Ok.

Ix = Ib*(β+1) and Ib = Vx/Rx we have

Ix = Vx/Rx*(β+1) so

Ix/Vx = (β+1)/Rx and from there Rt = Vx/Ix = Rx /(β+1)
Or notice that Vx = Ib*Rx and also Ix = Ib*(β+1) ---> Vx/Ix = (Ib*Rx)/(Ib*(β+1)) = Rx/(β+1)
And finally Ro = Re||(Rx/(β+1))

 

I'm sorry but I'm very confused!

Is on this diagram that you wrote those equations, ignoring temporarily Re?



Edited;

After looking to the diagram and to your equations I was able to clear out some of them.

By Nodal Analysis, I understood:
Ix = Ib + hfe*Ib = Ib*(hfe + 1)

And because Vx is applied to Rx, I can also understand that Ib = Vx/Rx

But what was the point on writing Ix as:
Ix = Vx/Rx*(hfe + 1) ???

Also at your post #17 you introduce a new parameter, Rt. What is this Rt?

 

Yes, I temporarily omitted the Re resistor because Re is in parallel with Vx.

 

But we do not have to do that
Ix = Vx/Rx + hfe*Vx/Rx + Vx/Re = ((Re + hfe*Re + Rx)*Vx)/(Re*Rx) ---->Ro = (Re*Rx)/(Re + hfe*Re + Rx) = (Re*Rx)/(( hfe+1)*Re + Rx)
But for me this expression is too complicated and this is why I omitted Re and find Rx/(hfe+1) and finally Ro = Re||(Rx/(β+1)). And this equation looks more pleasant don't you think so ?

 

But we do not have to do that
Ix = Vx/Rx + hfe*Vx/Rx + Vx/Re = ((Re + hfe*Re + Rx)*Vx)/(Re1*Rx) ---->Ro = (Re1*Rx)/(Re + hfe*Re + Rx) = (Re1*Rx)/(( hfe+1)*Re + Rx)
But for me this expression is too complicated and this is why I omitted Re and find Rx/(hfe+1) and finally Ro = Re||(Rx/(β+1)). And this equation looks more pleasant don't you think so ?

Indeed, but it's still a little hard to me to understand your method at the first time I have contact with it! I need to look to it several times before I can get into your mind and understand it the way you do!

We are not used, in classes, to do it that way. I mean to simplify the circuit and split the problem in smaller problems! Also, to do this, can lead us into simplification errors on the circuit. But I agree that if we were taught to approach it this way, probably it would be easier!

Yet about Ro, I would like to be able to understand it the way we do it in classes.

For instance, for Vi, we have wrote an equation of a closed loop (or net) that included voltage drop at hie and the voltage drop at Re//R_Load, right?

For Ro, we replaced R_Load by a voltage source Vx and we also shorted Vsig. From here, can I try to write an equation as we did for Vin? An equation of a closed loop (or net) that includes the voltage drop at Re, the voltage drop at hie and the voltage drop at Rb1//Rb2???

 

But what was the point on writing Ix as:
Ix = Vx/Rx*(hfe + 1) ???

Because now we can solve for Vx/Ix directly

Ix = Vx/Rx*(hfe + 1) / divided by Vx

Ix/Vx = 1/Rx * (hfe + 1) = (hfe+1)/Rx but now our result is in Siemens (1/R = conductance in Siemens ) not in Ohms. So to get the result in Ohms we take the reciprocal of the expression Vx/Ix = Rx/(hfe+1)

Also at your post #17 you introduce a new parameter, Rt. What is this Rt?

Do not bother about Rt it is only an auxiliary variable.

For instance, for Vi, we have wrote an equation of a closed loop (or net) that included voltage drop at hie and the voltage drop at Re//R_Load, right?

For Ro, we replaced R_Load by a voltage source Vx and we also shorted Vsig. From here, can I try to write an equation as we did for Vin? An equation of a closed loop (or net) that includes the voltage drop at Re, the voltage drop at hie and the voltage drop at Rb1//Rb2???

Yes, for sure you can use the same method for finding Ro that you have been using for finding Rin. And good luck this, "high entropy expression" guaranteed.

 

Also I want to point out that your voltage gain expression do not show/cover the hole story. What about Rsig ?? Rsig also have a big influence on the voltage gain.

 

Also I want to point out that your voltage gain expression do not show/cover the hole story. What about Rsig ?? Rsig also have a big influence on the voltage gain.

I saw that you have been including somehow Rsig, but we have not done any example in classes including Rsig. We have only included Rsig on some cases for Ro, not for Vi or Vo or Ri or input or output currents.

 

So why you drew a Rsig on the diagram ? Or maybe your teacher said to you that Rsig = 0Ω.

 

Ok, so gathering all info I get:

Ri
\(\displaystyle {R_{i}= \left (R_{b1}\parallel R_{b2} \right )\parallel \left ( hie + \left ( hfe + 1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right ) \right )}\)

Vi as we use to do in classes (not including Rsig):

\(
\displaystyle {\begin{matrix}
V_{i}= I_{b}\cdot hie + I_{e}\cdot \left ( R_{e} \parallel R_{Load}\right )\\
\Leftrightarrow V_{i}=I_{b}\cdot \left ( hie + \left ( hfe + 1 \right ) \left (R_{e} \parallel R_{Load} \right )\right )
\end{matrix}
}
\)

Vo
\(
\displaystyle {\begin{matrix}
V_{o}= I_{e}\cdot \left ( R_{e}\parallel R_{Load} \right )\\
\Leftrightarrow V_{o}=I_{b}\cdot \left ( hfe + 1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right )
\end{matrix}
}
\)

Av
\(
\displaystyle {\begin{matrix}
A_{v}=\frac{I_{b}\cdot \left ( hfe + 1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right )}{I_{b}\cdot \left ( hie + \left ( hfe + 1 \right )\cdot \left (R_{e} \parallel R_{Load} \right )\right )}\\
\\
A_{v}=\frac{\left ( hfe + 1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right )}{\left ( hie + \left ( hfe + 1 \right )\cdot \left (R_{e} \parallel R_{Load} \right )\right )}\approx 1
\end{matrix}
}
\)

Still editing for Ro.