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# [SOLVED][moved] converting button cells to AAA

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#### squirrel online

##### Junior Member level 1
i have a laser pointer that is powered by 3x1.5 button cells. testing the batteries results to 4.5+ volts and .14A.
i want to convert it to use 3xAAA that has 1.2v and 800mA

what i've done so far

used 3xAAA with 1.2v = 3.6+ volts and 800mA
it lights up for a few seconds only
removed the power source for a minute
it lights up again for a few seconds

used 4xAA with 1.2v = 4.8+ volts and 2000mA
it blinked and did not light up again

what do i need to make it work? thanks

Last edited:

Hi,

what do i need to make it work? thanks

It seems you killed it by overpowering.

******
I don't know how you get the current values from your batteries. Did you eventually short circuit them?

If you short circuit LiIon batteries they may explode. Others too.

Look for "Ohm's law" and "how to calculate current limiting resistors"

Klaus

Hi,

Buy a new one, It seems you killed it by overpowering

I don't know how you get the current values from your batteries. Did you eventually short circuit them?

Klaus

i sure will, and will not repeat the 4 AA batts again.

i used a digital multi-meter to get the volts and amps values of the 3 button cells. as for the AAA's and AA's volts and amp it's written on its body.

this is where i got stuck
a resistor decreases the volts and not the current.
and from my trials i assume that the current killed it?

Hi,

i sure will, and will not repeat the 4 AA batts again.
good idea.

i used a digital multi-meter to get the volts and amps values of the 3 button cells.
Voltage measurement is OK.
If you try to measure current of a battery by just connecting the two wires of the DVM to the battery, then you short circuit the battery. This may hurt your DVM, your batteries, and yourself if the batteries catch fire or explode.
==> Current measurement is in the very most cases with a (defined) load in series.

as for the AAA's and AA's volts and amp it's written on its body.
I doubt there is a "A" value written on the batteries (Although on starter batteries it sometimes is).
Usually there is an "Ah" value on the battery. This is different. It is not the current but the electrical charge in "ampere hours".
(It´s similar to km/h and km. While km/h is speed, km is distance.)
800mAh means you can draw 1 mA for 800h, or 8mA for 100h, or 80mA for 10h until battery is drained out.

a resistor decreases the volts and not the current.
Ohm´s law is true. With constant volts you can adjust current (by varying the resitance). With constant current you can adjust voltage (by varying the resitance). And anything inbetween.
And the phrase "current limiting resistor" is valid also.

and from my trials i assume that the current killed it?
Yes and no.
Current without voltage will not hurt anything. Voltage without current will not hurt anything, too.
Voltage * current gives power. Power generates heat. But in detail power it self will also not hurt anything.
You need a special time (depending on power) for something to heat - or better say to overheat. With enough power this may take microseconds.
With less power it may take hours for something to overheat.

***
I see you are learning. This is good. And failures will teach you.
Everyone of us made mistakes. And everyone of us still makes mistakes. So don´t be afraid.

Use some resistors and a power supply. Measure voltage, measure current and feel the generated heat. Make it explode (in a safe way! don´t burn your house!) and see what happens.
That´s the way to learn and get a feeling of electricity.

Klaus

squirrel online

### squirrel online

Points: 2
Some cheap laser pointers are designed to use the battery internal resistance for current limiting. Operating them with a different power supply can be tricky.

Unless you plan to make a switched mode converter, you are stuck to use at least 4 1.2V accus. The most simple solution would be a series resistor calulated for 4.5V/0.14A output (presumed your measurement is correct). A constant current circuit would be better, but difficult with only 0.3 V margin.

squirrel online

### squirrel online

Points: 2
wow!
from your inputs it seems that i have asked the wrong question and assumed the wrong values.

as the image below shows
the button cells works
the the 3 AAA lights it up, then not, then lights it up again
and the 4 AA killed it.

good thing i have another extra

my learning objective for the coming days is to use the 3 AAA as the power source.
will now then explore the world of resistors.

A rechargeable battery cell is labelled with the mAh rating which is how long it can supply some current, not its maximum current.
My Name brand (Energizer) AA Ni-MH cells measure 11A when shorted with my multimeter. They are 1.45V when fully charged and about 1V when nearly dead so their average voltage is about 1.2V. Since the current is 11A then the internal resistance of the cell plus the resistance of the leads and meter is a total of about 0.11 ohm. These cells have a capacity printed on them of 2300mAh. When the cell is worn out its shorted current when fully charged measures only 300mA or less.

