Hello everyone. Would it be possible to check my analysis for my attached circuit?
Here is my analysis:
The question states that B is 200 and that Vbe = 0.7v.
From the graph, vb = 0v, because it is grounded. Therefore:
Vbe = Vb - Ve
0.7 = 0-Ve
Ve = -0.7V
Now to find IE. using ohms law I have:
Ve - (-10v) = IE*RE
(-0.7)+10 = IE *10K
IE = 0.93mA
finding IC by using the formula IC = αIE
α=β/(β+1)IE
α = 200/201* 0.93mA
IC = 0.925mA
Finding Vc using omhs law again
Vcc -Vc = ICRC
10-Vc = 0.925mA * 10k
Vc = 0.75V
I would just like to know if my method is correct, because my teacher has completely different values! Thanks.
Here is my analysis:
The question states that B is 200 and that Vbe = 0.7v.
From the graph, vb = 0v, because it is grounded. Therefore:
Vbe = Vb - Ve
0.7 = 0-Ve
Ve = -0.7V
Now to find IE. using ohms law I have:
Ve - (-10v) = IE*RE
(-0.7)+10 = IE *10K
IE = 0.93mA
finding IC by using the formula IC = αIE
α=β/(β+1)IE
α = 200/201* 0.93mA
IC = 0.925mA
Finding Vc using omhs law again
Vcc -Vc = ICRC
10-Vc = 0.925mA * 10k
Vc = 0.75V
I would just like to know if my method is correct, because my teacher has completely different values! Thanks.
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