modified 11 years ago

NPN driving Load - Normally off - Active on

01:20:00

Vcollector=Vout
Vin = 0V Then Vout = '1' when load is 'off', transistor is 'off'.
Vin = 5V Then Vout = '0' when load is 'on', transistor is 'on'.
Potential divider network, R/(RB +R) = VBase.
-To turn 'on' transistor Vb > Ve by 0.7V
-Make Vb=1V when Vcc=5V
-Therefore Rbase=4K and R=1K then 1/(1+4)=0.2 x Vin
-When Vin = 5V, Vbase = 0.2 x 5 = 1Volt. Target reached.
Rload=RLed= 250R for 20mA with 5V Vin

published 11 years ago

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