The above equation states that if you begin with the voltage applied to the circuit, 5V in this case and then you subtract the voltage drop across the resistor Rb and then you subtract the voltage across the base-emitter junction, you should end up with 0 volts.Let Rb= 1k , Rc=100 and Vcc= 5v
Can anybody simplfy this circuit and explain why is Vcc-V(Rb)-V(Be)=0
Thanks
I already know that. But I'm wondering if we used KVL why ignored V(Re) as in the left loopThe above equation states that if you begin with the voltage applied to the circuit, 5V in this case and then you subtract the voltage drop across the resistor Rb and then you subtract the voltage across the base-emitter junction, you should end up with 0 volts.
hgmjr
The loop is the one you posted first, Vcc-V(Rb)-V(BE)=0.can you tell my which loop are you loooking at
by Luke James
by Jake Hertz
by Luke James