Good job Sacha !
Now the LED problem is definitely solved...

Vasile

On Fri, 19 Jul 2002, Mircea Chiriciuc wrote:

> Use this. With a 1V voltage drop across the resistor the current is also
> limmited at about  30mA.
> The voltage drop parammeter for a zenner diode states for the voltage drop
> across the diode when reversed biased. So in the drawing attached the diode
> is "keeping" the voltage at a maximum of 5.1V. If the voltage tends to
> increase the zenned will permit current to flow trough it so the voltage
> drop will remain aprox. constant untill the current limit trough the diode
> is passed and the diode will becone a wire. If the diode becomes a wire, the
> resistor will get the full current of your batteries I=U/R=6/33=181mA. The
> power the resistor will try to excange in heat will be P=U*I=6*0.181=1.09W.
> If you use a 1/4W resistor it will provide also protection for your circuit
> as it will burn and open the circuit if the current will exceed 50mA.
>
> Good Luck,
> Mircea Chiriciuc.
>
>

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