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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//chapter5
+//example5.3
+//page85
+
+// for Rl=50, Av=30
+//for Rl=85, Av=34
+
+// Av=mu*Rl/(rp+Rl)
+// thus
+// Av*rp-mu*Rl=-Av*rl
+// substituting for Rl=50 and Rl=85 we get the following linaer equations
+
+// 30*rp-50*mu=-1500 and
+// 34*rp-85*mu=-2890
+// solving by matrix
+
+a=[30 34 ; -50 -85]
+b=[-1500 -2890]
+solution=b/a
+mu=solution(1,2)
+rp=solution(1,1) // in kilo ohms since RL was in kilo ohm in the equations
+
+gm_kilo_mho=mu/rp
+gm=gm_kilo_mho/1000
+printf("mu = %.3f \n",mu)
+printf("rp = %.3f kilo ohm \n",rp)
+printf("gm = %.4f mho \n",gm)