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Reversing amplifier

1. Definition

The inverting amplifier assembly has a differential linear amplifier and two resistors

Reversing amplifier

2. Simulation

2.a. Transfer function

The amplifier is modeled as a linear quadrupole, with an open loop gain of the following form: g (s) = g01 + sw0

We use the Mathematica module:

Get["../simulin/simulineaire.m"];

First, the circuit is defined in the following function:

inverseur[r1_,r2_,g0_,w0_]:=Module[{n,A,B,g},
    n=4;
    A=Table[0,{n},{n}];
    B=Table[0,{n}];
    A=ajouterResistance[A,1,2,r1];
    A=ajouterResistance[A,2,3,r2];
    g=g0/(1+s/w0);
    {A,B}=ajouterSourceTensionSTCT[A,B,3,4,4,2,g];
    {A,B}=ajouterMasse[A,B,4];
    {A,B}=definirEntree[A,B,1];
    Return[{A,B}];
]
            

Here is the transfer function

{A,B}=inverseur[r1,r2,g0,w0];
h=transfert[A,B,3]/.s->I*omega
            

-g0 r2 w0g0 r1 w0+i ω r1+i ω r2+r1 w0+r2 w0

2.b. Bode diagram

The open loop gain is fixed at g0 = 105 and the cutoff pulse w0 = 20π (10 Hz). The resistors are chosen so as to obtain a gain of 10 at low frequency:

{A,B}=inverseur[10^3,10^4,10^5,20*N[Pi]];
h=transfert[A,B,3];
                
bodeGain[h,0,6,0,60]
plot1.png
Bode diagram

Let’s increase the gain by a factor of 100:

{A,B}=inverseur[10^3,10^6,10^5,20*N[Pi]];
h=transfert[A,B,3];
                
bodeGain[h,0,6,0,60]
plot2.png
Bode diagram
bodePhase[h,0,6]
plot3.png
Bode diagram

There is a reduction in the bandwidth of 2 decades.

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