# Change button battery calculator to AA power?

Discussion in 'Power Electronics' started by NuLED, Feb 18, 2013.

1. ### NuLED

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Jan 7, 2012
Hello!

I have an old HP 11-C calculator which is powered by those button batteries. (In fact, I even have a newer model that is powered by 3V lithium cells also, but I want to deal with the older model first.)

I can see the positive and negative connectors for the series batteries (it has 3 button batteries, so I guess that is 3 x 1.5 = 4.5 volts)

My question is, can I rig up three AA batteries in series (4.5 volts also) and adjust the resistance with resistors to provide the same power level to the calculator? (Attach wires to the battery connectors on the device itself).

It seems pretty straightforward but of course I am assuming it is deceptively simple and that is not as easy as it looks? (I would probably try to make some measurements of current using a DMM with the original button batteries in series).

Mostly I am not sure how to make sure the current is not too strong, but when I think about it, if I increase resistance, the voltage will drop also.

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
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Jan 21, 2010
Why? These would often run for a decade on a set of batteries.

3. ### donkey

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Feb 26, 2011
I agree with steve first and foremost. its ALOT easier to just leave it how it is and buy 1 (or 3 in your case) batteries every few years.
anyway, why the heck would you put a resistor on it? most (not all) devices have their own in built current limiter or to put it easy they only draw what they need. a quick way of showing this is a car battery. the amps needed to start a car is 40+ the amps needed for a car stereo is alot less, if the stereo was just going to hog all the power it would fry in 2 seconds but it only draws what it needs. same with your calculator
another thing is make sure they are 1.5volt andnot 3 volts button cells.
last thing is if you hook + to + and - to - it should work... having said all of that,I would grab the button cells and not worry about it too much or switch to a solar calculator

4. ### CocaCola

3,635
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Apr 7, 2012
BTW if the OP is annoyed with local button cell prices (they do annoy me) hit up Ebay... Instead of paying \$5 for 1 or 2 buttons cells at the local store you can get 50-100 delivered to your door from China... They might not be name brand but you kinda get a few extras that more than make up for them being a little inferior in quality...

2nd BTW to the OP, since this is a vintage calculator it originally used NR44 mercury batteries, but LR44 or AG13 or 157 alkaline button cells will work just fine as a replacement, and can be found everywhere...

Last edited: Feb 18, 2013
5. ### NuLED

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Jan 7, 2012
OK thanks guys - I agree with what you all said, and in fact I did buy a few button batteries, but aside from the possibility of using AA batteries, I did want to understand the theory and practice behind it.

I guess I wasn't so sure that there *was* a current limiting thing in such devices; it just seemed like I might fry circuitry by providing too much current (case in point, we are supposed to provide resistors to LEDs instead of let them run straight off a voltage source). I guess I am a bit confused over this aspect.

I understand battery CAPACITY as amp-hours (or in the case of microelectronics, milliamp hours, such as in rechargeable battery capacity) but I am not really sure I understand how to ensure I provide the right current at a specific voltage, for specific devices that operate at those current levels.

Put another way, if I hook up the aforementioned calculator, with 3 x 1.5v batteries in series, will it work or not, or will I damage the unit? And along this line, if I provide a solar power source, delivering 4.5 volts, will it also work?

Thanks!

6. ### CocaCola

3,635
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Apr 7, 2012
If it runs of 4.5 Volts DC than any 4.5 Volt source of equal or greater current than it requires will work... In this case pretty much any 4.5V source will work since the button cells only supply a few mA of current, so it's current requirements are nil...

7. ### davennModerator

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Sep 5, 2009
There isnt a current limiter as such ... as in a specific piece of circuitry ... donkey was throwing your a bit of a "curve ball" with...
but he did start to redeem himself with ....
Any DC circuit is a resistive load across the power supply. And the value of that resistive load determines what current will flow through that circuit for a given voltage.

cheers
Dave

8. ### NuLED

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Jan 7, 2012
Yes, exactly thanks! Based on Ohm's Law, that is what I understood too. I was always confused over the voltage and current. From reading various other materials, I got the impression that for a given voltage, there are varying currents. Yes, based on the resistance at that voltage.

Now, the question then becomes, do I need to replicate the internal resistance of the original battery series (whatever it might be) before connecting the device?

That is, I suppose I would infer the internal resistance by measuring the current across the series batteries, with a DMM, and then measure the voltage again, then calculate the internal resistance? (But of course, I would also be calculating the DMM's own internal resistance).

