Hello all,
I have a question regarding the BJT configuration that I have posted. The values I had to find were the current Ie and the voltage V3, which is the same as Vc.
The answers are: Ie = Ve - (-10v) / 4.7K = 2mA
Vc = 10V - IcRc = 10V - 2MA*(3.3K) = 3.4V
these answers are correct. But what I am confused about is the following:
When I first look at this simple circuit, I see that the base is tied to ground. So to me, that mean that there is no current going into the base, therefore the transistor is in cut-off. If it is in cut-off, then the transistor behaves as an open switch and there-for no current through the collector or emitter.
i know I am wrong here, but this is my first assumption when I first look at it. Can someone shine some light for me here. Thanks
I have a question regarding the BJT configuration that I have posted. The values I had to find were the current Ie and the voltage V3, which is the same as Vc.
The answers are: Ie = Ve - (-10v) / 4.7K = 2mA
Vc = 10V - IcRc = 10V - 2MA*(3.3K) = 3.4V
these answers are correct. But what I am confused about is the following:
When I first look at this simple circuit, I see that the base is tied to ground. So to me, that mean that there is no current going into the base, therefore the transistor is in cut-off. If it is in cut-off, then the transistor behaves as an open switch and there-for no current through the collector or emitter.
i know I am wrong here, but this is my first assumption when I first look at it. Can someone shine some light for me here. Thanks