CE Voltage Amplifier - Negative Feedback

Thread Starter

elec_eng_55

Joined May 13, 2018
214
I am going to understand this stuff and if it kills me.

I have uploaded a simple CE voltage amplifier schematic as modified by Jony.

I believe the DC gain is 2.7K / 380 ohms = 7.1 and before C3 and R5 are added,
the AC gain is also 7.1. Although the simulation says the AC gain is 4.5.

Once C3 and R5 are added, I believe that the AC gain is 2.7K / [(380 * 9.1) / (380 + 9.1)] = 303.
However, the simulation says that the AC gain is 100.

With this type of negative feedback, is AC gain = Rc / Re//Rfeedback?

I am confused.

David.
 

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Jony130

Joined Feb 17, 2009
5,023
The voltage gain will be equal around

AV = (Rc||RL)/( re + R4||R5) = (2.7kΩ||5.6KΩ) / ( 8.6Ω + 380Ω||9.1Ω) ≈ 104V/V

Where re = 26mV/Ie

And this is a current - series feedback.

Also, notice that now when re << (R4||R5) the re variations in the rhythm of the input signal do not count much now.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
The voltage gain will be equal around

AV = (Rc||RL)/( re + R4||R5) = (2.7kΩ||5.6KΩ) / ( 8.6Ω + 380Ω||9.1Ω) ≈ 104V/V

Where re = 26mV/Ie

And this is a current - series feedback.

Also, notice that now when re << (R4||R5) the re variations in the rhythm of the input signal do not count much now.
Many thanks, Jony.

David
 

Jony130

Joined Feb 17, 2009
5,023
C3 ≈ 1/(2 * pi * F1 * R5) ≈ 0.16/(F1 * R5) ≈ 0.16/(10Hz * 9.1Ω) ≈ 1758μF ≈ 2200μF

Or if you want to F1 to be set at 100Hz

C3 ≈ 0.16/(100Hz * 9.1Ω) ≈ 175.8μF ≈ 220μF

In reality C3 see this resistance R5 + (re + (R1||R2)/(β+1) )||R4
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
C3 ≈ 1/(2 * pi * F1 * R5) ≈ 0.16/(F1 * R5) ≈ 0.16/(10Hz * 9.1Ω) ≈ 1758μF ≈ 2200μF

Or if you want to F1 to be set at 100Hz

C3 ≈ 0.16/(100Hz * 9.1Ω) ≈ 175.8μF ≈ 220μF

In reality C3 see this resistance R5 + (re + (R1||R2)/(β+1) )||R4
Thanks again.
 

Jony130

Joined Feb 17, 2009
5,023
Next time if you want to know how to find the capacitor value. You simply need to find the circuit time constant.
So, all you need to do is to find the effective (equivalent) resistance seen by the capacitor.

That means you have to consider the circuit as seen from the capacitors side.

For example, C2 will see Rc in series with RL if we ignore the transistor output resistance.

t = C2 * (Rc + RL) and the corner frequency is

Fc = 1/(2 * pi * t) = 1/( 2 * pi * C2*(Rc+RL) )
 
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