hi Psy,
Have you read the link in my post #17.?
Regarding Vce
E

Yes I have. My last post was about that when I say that in equation 1 they are ignoring Re*Ie.

 

hi Psy,
Have you read the link in my post #17.?
Regarding Vce
E

EDIT:


Not really.
Think about what you are trying to determine for the Q point ie: Vcollector at the steady state collector current Ic.

So why isn't Re*Ie in the equation 1???

I have written the output net equation as Vcc = Rc*Ic + Vce + Re*Ie <=> Vce = Vcc - Rc*Ic + Re*Ie. So, I'm no really understanding why Re*Ie is not included in the equation!

 

hi,
Its considered that the Q-point is specified by the DC collector current and the DC collector-to-emitter voltage (in the case of a BJT).
E

 

Well, because a few posts back, Mr. EricGibbs said that Vce couldn't be 1.4V. He did the following: Ic*Vc = 4.18*2.2 = 9.6V. Then Vcc - Vc = 12 - 9.2 = 2.8V so it could not be 1.4V.

I don't know what to say.

Vce = Vc - Ve right ??

Vc = Vcc - Ic*Rc ≈ 12V - 4.2mA*2.2kΩ ≈ 2.8V right??

Ve = Ie*Re ≈ 4.2mA *0.33kΩ ≈ 1.39V right??

So Vce = 2.8V - 1.39V = 1.41V right??

 

You asked why: So why isn't Re*Ie in the equation 1???

I posted my reply in post #23.

Psy,
I think there has been some misunderstanding in the terms regarding Vc and Vce, where you kept including the IeRe value.
I consider Vc as the voltage between the collector and the 0V line, Ve the voltage from emitter to the 0V line and Vce as the voltage 'across' the transistor which is the value that is used for the Q point calculation.

Hope we are now both on the same page.
E

 

I don't know what to say.

Vce = Vc - Ve right ??

Vc = Vcc - Ic*Rc ≈ 12V - 4.2mA*2.2kΩ ≈ 2.8V right??

Ve = Ie*Re ≈ 4.2mA *0.33kΩ ≈ 1.39V right??

So Vce = 2.8V - 1.39V = 1.41V right??

Well, I'm very sorry if I'm making your time a hard time. But we are not used to work with these values. We are used to work with the input and output net equations as I did.

And as we use to do in our classes, we write the input and output equations, and calculate Ic and Vce from there. And that's it. In the end we just need to check if our initial assumption (TR working region) was correct or not! We have never considered Vce = Vc - Ve nor Vc = Vcc - Ic*Rc. And when you say that Ve = Re*Ie and use 4.2mA for Ie, you're considering Ic = Ie, right?


EDITED;

I posted my reply in post #23.

Psy,
I think there has been some misunderstanding in the terms regarding Vc and Vce, where you kept including the IeRe value.
I consider Vc as the voltage between the collector and the 0V line, Ve the voltage from emitter to the 0V line and Vce as the voltage 'across' the transistor which is the value that is used for the Q point calculation.

Hope we are now both on the same page.
E

Hum, ok. I got that point.

So, are my calcs correct or not when I say Q-point = (1.404V; 4.187mA)???

 

Anyway, I think my results are fine. Already made my calcs to the DC and AC Load Lines, and also calculated the correct Ic curent to have Maximum Possible Compliance. Now I need to re-calculate the correct Rb1 for this current.

To do this I have taken the input net equation and replaced Vth with (Rb2*Vcc)/(Rb1 + Rb2) and Rth replaced by (Rb1*Rb2)/(Rb1+Rb2). This way I get an equation of only one unkown which is the one I need to know for the "optimal Ic current" that will give me what we call "symmetric excursion" or "Maximum Possible Compliance".

Is thing thought correct? I've drawn both Load Lines and Q-point for the initial Ic current and the new AC Load Line and new Q-point and I got the "symmetric excursion/Maximum Possible Compliance" so I'm assuming that I did calcs and everything the right way!

 

Yes, your Q-point is poorly chosen. You do not get maximum symmetrical voltage swing at the load.
And for your amp the optimal Ic for maximum voltage swing is equal to

\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2Rc+Re} \approx \frac{12V}{4.73k \Omega} \approx 2.53mA\)

 

The formula my teacher gave us to calculate that ideal Ic was Ic=Vcc/(Rac + Rdc) where Rac is the output impedance for the AC circuit analysis and Rdc is the output impedance for the DC circuit analysis! I found that Ic to be 3.306mA and after re-drawing the Q-point and the new AC Load line, I think I got that maximum possible compliance!

