The gain of the first stage is multiplied by the gain of the second. The gain of the first stage is equal to the ratio of the resistors (approximately). The gain of the second stage is equal to the ratio of resistors + 1 (approximately).

 

The gain of the first stage is multiplied by the gain of the second. The gain of the first stage is equal to the ratio of the resistors (approximately). The gain of the second stage is equal to the ratio of resistors + 1 (approximately).

first stage gain is 28 or 24?

how about 2nd stage? i tried but answer not multiplying up

gain for second stage : R6 in parallel with R7 and R8 ? gives answer but why like that?

first stage = (30 + 180)/7.5 ?

2nd stage = (180||36||30)/0.3 ?

 

first stage < 180/7.5=24
2nd stage < 30/0.3+1=101
GainFB =Gain/(1+beta*Gain), Gain-gain without feedback.
For more accurate calculations there is a Spice. I was engaged in such calculations 35 years ago, when I did not have modern programs. For calculations, it is necessary to know the laws of Ohm and Kirhoff, as well as equivalent transistor circuits.
I'm not interested to carry out detailed calculations. I consider it a waste of time, because have a spice. It is enough to be able to do approximate calculations.
What does your teacher do?

 

first stage < 180/7.5=24
2nd stage < 30/0.3+1=101
GainFB =Gain/(1+beta*Gain), Gain-gain without feedback.
For more accurate calculations there is a Spice. I was engaged in such calculations 35 years ago, when I did not have modern programs. For calculations, it is necessary to know the laws of Ohm and Kirhoff, as well as equivalent transistor circuits.
I'm not interested to carry out detailed calculations. I consider it a waste of time, because have a spice. It is enough to be able to do approximate calculations.
What does your teacher do?

oh, did that at first but did not multiply to 1400...
I have so many teachers for now, Jony, Bordodynov, Yill, Crutschow, just to name a few..which of them you talk about?

 

I thought you were a student and meant a teacher. And about the mismatch of the gain, I did not in vain put a sign less than (<). The gain of the first stage depends on the input resistance of the transistor (~ 60 kΩ) and the base-emitter resistor (36 kΩ). The gain drop is 7.5 / (36 || 13 || 7.5) = 7.5 * (1/36 + 1/13 + 1 / 7.5) = 1.785 times. Also, the gain of the second amplifier is less than 101. I have given the formula for calculating the gain of the feedback circuit.

 

I thought you were a student and meant a teacher. And about the mismatch of the gain, I did not in vain put a sign less than (<). The gain of the first stage depends on the input resistance of the transistor (~ 60 kΩ) and the base-emitter resistor (36 kΩ). The gain drop is 7.5 / (36 || 60 || 7.5) = 7.5 * (1/36 + 1/60 + 1 / 7.5) = 1.333 times. Also, the gain of the second amplifier is less than 101. I have given the formula for calculating the gain of the feedback circuit.

Student drop out..Back in few months..Anyways, thanks.

 

why current mirror will not work?

Look very closely at your schematic
https://forum.allaboutcircuits.com/attachments/screen-shot-2019-06-01-at-5-59-43-pm-png.178764/

Notice that Q2 base-emitter is connected between Q3 collector and Vcc ( Q2 base-emitter is in parallel with Q3 Collector-Emitter + R7).
And the parallel connected components experience the same voltage.

Vbe2 = Vec + Ie*R7
0.7V = Vec + 100μA*12kΩ
0.7V = Vec + 1.2V

As you can see we have inequality and this is why the circuit doesn't want to work.

 

Look very closely at your schematic
https://forum.allaboutcircuits.com/attachments/screen-shot-2019-06-01-at-5-59-43-pm-png.178764/

Notice that Q2 base-emitter is connected between Q3 collector and Vcc ( Q2 base-emitter is in parallel with Q3 Collector-Emitter + R7).
And the parallel connected components experience the same voltage.

Vbe2 = Vec + Ie*R7
0.7V = Vec + 100μA*12kΩ
0.7V = Vec + 1.2V

As you can see we have inequality and this is why the circuit doesn't want to work.

Just reading...and in circuit there was a wire joining Q4 collector and base but I don't think this is important for your explanation
If understand correctly then have to add resistor on Q2's emitter or reduce the resistors on Q3 and Q4 to about 200 ohms?
What did I miss?

Have new circuit to test but rn just with phone only, can't post

 

Just reading...and in circuit there was a wire joining Q4 collector and base but I don't think this is important for your explanation

Well, this wire is necessary if you want to have the current mirror.

If understand correctly then have to add resistor on Q2's emitter or reduce the resistors on Q3 and Q4 to about 200 ohms?
What did I miss?

Not much...Haha.

 

Well, this wire is necessary if you want to have the current mirror.


Not much...Haha.

Yes it is necessary. I omitted in first diagram posted. You have first schematic

Hmm... Sad then... Will retry because think I see mistake now