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reducing voltage via potentiometer

izzy007

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no?
then what will it be. it supposed to make the pot see 1.25v isnt it. thats you do 1.25/resistance.
 



chimo

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izzy007 said:
no?
then what will it be. it supposed to make the pot see 1.25v isnt it. thats you do 1.25/resistance.
You asked if the output of the LM317 was 1.25V.  No, it will not be.  The voltage drop from Vout pin to the Adj pin will be 1.25V.  

The voltage drop across the LM317 will depend on a variety of factors.  You are creating a current source so you do not have to worry about the voltage - other than ensuring your supply voltage is high enough to make up for the voltage drops across the components connected in series to the load (LM317 and sense resistor).

The voltage drop across the LD will depend on the current and the temperature of the LD.  As it heats up, if the current is held constant, the voltage will drop will be less.

V(batt) - V(LM317) - V(Rsense) - V(LD) = 0

If conditions are correct, the LM317 will try to maintain 1.25V between the Vout and Adj pins. As a basic explaination, it will do this by acting like a variable resistor between its Vin and Vout pins.


 (+)--->(Vin)LM317(Vout)--->Sense Resistor(Adj)----------|
   |                   |_________________________|              |
Battery                                                                     Laser Diode      
   |                                                                                |
  (-)                                                                              |
    ----<-----------------------------------------------<----------
 




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