1. Definition
The inverting amplifier assembly has a differential linear amplifier and two resistors
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]
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]
bodePhase[h,0,6]
There is a reduction in the bandwidth of 2 decades.
