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

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

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.

Quote:

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The 3468A has an HP-IL interface, so it was commonly used with the HP-41, HP-71, and HP-75.

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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.

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.

Thank you, Katie!



I believe that your suggestion is the most sophisticated way to resolve the problem.



I know that the item is somewhat OT as it was stated by Ren. However, as Eric remarked, the multimeter has an HP-IL interface and may be considered as a peripheral of those HP calculators which support that interface. Therefore, I thought that it was not too remote to believe that someone in this forum could give some advice. Indeed, I purchased a used HP-3468A because I was curious to use it with my HP-41.

It works fine currently, and a loop together with a IL-printer can be established which then serves as a logger. The issue with the battery does not urge. I use to check the voltage yearly and I planned to replace the battery when the voltage drops.

As I am only a medic whith no in-depth knowledge about electrical engineering I just wanted to be prepared for the day X. Also, I do not have the equipment for a new calibration, and because of that an in vivo replacement of the battery seemed to be a good option.

Apologies if I may have caused discomfort.



Best regards

Frido

Frido,
I wasn't opposed to your posting, I wondered if there wasn't a better place to get the information you sought. Katie has certainly shown me otherwise!

Ren

dona nobis pacem

Just a small note: That 0.1 uF capacitor will not hold the memory contents, not even for a second.
Katie's suggestion of a parallel through a diode is certainly the way to go.