Both images are false. When you combined the two resistors (Ry + hie4) You have hidden "base" inside united resistor. Where the base you do not care. The importance of the base current.

 

Both images are false. When you combined the two resistors (Ry + hie4) You have hidden "base" inside united resistor. Where the base you do not care. The importance of the base current.

So why does @Jony130 says that the Base is grounded?

 

You guys are geniuses mate

 

But I do not understand your problem. As Bordodynov we do not care where the base is. The base current is what we should care about.
Also after we drawn a small-signal representation we now can forget that we are dealing with transistor circuit.



From now on we are dealing with a circuit that contains a dependent source.

For example we can use a nodal analysis to solve for Ix.

Ix = Ve/(RE||(hie+Ry))

And the nodal equation for Ve is:

\(hfe*Ib +\frac{Vx-Ve}{ro}=\frac{Ve}{RE}+\frac{Ve}{hie+Ry}\)

And one additional equation for Ib =-Ve/(hie+Ry)

And solve the circuit.

\(Ve = \frac{RE (hie + Ry) Vx}{hie (RE + ro) + ro*Ry + RE (ro + hfe* ro + Ry)}\)

and Ix

\(Ix = \frac{Ve}{\frac{1}{\frac{1}{RE} + \frac{1}{hie+Ry}}}=\frac{(hie + RE + Ry) Vx}{hie (RE + ro) + ro*Ry + RE (ro + hfe*ro + Ry)}\)

And
\(Rout =\frac {1}{Ix} = ro +\frac{ RE*(hie + hfe*ro + Ry)}{(hie + RE + Ry)}\)

 

I'm going through your equations, but the one for the currents at node E, according to the signs that you used, suggests that IB and IE current are flowing out of the node E, and hfe*Ib and I_ro are flowing into node E, right?

If my thought is correct, the Ib current direction you drew in the schematic should be out of node E into the node GND, no? Because the very next equation you say that that same current have the minus sign!

I mean:



The one in red is Ib, right???

Then you just say that


So, shouldn't the first equation be:

 

I'm going through your equations, but the one for the currents at node E, according to the signs that you used, suggests that IB and IE current are flowing out of the node E, and hfe*Ib and I_ro are flowing into node E, right?

Yes, exactly

If my thought is correct, the Ib current direction you drew in the schematic should be out of node E into the node GND, no? Because the very next equation you say that that same current have the minus sign!

No. In my diagram the CCCS current flow from top to the bottom ( into node Ve) only if Ib is flow in the direction I drew on the diagram.
So, if you change Ib direction, you also need to change the CCCS current direction.

And now the circuit looks like this

 

Rout=Vx/Ix not Rout=1/Ix !
But the final result is true. When I make these calculations, I instead of unknown value (Vx) using 1V (Ux = 1). This method eliminates of mathematical expressions (equations) one variable.

 

Rout=Vx/Ix not Rout=1/Ix !
But the final result is true. When I make these calculations, I instead of unknown value (Vx) using 1V (Ux = 1). This method eliminates of mathematical expressions (equations) one variable.

I also use Vx = 1V and this is why Rout= 1/Ix is true

 

So why does @Jony130 says that the Base is grounded?

It is grounded through hie4 and Ry. There's no signal (AC signal, that is) coming from Q3, so the base of Q4 is effectively grounded through hie4 and Ry.

You could have solved this mirror circuit more simply and accurately by just using nodal analysis to solve the whole thing.

There are 4 essential nodes in your circuit. Number them like this:

Node 1 is the collector of Q3 and bases of Q3 and Q4; voltage here is v1
Node 2 is the collector of Q4; voltage here is v2
Node 3 is the emitter of Q3; voltage here is v3
Nove 4 id the emitter of Q4; voltage here is v4

You have only to write 4 nodal equations, and since you only want impedances at the nodes, you don't have to provide a stimulus vector.

For example, the equation for node 1 is:

(1/hie3 + hfe/hie3 + 1/hie4 + hoe + 1/Rc3)v1 + (0)v2 + (-1/hie3 - hfe/hie3 - hoe)v3 + (-1/hie4)v4 = 0

You can write these equations by inspection of the circuit. The equations can be put in the form of a Y matrix. Then the matrix is inverted to become a Z matrix. The impedances of the 4 nodes are on the main diagonal of the Z matrix. The computer is doing all the hard algebra when it inverts the Y matrix. Here's the result:



This method makes no approximations, and the symbolic results can be complicated. But, once you substitute numerical values, the complicated expressions for the 4 impedances become single numbers.

The value of Ry from this method includes not just impedances associated with Q3, but also the effect of hie4 and RE4; this is slightly different from what Jony130 did. He includes hie4 and RE4 in his final solution for Ro:



You also get the impedances at the two emitters for free even though you didn't need them.

