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HP Forum Archive 19

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HP-35S : do the battery last that long?
Message #1 Posted by Jean-Michel on 8 Mar 2010, 12:34 p.m.

Hi folks,

My HP-35S I bought new in nov' 2007 and barely used since needed some fresh batteries to power on today! (Not a problem of contrast setting, I checked this).

My question: did you guys notice that the lifetime of the two CR2032 cells was that short in this calc?

Just as an information: my 26 years old CASIO fx-602P needs much less similar batteries. I don't remember when I bought some new, but it's much more than two years ago!

Any opinions appreciated.

Kind regards.
Jean-Michel

      
Re: HP-35S : do the battery last that long?
Message #2 Posted by Michael de Estrada on 8 Mar 2010, 1:23 p.m.,
in response to message #1 by Jean-Michel

Yes, that is normal for the HP-35s. I only get an average 18 months use out of a pair of batteries. The calculator drains power even when turned off to maintain the contents of memory. The HP-33s is even worse.

            
Re: HP-35S : do the battery last that long?
Message #3 Posted by Jean-Michel on 8 Mar 2010, 5:22 p.m.,
in response to message #2 by Michael de Estrada

Michael,

many thanks for your prompt answer. I'm rather sad to know that this calc of the 21st century needs that much power to maintain the memory whereas calcs over twenty years old don't...I was far from thinking of this. Is this what we call progress?

One last question (probably stupid, but who cares?) : would it drain the same power to maintain the content of the memory if this one would be totally empty (variables and prgm lines)? Since my wife uses it quite exclusively now without ever putting anything into memory, I could leave it empty if this would help to increase the lifetime of batteries, but I doubt! If one of you have an idea...

Kind regards. JM

                  
Re: HP-35S : do the battery last that long?
Message #4 Posted by Michael de Estrada on 8 Mar 2010, 5:53 p.m.,
in response to message #3 by Jean-Michel

I am certainly no expert with electronics, however, I doubt that the power consumption of the non-volatile memory is in any way related to it's state. OTOH, when the calculator is performing a calculation or running a program, it will consume more power than when it is idle. I believe the reason that the power consumption of the HP-33s is greater than the HP-35s is because its CPU runs at a faster speed. Because the memory of the HP-35s is much larger than older vintage calculators, it probably requires more power to maintain its contents when the calculator is powered off. If you use this calculator only occasionally and don't need to maintain the contents of memory, you might consider simply removing the batteries during storage,since it is extremely easy to remove and install batteries in this model.

                        
Re: HP-35S : do the battery last that long?
Message #5 Posted by Katie Wasserman on 8 Mar 2010, 9:24 p.m.,
in response to message #4 by Michael de Estrada

The 35S has a fairly high off-state current draw. I measured it on at around 10ua per cell. A typical CR2032 cell has a maximum rated capacity of around 220mah so you could expect about 2.5 years of battery life if you never use the calculator.

The amount of memory is insignificant to the off-state current draw for most calculators. What is stored in memory has zero effect on current draw in CMOS technology. What counts is:

1) circuit design -- cells in parallel with no blocking diodes is bad (the 12C+, 20b and 30b all have this problem).

2) silicon technology -- the leakage current on each gate. The Voyager series used chips that had near zero leakage current and the silver oxide button cells lasted for many, many years.

3) firmware design -- if the processor needs wake up often to scan the keyboard, update a RTC or do other maintenance functions, battery life is going to suffer.

-Katie

                              
Re: HP-35S : do the battery last that long?
Message #6 Posted by Jean-Michel on 9 Mar 2010, 3:50 p.m.,
in response to message #5 by Katie Wasserman

Hello Katie,

let me thank you for your very complete and technical explanations. I couldn't have any better answer.

Kind regards.

Jean-Michel.


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