Two-Stage Negative Feedback

Thread Starter

newbie2019

Joined Apr 5, 2019
95
This two-stage swamped amplifier has an overall voltage gain of 502 without two-stage negative feedback.
The input impedance is 1.32KΩ and the output impedance is 3.6KΩ.

When I implement two-stage negative feedback (schematic attached), the voltage gain drops to 210.

Av = 502
β = RE11 / RE11 + RF = 50Ω / 50Ω + 18KΩ = 0.00277
ACL = Av / (1 + (Av * β)) = 502 / (1 + (502 * 0.00277) = 210

I understand from my research that in order for ACL to be stable within +/- 1%, Av should be equal to or
greater than 100 * ACL. That would mean that either Av would have to be 21,000 or ACL would have to
be ACL would have to be 5.02. What am I missing?

Newbie2019
 

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Jony130

Joined Feb 17, 2009
5,163
What am I missing?
You did not miss anything. Maybe except the "loading effect" of a feedback network on an open loop gain (AOL).

If you would include the "loading effect" on the AOL the open loop gain will be around 439.5V/V , therefore the closed loop gain is

ACL = 439.5/ (1 + (439.5 * 0.00277) = 198.2V/V
 

Thread Starter

newbie2019

Joined Apr 5, 2019
95
You do not miss a thing. Maybe except the "loading effect" of a feedback network on an open loop gain (AOL).

If you would include the "loading effect" on the AOL the open loop gain will be around 439.5V/V , therefore the closed loop gain is

ACL = 439.5/ (1 + (439.5 * 0.00277) = 198.2V/V
Thanks Jony. How would I calculate the input and output impedance with negative Feedback?
 

Thread Starter

newbie2019

Joined Apr 5, 2019
95
Yes, and this is why the Rin will increase by this factor (1 + (439.5 * 0.00277) but Rout will decrease by the same factor.
Thanks Jony. Could you please guide me as to how you applied the "loading effect" that brings the
voltage gain down from 502 to 439.5. I have been running numbers but can't figure it out.

One last thing, is how can Av >= 100 * ACL?

Thanks,

Newbie2019
 

Jony130

Joined Feb 17, 2009
5,163
Sorry, I am not sure what reply #8 means.
To be able to find the open loop gain I create an equivalent circuit. So I disable the feedback network and calculating the open loop gain with the equivalent feedback loading ( show in red). For the voltage type of feedback. To find resistance seen by Q1 emitter I simply disconnect the feedback network from the output and we short it to ground. and this is why we have we have RE11||RF at the Q1 emitter.
But from the output node perspective, I disconnect T1 emitter. And now our RE11 and RF resistors appear to be in series connection.

And this is why I end up with this equivalent circuit.

10.PNG

So now I can find AOL.

See another exampel for a different type of feedback topology.
 

Thread Starter

newbie2019

Joined Apr 5, 2019
95
Thanks Jony. Here are my calculations for AOL, ACL, Input and Output Impedance.
Based on

Assumption: 2N3904's hie = 3KΩ and hfe = 100

RL'1 = RC1║RB21║RB22║hie = 3.6K║10K║2.2K║3K = 2.65K║2.2K║3K = 1.2K║3K = 0.857K
RL'2 = RC2║(RF + RE11_2) = 3.6K║18.05K = 3KΩ

Av1 = - hfe * RL'1 / hie = - 100 * 0.857K / 3K = -28.57
Av2 = - hfe * RL'2 / hie = 100 * 3K / 3K = -100

Av = Av1 * Av2 = -28.57 * -100 = 2,857

β = RE11 / RE11 + RF = 50Ω / 50Ω + 18KΩ = 0.00277

Av * β = 2,857 * 0.00277 = 7.91

Av(feedback) = Av / (1 + (AV * β)) = 2,857 / 8.91 = 320.65

Ro = RC║RF = 3.6K ║18K = 3K

Ro(feedback) = Ro / (1 + (Av * β)) = 3K / 8.91 = 0.337K

Rin = hie = 3K

Rin(feedback) = (1 + (Av * β)) * Rin = 8.91 * 3K = 26.73K
 
Last edited:

Jony130

Joined Feb 17, 2009
5,163
Can I ask you one question?

What is the voltage gain of this CE amplifer?

1.PNG


Assumed that the Ic = 1mA and β = 100 if needed.
 

Jony130

Joined Feb 17, 2009
5,163
rc = RC1║RL = 1K║1K = 500Ω
re = RE1 = 100Ω

IE ≈ IC
r'e = 25mV / IE = 25mV / 1mA = 25Ω

Av = rc / (re + r'e) = 500Ω / (100Ω + 25Ω) = 4
Very good. In this case, can you tell me why did you omit the RE11 and RE21 in your calculation for an open loop gain in post #14?
 
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