As I said in my previous post your diagram is a small-signal model for AC signals only. And this model is just a abstract concept.
And the task of this abstract model is to help us analyze this nonlinear circuit using only traditional "linear abstract models" and network theories (Ohms law, kvl, kcl, mesh, nodal etc) based on linear algebra. Also BJT must be properly biased because AC signal will "modulates" this Q point in the "rhythm" of AC input voltage.
See the example
http://forum.allaboutcircuits.com/threads/class-b-push-pull-amplifier.85631/#post-614844

As for your dif amp as input signal voltage rise the Ie1 will also rise but Ie2 will decrease by the same amount.
So we can say that Ie2 is "negative" because Ie2 is now smaller than the Ie2 quiescent current.
http://forum.allaboutcircuits.com/attachments/1-png.30861/ (Q point is Ie1 = Ie2 = 1mA)
But Ie1 and Ie2 will still flow in the same way (into Ree).

 

Ok, now about node 1 to be considered as grounded! In my teacher's similar circuit (without Re1 and Re2), he says that that node is as if it is grounded!

Can you explain why?

 

Because the AC component of a Iee current is 0A. So, the voltage at node 1 must be constant.

 

Ok, so in my circuit, that same node will be also grounded?

Is my equivalent model correct? And if that node is grounded, can I do the Differential mode analysis as if there was no Re1 and Ree???

 

Ok, the previous post is wrong and I've re-worked that stuff!

Now I have this equivalent simplified Hybrid-Pi model and the corresponding Diff Mode Gain... I would like for confirmation on the result!


Errata:

Om the attachment, at the dependant current source, where is hie*ib should be hfe*ib!

 

Ok, I tried to evaluate the formulas to the Common Mode gain.

I attached the equivalent simplified Hybrid-Pi circuit and the formulas I used.

Please someone help me to confirm my calcs!

 

Both of your formulas looks good.

 

Oh, great!!

Ok, I've also been reading the book I bought, H&H - 3rd Edition but they equations they use there are not even cloe to the ones I got, at least at a first glance!

I attached a photo of that page and surrounded the equations of interest in green and red...
Can you give an insight of the differences between my equations and the ones used in the book?

Here is the image, full size
http://imgur.com/9y9kEo4

 

While nobody answers to my previous post, I'm going to advance in the assignment.

We are also asked to evaluate the CMRR. We were taught that CMRR = Adm/Acm.

But the fact is that I'm finding that the result is probably not matching the expectations! So I'm going to show you my work so that you can tell me if it is correct or not!

Knowing that hfe = 255, Ib = 8.07μA






but using this CMRR formula I get a value about 40 which I think it's low because I was expecting an higher value but I might be wrong about my expectations!

 

Oh, great!!

Ok, I've also been reading the book I bought, H&H - 3rd Edition but they equations they use there are not even cloe to the ones I got, at least at a first glance!

I attached a photo of that page and surrounded the equations of interest in green and red...
Can you give an insight of the differences between my equations and the ones used in the book?

First of all the equations from AoE are for asymmetric output (single-ended output) and they use re instead of hie. And I also prefer re (T- model)
http://forum.allaboutcircuits.com/t...citor-in-common-base.90570/page-2#post-658421


And your formula for Adm is for symmetric output.
So to get equation for asymmetric output simply divide your Adm equation by 2.

\(Aam =\frac{Adm}{2} = \frac{hfe*Rc}{2 (hie + (1 + hfe) Re1)} \)

Now if we replace hie with this:

\(hie = (hfe+1)*re\)

Where :
re - small signal resistance between base and emitter looking into the emitter

\(re =\frac{dV_{be}}{dI_{e}} = \frac{Vt}{Ie}\approx \frac{26mV}{Ie}\)

\(Aam = \frac{hfe*Rc}{2 (hie + (1 + hfe) Re1)} = \frac{hfe*Rc}{2 ((hfe+1)*re + (1 + hfe) Re1)} = \frac{hfe*Rc}{2 ((1 + hfe)*(Re1+re))}\)

Now notice that hfe/(hfe+1) = Ic/Ie = alpha ≈ 1 we finally have this :

\(Aam = \frac{Rc}{2(Re1+re)}\)

\(Adm = \frac{Rc}{Re1+re}\)

We are now getting exactly the same equation as in AoE.

Aam = 2.2kΩ/2*(100Ω+12.1Ω) = 9.81V/V

Adm = 2.2kΩ/(100Ω+12.1Ω) = 19.62V/V

Acm = Rc/(Re1+re+2Ree) = 2.2kΩ/(100Ω+12.1Ω+4.4kΩ) = 0.488V/V

And CMRR = 19.62/ 0.488 = 40.2

 

Now if we replace Hie with re we get this

I didn't understood it here...

Replace hie by re, where? I don't know that formula for hie. When you write 'Hie', is it the same as 'hie'

Does the 'd' before Vbe and Ie stands for derivative???
I only know the formula for 'hie' as hie = Vt/Ib... (Quiescent Ib).

