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Looking for a red laser diode circuit (controlable)

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harryd

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hi, i have a red laser diode and i want to build a laser with a max of 5mw output but i want to control the output using a voltage/current supply. so basically what i am looking for is a simple reliable circuit that makes the laser diode works and i can control the output by changing the voltage. Can u help me plz???
 
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sorry meant 5mw


can i just use one of the those simple Driver circuits and change the pot value, hence, the output power changes ?
 
well what circuit are you refering to?
and you cant make a 5mw diode any better than 5mw without making your diode last for about 30 seconds. your driver will put to much current through your diode and blow it.
with the driver i cannot use the diode thats linked below unless i modify it becuase the driver nneedes 55milli amps und up thus meaning you can adjust the laser driver to your diode.
i dont really understand what you want, do u want to make a 5mw laser 10mw? or even burn to burn u need over 100mw
cheers













10mW Laser Diode - Jaycar Electronics check out this aussie website for your parts.
 
ok i have a 1x laser diode. i need a circuit for the laser diode to work and i can control the value of the output. so i want the max to be 5mW, because i will use it in a simple experiment ( measuring the beam width of the beam). also, i want to measure with 3 or 4 different output power, lets say, the 3 different output power are 1.5mW, 2.5mW and 3 mW.

i was reading some topics before, and i came across the design attached below.

as i can see the pot is used to change the overall resistance of the circuit, which controls the output. so if i can change the value of the pot the value of the output changes as as well, right ?
 

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harryd said:
ok i have a 1x laser diode. i need a circuit for the laser diode to work and i can control the value of the output. so i want the max to be 5mW, because i will use it in a simple experiment ( measuring the beam width of the beam). also, i want to measure with 3 or 4 different output power, lets say, the 3 different output power are 1.5mW, 2.5mW and 3 mW.

i was reading some topics before, and i came across the design attached below.

as i can see the pot is used to change the overall resistance of the circuit, which controls the output. so if i can change the value of the pot the value of the output changes as as well, right ?
Click to expand...

That is the circuit that is known as the DDL Current regulated Laser
Driver I linked to above... (notice (AKA DDL) on the Drawing)


Jerry
 
thanks for helping man :D
it is just i dont have much time and i have alot to do.

if i build the same circuit, and change the value of the pot, will the output of the laser diode increase?
 
well i have an optical power meter in the lab, i can measure it and make sure i don't reach the maximum right ?

ok now for the values how do i set the max of the laser diode. i mean, some people make lasers more powerful than other, how do i make the max. 5mW
 
well on the diode it will tell you the max mili amps and the max voltage in put just dont exeed that? if you want a 5mw to be more powerful just get a 10mw.
may i ask y you want to make a a5mw powerful? it definatly wont burn when u change it
 
i dont want to make a 5mw powerful . all i want is to get 3 different output power from the laser. so i can do the experiment 3 times with different power. i dont want to exceed 5mw , because after that i think i am gonna have to wear eye protection and i want to keep it as simple as possible
 
harryd said:
i dont want to make a 5mw powerful . all i want is to get 3 different output power from the laser. so i can do the experiment 3 times with different power. i dont want to exceed 5mw , because after that i think i am gonna have to wear eye protection and i want to keep it as simple as possible
Click to expand...

Unless the diode has an APC circuit integral, the amount of current corelates with the laser's output. Turn up the current, the output goes up. Turn down the current, the output goes down. Since you have a meter available (the only way to measure output), set up the diode/driver, shine it on the meter and VERY CAREFULLY adjust the current.

Remember to give it enough voltage at the driver. You need whatever voltage the diode needs (probably 3V) PLUS whatever voltage drop-out the driver has (IIRC the DDL = 2.2V). So, you would need >5.2V at PSU for the driver to regulate the current. If you drop below that voltage threshold, the driver will not regulate the current effectively and the current will begin to drop.

It is unclear whether you have any focusing optics included in your experiment. Be aware that diodes do NOT produce the type of "beam" that you see in our pictures. The beam from diode lasers diverges very quickly. They must be focused through optics to get a "beam."
If you need a "beam" without optics, you will have to go with some sort of gas laser. At the low outputs you are talking about, a helium/neon (HeNe) will do.

Peace,
dave
 
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thanks a lot man for the advice. actually what i am trying to do is measure the beam width of the laser, using the same setup as the knife edge technique but only using an optical fiber to couple the laser beam.

i have three questions
1- the laser diode works on 35mA and max of 50m A and according to the equation 1.25/current= resistance. 1.25/ 50mA= 25 ohm. does this mean i should have a total of 25 ohm ( 2x 10 ohm=5 ohm + the value of the pot=20 ohm ) and that would be as far as i can go with the pot, right?.

2-about the dummy circuit, when i measure the voltage across the 1 ohm resistance..... then what?. i mean what does this value imply.

3- about the input voltage would a 9v battery get the job done ?


again thanks man, u are really helping me :D
 
For 35 mA you need 35 ohms, for 50 mA you need 25 ohms resistance. So the logical thing to control that range is using a fixed 25 ohm resistor in series with a 10 ohm potmeter. Take a 22 ohm (standard value) in series with a 20 ohm pot to get some more range.

On question 2: you measure the current in mA as mV on the multimeter.

A 9V battery would be fine.
 
ok thanks
on your answer to question 2: after i measure the current in mA... what is this value, is this the value of the output current if i the load is replaced with the laser diode ??

and how do i convert this current to power, is there any equation for that ?
 
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