Does anyone know why this message is molded into the battery cover of the 10BII+ :

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Batteries must not be removed at the same time. Please replace batteries one at a time.

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It sounds ominous. I wasn't planning on replacing batteries at this time yet thought I'd ask.

Nothing ominous here. Replacing One batt at a time helps retaining calc memory;-)

HTH

Ray

The processor works at three volts, not at six. So the batteries are wired in parallel. Older designs used to need higher voltages and had the batteries wired in a serial configuration. The two main cells were often accompanied by an extra cell for memory retention.

Sounds ominous to me. Does it not have a capacitor to retain memory during battery changes like all other HP's since - what, the Voyagers?

Nothing new there. Even the HP 30b cover warns: "Replace batteries one at a time."

This where my knowledge comes from.

There's no capacitor in there to retain memory. None of the HP ARM-based calculators have a capacitor for memory retention. You've got at best a few seconds before you'll lose memory if you remove both coin cells in any of these machines.

Here's a pic of 10bii+ pcb:

-Katie

The difference with this one is the stronger message in the first sentence "Batteries must not ...".

Well, if it only has to do with loosing settings or data stored in memory then there's little to worry about since we can't program it. With programs that you don't really want to lose is when you'd want to be more cautious.

Thanks all.

I wonder why the Y^x does not work like a"Two-number function"? for instance a % change works:

number 1
<INPUT>
number 2
<% Chg>

However the Y^x works diferently using the "=" sign:

number 1
<Y^x>
Number 2
<=>

% Change works like an RPN machine, but Y^x work like an algebraic machine. It is like using two different logics with "two-number functions". I would have called a Y^x a "two number function" and treat it like the % change. It's little confusing at first until one remembers they function differently...

Hello,

Please see page 37 on your manual where it talks about which functions function in the dual way. They are perm, comb, %chg, ddays, date, t, and inverse t.

The reason why not all are supported is that it needed to be able to support the fact that INPUT does an = for any pending operators. If something like * or ^ worked the same, if you wanted to do 4*5 INPUT then 4*2 %CHG, it would actually do 4*5*4 when you pressed the second * key.

TW

Excuse my (electronic) ignorance, but if there in no capacitor, why would you get even a few seconds?

If there were enough small valued (0.1uF) capacitors used for bypassing they might retrain enough of a small charge long enough, depending on the current drain in the circuit, to provide current to circuits to keep them active.

exactly

I like that the 10bii+ allows the number1 FUNCTION number2 = key sequence on two-argument functions (%CHG, DAYS, COMB, etc...).

The sequence is straight forward and consistent with the two-argument arithmetic operators.

It's a small thing, but I'm still not getting this:

@Katie:

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There's no capacitor in there to retain memory...You've got at best a few seconds before you'll lose memory if you remove both coin cells

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...if there in no capacitor, why would you get even a few seconds?

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If there were enough small valued (0.1uF) capacitors used for bypassing they might retrain enough of a small charge...

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exactly

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There's no capacitor in there to retain memory...

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Katie:

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"None of the HP ARM-based calculators have a capacitor for memory retention."

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Jim:

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If there were enough small valued (0.1uF) capacitors used for bypassing they might retrain enough of a small charge...

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You:

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Although there is not a large capacitor for long-term memory retention the small valued bypass capacitors might retain the memory for a short while. They are the small SMD components (CXX) around the black blob IC and elsewhere in the printed circuit board.

More about bypass capacitors here. The first two paragraphs and the Figure 4 might give you an idea.

Edited: 25 Apr 2011, 4:57 p.m.

Let me try to explain.....

Whenever an IC is used on a printed circuit board you will typically use a "bypass capacitor" across the power terminals as close to the IC as possible. The reason for this is to create a very small reservoir of current just for that IC. When IC needs to momentarily draw a larger amount of current it will have that available without the resistive drop created by the thin traces on the circuit board that would happen if the current had to come from further away on the board. This is particularly true for microprocessors that have high momentary current needs. Really any CMOS circuit with a lot of transistors has this issue.

It's pretty much a given that there's going to be a bypass capacitor for the ARM chip. Bypass capacitors are typically 0.1uF -- not very much capacitance. Depending on the off state current draw of the ARM chip (maybe .1uA, just a guess) you might get a few seconds (roughly: time=V*F/A; 3*.1uF/.1uA = 3 seconds) of memory retention from the bypass capacitor. A memory retention capacitor might be 100uF (often much more) depending on the off state current draw.

-Katie

Nice article. It's obvious that nobody would think of distributing 76 64nF caps around that one device and normally a much smaller number of 0.1uF and 0.01uF caps around that "one" device being used in the article.

Some manufactures (i.e., Intel - since I use to work there as a PCB designer) would design in the number of caps they thought would be needed and then when the board was working satisfactorily would remove components to reduce costs but not so much as to exceed the amount of noise they needed to stay under to pass tests.

Never-the-less, thanks.

The capacitors I am more familiarized with are somewhat bigger and are rated in kvar instead of nF or uF, but their smaller siblings are always interesting :-)

Regards,

Gerson.

Thanks. You splained it to me.