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#1 flyfishingrules2

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Posted 14 April 2006 - 09:35 PM

I know I sound like a broken record on the this, but the first thing I do when I troubleshoot and an older computer (2 or 3 years old or older) is replace the motherboard battery, sometimes referred to as a CMOS battery. They are cheap, less than $3, and cure a world of ills. Symptoms of a bad battery: --takes too long to boot because it is re-creating BIOS settings, locations of hard drives, memory amount, etc. --does not hold BIOS settings; computer does not run at correct speed, for instance --Windows time and date are wrong or keep changing on reboots --any CMOS error or System Configuration Lost error --error 201 (memory) --speakers or keyboard do not work Not to get too technical but BIOS stands for Basic Input/Output System; CMOS stands for Complementary Metal-Oxide Semiconductor, pretty fancy for what we would call, umm, you know, a battery. The battery in the U.S. is usually a CR2032 lithium. The battery is about dime-sized, round, and silver. Care should be taken to remove and replace it, but don't be intiminated. Ground yourself by touching the metal on your case and carefully press the little metal clips that hold it. Replace and enter BIOS to check or change settings. Also if you know the battery is good, popping it out for a few minutes resets BIOS settings to defaults, sort of like clearing out all the cobwebs, a good thing. If you are familiar with your motherboard, the same thing can be accomplished by moving the BIOS jumper from 1-2 to 2-3 and back. Check your motherboard manual for details. Happy computing. :)

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#2 IntelGuy

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Posted 14 April 2006 - 10:15 PM

Great tip Fly! I've actually had a half-dozen instances where the computer wouldn't boot at all due to a low or dead battery .

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#3 DK64_MASTER

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Posted 16 April 2006 - 12:12 AM

Great tip Fly! I've actually had a half-dozen instances where the computer wouldn't boot at all due to a low or dead battery .


Instead of buying a battery, could you test the voltage of the battery (should be 3.3 volts) and determine if it's dead according to that?

Cheaper, and faster solution :).

#4 tatianna

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Posted 31 May 2006 - 12:13 AM

Reading the battery voltage may work sometimes...however the battery voltage has little to do with the battery current. In other words, you may have a battery reading 3.3 volts but it may have little or no current available to operate the device. :)

#5 DK64_MASTER

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Posted 31 May 2006 - 01:44 AM

Reading the battery voltage may work sometimes...however the battery voltage has little to do with the battery current. In other words, you may have a battery reading 3.3 volts but it may have little or no current available to operate the device. :)


Ohm's law: V=IR, and in this case, R is material dependent, and should be relatively same for a set of batteries. If you can't sink enough current, then I'm sure your voltage would would reflect that. :)

#6 IntelGuy

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Posted 31 May 2006 - 06:20 AM

Ohm's law: V=IR, and in this case, R is material dependent, and should be relatively same for a set of batteries. If you can't sink enough current, then I'm sure your voltage would would reflect that. :)


Actually, to get an accurate reading of the amps, an ammeter should be connected in series with the battery, and the battery should be under a load. I've had several instances with automobile batteries where the volts read 13.5 (on a 12V system) but it wasn't pushing enough amps to light up the dome light.

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#7 DK64_MASTER

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Posted 31 May 2006 - 09:50 AM

Actually, to get an accurate reading of the amps, an ammeter should be connected in series with the battery, and the battery should be under a load. I've had several instances with automobile batteries where the volts read 13.5 (on a 12V system) but it wasn't pushing enough amps to light up the dome light.



Ah crap, forgot about that. And I'm an electrical engineer :nono: :hammer:.

I retract my post.

#8 TomGL2

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Posted 31 May 2006 - 11:23 AM

Instead of buying a battery, could you test the voltage of the battery (should be 3.3 volts) and determine if it's dead according to that?

Cheaper, and faster solution :).


Batteries are usually tested by measuring the voltage while a load is across the terminals. For a 2032 lithium cell, the load is typically a few kilohms to produce a current of about 1 milliampere. The end of the cell's service life would be reached in 150 to 200 hours if the load were continuous.

The current required by RTC/NVRAM chips is minute by comparison, however, typically about 500 nanoamperes, equivalent to a 600 megohm load. This is much greater that the resistance of even a very sensitive voltmeter, the result being that the meter imposes a greater load than does the component the cell powers.

It is therefore not necessary to place an additional load on the cell, bearing in mind that that is true only for micropower applications such as this. RTC/NVRAM parts typically have a 10% voltage tolerance, so the cell should be replaced when its voltage approaches 2.7 volts.




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