Notice that Jony130's result is equal to mine to 7 decimal places. The tiny difference is due to the fact that he didn't take hoe into account in his expression for Ry. It makes essentially no difference to ignore hoe in the computation of Ry, but it is crucial to include it in the computation of Ro

 

Notice that Jony130's result is equal to mine to 7 decimal places. The tiny difference is due to the fact that he didn't take hoe into account in his expression for Ry. It makes essentially no difference to ignore hoe in the computation of Ry, but it is crucial to include it in the computation of Ro

Of course you hoe3 have in mind ?

 

We could use the same kind of nodal analysis to determine Ry considered only due to Q3. We only need two nodal equations in this case:

 

Of course you hoe3 you have in mind ?

Of course.

And, of course, if it isn't obvious to everybody, I just picked an arbitrary value of 1/100000 for hoe (which includes hoe3 and hoe4 which weren't distinguished).

 

Also any comments on #85? And confusion about base current direction. Because I don't know how this can be explained to PsySc0rpi0n in simple words.

 

Ok, I appreciate the help and all he hard work you guys are having with me!

I'm not sure if I fully understood that Ib current matter, but I need to move forward. I understand that Ib should flow into node E because it's the way it works on the transistor. Ib flows from Base to Emitter! Period! I understand the nodal equation @Jony130 wrote for that node if Ib current was flowing out of node E to the GND node!

Anyway, I need to move forward and I'll ask my teacher to explain that to me, face to face!

Ok, what about the equation you wrote for Ve.
Have you found that equation by solving the nodal equation at node E for Ve? Or have you used another method?

 

I just simply solve for Ve using this nodal equation


Also, which program you are using to write your equations ?

 

I just simply solve for Ve using this nodal equation
View attachment 96780

Also, which program you are using to write your equations ?

I use Code Cogs... Sometimes I paste the latex code here, sometimes I download the generated gif of the formula and upload it here in the forum!


I'm going to try to solve the equation for Ve!

 

Ok, I wasn't able to find the same result as you @Jony130 . At least apparently...

Can you check my calcs:



Step by step explanation of my work with the expression
2nd line - split (Vx-Ve)/(ro) into 2 fractions and moved Vx part to the other side of the equal sign, moved all terms with Ve into the left side and multiplied all terms by (-1)

3rd line - highlighted Ve/(hie+Ry) out of those 2 terms and Ve out of the other 2 terms, made 1/ro and 1/Re only 1 fraction.

4th line - highlighted Ve again

5th line - summed the 2 left fractions converting them into one fraction and moved the result to the other side of the equal to divide Vx/ro by this term.

6 th line - converted a fraction of a fraction into a single fraction!

Latex code for edition if needed, using that same site!

Spoiler: Latex Code

\displaystyle {\begin{matrix}
-hfe\cdot \frac{V_{E}}{hie + R_{y}}+\frac{V_{x}-V_{E}}{ro}=\frac{V_{E}}{R_{E}}+{\frac{V_{E}}{hie+E_{y}}}\\
\\
\frac{hfe\cdot V_{E}}{hie+R_{y}}+\frac{V_{E}}{r_{o}}+\frac{V_{E}}{R_{E}}+\frac{V_{E}}{hie+R_{y}}=\frac{V_{x}}{ro}\\
\\
\left ( hfe+1 \right )\cdot \frac{V_{E}}{hie+R_{y}}+V_{E}\cdot \left ( \frac{R_{E}+r_{o}}{r_{o}\cdot R_{E}} \right )=\frac{V_{x}}{r_{o}}\\
\\
V_{E}\cdot \left ( \frac{hfe+1}{hie+R_{y}}+\frac{R_{E}+r_{o}}{R_{E}\cdot r_{o}} \right )=\frac{V_{x}}{r_{o}}\\
\\
V_{E}=\frac{\frac{V_{x}}{r_{o}}}{\frac{\left ( hfe+1 \right )\cdot r_{o}\cdot R_{E}+\left ( hie+R_{y} \right )\cdot \left ( R_{E} + r_{o}\right )}{\left ( hie+R_{y} \right )\cdot \left ( r_{o}\cdot R_{E} \right )}}\\
\\
V_{E}=\frac{V_{x}\cdot \left ( hie+R_{y} \right )\cdot \left ( r_{o}\cdot R_{E} \right )}{r_{o}\cdot \left [ \left ( hfe+1 \right )\cdot r_{o}\cdot R_{E} + \left ( hie+R_{y} \right )\cdot \left ( R_{E}+r_{o} \right )\right ]}\\
\end{matrix}}

 

Your equation is correct. Now divide is by RE||(hie+Ry)

 

Your equation is correct. Now divide is by RE||(hie+Ry)

That operation is only for simplification or is it to calculate something else??? I got lost in the meantime!

Eited;

Is that for Ix???

 

Is that for Ix???

Yes, for Ix.
Ix = Ve/RE||(hie+Ry) also Vx = 1 therefore Rout = 1/Ix