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diff --git a/181/CH7/EX7.27/example7_27.sce b/181/CH7/EX7.27/example7_27.sce
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+// Find Id,Vds,Vgs,Av
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 7-27 in page 331
+
+clear; clc; close;
+
+// Given data
+Idss=3; // Drain-source current in mA
+Vp=-2.4; // Pinch off voltage in volts
+
+// Calculation
+printf("Id^2-6.73*Id+5.76=0\n");
+Id1=(6.73+sqrt(6.73^2-4*1*5.76))/2;
+Id2=(6.73-sqrt(6.73^2-4*1*5.76))/2;
+printf("Id = %0.2f mA or %0.2f mA\n",Id1,Id2);
+printf("(a)The possible value is 1.01 mA\n");
+Vgs=-Id2*1;
+Vds=20-(1.09*(1+10));
+printf("Vgs = %0.2f V\nVds = %0.2f V\n",Vgs,Vds);
+gm=(-2/Vp)*sqrt(Id2*Idss);
+Av=gm*10;
+printf("(b)Voltage gain Av = %0.1f",Av);
+
+// Result
+// Id = 1.01 mA
+// Vgs = -1.01 V
+// (a) Vds = 8.01 V
+// (b) Av = 14.5
+\ No newline at end of file
diff --git a/181/CH7/EX7.27/example7_27.txt b/181/CH7/EX7.27/example7_27.txt
new file mode 100755
index 000000000..f879c941c
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+Id^2-6.73*Id+5.76=0
+Id = 5.72 mA or 1.01 mA
+(a)The possible value is 1.01 mA
+Vgs = -1.01 V
+Vds = 8.01 V
+(b)Voltage gain Av = 14.5 
+\ No newline at end of file