Two laser pointers

[cannibal]

hello

Need help in how to connct 2 optical laser pointers togather in one circuit using one battry 9V with switch to awitch ON/OFF the lasers togather

??

thanks

 

[audioguru]

It looks like you have a cheap laser pointer. it operates at about 3V from 3 coin battery cells in series. The internal resistance of the battery limits the current to about 40mA.

A 9V battery will need to have a current-limiter or resistor in series with each laser pointer to limit the current to about 40ma. Then two-thirds of the power of the battery will be wasted in the current-limiter because the battery voltage is too high.
The 9V alkaline battery will last about 6 hours.

 

[mneary]

Connect them in series, with a resistor to limit the current. Do you need help with how to connect a switch?

 

[cannibal]

Hi

So what is the best way to avoid the wastage of the voltage?

The problem is I can't find a holder for 3 coin battery cells in series


I attached image of the parts please provide me with the best solution to connect them to keep the system running for longest possible time (without using power supply)

 

[audioguru]

Little coin cells don't have enough power to light a laser for much time. They don't have enough power to cause damage from too much current.

Larger AAA or AA battery cells can easily blow up a laser due to too much current so a current-limiting resistor MUST be used in series with the laser.

 

[cannibal]

Could you please show me how to connect since my skill is not that in electroics

 

[audioguru]

Wire the thing like this.
The resistor's value is calculated with Ohm's Law:
The voltage of the battery minus the voltage of both lasers in series divided by how much current you want in the lasers. Too much current and they blow up.

 

[cannibal]

I used Ohms law

I got 1.5 K ohms

6 / 4X10^-3 = 1.5K ohms

hope I did good
I can't see my lasers blow up infront of me

by the way I may need to know how to add LED to show that the system is running

 

[audioguru]

No.
Ohm's Law says that the value of a resistor is calulated by the voltage across the resistor divided by the current through it. The voltage across the resistor is much less than the supply voltage because the two lasers are in series with it.
How many volts for each laser? 2V? Then the resistor has only 2V across it.

A laser doesn't work at the very low 4mA that you used. It needs at least 20mA or more. But too much current will burn it out. You need to measure the original laser circuit to determine the voltage and the current.

Add an LED in series with a current-limiting resistor. If the supply is 6.0V, the LED is a 2V red one and its max allowed current is 30mA then the resistor has 4V across it and 25mA through it which calulates to a 160 ohm resistor. A 180 ohm resistor makes a current of 22.2mA.

 

[cannibal]

Hello

I decided to use power supply for this circuit

Could you please show me how to combine the connection of this circuit with the two laser pointers

https://www.aaroncake.net/circuits/supply6.asp

 

[audioguru]

Could you please show me how to combine the connection of this circuit with the two laser pointers

You forgot to measure the original circuit:
"You need to measure the original laser circuit to determine the voltage and the current."

If you don't, then the lasers either won't work or they will burn out.

 

[cannibal]

the voltage is 4.5 V
the current is 315mA

for one laser

 

[audioguru]

$.5V at 315mA is 1.5W That is an awful lot of power for a little laser pointer. It can do serious damage. Its battery would need to be pretty big to supply 315mA.

I think it is about 3V but the 3 tiny button battery cells drop their voltage with the laser as their load. I think the current of the laser is about 40mA.

 

[cannibal]

I saw the data sheet of small battery of the laser pointer

it shows 1.5V and 105mA

I think you are right maybe it's 3V with 40mA

so if I want to use the
https://www.aaroncake.net/circuits/supply6.asp

for supplying two laser pointers

what output voltage I should select
and what should be the value of U1 and T1
and it appears that all values of T1 give 1.5A

so please I need also help in selecting the right resistor and also inclouding one LED in the circuit that shows the system ON

 

[audioguru]

The laser uses current to glow, not the battery. The laser will burn out if its current is too high, not the battery.

The battery is NOT rated for 105mA. That is current. The battery is rated for 105mAh, which is how long it can supply 1mA (105 hours) or less.

The power supply circuit is able to supply from no current to 1.5A, or up to 0.7 times the current rating of the transformer. The load takes as much current as it needs.

You must measure the current in the original laser circuit for you to be able to calculate the current-limiting resistor.

 

[cannibal]

Hello

I opened the laser pointer and I found this cirucit

I need help to measure it

the laser came with lamp that was connected to blue and red wires
I cut it because it is not needed

please see the attachment

 

[audioguru]

I have a little laser pointer.
It uses 3 alkaline button battery cells (a total of 4.5V without a load) for power. The internal resistance of the battery cells reduces their current to about 30ma to 40mA to the laser that is about 3V.

I don't know the voltage of your laser diode and don't know how much current will burn it out.

The voltage is low so it is difficult to measure the current. A current meter will reduce the current.

 

[cannibal]

May I know where is the + and - in the laser so I can test without the switchs

 

[tom3000]

i dont know where you get your lasers from, but mine have additional resistor behind the laser diode, i ripped about 6 of them appart (they're cheap here $1, 3 batteries included)

all i do to mine is feed 4.5V directly from my LM317 and that's it, 3 AA will do fine

 

[audioguru]

all i do to mine is feed 4.5V directly from my LM317 and that's it, 3 AA will do fine

Button battery cells have an internal resistance that limits their current so the laser is not destroyed. A laser needs about 3V at about 40mA. Then the 4.5V worth of button battery cells have a total internal resistance of 37.5 ohms. Your 4.5V LM317 or 3 AA cells will burn the laser out.

Maybe the extra resistor is in series with the laser to help limit the current.