# Current Limiting Resistors

So I have an old calculator (an HP11C if you must know) that runs off LR44 batteries. But these batteries have gone the way of the dodo bird in my area... can't find them. So I looked up up the LR44 battery. Each one is 1.5V@150mAh. 3 were in the calculator. If I use 3 AAA to power it, I'll have my 4.5V. As for the current, can I hook up a 10? resistor inline with the batteries?

V=IR; 4.5=.45 * R; 4.5 = .45 * 10

Will this work? It was a very nice calculator...

Thanks.

Batteries are a current source. Your calculator is a current consumer. The amp-Hour rating of a battery is its total capacity, not the amount of current it will "push" into a device.

The calculator will only draw the amount of current it needs from the batteries, regardless of what their capacity.

There is no need for resistors in this situation. Also, why not just buy new LR44?

But then why does an LED short if you send >20ma through it?

You are comparing apples and oranges.

The reason for a current limiting resistor with a LED is because of a special property of the diode. When enough voltage (the forward voltage) is dropped across an LED, it effectively turns into a short circuit. It will draw all the current that is available. The current limiting resistor does two things: 1) it limits the amount of current that can flow through the LED and 2) it keeps the voltage drop of the LED constant. (This is Ohm's law.)

You'll notice in the process of calculating a current limiting resistor for a LED, you do not take into account the current source's capability, right? You simply calculate based on the total Voltage across the LED + Resistor and how much current you want to flow through the two.

But seeing as this calculator is 1E6 years old, shouldn't I take some precautions with it and put in a resistor? Will it hurt?

The resistor will add unnecessary complication, drop some of the voltage, waste power, etc. If the calculator is damaged then applying power either with a resistor or without won't matter much, will it?

Again, I would think the smarter approach would be to order the proper batteries from an online source and go about it that way.

Alright.

Picture it like this:
The three LR44 batteries are the equivalent to a glass full of water.
The three AAA batteries are the equivalent to a bottle full of water.
The calculator sucks the water with a straw.

If you suck water with a straw from a bottle are you sucking a higher volume of water per second than if you sucked from a glass of water?
No. It's the same thing. The only difference is that the bottle has more water, ie, there's more for you to suck (ba dum tish!).

The same applies to the calculator. The only difference between the batteries is that if you use LR44 batteries the charge will last 10 days and if you use AAA batteries the charge will last a century.

So yeah, you could even use 3 D batteries.

But 3 D batteries couldn't easily be attached to the back of the calculator with black electrical tape like AAAs can.

The LR44 is the same as an energizer 357. I'm really surprised you cannot find any. They are everywhere for me. They are what my digital calipers use and I need to replace them every year.

Attaching a resistor will only drain the batteries faster... is buying a new calculator out of the question? I know this is silly, I didn't put away my Yashica when they stopped making batteries for it (and found a new version ten years later half way around the world) but you can probably get a just as powerful calculator for less than that has costed you then.

Attaching a resistor will only drain the batteries faster

In parallell with the calculator, yes. But as far as I got he meant to use it in series with it. Not that it would work as intended as discussed above.

... is buying a new calculator out of the question? I know this is silly, I didn't put away my Yashica when they stopped making batteries for it (and found a new version ten years later half way around the world) but you can probably get a just as powerful calculator for less than that has costed you then.

I don't doubt there are better calculators out there now, but one thing that puzzled me was how the price for calcs was pretty much the same for years. I had a FX4000P, I am not sure where it is now, but its price never dropped from I got it mid-late 80's until, well, at least a decade later, probably more I don't remember exactly (Of course the real price was probably less anyway, with inflation and all).

XKCD is pretty much spot-on here

But a netbook with wxMaxima, GNUplot and similar software beats any calculator anyway these days. Plus you can see movies on them

EDIT: And of course the main point I somehow managed to not write was that the calculators didn't evolve Not just the price.

But seeing as this calculator is 1E6 years old, shouldn't I take some precautions with it and put in a resistor? Will it hurt?

Don't mess with your HP11C. You might be better off selling the calculator on eBay to someone who will appreciate and treasure it. It is one of a family of the best and most sought after (by those who appreciate the ease of use of 'Reverse Polish Notation') scientific calculators that has ever been manufactured.

Don

Don't mess with your HP11C. You might be better off selling the calculator on eBay to someone who will appreciate and treasure it. It is one of a family of the best and most sought after (by those who appreciate the ease of use of 'Reverse Polish Notation') scientific calculators that has ever been manufactured.

I appreciate and treasure it. That's why I'm asking before powering it with AAAs. And yes, I know how to use RPN. My current calculator, an HP48SX, uses RPN (duh). But it is sort of like using a nuclear power plant to power an ATtiny-- its over-necessary. I like the 11C because of its relative simplicity.

And I found the 357 at walgreeens... didn't realize the batteries had so many different names. Thanks all!

raron:

Attaching a resistor will only drain the batteries faster

In parallell with the calculator, yes. But as far as I got he meant to use it in series with it. Not that it would work as intended as discussed above.

There would still be a voltage drop on it...

There would still be a voltage drop on it...

By how much?

[I found the 357 anyway, so I'll just get 3 of those.]

bubulindo:

raron:

Attaching a resistor will only drain the batteries faster

In parallell with the calculator, yes. But as far as I got he meant to use it in series with it. Not that it would work as intended as discussed above.

There would still be a voltage drop on it...

Attaching a series resistor don't usually drain more current through a circuit.

I = V/R1 > V/(R1+R2)

But yes, there will be a voltage drop over the resistor, and lowered one for the calculator, assuming it is linear and still works with it...