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I have an hp-41CV, and when you turn it on it displays "MEMORY LOST", and that is it. Any ideas what this means?

Hi;

I have seen exactly this in my HP41C when it was about to complete 1 year with me. At that time (Dec. 1983? Jan. 1984?) I was not aware of anything but using it. After about 5-6 years I was delving into its guts and I got a better understanding about its electronics. With another HP41C I successfully simulated this particular situation.

The fullnut (almost all HP41C's, some HP41CV, a f

Hi;

I have seen exactly this in my HP41C when it was about to complete 1 year with me. At that time (Dec. 1983? Jan. 1984?) I was not aware of anything but using it. After about 5-6 years I was delving into its guts and I got a better understanding about its electronics. With another HP41C I successfully simulated this particular situation.

The fullnut (almost all HP41C's, some HP41CV, a few HP41CX) has a bunch of RAM IC's, basically one that contains the "control" registers (ALPHA, Stack and those one can access with Synth commands) and the others that will contain the "user" registers. If an HP41C, only two RAM IC's: one of each. If an HP41C/CX, six RAM IC's: the first one for the control registers and five 64-registers RAM IC's, in a total of 320 registers (one is always with the permanent .END., so we can only access 319 for data registers and 4 bytes for program OR 2237 bytes of program and no one register).

In my particular case (finally!) the first RAM IC was dead. I used this IC in two other 41C's and the resulting MEMORY LOST was observed, too. The original "amnesia" HP41C worked fine with the others' control RAM IC.

The bad news is that I cannot tell you this is the only one cause for the MEMORY LOST. This is the one I know!

The codes for the IC's are:

1LA701-xxxxx - first RAM: 16 registers
1LA702 to 705 - block in the first HP41C's: 4 x 16 registers
1LE701 to 705 - block with 5 x 64 registers (320)
In the later HP41CV's (fullnuts), the 6 RAM IC's were soldered in the mainboard as three blocks of two IC's; the mainboard had three positions only. If you have an HP41C that uses this type of maninboard, you'll see two electrolytic capacitors mounted outside the PCB's boundary (left side) and, aligned with them in the right, the three positions for the RAM IC's. There is no specific position for each RAM, the three DIP connectors are in parallel with each other. If you have an HP41C, only two IC's will be found, and one DIP is free.
Hope this helps.

I MUST thanks to many guys in here that provided me all of the final information (mostly the path for downloading the 41C's schematics) so I can post it here with confidence.

Cheers.

Right ...
Try those two things.
Press & maintain ENTER key, then ON. Realease. 0.0000?
Press & maintain <- key, then ON. Release. MEMORY LOST?
Else I have the Service Module (please see http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/forum.cgi?read=23586) and spare boards for HP-41.