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// Exa 2.4
clc;
clear;
// Given data
// Referring circuit given in Fig. 2.7(a)
R1= 5*10^3; // Ω
Rf=20*10^3; // Ω
Vi=1; // Input voltage(V)
RL=5*10^3; // Load resistor(Ω)
// Solution
Vo= (1+(Rf/R1))*Vi; // Output voltage(V)
printf('The output voltage i.e vo = %d V. \n',Vo);
AcL=Vo/Vi; // Closed loop Gain
printf(' The closed loop gain i.e Acl = %d. \n',AcL);
IL=Vo/RL; // Output current(A)
printf(' The load current i.e iL = %d mA. \n',IL*1000);
I1=Vi/R1; // Input current(A)
Io=IL+I1; // Total current(A)
printf(' The output current i.e io = %.2f mA. \n',Io*1000);
disp("The op-amp output current Io flows outwards from the output junction.");
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