Why I shoudn´t use a resitor + 2 NiMH ?

[Fred_DASP]

I did the DDL and it´s ok. Nothing to say about it.
But once time I got the mA an Volts to drive my laser, can I use 2 NiMH (AA or AAA) and a resitor in series with diode?

I almost think that answer is not. but WHY if the Ohm´s Law is the LAW ?

My diode is running at 2.5 Vots and 100 mA (an 8x DVD)
With those info, what resistor I need?
V=RI -> R=I/V -> R= 0,01A/2.5Volts = 0,04Ohms
It´s very little... not a comercial value.
Is there something wrong?

A just want do a portable laser pen, and I don´t have the bat. CR123 and chagers for it.


I´m really confused!

 

[positron]

I did the DDL and it´s ok. Nothing to say about it.
But once time I got the mA an Volts to drive my laser, can I use 2 NiMH (AA or AAA) and a resitor in series with diode?

I almost think that answer is not. but WHY if the Ohm´s Law is the LAW ?

My diode is running at 2.5 Vots and 100 mA (an 8x DVD)
With those info, what resistor I need?
V=RI -> R=I/V -> R= 0,01A/2.5Volts = 0,04Ohms
It´s very little... not a comercial value.
Is there something wrong?

A just want do a portable laser pen, and I don´t have the bat. CR123 and chagers for it.


I´m really confused!

well, the resistor has to drop the voltage that is the difference between the supply and the diodes voltage drop. since your batteries will be at more than 1.25V when fully charged and much less when getting flat, that makes it pretty dimm pretty quick since you would need to set your resistor at the appropriate value to drop the right voltage when the batteries are at full charge. I hope that makes some kind of sense

 

[Fred_DASP]

PLEASE, DELETE THIS TOPIC

 

[positron]

PLEASE, DELETE THIS TOPIC

That is like being snubbed! (maybe I'm too sensitive )

Did I misunderstand the question? Was my reply more confusing than no reply? Is it just a language/cultural barrier that got in the way? Asking for nothing more than the thread to be deleted is a weird reply.

Anyway, have a lovely day.

 

[Fred_DASP]

You got the question. And I the answer
I'm embarrassed about my own question.
I thank you for opening my mind.
have a nice day u 2

 

[woop]

I don't think we have any mods, so this topic is staying where it is.
its a lesson for others anyway.

to put the answer another way, Ohms law is the law, but batteries are not a very constant voltage source, basically its a sucky way of powering your laser, and there are so many better ways.
however, there is no reason why you can't do it, you will just not get a very constant light output, and you run the risk of blowing your laser after you charge the batteries or turn on the laser from a cold start

oh yeah and thats not the way you use ohms law in this situation, you want a resistor to drop the excess voltage from the battery.
so v= Vbat-Vld = around zero for 2AA ni-mh. this is why some people run burners directly off 2 AA's. you might want to put a 10ohm pot in series with the laser if you are going to try this, just in case, then you can adjust it according to the battery charge.
by the way, i have never run a laser directly off 2 AA's, i didn't want to risk it... and i don't recommend it either

 

[IgorT]

You got the question. And I the answer
I'm embarrassed about my own question.

You shouldn't be embarrassed about this question. Many people are wondering about the same thing, and then they ask why their diode is dead.

Others extract the diode, and then test it by connecting it directly to a Li-PO with more than 3.6V (a full Li-PO has 4.2V), and then don't understand why this is not a good idea. I don't even see the point of such a "test". It's not really much of a test if the only thing it does is verify that the diode did work, before you "tested" it. Now that would be an embarrassing question... And yes, it was asked.


You already got most of the answers, but another reason why you wouldn't want to connect the laser diode directly to the batteries (even if they could provide constant voltage) is because the diodes drop in resistance as they warm up. A proper driver will just lower the voltage a bit to keep current constant. A voltage source would push more current through the diode, resulting in more heat, resulting in even more current, ultimately resulting in a dead diode (thermal runaway).

With a driver you get the same power every time you turn the laser on (until the batteries are empty) and if you put different batteries in, or even batteries with a higher voltage, the diode will still see the same current and be safe.

You can push a diode further with a driver, and it would still be safer than without.