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# Questions about mAh [HELP A NEWBIE AS ME]

#### megha

##### New member
Hi,
I have to supply my new laser with a maximun of 12V @ 180mAh. (6x diode, 405nm, not GGW, maybe it`s the other diode, as SF-AW210)

Which batteries do you suggest ?

4 x 3.0V @ 1000 mAh (CR123A)
3 x 3.7V @ 1300 mAh (CR123A)
3 x 3.7V @ 880 mAh (LIR123A)

What I don`t understand about mAh is... when I know that I have to "give" or "supply" 180mAh to my laser, we are talking about how much electricity or power is going from the battery to the laser ? and when a rechargeable battery says "1300mAh" is the capacity to be charged or to give power? Because I`ve read some alkaline batteries that also say "nominal power 50mAh" and there is when I don`t know when we are talking about power supply capacity or charging (holding) capacity.

Another question that you have to laugh about: My laser is regulable, that is where I set the quantity of mAh will the circuit accept ? so it will determinate the mW of output?

Sorry for my bad english and my stupid questions, but I`ll burn it if I do a disaster... thanks for your time

#### Cyparagon

##### Well-known member
I think you're a bit confused. What driver are you using? 12V is absurdly high.

#### iskor12

##### New member
mA is the amount of current that is flowing.
mAh is the capacitance of the battery.

if you circuit is running at 180mA and you use a 2400mAh battery.....it should run for about 13.33 hours before the battery dies.

2400mAh/180mA=13.33hours

Also, if you are going to use rechargable batteries....make sure you know what the voltage is.

rechargeable
3.0 V battery when fully charged = 3.6 V x 3 = 10.8 V Total
3.6 V battery when fully charged = 4.2 V x 3 = 12.6 V Total

You will likely fry your driver and diode if you use 4 of either battery. Stick to 3 batteries max.

12 V is the max that the driver can handle. You would be more than ok if you only used 9 V.

#### HIMNL9

##### New member
Hi, Iskor, sorry if i correct you, but the driver can handle more than 12V (in fact, the LM317 can handle a maximum VI - VO difference of 37V )

Then, no problem using 4 batteries ..... the problem, in fact, can be another one.

The regulator, for keep the current constant, must reduce the voltage, and this mean that it need to dissipate the corresponding energy in form of heat ..... and, more voltage you give it, more heat it need to dissipate ..... so, just as example, if you use 11/12 V, and it need to dissipate half watt, using 24V, it still regulate the current in the same way, but need to dissipate like 4 or 5 W ..... and cause the TO220 case is designed for a common 1 to 1,5W dissipation, it go quickly in overtemperature protection (the driver shut down itself when the junction temperature exceed 120 degrees)

For this reason is better to limit the power supply to 3 batteries, but you don't risk to burn it, using 4 or 5 of them ..... just risk to have a driver that work intermittently (and that burn your fingers, if you touch it when the case is around 100 C )

#### eetlotsgloo

##### New member
mA is the amount of current that is flowing.
mAh is the capacitance of the battery.

if you circuit is running at 180mA and you use a 2400mAh battery.....it should run for about 13.33 hours before the battery dies.

2400mAh/180mA=13.33hours

Also, if you are going to use rechargable batteries....make sure you know what the voltage is.

rechargeable
3.0 V battery when fully charged = 3.6 V x 3 = 10.8 V Total
3.6 V battery when fully charged = 4.2 V x 3 = 12.6 V Total

You will likely fry your driver and diode if you use 4 of either battery. Stick to 3 batteries max.

12 V is the max that the driver can handle. You would be more than ok if you only used 9 V.

I just feel the need to nitpick. mAh is not capacitance, it is capacity. Capacitance is measured in Farads (F). Also, your runtimes are just a rough estimate. As voltage drops, current draw increases, depending on the driver. Although with the LM317 I think efficiency increases instead. Basically what I'm saying is, runtimes aren't quite that easy to calculate

@OP:
I think that driver should be able to deliver full power to the laser over 7.2v. That might be for red diodes though. That is the low end of voltage for 2x li-ion cells. Once they hit 3.6v/cell, voltage drops real quick. It is also recommended to charge them around this point, for increased cell life.

D

mah = ma/h

better now?