BJT operation

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Ratchit said:
Why don't you refer to it as Rc as the book does? Then there won't be any confusion.

Ratch
Click to expand...

Okay. Anyway, the book uses capitalized subscripts for DC values. When you wrote "Rc", you used lower case "c" subscript. For example, the book says, Vc=Ic*RC where both "Vc" and "Ic" are AC values while "RC or R_C" a DC value.
 
PG1995 said:
Okay. Anyway, the book uses capitalized subscripts for DC values. When you wrote "Rc", you used lower case "c" subscript. For example, the book says, Vc=Ic*RC where both "Vc" and "Ic" are AC values while "RC or R_C" a DC value.
Click to expand...
I don't know how to do subscripts in this forum''s text. Anyway, do you understand now how those equations were derived?

Ratch
 
Everything looks OK to me in the book. Perhaps someone else can explain to you what it means.

Ratch
 
Thank you!

I think I do understand the equations. In Question #2, you have Vin=Vs+V_BB. Actually Vs is riding on top of DC supply V_BB.

Question #1:


As the input signal, Vin, increases, so does Vb , and Vc is an inverted output. Av=-Vc/Vb=-(Ie*R_C)/(Ie*r'_e)

Thanks.
 

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Hi again,

I thought that I should confirm this with you. From our previous discussion, it was said and concluded that both voltmeters, ckt_bjt111, would read the same, right? Thanks!
 

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PG1995 said:
Hi again,

I thought that I should confirm this with you. From our previous discussion, it was said and concluded that both voltmeters, ckt_bjt111, would read the same, right? Thanks!
Click to expand...
The sum of the voltages across the transistor and Rc equals Vcc. Therefore, if the voltage is large across the transistor, it is small across Rc and vice-versa. That makes the amplitude of the voltages unequal.

Ratch
 
PG1995 said:
Thank you!
Please have a look on this attachment, bjt_amp1a.

Question 1:
Shouldn't it be Vc = - (Ic*R_C)?

Question 2:
Shouldn't it be Vin instead of Vs?
Click to expand...

Question 1: Nope. It's a positive voltage. Think of the voltage divider rule, with the transistor acting like the lower resistor.

Question 2: No because the book is talking about Vbe, which is the voltage drop between the base and the emitter.
 
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