modified 9 years ago

rezystor do tranzystora

00:24:00

Ic=B*Ib                      // I Kolektora
Ib=Ic/B                      // I Bazy

B= 100A/A   // wzmocnienie tranzystora
Ic= 20mA = 0.02A (led) 

Ib= Ic/B = 0.02/100=0.0002A=200uA

R=U/I         //Rezystor
U=VCC-0.7        //napięcie na rezystorze
R=U/I=4.3V/0.0002A=21500ohm

published 9 years ago

GingerKing

9 years ago

Nice math but this is a bad idea. B increases with temperature which is not simulated in EC. In other words, you could potentially destroy your led with this circuit on a hot day. Im not saying that the circuit wont work. Just be careful. Good luck

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