Question about the resistor value

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    Resistor Value
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SUMMARY

The ideal resistor value for a laser module operating at 3V and 45mA with a 3.3V power supply is calculated using the formula R = (V_supply - V_laser) / I, resulting in 6.6 ohms. However, using a constant current source is recommended due to the risk of exceeding the laser diode's current rating if the supply voltage fluctuates. A variable voltage supply with a fixed resistor, such as a 100-ohm resistor in series, allows for safe adjustments while monitoring current to prevent damage to the laser diode.

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anita1984
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Hello Forum ,
I would like to ask you about the resistor value ideal for a laser module 3V @45mA and the power supply is 3.3V , R= (3.3-3)/45= 6.6 ohm ? it sounds a little strange to me.
Have a nice day and thank you in advance.
 
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yes, that's one of the reasons why you need supply power to a laser diode using a constant current source.
 
It seems risky to have the power supply voltage so close to the laser voltage, though, unless you are sure it isn't going to change.

For example, if the voltage went up to 3.6 Volts, the current would go up to 90 mA. That is double the current for a 9% increase in voltage.

I read up on this a little and it seems you should start with a variable voltage and a fixed resistor in series and gradually increase the supply voltage until the laser starts "lasering".
No idea what that means but I guess it is an obvious change in the light output.
The article I read suggested a 0 to 10 volt supply and a 100 ohm resistor in series, but with a current meter in circuit to see that you don't exceed the laser diode ratings.

If it is a visible laser, you should see a change in behaviour, but if it is infra red, I don't know how you would tell where the best current was.
 

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