Re: Multimeter HP 3468A, CMOS RAM battery replacement Message #5 Posted by Katie Wasserman on 12 Jan 2011, 12:49 a.m., in response to message #4 by Martin Pinckney
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.
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