I have a question that asks to calculate all currents and voltages of a common base BJT transistor amplifier circuit with the emitter resistor removed making it an open circuit. I would think it should just be like getting rid of the entire left side of the circuit, and current would flow through BC going clockwise around the second part, but I'm not sure what happens with Vee still connected. In the question Vee = -3v and Vcc = 12v. Would I just subtract Vee from Vcc and pretend it's just like having one voltage source of 9v?

 

Maybe the teacher meant remove the resistance (not the resistor) in the emitter circuit?

 

I have a question that asks to calculate all currents and voltages of a common base BJT transistor amplifier circuit with the emitter resistor removed making it an open circuit. I would think it should just be like getting rid of the entire left side of the circuit, and current would flow through BC going clockwise around the second part, but I'm not sure what happens with Vee still connected. In the question Vee = -3v and Vcc = 12v. Would I just subtract Vee from Vcc and pretend it's just like having one voltage source of 9v?

If the emitter resistor is removed,the base-emitter junction is turned off,the CB junction is reverse biased,so what do you think happens to the collector current?

What does that imply about Vout?