Hello, I want to design a common emitter 2 stage amplifier with the 2n3904 and 12v vcc but I have not found a good reference to calculate the values for the following circuit.

 

Hello, I want to design a common emitter 2 stage amplifier with the 2n3904 and 12v vcc but I have not found a good reference to calculate the values for the following circuit.

try https://leap.tardate.com/electronics101/bjt/twostagecommonemitteramplifier/

ADDED:
read this thread
https://forum.allaboutcircuits.com/...ge-discrete-bjt-amplifier.193939/post-1823460

 

And... what's your question? What are the specs you are trying to meet? What gain? What frequency range? What input impedance, What input voltage range? What output voltage range? What output impedance?

Is this some sort of homework project/assignment?

 

This doesn’t look like a power circuit and hence should not be in the power electronics forum.

 

And... what's your question? What are the specs you are trying to meet? What gain? What frequency range? What input impedance, What input voltage range? What output voltage range? What output impedance?

Is this some sort of homework project/assignment?

It is a project, they told me that I choose the values but I did not understand very well what he was referring to, what I have searched for on the internet does not have what I am looking for for this design

 

try https://leap.tardate.com/electronics101/bjt/twostagecommonemitteramplifier/

If my circuit does not require these resistors, how are the equations modified?

 

MOD NOTE: Moved from Power Electronics to Homework Help.

 

If my circuit does not require these resistors, how are the equations modified?

((VCC-0.7-(0.5*VCC/RC1)*RE1)/RB2)*B@(0.5*VCC) = (0.5*VCC)/(RC1+RE1)
And in this case you do not need RE1b and CE1.
If your amplifier will work without any load (it is nonsense), simple remove RL.
EDIT: slightly changed equation.

 

Your first circuit does not have the negative feedback resistors RE1 and RE2 shown in the last circuit.

 

If my circuit does not require these resistors, how are the equations modified?

View attachment 297147

Hello,

You should probably try to figure out why the resistor values are what they are in that schematic.
For example, the resistors associated with Q2 may be much too large for your application, but you would need to specify the output load impedance or type. It the load a speaker, or another stage, or something else?

 

Well, I was guided by this pdf https://learnabout-electronics.org/Downloads/amplifiers-module-02.pdf beta 100 and I simulated it with an input of 0.1Vp 1kHz with the 2n2222 and it worked
and i have a question why the output of stage 1 scrolled up?

 

and i have a question why the output of stage 1 scrolled up?

What do you mean by scrolled up?
Are you asking why the waveform has been shifted upwards by 2V?

CH2 - cyan (light blue) colored trace is the output at C5.
This output is a C-R load such as this, where C is C5 and R is input impedance of the oscilloscope which could be 1MΩ



CH1 - yellow trace, is the signal after C2.
This signal is loaded by the equivalent of R2, R7, base-emitter junction of Q2, R6, and the input impedance of the oscilloscope.
The major contributing factor is the voltage divider combination of R2 and R7 which presents a bias voltage of about 2V at the base of Q2.
Thus, to answer your question, the 2V shift is the base bias on Q2 presented by R2 and R7. It has nothing to do with the signal coming from C2.
The signal from C2 is simply superimposed (i.e. added) on top of the DC bias voltage. Or you can say it the other way around. The signal from C2 is presented to the base of Q2 with a 2V DC bias superimposed (added) on to the signal.

Notice also that the signal on CH1 is out of phase with the input at C1 and the output at C5.

 

Remember that the total actual voltage at any point in your circuit is the superposition of the large-signal (DC) response and the small-signal (AC) response.

What is the DC response at the node that you have connected to the scope?

You AC response is going to be laid on top of that DC response.

 

You cannot buy a transistor with a beta of 100. A transistor part number has a wide range of beta so the base resistor to ground resistor is important unless you measure the beta of Q1 and calculate R1 to match it, then repeat if Q1 is ever replaced.

Your new schematic has a problem with C3 and C4:

 

Well, I was guided by this pdf https://learnabout-electronics.org/Downloads/amplifiers-module-02.pdf beta 100 and I simulated it with an input of 0.1Vp 1kHz with the 2n2222 and it worked
and i have a question why the output of stage 1 scrolled up?
View attachment 297627

Hi,

Take a look at how Q2 is biased.

 

In my simulation here, Q2 is clipping because it is biased for a transistor with higher beta, I increased R2 to bias it better.
The output has the top of the waveform squashed because there is no negative feedback and because C2, C3 and C4 are too small which cuts the 1khz fundamental frequency but gives full gain at high harmonic distortion frequencies. .