First i must say that i don't really understand the meaning of the value of the resistors and capacitors here (different from my book), to make my signal passed through the amplifier or not.
The role played by Ce is also something i don't really understand. The capacitor Ce get charged when the transistor let pass the current, and when he doesn't, the capacitor discharged through Re i think, but what role it play on the signal i try to amplify?
I gonna resume what I seem to understand (with my basic words) : we use the transistor on the range of possible voltages he could amplify before he get saturated. So the goal is to "squeeze" the signal between the lower voltage where the transistor let the current passed and the voltage where he get saturated. On that range, the voltage collector-ground "follow the shape" of the voltage base-ground and we can take back our signal amplify by the 12V power supply, then we make it passed to a capacitor to filter the DC current. I don't really see the meaning of RL also.
If i'm totally wrong, please tell me! Thanks for all you help, and i'm definitely gonna use a simulation software! If you have any to suggest for linux/ubuntu?
I started to write out a lot of stuff from a historical perspective,but after I read the first few paragraphs,ny eyes started to glaze over,so I'll try to make it a bit simpler.
Going back to your original circuit,imagine replacing the transistorT1 with a resistor Rt from collector to emitter,& R4 with a short circuit ( the base connection now goes nowhere).
You now have a voltage divider consisting of R3 & Rt across your 5v supply.
The voltage ( with respect to the negative PSU terminal) at the junction of R3 & Rt (pt "c") is now +5( Rt/(Rt +R3) volts.
Now make Rt a variable resistor.
You can see that you can vary the voltage at point (c) by varying the value of Rt.
Now,imagine you have superpowers & can look at a signal incoming & move the knob of the variable resistor fast enough to keep up with it.
You will now have a dc voltage at point (c),varying at the signal rate.
This
ac component will pass through C2,& appear at the "sortie" terminals.
After a while,you will discover (assuming Rt is a real-world component) that for best distortion of the output signal,at expense of voltage level,you will not use all of the adjustment range of the variable resistor.
The transfer function of Rt is not completely linear across the full range,so you have applied a mechanical "bias " to get around this.(we'll come back to this)
Now restore R4 --the voltage at (c) now becomes:- +5( Rt+R4/(Rt +R3+R4)
Varying Rt now varies the total value of the "bottom" part of the divider by a lesser proportion,& hence the level of variation of the current through the divider is less,the voltage at (c) has a smaller ac component,& the ac output at the "sortie" terminals is less.
At this point,Microsoft has revoked your superpowers,& you have to restore T1.
T1 needs a dc bias to place it in the linear portion of its "transfer function",just like Rt needed a mechanical bias.
In the case of a BJT,this is a current bias
R1,R2,R4 are all part of the circuit to provide this bias.
So we are pretty much stuck with R4,but in this circuit,the gain is lousy.
The formula I learnt back in the Dark Ages for CE amplifiers with unbiased emitter resistors was A =gmRc/1+gm Re
gm is usually >>1,so this is near as dammit A= Rc/Re,which,for the original circuit ,is a gain of about 4.
There is,a "fiddle" though--if you use C3,for the ac component of emitter current it looks like a short,so Re disappears from the gain formula
A= gmRc/1 ( as gm times zero
is zero),or just,gm Rc.