Maybe this will help:
First notice that the voltage at Q4 emitter is
Ve = Ix * (Re4||(hie+Ry)) and if substitute Rx = (Re4||(hie+Ry)) we have Ve = Ix*Rx
Next we notice that I_hoe = Ix - (-Ib*hfe) and Ib is (Ix*Rx)/hie we have our KVL equation
Vx = Ix*Rx + [Ix - ((-Ix*Rx)/hie*hfe)]*hoe
And now all you need is to do it solve for Ix and Ro = 1/Ix

 

By Rx you mean the output impedance of this right side block, correct?

Also I didn't understood why I_hoe4 = Ix - (- IB*hfe)...

I would have considered that at node C, we have hfe*IB flowing out of the node C into node E and I_hoe flowing out of the same C node into E node, so I would write that Ix = hfe*IB + I_hoe, therefore I_hoe = Ix - hfe*IB.

Also I didn't understood how Ib = (Vx*Ix)/hie.

 

Rx = (Re4||(hie+Ry))

 

Rx = (Re4||(hie+Ry))

Ah ok... Sorry!

I've added some text to my previous post after hitting the "Post Reply" button

Also I didn't understood why I_hoe4 = Ix - (- IB*hfe)...

I would have considered that at node C, we have hfe*IB flowing out of the node C into node E and I_hoe flowing out of the same C node into E node, so I would write that Ix = hfe*IB + I_hoe, therefore I_hoe = Ix - hfe*IB.

Also I didn't understood how Ib = (Vx*Ix)/hie.

 

Also I didn't understood why I_hoe4 = Ix - (- IB*hfe)...

I would have considered that at node C, we have hfe*IB flowing out of the node C into node E and I_hoe flowing out of the same C node into E node, so I would write that Ix = hfe*IB + I_hoe, therefore I_hoe = Ix - hfe*IB.

Notice that Ib flow in opposite direction the the Ix. And this is why we must have a minus sign.
http://forum.allaboutcircuits.com/attachments/qrms9oo1-png.96632/

Also I didn't understood how Ib = (Vx*Ix)/hie.

Ah you are right. I completely forgot about Ry

Ib = (Vx*Ix)/(hie + Ry)

 

Notice that Ib flow in opposite direction the the Ix. And this is why we must have a minus sign.
http://forum.allaboutcircuits.com/attachments/qrms9oo1-png.96632/


Ah you are right. I completely forgot about Ry

Ib = (Vx*Ix)/(hie + Ry)

Ok, I got it about the minus signal... I was thinking about hfe*IB current direction...

But about IB, I'm not sure I get it!
Im trying to use units to try to track your thought!

Vx = Volts
Ix = Amps
hie + Ry = Ohms

So we have ( V/Ω )*A and V/Ω = A, so the result of that equation should be A^2... I'm not understanding. I'm sorry!

 

Forgive me, I'm sick today and I have a fever.
Look at my original post #16 and you will see that Ib is (Ix*Rx)/hie. So the correct equation for Ib is :
Ib = (Ix*Rx)/(hie + Ry). Now everything looks ok. I hope so.

 

Ok, I think I got it...

I think you have mistype IB... Should be IB = ( Ix*Rx ) / (Ry + hie4), right?

 

Forgive me, I'm sick today and I have a fever.
Look at my original post #16 and you will see that Ib is (Ix*Rx)/hie. So the correct equation for Ib is :
Ib = (Ix*Rx)/(hie + Ry). Now everything looks ok. I hope so.

Yeah... I'm sorry for your health condition! I hope you get better soon. Festivities are about here and would suck that you have to go through them sick!

 

Yes, you are right. THX for kind words

 

Yes, you are right. THX for kind words

I'm sure you'll get better quick...

About the IB equation I'm still trying to understand it!

I've attached a picture. I'm not understanding the relation you're using to write that I_hoe = Ix - (-hfe*IB)...


At that Emitter node we have that

hfe*IB + I_hoe + IB = IE

<=> I_hoe = -hfe*IB - IB + IE
<=>I_hoe = IB*( (hfe + 1) -1 - hfe)
<=>I_hoe = IB*0 = 0, which, obviously, is wrong!!!

I'm not following your thought!

 

But in your small-signal model the base terminal is at ground. This mean the that if we assume that Ix is positive, the Ve will also be positive and Ib will flow in the opposite direction. And this is why a give this minus sign. Because Ix flow in opposite direction to the Ib current.

Is that not a standard procedure ??
http://forum.allaboutcircuits.com/t...uit-with-dependent-source.104152/#post-788871

 

@Jony130 is this the figure that you're referring to as the small-signal model? If so, isn't the Q4 base connected to Ry which was the Q3 input impedance?

 

Yes. But what is the question?

 

Yes. But what is the question?

You were saying that the base was grounded in the small-signal model! I didn't understood that! The Q4 base is connected to Ry...

 

I treat hie4+Ry as a single resistor connected between base and ground.

 

After doing all the math I get this result:

\(Ro = h_{oe4}+R_{E4}+\frac{(h_{fe4}* h_{oe4} - R_{E4}) *R_{E4}}{h_{ie4} + R_{E4} + R_y}\)

 

After doing all the math I get this result:

\(Ro = h_{oe4}+R_{E4}+\frac{(h_{fe4}* h_{oe4} - R_{E4}) *R_{E4}}{h_{ie4} + R_{E4} + R_y}\)

hoe4 is a conductance, not a resistance, so where you have hoe4 shouldn't it be 1/hoe4?

 

Yes, of course you are right. I do not know why but I always think is term of resistance ro. Additional I turn OFF my thinking process today.

 

I treat hie4+Ry as a single resistor connected between base and ground.

Ok, I understand that, but I don't understand why you say that the base terminal is grounded! Can you please try to explain me again?

Which version are you considering between the 2 pics attached?