Now notice that hfe/(hfe+1) = Ic/Ie = alpha ≈ 1 we finally have this :

Does that relation comes from the active zone relations we know between Ic, Ib and Ie??? That alpha parameter is that parameter that stipulates the beta of a transistor?

Finally, when you wrote Adc at the end of your post, did you meant Acm (Common Mode Gain)?

 

Replace hie by re, where?

For example here
Adm = (hfe * Rc)/(hie + (hfe+1)*Re)

hie = (hfe+1)*re

Adm = (hfe * Rc)/((hfe+1)*re + (hfe+1)*Re) = Rc/((re + Re) * hfe/(hfe+1) ≈ Rc/(re + Re)

I don't know that formula for hie.

Now you know

When you write 'Hie', is it the same as 'hie'

Yes I made mistake, I should use a lowercase letter.

Does the 'd' before Vbe and Ie stands for derivative???

d = derivative

Does that relation comes from the active zone relations we know between Ic, Ib and Ie??? That alpha parameter is that parameter that stipulates the beta of a transistor?

Alpha is a current gain for a common base amplifier.
Current gain = Iout/Iin = Ic/Ie and Ic = Ib*hfe and Ie = Ib*(hfe+1) and this is why Ic/Ie = (Ib*hfe)/(Ib*(hfe+1)) = hfe/(hfe+1)

Finally, when you wrote Adc at the end of your post, did you meant Acm (Common Mode Gain)?

Yes.

 

Ok, I finished the part of Gains and CMRR and all match LTSpice closely. I have simulated the circuit separately for Common Mode and Diff Mode.

Now teacher is asking what would be the range of the differential voltages! I don't know what teacher means with this! Does anyone has a clue?

 

Ok, I got the question about differential voltages rage ...

I'm starting to analyse an improved version of that circuit, using a current source built with 2 transistors and placed at the node 1 as the attached asc/png file shows.

I just don't know where to start to evaluate the transistors Q-Point!

Any hint to start?

 

Start at Q3 , Rc3 and Re3. Simply write a KVL for this part of a circuit.

 

Start at Q3 , Rc3 and Re3. Simply write a KVL for this part of a circuit.

So if I'm not mistaken the equation is:

Vcc = Rc3*Ic3 + Vce + Re3*Ic3*( (β+1) / β)

I pictured a circuit like this:


Can you confirm, please?

 

Your equation looks good. So Vce3 = ? and Ic3 = ?

 

Your equation looks good. So Vce3 = ? and Ic3 = ?

Ok, I think I got this Q-point too with the help of a fellow net friend!

It was a long way but I got there, I guess.

The circuit schematic and loops can be viewe at the attachments!

For the Q3 I found Q-point as Q-p (0.67V; 4.24mA). Vce = Vbe as they are shorted...

The I had some help from some other firends (old IRC) and I think I managed to sort out the whole circuit to find out the new Q-point of all transistors.
It was a long night until 4:00 am and I'm not sure if everything is OK, so I'm going to post here all my calcs, properly commented, so that someone can check it.

Before that, in that current source (mirrored, as far as I know), is the Q3 Q-point equal to Q4 Q-point???? I got 2 different Q-points for them!


Ok, firstly, used the loop marked as 1 to find IC4, knowing that VCE4 = VEB4 because these 2 nodes are shorted, so:
I'm sorry the image sizes... I can't control them here directly in the forum.


So, Q3 Q-point should be Q3-point (0.67 V; 4.24 mA)

As this is a mirrored current source, IC3 = IC4 = 4.24 mA

From here I can find IC1 because:



and therefore



Now I need V1, which can be evaluated by:


This V1 matches VC4, that will be needed to find VCE4


Now I'll need IE4 to find VE that will also be needed to find VCE4




Now, we can evaluate VE4 as:




And after that, I find VCE4 by means of:




and finally to find VCE1 I'll use the loop 2 equation:




I found that:

Q1 Q-point (6.01 V, 2.11 mA)
Q2 Q-Point = Q1 Q-point
Q3 Q-point (0.67 V; 4.24 mA)
Q4 Q-point (8.286 V; 4.24 mA)

But I'm not sure about 2 things:

1 - if Q1 Q-point is really exactly the same as Q2 Q-point

2 - if Q4 Q-point should be equal or not to Q3 Q-point!

 

AD1 - If transistors are identical and Vd = 0V then Ic1 = Ic2.
AD2 - Because you prefer "pedantic calculations" the answer is no. Ic3 is not equal to Ic4.

 

AD1 - If transistors are identical and Vd = 0V then Ic1 = Ic2.
AD2 - Because you prefer "pedantic calculations" the answer is no. Ic3 is not equal to Ic4.

I forgot to add a screen of the schematic... Added now!

Transistors are supposed to be a match as I chosen 4 2N2222A from LTSpice library!

Sorry, but what you mean by AD? Is it Advice???

And what is that Vd that you're referring to?

What means pedantic calcs?

I think I have found a small difference between IC3 and IC4. In fact I_RC3 = IC3 + IB3 and IC4 might be different because of that... I probably assumed, wrongly, that IB3 = 0A. But I'm not sure if I can do that!


Edited...
My old Longman Dictionary told me what pedantic means!