But in case when the capacitors are not short circuits, the gain is equal
Kusf=Ku1*ku2

Ku1=[ R2 || (1/Hoe) || (Xc2+R5) ] / [re+(R4 || Xc3)]
Ku2=Rin/(Xc1 + Rin)=(ω*Rin*C1) / √( 1+(ω*Rin*C1)^2 )

Rin=R1 || R3 || ( (β+1) * [re+(R4||Xc3) ] )

In mishu.daniel circuit for F=6Khz and hfe=100; Ube=0.69V
Then the base current is equal
Ib=71uA and Ic=7.1mA; re=3.6Ω
The gain of transistor is equal:
Ku1=[ R2 || (1/Hoe) || (Xc2+R5) ] / [re+(R4 || Xc3)]
Re=( R4 || Xc3 )=199=200Ω
Ro=(Xc2+R5)=10.3KΩ
Rc=(R2 || 1/hoe)=675Ω

And now the gain of BJT is equal
Ku1=( Rc|| Ro )/( re+Re )=633/203=3.1[V/V]

And now the gain of all amplifier is equal:
We need to take into account influence of C1.
C1 and Rin form a voltage divider whit "gain"
Ku2=Rin/( Xc1 + Rin )=( ω*Rin*C1 ) / √( 1+(ω*Rin*C1)^2
For F=6Khz we get
Rin=R1 || R3 || ( (β+1) * [re+(R4||Xc3) ] )=2.45KΩ
C1=10nF
and
Ku2=( ω*Rin*C1 ) / √( 1+(ω*Rin*C1)^2=0.687[V/V]

Add the gain
Kusf=3.1*0.687=2.1[V/V]
And of course C2 and R5 is voltage divider to, but for F=6Khz the "gain" is equal
y=( ω*R5*C2 ) / √( 1+(ω*R5*C2)^2 = 0.966[V/V]
and finally
Gain =2.1*0.966=2[V/V]

And spice confirm this result​

 

U know there is a simple explanation for the emitter cap in CE amp circuit.
The capacitive reactance shows a low resistance path to AC so that all the amplified signal develops at the collector.
The capacitor does not effect the biasing circuit.
So if you do a DC analysis of the CE amp, the capacitors are ignored.
It comes into play only in AC analysis, hence the gain of the amp.

Rifaa

 

Gain =2.1*0.966=2[V/V]

What about the phase angle of the gain?

 

Well, as everyone know the phase angle for the CE amplifier is -180°. But in this case the phase angle will be less then -180°
The phase angle of C1 and Rin for F=6KHz is equal 47.2°
And for C2, R5 is 14.8°
So overall phase angle will be -118°
That phase angle will give as a time shift Δt, for the period of F=6KHz---> T=166us
Δt=(-118°/360°)*166us=-54.6us

 

So overall phase angle will be -118°

Did you forget the effect of R4-C3?

 

No, I don't forget, but the effect of R4-C3 will by "negligible", from a practical point of view.
F=0.16/RC=0.16/(200*10nF)=80KHz
So for F=6KHz the phase shift will be arc tg(ω*R*C)=4.3°
And this will give the overall phase shift something around -113.7°
Spice show -114.77°

 

I'm not familiar with this solving method Jony130. I have applied nodal method on the amplifier input and gave me the same result as spice model 2.12V. And then from the output side of amplifier I manage to obtain the gain as in spice simulation.
Why do you use this method Jony130? Don't you think is more complex than the other one?
I found some tutorials related to my topic:
http://www.electronics-tutorials.com/amplifiers/negative-feedback.htm
I saw there a trick about replacing the load resistor with an LC circuit. I know that RF stuff is more complex so I tried to understand what happens there in that circuit. It is unclear for me the formulas there "2 * Po = [Vcc - Ve][SIZE=-1]2 [/SIZE]/ R " and also how is behaving this LC circuit there. The Q factor is very blurry in my mind, I know how LC is behaving in theory, I cannot "feel it", I also know that Q=X/R but I saw also Q=R/X (???) and how the gain is obtained from this circuit. Could you help me with this?
Thanks.

 

No, I don't forget, but the effect of R4-C3 will by "negligible", from a practical point of view.
F=0.16/RC=0.16/(200*10nF)=80KHz
So for F=6KHz the phase shift will be arc tg(ω*R*C)=4.3°
And this will give the overall phase shift something around -113.7°
Spice show -114.77°

Do you have a good tutorial about this method? I found it in several books but I think that authors are skipping some parts which I consider useful and I've quit each time ...
Thanks.

 

No, I don't forget, but the effect of R4-C3 will by "negligible", from a practical point of view.
F=0.16/RC=0.16/(200*10nF)=80KHz
So for F=6KHz the phase shift will be arc tg(ω*R*C)=4.3°
And this will give the overall phase shift something around -113.7°
Spice show -114.77°

If this were a design for a production amplifier, then we could say that with 5% tolerance resistors, 10% tolerance capacitors, β with variation much larger than for the passive components, etc., we could consider the effect of R4-C3 "negligible, from a practical point of view".

But this is an exercise, and we can give the circuit elements "exact" values for the purpose of deriving a mathematical solution for the circuit.

mishu.daniel wanted a solution for the case where the capacitors are not assumed to be AC short circuits. A proper solution should include all the capacitors. And since the capacitors are reactive elements, the gain involves both magnitude and phase angle.