 

Is it normal for a BJT to have an hie = 2.7kΩ?

 

Is it normal for a BJT to have an hie = 2.7kΩ?

You should find this PDF useful for all your BJT studies.
Refer page #392, Example 6.4.

E

 

You should find this PDF useful for all your BJT studies.
Refer page #392, Example 6.4.

E

I'm going to buy a book that someone advised me to buy. It's H & H - Art of Electronics 3rd Edition. Hope it's a good book.

Anyway, my teacher told me that to find hie I should divide 25mV (I don't know where this value comes from but I already asked him) by the base current! That's what I did and it gave me 2.7kΩ.

I did:

Ib = Ic/β = 4.187/458.7 = 9.128μA

Then 25mV/9.128μA = 2.7kΩ

 

Anyway, my teacher told me that to find hie I should divide 25mV (I don't know where this value comes from but I already asked him)

Page #2 of this PDF explains the 25mV

 

Page #2 of this PDF explains the 25mV

Ok, I got it. What about hie value that I calculated? Looks normal to you?

 

I'm going to buy a book that someone advised me to buy. It's H & H - Art of Electronics 3rd Edition. Hope it's a good book.

Art of Electronics is a really good book, but this book is not for you.

Ok, I got it. What about hie value that I calculated? Looks normal to you?

Looks good.

 

Art of Electronics is a really good book, but this book is not for you.
...

Why you say it's not good for me? Is it too advanced for my level of knowledge (which is zero or close to that)? I already bought it!

Editing (already edited);

For that circuit I've calculated the following:

aV - Voltage Gain
aI - Current Gain
rIN - Input Impedance
rOUT - Output Impedance (not calculated yet)

hie = 25/9.128μA = 2.7kΩ

aV = Vout/Vin = -(hfe*Ib*Rc)/(Ib*hie) = -(hfe*Rc)/hie = -373.76 => ~374
aI = Iout/Iin = (hfe*Ib)/Ib = hfe = 458.7

rIN = (Rb1//Rb2)//hie = (1.67kΩ*2.7kΩ)/(1.67kΩ+2.7kΩ) = 1.03kΩ

rOUT, I need to check my classes notes because I can't remember out to do it. I know that we replace Vsig for a short (i guess) and replace the Rc for a Vx so that we can have an Ix but there is something else that I can't remember now!


Edited 2;

I was reading this link and at some point at "Common Emitter Votlage Gain" section, they say that Re = 25mV/Ie. Is this the same that what I call hie??? If so, why they calculate this as 25mV divided by Ie and my teacher told me that is 25mV/Ib???? Ib and Ie are quite different!!!

 

Why you say it's not good for me? Is it too advanced for my level of knowledge (which is zero or close to that)? I already bought it!

Editing;

Yes, this book is not for beginners.

aV - Voltage Gain
aI - Current Gain
rIN - Input Impedance
rOUT - Output Impedance (not calculated yet)

hie = 25/9.128μA = 2.7kΩ

aV = Vout/Vin = -(hfe*Ib*Rc)/(Ib*hie) = -(hfe*Rc)/hie = -373.76 => ~374
aI = Iout/Iin = (hfe*Ib)/Ib = hfe = 458.7

Rin = (Rb1//Rb2)//hie = (1.67kΩ*2.7kΩ)/(1.67kΩ+2.7kΩ) = 1.03kΩ

Rout, I need to check my classes notes because I can't remember out to do it. I know that we replace Vsig for a short (i guess) and replace the Rc for a Vx so that we can have an Ix but there is something else that I can't remember now!

Why did you not include RL in your calculations for the Av ?? Also in Ai why do you not include RL and Rth ??

 

...
Why did you not include RL in your calculations for the Av ?? Also in Ai why do you not include RL and Rth ??

Editing;

For aV I'm considering the voltage drop across hie, so, as we have a parallel setup between Rth and hie, the voltage drop at Rth is equal to the voltage drop across hie, so I've used the hie voltage drop. Regarding not including R_load, probably I am wrong, however in this link in the 1st example they have for "CE stage with Re bypassed" they also don't consider R_load!

 

So what about RL ?? And I always thought that Vo is a voltage across load resistance ??

 

So what about RL ?? And I always thought that Vo is a voltage across load resistance ??

I know precisely what you mean but I'm afraid I don't know how to answer you! What you're saying makes all the sense but I was following that link! I'll talk to my teacher later today...

And once more, voltage drop across R_load is equal to voltage drop across Rc because they're also at a parallel setup, right?