BJT operation

[Ratchit]

Thanks.

It's collector resistor; the resistor connected to collector terminal of transistor.

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

Ratch

 

[PG1995]

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

Ratch

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]

Re: post #18, https://www.electro-tech-online.com/threads/bjt-operation.155724/post-1342003

I don't think that the book is correct. The output is inverted therefore there should be a minus sign to show that inversion for the amplification. As the voltage across transistor drops/rises, there is an equivalent rise/drop, respectively, in voltage across R_C. Hence, Vc=-(Ic*Rc).

Also, I do believe that it should have been "Vin" instead "Vs".

 

[Ratchit]

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.

I don't know how to do subscripts in this forum''s text. Anyway, do you understand now how those equations were derived?

Ratch

 

[Ratchit]

Re: post #18, https://www.electro-tech-online.com/threads/bjt-operation.155724/post-1342003

I don't think that the book is correct. The output is inverted therefore there should be a minus sign to show that inversion for the amplification. As the voltage across transistor drops/rises, there is an equivalent rise/drop, respectively, in voltage across R_C. Hence, Vc=-(Ic*Rc).

Also, I do believe that it should have been "Vin" instead "Vs".

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

Ratch

 

[PG1995]

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.

 

[PG1995]

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!

 

[Ratchit]

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!

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]

Thank you!

So, what was Vc in the link below? Is it the voltage across transistor, or the voltage across resistor? I'm sorry but I need to confirm this. I had thought that Vc is voltage across the transistor.

https://www.electro-tech-online.com/attachments/amp432-jpg.115825/

 

[Ratchit]

Thank you!

So, what was Vc in the link below? Is it the voltage across transistor, or the voltage across resistor? I'm sorry but I need to confirm this. I had thought that Vc is voltage across the transistor.

https://www.electro-tech-online.com/attachments/amp432-jpg.115825/

If a AC meter is used, the Vcc will be read as zero AC, and the AC voltage across the transistor or Rc will be equal.

Ratch

 

[PG1995]

Thanks.

In post #27 I meant to say AC voltmeters.

 

[MichaelaJoy]

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?

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.