Hello,
the multimeter (HP-3468A) has a 3V lithium battery to prevent the deletion of the CMOS RAM while turned off.
This is crucial because otherwise the calibration data would be erased.
My point now, is to ask whether someone has ever replaced this battery while maintaining the calibration data.
According to the service manual, the CMOS RAM gets its supply voltage from the grid while the machine is turned on. So it would be obvious to do some soldering and replace the battery when the multimeter is turned on.
Another idea would be to keep the old battery. Then, to connect the new battery in parallel, and eventually, to disconnect the old battery.
However, I am afraid to produce some shorts or whatsoever which could drain the contents of the RAM anyway, or even worse, damage some other part of the multimeter.
Does anyone have some ideas or comments on this?
Any help is appreciated.
Best regards
Frido
Multimeter HP 3468A, CMOS RAM battery replacement
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Post: #9
01-11-2011, 03:09 AM
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Post: #10
01-11-2011, 09:06 PM
Although I have calibrated and repaired HP multimeters in the the past, I am unfamiliar with that model or any that contain a lithium battery. You may have posted this in desperation, or were you unaware this forum is for HP calculators? I suggest posting your question the to sci.electronics.repair newsgroup, or Google search Sincerely, Ren dona nobis pacem
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Post: #11
01-11-2011, 09:24 PM
The 3468A has an HP-IL interface, so it was commonly used with the HP-41, HP-71, and HP-75. (Meters with an HP-IB interface could be used, but required the 82169A IL/IB translator.) The 3468A calibration is stored in battery backed RAM. If the battery dies, you lose the calibration. Unfortunately mine had a dead battery when I received it. ▼
Post: #12
01-11-2011, 11:07 PM
Quote:Does this then, allow an "in" for Frido to ask the question here? I know nothing about this unit, but just on general principles, it would seem that soldering a new battery in parallel (with the unit off), then removing the old would be preferable to doing any modifications while the unit is turned on. ▼
Post: #13
01-12-2011, 12:49 AM
I know nothing about this meter either, but found the service manual here. The schematic shows that the memory is powered by diode steering logic going to the +5V supply and the 3.6 volt battery. Either will power memory. There's also a .1uf capacitor at the junction so even with no battery and the +5 volt supply off, memory will be preserved for at least a short while. If I were going to replace the 3.6 volt battery I'd: (1) disconnect it from the AC power; (2) hook up an external 3.6 volt power supply thru a third (temporary) diode to this junction point using some good micro-clips; (3) remove the old battery by just clipping the leads (it's faster than desoldering); (4) solder in the new battery to those cut leads. The reason for using 3.6 volts and going thru a third diode is just in case the your external power supply quits or is shorted out, you'd still have the .1uf cap to hold memory for a bit.
Edited: 12 Jan 2011, 1:15 a.m. ▼
Post: #14
01-12-2011, 04:32 AM
Thank you, Katie! ▼
Post: #15
01-12-2011, 08:36 AM
Frido, Ren dona nobis pacem
Post: #16
01-13-2011, 03:46 PM
Just a small note: That 0.1 uF capacitor will not hold the memory contents, not even for a second. |