# CE Voltage Amplifier - Negative Feedback

#### 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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Joined Mar 10, 2018
3,630
Looks like G = 110 based on sim.

See attached.

Regards, Dana.

#### 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.

#### elec_eng_55

Joined May 13, 2018
214
Looks like G = 110 based on sim.

See attached.

Regards, Dana.
Gain = 0.100 Vout / 0.001 Vin = 100 based on my sim.

#### 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

#### elec_eng_55

Joined May 13, 2018
214
See
gain: 1/(1/2700+1/5600)/(9.1+25.8/3+0.1)=102.342, but C1 and C3 reduse to 100.
View attachment 158525
Thanks Bordodynov. How does one arrive at the value for C3? I tried using

Xc = 0.1 * Re = 0.1 * 390 ohms = 39 ohms

C3 = 1 / 2 * pi * f1 * Xc = 1 / (2) (3.14) (100Hz) (39 oms) = 40.83 uF

which doesn't yield 2200 uF.

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

#### 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) )