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Johnh · 09-29-2024, 03:43 AM

Has anyone changed batteries on an HP15C CE?

Mine is a year old, running fine, but I identified the following issue when I opened up the battery compartment yesterday to (successfully) do firmware and VoyagerSave updates. This was actually the first time opening the battery compartment.

The idea per the manual is that you can switch off, replace the batteries one at a time, and when you switch on, data is still preserved. On mine, if I remove and quickly replace the battery nearest the centre, all data is lost, like a full reset. The outer battery near the corner is not providing charge to maintain data. This is with new, tested batteries. I can however, remove the corner battery and actually run the machine and it works ok

Testing without batteries with a meter reading resistance, there is no connection from the outer battery’s lower ‘-ve’ connection, to that of the centre battery, (test range 2M Ohms), and with one battery in place in the centre position, no voltages are detectable at the corner battery slot, and using alligator clip and leads to connect the second battery, no current measured flowing through it either.

I think my corner battery is just not connected to anything? And its not just lack of contact battery to calculator. The calc runs fine on one battery, but this can't be replaced without data loss.

Can anyone comment or confirm?

ive also posted about this while getting the cable connection to work, on this thread:
https://www.hpmuseum.org/forum/thread-22103.html

But this battery issue is more basic I think.

Steve Simpkin · 09-29-2024, 05:46 AM

It is possible that there is just a broken wire. Each battery has its own two wires connected to the main PCB.
https://drive.google.com/file/d/11AjaPCl...sp=sharing

AnnoyedOne · 09-29-2024, 03:50 PM

(09-29-2024 03:43 AM)Johnh Wrote: Can anyone comment or confirm

See your other thread.

https://www.hpmuseum.org/forum/thread-22...#pid192509

A1

Johnh · (This post was last modified: 09-30-2024 07:15 AM by Johnh.)

Thanks for all the input, on two threads. Based on that, I have a good workaround, and its fully reversible.

It's just a short length of insulated stranded wire, solder- tinned at each end, tucked under each negative connector, to squeeze it against the case. Batteries fit and both connect (at a slight tilt). And the cover slides back as normal. The bush around the screw-hole even eases the wire out of the way so the screw works fine.

Now Im able to remove either battery and it still runs, and data is preserved. So the batteries should share the load now, and I can check and change them when needed. This with VoyagerSave now going, will allow me to add and develop my programs! Smile

Pictures below.

RPNerd · 09-30-2024, 11:29 AM

(09-30-2024 07:03 AM)Johnh Wrote: Thanks for all the input, on two threads. Based on that, I have a good workaround, and its fully reversible.

I'd reverse it now if I were you and if you don't want batteries to leak and/or swell in the calculator.

If both batteries don't have exactly the same charge level then one is going to try and charge the other. In circuit, the positive terminals of the batteries each go to the anode of a diode and the cathodes of those diodes are connected together to form the positive rail at about 2.5V. This prevents discharge from one battery to the other and allows for load balancing.

If your calculator went dead when you removed one battery then it simply means that the other one didn't have enough juice left to power it.

AnnoyedOne · (This post was last modified: 09-30-2024 01:13 PM by AnnoyedOne.)

(09-30-2024 11:29 AM)RPNerd Wrote: If both batteries don't have exactly the same charge level then one is going to try and charge the other.

True. Which is why the manual says to replace both with new/same brand ones. Effectively the same voltages so no current flow between them. Plus they should discharge at the same rate.

Diodes aren't needed.

https://www.hpmuseum.org/forum/thread-22...#pid192536

A1

Johnh · 09-30-2024, 01:24 PM

The issue that my calculator had was unchanged with two brand-new identical batteries measuring the same voltage. I can also confirm from measurement, that the positive sides are directly connected to each other. I had wondered about diodes, as in the past linked above, but I also see many references online to direct parallel connection being ok with these batteries due to their internal resistance, based on batteries of the same type and age.

There's a fault inside, but it's working OK now and I decided not to open it all up further to investigate further. But I'll check the batteries again soon.