Here is the crux of where I am "stuck" in trying to understand this aspect of the whole thing.

Thanks.

9. ### davennModerator

13,813
1,945
Sep 5, 2009
There isnt a current limiter as such ... as in a specific piece of circuitry ... donkey was throwing your a bit of a "curve ball" with...
but he did start to redeem himself with ....
Its a common misunderstanding that if a power supply of hi current capability is connected to a piece of gear that requires a lower current, that it will fry the gear

This doesnt happen

the current flowing in the piece of gear is set by the the load resistance of that gear and the voltage of the supply

have a look in this thread for some Ohm's Law explanations I did for some one else regarding voltage resistance and current --- posts # 7 and 8

cheers
Dave

Last edited: Feb 18, 2013
10. ### NuLED

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Jan 7, 2012
ok thanks; will do now

11. ### davennModerator

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Sep 5, 2009
click on the this thread link in the last post
it will take you through a bit of Ohms law

DAve

12. ### CocaCola

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Apr 7, 2012
So riddle yourself this, in said calculator, that has a fixed 'internal running resistance' and is supplied with a fixed 4.5V what do you know about the current used based on Ohm's law?

No, it's insignificant in this instance since the vast majority of the 'resistance' is from the device itself... The batteries resistance only becomes a big issue when you are draining it beyond it's output capacity, this can be exploited to limit current but it's not advisable as you are basically causing the circuit to fail to maintain equilibrium...

13. ### NuLED

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Jan 7, 2012
OK, good question, so basically now I infer the following, which was non-obvious to me, and which I am not yet 100% sure is correct:

A device that draws a small current, such as measured in microamps or milliamps, actually has a very HIGH resistance, compared to something that draws a very high current, at a given voltage for both cases. And because many devices drawing milliamps are usually microelectronic, miniature devices, there arises a misunderstanding that these little devices have low resistance and are fragile against high current.

Is that a correct understanding?

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Typically, batteries are chosen so that their internal resistance is negligible compared with that of the circuit.

Said another way, the current demanded from the batteries is well within their capabilities.

What this means is that the internal resistance of the batteries can be ignored.

If you want to use a smaller battery, their internal resistance may be higher, it may become significant, and you may have problems.

Going the other way, a smaller internal resistance will be even less significant, so it won't be a problem.

Having said that, there are *some* circuits where the battery's internal resistance IS used to limit current. An example is a "LED Throwie". In this case the LED is placed across the battery and the battery's internal resistance limits the current to a safe value. It's a poor circuit; it can damage the LED, and it won't get the most from the battery, but you don't care. If you were to replace that battery with a larger one, you might destroy the LED very quickly, or you could even cause the battery to explode. Very few circuits are like that. The 1980's vintage HP calculator is certainly NOT in this category.

15. ### NuLED

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Jan 7, 2012
Mmmmm... so just clinging on the cliff of confusion with fingernails... I'd like to ask WHY is the LED situation different; what is different about it? the LED has no resistance to speak of because it is designed to be a good directional conductor?

16. ### CocaCola

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Apr 7, 2012
Read Steves sticky on LEDs I believe it covers the LED (diode) phenomenon... LEDs do in fact have 'resistance' but unlike many other things it's FAR from linear and other factors and rules need to be applied, a very small voltage change or change in environments (aka heating up) can DRASTICALLY change that 'resistance' and you can get a current runaway phenomenon...

There is of course A LOT more to it than that, read Steve's sticky and then Google up how LEDs and even diodes act differently than say a resistor, there are pages and pages of documentation about this, more than I want to type

17. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
A LED when powered from a voltage source (i.e. a perfect battery) *REQUIRES* something to limit the current through it.

In most cases you will use a resistor.

The internal resistance of a battery is like a resistor in series with a perfect battery (kinda -- it's more complex, but that covers 90% of it).

If the internal resistance of the battery is close to the resistance you would require to place between a perfect battery of that voltage and a LED then... the internal resistance will act to limit the current.

If you get a battery with a lower internal resistance, too much current will flow. This can destroy the LED, but it can also cause the battery to get hot and (in extreme cases) catch fire or explode.

See here for heaps of stuff about LEDs (it also explains why LEDs need current limiting)

18. ### NuLED

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Jan 7, 2012
Alright, thank you everyone. Am reading further.