Assuming re = 3.6, β = 100 and hre = 0, a solution using the nodal method where the entire circuit is solved at once (using 12 digit arithmetic throughout), instead of doing sections one at a time, gives the result Av = 2.0241 < -114.874°

Assuming re = 3.3, β = 100 and hre = 0, Av = 2.02687 < -114.8637°

Doing sections one at a time allows numerical rounding errors to accumulate, which is probably why your computed result is not as accurate as your spice result.

 

I found some tutorials related to my topic.
Could you help me with this?
Thanks.

mishu.daniel, PM me with a valid email address, and I'll email you some tutorials.

 

For me "My" method is simpler. I feel intuitively the behaviour of BJT, and don't need to draw any equivalent circuit to determine the gain or any other parameter. Of course this "method" is not so accurate, but for me is more intuitive and easier. In "The Art of Electronics" use very similar method to analyse various circuit.
And I'm a hobbyist without any electrical engineering education.
And all I need to know is to intuitively understanding the basics law and remember some equation.
eg:
The gain of BJT is equal
Au=Rc/(re+Re)
where
Rc- Is all impedance looking away of the collector
Re - all impedance looking from emitter to the ground.

 

The gain of BJT is equal
Au=Rc/(re+Re)
where
Rc- Is all impedance looking away of the collector
Re - all impedance looking from emitter to the ground.

Consider the circuit at the beginning of this thread. Now connect a 10k resistor from collector to base.

Does "Rc- Is all impedance looking away of the collector" give the right answer in your formula for this modified circuit?

What is the correct gain with the added 10k resistor?

 

Consider the circuit at the beginning of this thread. Now connect a 10k resistor from collector to base.

Does "Rc- Is all impedance looking away of the collector" give the right answer in your formula for this modified circuit?

Yes, you can get the "correct" answer.
Using Miller's claims (principles) we know that we can add a resistor parallel to
R3.
Rm1=Z/(1-Ku)
And second resistor connected between collector and the gnd.
Rm2=10K

Gain of the BJT
Ku=(Rc||Ro||10K)/203=-2.9[V/V]
and Rm1=2.5KΩ so imput impedance will be lower.
Rin=2.45K||2.5K=1.2kΩ
And this will result that the gain of entire amplifier will be reduce , mostly by a effect of a voltage divider C1, Rin.
Gain=1.2[V/V]

 

Yes, you can get the "correct" answer.
Using Miller's claims (principles) we know that we can add a resistor parallel to
R3.
Rm1=Z/(1-Ku)
And second resistor connected between collector and the gnd.
Rm2=10K

Gain of the BJT
Ku=(Rc||Ro||10K)/203=-2.9[V/V]
and Rm1=2.5KΩ so imput impedance will be lower.
Rin=2.45K||2.5K=1.2kΩ
And this will result that the gain of entire amplifier will be reduce , mostly by a effect of a voltage divider C1, Rin.
Gain=1.2[V/V]

But now you're no longer using the formula you gave, and to which I referred:

The gain of BJT is equal
Au=Rc/(re+Re)
where
Rc- Is all impedance looking away of the collector
Re - all impedance looking from emitter to the ground.

You've transformed the feedback resistor into 2 other resistors; this transformation wasn't described as part of your formula. Somebody looking at your simple formula might not know that they should do that.

Perhaps you meant that the formula was only for a simple BJT circuit without feedback.

The result you've gotten for the case with the 10k feedback resistor is fairly accurate, but when I compare to the result from nodal analysis I see some small discrepancy. The differences get larger when there is more feedback, e.g., with 1000Ω feedback, for example.

Try this case: using the original circuit from post #1, add a 1000Ω resistor from collector to emitter.

Now you have an additional 1000Ω looking away from the collector and from the emitter. What does your formula give for the gain in this case?

 

Haha,
But that can be say on any "formula" in the textbook.

 

Did you do this yet?

Try this case: using the original circuit from post #1, add a 1000Ω resistor from collector to emitter.

Now you have an additional 1000Ω looking away from the collector and from the emitter. What does your formula give for the gain in this case?

 

But I don't need to calculate the new gain. I know that this formula will give the wrong answer.
You have to remember that every equation has his own limitation.
Even the Ohms law don't hold in every situation, or for example
Ic=β*Ib.
Similarly the textbook equation for gain of CE amplifier.
Au=( Hfe*Rc ) / ( H11 ) = ( Hfe*Rc) / ( hfe*re ) = Rc/re = Rc /(Ut/Ic)=(Rc*Ic)/26m = gm*Rc = 40*Ic*Rc.
Doesn't take into account every real life situation, and has his limited application.

 

But I don't need to calculate the new gain. I know that this formula will give the wrong answer.

Why does it give the wrong answer? If you wanted to solve the circuit (without using SPICE), what would you do?

In post #31 you said:

And all I need to know is to intuitively understanding the basics law and remember some equation.

I'm trying to suggest that for the purposes of solving linear networks, it might be better to understand more general methods, rather than remembering some equations. Simple equations (and intuition) may fail for all but the simplest circuits. Just adding a single resistor from collector to emitter caused failure of this method:

The gain of BJT is equal
Au=Rc/(re+Re)
where
Rc- Is all impedance looking away of the collector
Re - all impedance looking from emitter to the ground.

This is not to say that for very simple topologies, simple formulas may not be worth remembering, but their limitations should be understood. For a simple circuit, the more general methods are also very quick and easy, and no formulas for a particular circuit need be remembered; they can be derived as needed.

By the way, the formula Au=Rc/(re+Re) is slightly in error. It should be:

Au=(β/(β+1))*Rc/(re+Re)

otherwise you would have the impossible situation that Au would be non-zero when β is zero.