My very cheap Chinese solar garden lights came with AAA Ni-MH cells rated at 300mAh and 600mAh. All the ones rated at 600mAh measure only about 100mAh. Most of the 300mAh cells measure a little more than 300mAh and produce 6A when shorted with my multimeter.

A button cell is tiny so it has some internal resistance that limits its maximum current to 0.14A. Your three AAA rechargeable cells (they might produce 6A) and four AA rechargeable cells (they might produce 11A) produced a current in the laser diode that was much too high and destroyed it.

squirrel online

### squirrel online

Points: 2

i have a question regarding the following pix

with the led bulb there was a little voltage drop

with the laser diode there was a big voltage drop

why did this happen?

I'm not aware of a LED with 4.5V forward voltage. Wrong polarity?

squirrel online

### squirrel online

Points: 2
now the led bulb is busted

weird though, i used the same led bulb i removed from the flash light.

Hi,

think your given voltages in the pictures of post #8 are wrong. The values in the text semms to be OK.

LED: 2.65V, input 4.66V, 27Ohms
Input voltage is 4.66V, LED voltage = 2.65V, remaining voltage at the resistor is 4.66V-2.63V = 2.03V.
Now you know the voltage across the R: = 2.3V and the R = 27 Ohms. So you can calculate I: R=U/I ==> I=U/R = 2.03V/27Ohms = 0.075A = 75 mA.
The same current runs through your LED. Too much for a standard LED. It burns.

***
LASER: 4.52V, input 4.65V, 27Ohms
Input voltage is 4.65V, LASER voltage = 4,52V, remaining voltage at the resistor is 4.65V-4.52V = 0.13V.
Now you know the voltage across the R: = 0.13V and the R = 27 Ohms. So you can calculate I: R=U/I ==> I=U/R = 0.13V/27Ohms = 0.0048A = 4.8 mA.
The same current runs through your LASER. Maybe too low for your LASER to work correctly.

***
The way to go:
You need to know:
* Input voltage. Lets say 4.66V.
* Nominal LED voltage from datasheet. Lets say 2.1V.
* Nominal LED current from datasheet. Lets say 20mA
--> calcualtion of R:
Input voltage is 4.66V, LED voltage is 2.1V --> voltage remaining for the R: = 4.66V -2.1V = 2.56V
Current thourgh R is the same like current throughLED = 20mA = 0.02A
Resistor: R = U / I = 2.56V / 0.02A = 128 Ohms. Try to find the next bigger one in your stock. Maybe 150 Ohms.

You may find it difficult, but everyone of us has to do the same. Maybe hundreds of times for a single project. That´s life.

Klaus

squirrel online

### squirrel online

Points: 2
is my multi meter defective?
or i tested it wrong

the 4.68 to 4.68 is where the busted led used to be
the 4.28 to 2.65 has a laser diode installed

I have made hundreds or thousands of LED circuits but I have never burned out an LED because I calculate the resistor value first, not last.

squirrel online

### squirrel online

Points: 2
Hi,

with the led bulb there was a little voltage drop
with the laser diode there was a big voltage drop
How does this meet the values of the picture?

Btw: in both pictures the LED is disconnected.

So I don't understand what you measure at all.

Klaus

Last edited:
squirrel online

### squirrel online

Points: 2
oh, i think im wrong again. now i know what im going to do on the next one. ill be back after a few days. thank you all for the replies.

success!!!! its been lighted up for 30minutes

thank you klaus for making everything clear
thank you Audioguru for pointing out that everything must be computed
thank you FvM for the series layout

i just hope that it will still work if there are laser pointers connected to the 3 x AAA batteries.

bought a new laser pointer
got off initial values of .02A, input 4.51v, forward 3.78, computed resistance is 36.5ohms
but for now i got a 47.9ohm and i tried it
and it works.

question. do i still need to use 36.5ohms? or the 47.9 will do? is one better than the other since computed resistance is 36.5.

You computed a current limiting resistor value of 36.5 ohms to produce a current of only 20mA only when the battery is brand new. Its voltage and the current will drop as the battery is used.
But without seeing the spec's for the laser diode you have no idea if it will survive or whether it can safely use much more current.
The 47.9 ohm resistor you are using reduces the current to 15.2mA when the battery is new.

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