BJT question

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jack23

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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
 

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If the emitter was connected to 0V then what you are saying would be true as it isn't at 0V but -10V. The base is at 0V and the emitter is connected to -10V via a resistor meaning current is flowing from 0V to -10V.
 
Hero999 said:
If the emitter was connected to 0V then what you are saying would be true as it isn't at 0V but -10V. The base is at 0V and the emitter is connected to -10V via a resistor meaning current is flowing from 0V to -10V.

So, it does not matter if the base is connected to 0V, as long as the Base-Emitter junction is forward biased. Which in this case it is because the N-type emitter is at a lower voltage potential than the P-type Base. Correct?

Of course, the above statement is assuming the Collector-Base junction is reverse biased.
 
Ground is just an arbitrary reference point. Look at the 3 circuits below. The transistor quiescent point is exactly the same in all three.
 

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It makes me dizzy to see the ground bouncing around like that. Earthquake?
 
audioguru said:
It makes me dizzy to see the ground bouncing around like that. Earthquake?
Yep. It all started like this...
 

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