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+// Exa 4.5
+format('v',7);
+clc;
+clear;
+close;
+// Given data
+R_D= 5;// in kΩ
+R_D= R_D*10^3;// in Ω
+R_SS= 25;// in kΩ
+R_SS= R_SS*10^3;// in Ω
+I= 0.8;// in mA
+I=I*10^-3;// in A
+i_D= I/2;// in A
+unCox= 0.2;// mA/V^2
+unCox= unCox*10^-3;// in A/V^2
+WbyL= 100;
+// Formula i_D= 1/2*unCox*WbyL*V_OV^2
+V_OV= sqrt(2*i_D/(unCox*WbyL));// in V
+gm= i_D/V_OV;// in A/V;
+
+// Part (a)
+Ad= 1/2*gm*R_D;// in V/V
+disp(Ad,"Differential gain in V/V is : ")
+Acm= -R_D/(2*R_SS);// in V/V
+disp(Acm,"Common mode gain in V/V is ")
+CMRR= abs(Ad)/abs(Acm);
+CMRRindB= round(20*log10(CMRR));// in dB
+disp(CMRRindB,"Common mode rejection ratio in dB is : ")
+
+
+// Part (b)
+disp("Part (b) when output is taken differentially")
+Ad= gm*R_D;// in V/V
+disp(Ad,"Differential gain in V/V is : ")
+Acm= 0;
+disp(Acm,"Common mode gain in V/V is ")
+// CMRRindB= 20*log10(Ad/Acm) = infinite ;// in dB
+disp("Common mode rejection ratio in dB is : ")
+disp("infinite");
+
+// Part (c)
+disp("Part (c) when output is taken differentially but the drain resistance have a 1% mismatch.")
+Ad= gm*R_D;// in V/V
+disp(Ad,"Differential gain in V/V is : ")
+// delta_R_D= 1% of R_D
+delta_R_D= R_D*1/100;// in Ω 
+Acm= R_D/(2*R_SS)*delta_R_D/R_D;// in V/V
+disp(Acm,"Common mode gain in V/V is ")
+CMRRindB= 20*log10(abs(Ad)/abs(Acm));// in dB
+disp(CMRRindB,"Common mode rejection ratio in dB is : ")
+
+// Note: In the book, there is putting wrong value of Ad (20 at place of 10) to evaluate the value of CMRR in dB in part(c) , So the answer of CMRR in dB of Part (c) is wrong
+
+
+
+