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+clear;
+clc;
+
+//Caption:Design a video amplifier using MC1550
+//Given Data
+Avo=-25;
+Vagc=20;//in V
+Vcc=6;//in V
+hfe=50;
+rbb=50;//in ohm
+Cs=5;//in pF
+Cl=5;//in pF
+Ie1=1;//in mA
+ft=900;//in MHz
+Vt=26;//in V
+n=2;//eeta
+//re2 = infinity
+
+//Since Vagc=0 , transistor Q2 is in cut off region and collector current of Q1 flows through Q3....So
+Ie2=0;
+Ie3=1;//in mA
+re3 = (n*Vt)/Ie3;//in ohm
+disp('ohm',re3,'re3=');
+gm = (Ie1)/Vt;//in ohm^-1
+disp('ohm^-1',gm,'gm=');
+rbe=hfe/gm;
+disp('ohm',rbe,'rbe=');
+Ce=gm/(2*%pi*ft*10^-6);
+disp('pF',Ce,'Ce=');
+a3=1;//we make an assumption that alpha is one
+s=0;
+//Av0 = -((a3*gm)/(re3*rbb))*(1/(((1/rbb)+(1/rbe)+(s*Ce))*((1/re3)+(s*Cs))*((1/Rl)+(s*(Cs+Cl)))))
+//From here we can find Rl
+k = -((a3*gm)/(re3*rbb))*(1/(((1/rbb)+(1/rbe)+(s*Ce))*((1/re3)+(s*Cs))));
+Rl=Avo/k;
+disp('ohm',Rl,'Rl=');
+
+//C is in picoFarad so to compensate the whole equation some constants are multiplied
+f1 = 1/(2*%pi*Rl*(Cs+Cl)*10^-6);
+disp('MHz',f1,'f1=');
+f2 = 1/(2*%pi*Ce*10^-6*((rbe*rbb)/(rbe+rbb)));
+disp('MHz',f2,'f2=');
+f3 = 1/(2*%pi*Cs*re3*10^-6);
+disp('MHz',f3,'f3=');
+
+
+//end
+\ No newline at end of file