# 2 stage common emmiter with feedback

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#### waverunner

##### New Member
hello guys,
I would like to explain me how the feedback(R11,c2) affects the amplifier. I know about feedback from collector, but not from emmiter.
Thanks a lot

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#### audioguru

##### Well-Known Member
The emitter of the second transistor has exactly the same signal as its base which has exactly the same signal as the collector of the first transistor.

So R11 and C2 provide negative feedback.
But it is applied to the input which reduces the input impedance that was already low without the negative feedback.

#### waverunner

##### New Member
Thanks audioguru. I'll do calculations and if i have problem i post again.
Thanks

#### waverunner

##### New Member
I do the calculations for dc analysis, but in ac analysis how i find the gain?

#### audioguru

##### Well-Known Member
It is like an inverting opamp except its open-loop gain is only about 100 instead of 100,000 like an opamp. So the gain is a little less than R11/R1 (10k/150= 67) so the gain is about 50, except the 1k load on the 6.8k collector resistor of the second transistor reduces the gain to about 7.3.

I have the formula on my hard drive but I can't find it.

#### waverunner

##### New Member
I design the circuit in spice, and the voltage gain is 15,97... now i want to confirm that.

#### audioguru

##### Well-Known Member
I design the circuit in spice, and the voltage gain is 15,97... now i want to confirm that.

It only matters in school because the input impedance is so low that the circuit is not practical. The gain might be only 16 if the source impedance is zero. If the source impedance is more than zero then the circuit attenuates it and the gain is less.

#### The Electrician

##### Active Member
I design the circuit in spice, and the voltage gain is 15,97... now i want to confirm that.

I get an AC voltage gain of 16.13 by nodal analysis.

#### MrAl

##### Well-Known Member
I design the circuit in spice, and the voltage gain is 15,97... now i want to confirm that.

Hi there,

The input voltage is divided by 150 which puts some current into
the base of Q1, and that amplifies that a lot because of C4 so
the output at the collector WOULD be very large, except that that
signal is coupled to Q2 and Q2's emitter feeds the feedback network
R11 and C2 and so the gain of Q1 is reduced to about 10000/150
which is around 67. Unfortunately, it looks like Q2 is not set up
right to match the output impedance (1k) so that output impedance
causes a lot of attenuation. With a slight gain of very roughly 2
for Q2 and the attenuation of 7k to 1k the output gain would be
very roughly 19. The very low output impedance and Q2 bias
probably not in the best place you may also see quite a bit of
clipping of the output signal so you probably get a lot of distortion.
With a better circuit for Q2 you can probably do a lot better.

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#### waverunner

##### New Member

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