So, can I say that my Vi is either the voltage drop across Re or across hie, is this correct?

Yes, you are right.

If this is correct isn't Vi = ib*(hfe + 1)*Re or Vi = -hie*ib???

It may sound strange but this is wrong why? Well let as exam the simplest example:
I assumed Vin = 1V ; Re = 1kΩ ; hie = 1kΩ ; Rc = 1kΩ and β = 100.



I hope that now you see that the current through Re resistor is equal to IRe = Vin/Re =1V/1k = 1mA and also Ib = -Vin/hie = -1V/1k = -1mA.
Also notice that Iin = IRe + (Ib + Ic) = IRe + (β+1)*Ib. If you do not see it, well you are in big trouble.
But can you solve for Vout ?

 

I don't understand a few things.

1 - Why you have ignored/omitted RLoad?

2 - Why isn't Ire = Ie = ib*(hfe + 1)??? If we look to the transistor, Ie current flows to gnd through Re (on the original circuit).

3 - The horizontal line where you drew the Iin, the 1V mark and the E node, is the same node, the node E, right?
So, can I say the following, regarding currents?

If so, I would say that Ib+hfe*Ib = Iin + Ie, so I can't understand why you say that Iin is the only current getting into node E and that all the other 3 currents are getting out of node E.

 

1 - Why you have ignored/omitted RLoad?

Because in MY example I do not care about RL and at this point it is irrelevant detail.

2 - Why isn't Ire = Ie = ib*(hfe + 1)??? If we look to the transistor, Ie current flows to gnd through Re (on the original circuit).

Because we are talking about AC current situation and you forget about Vin and Iin.

3 - The horizontal line where you drew the Iin, the 1V mark and the E node, is the same node, the node E, right?

Yes, you have right.

So I can't understand why you say that Iin is the only current getting into node E and that all the other 3 currents are getting out of node E.

Because if we assume that the Vin = 1V then this automatically means that the emitter voltage is also at 1V. So from this, I already know that the current through Re resistor will flow from emitter to ground , the base current will also flow in the opposite direction.
And since the "control" current (Ib) is flowing in the opposite direction then hfe*Ib current must also flow in the opposite direction.
But maybe you're right I'm shouldn't mess with your head and using mental shortcut. And I should write this KCL for emitter node (look at my diagram in post 42) like this:
Iin + Ib + Ib*hfe = IRe from there we can find Iin:

Iin = -Ib - hfe*Ib + IRe

And since Ib =-1mA is negative as I already explained we end up with this expression.

Iin = -(-Ib) - hfe*(-Ib) + IRe = Ib + hfe*Ib + IRe

Iin = -(-1mA) - 100*(-1mA) + 1mA = 1mA + 100*1mA + 1mA = 102mA

 

Ok, I understood that and if I hadn't messed up my currents equation in my previous post, it would match yours. So, I agree with that!

So I should have wrote Iin + Ib + hfe*Ib = IRe. This says that there are 3 currents "going in" and one going out or the other way around!

But I think this equation is not the same as you wrote in your #41 post, I guess!

Iin = IRe + (Ib + Ic) = IRe + (β+1)*Ib

If I replace the values you used before in the above equation, are we gonna get the same results?

Edited;

Iin = 1mA + (-1mA - 100mA) = -100mA.

 

But I think this equation is not the same as you wrote in your #41 post, I guess!
Iin = 1mA + (-1mA - 100mA) = -100mA.

The equation from #41 post assumes that all the currents are positive which is true.

 

The equation from #41 post assumes that all the currents are positive which is true.

I can't understand that. If we take as reference, the gnd node, then we must consider or Ie negative and the other 3 currents positive or the other way around, Ie positive and all the others negative, right? How come you say that all currents are positive?

 

in = -Ib - hfe*Ib + IRe

And since Ib =-1mA

Iin = -(-Ib) - hfe*(-Ib) + IRe = Ib + hfe*Ib + IRe

But if you say that Ib = -1mA, then -Ib = 1mA.

So, when you write:

-(-Ib) this is the same as -(-(-1mA)).

I can't understand what you are trying to tell me with that "transformation".

If we assume your values:

Ie = 1mA and Ib = -1mA

and if we consider my equation, then I can understand it.

If you say that you consider all currents as positive, then I can't understand this statement nor the equation from your post #41.

PS: You say that I should read post #42. That post was mine. Do you really mean post #42?

 

All I'm trying to say is that if Vin is positive the current will flow in the way I show in post 47.

Ok, and that is true for any situation or only for small AC signal analysis?

 

It is true only when we analyse the CB amplifier using small signal analysis.
But for you it will be better if you will stick whit this starting equation.
Iin + Ib + Ib*hfe = IRe And finish the analysis using current direction show in #42. This will be less confusing/misleading for you.

 

It is true only when we analyse the CB amplifier using small signal analysis.
But for you it will be better if you will stick whit this starting equation.
Iin + Ib + Ib*hfe = IRe And finish the analysis using current direction show in #42. This will be less confusing/misleading for you.

Ok, so sticking to that equation, we are saying that IRe is getting out and that the other 3 currents are getting in.

So, to find Vi, as I already said, it is either the voltage drop across Re or voltage drop across hie, right?
And IRe \(\neq\)Ib*(hfe+1), right?

 

So, to find Vi, as I already said, it is either the voltage drop across Re or voltage drop across hie, right?

Yes

And IRe \(\neq\)Ib*(hfe+1), right?

Yes

 

Is this current representation correct to explain why \(I_{b}\cdot (hfe + 1) \neq I_{R_{e}}\)???

 

Yes, but can you tell me why do you still bother about that IRe current ?

 

Yes, but can you tell me why do you still bother about that IRe current ?

I'm just slowly trying to do something out of this circuit! Once I have already graduated at this class, it's no longer my major concern at the moment. However I still would like to get to some point wit this circuit!

 

RE is not Re.

 

RE is not Re.

I've seen that terminology but for currents and voltages. Using \(v_{ce}\), \(v_{CE}\), \(V_{ce}\) or \(V_{CE}\) is different. But I have never used it for impedances!

 

Morning Psy.
To check that you understand the operation of the Common Base circuit that you have posted in post # 52, ref attached image with component values added.

Are you able to calculate:

When Vs = 0v. the following: Vb, Ve, Vc, Ib, Ie and Ic

Also when Vs = 1v pk to pk at 1kHz,, what is the Vppk of Vout. ??

E

 

You mean when we make a small AC signal analysis? Shorting the caps and DC voltage supplies?

 

hi,
Use any method that you think is appropriate to answer the questions.
E

 

This thread is turning into a book...