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diff --git a/181/CH2/EX2.1/example2_1.sce b/181/CH2/EX2.1/example2_1.sce
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+// Calculate height of potential-energy barrier
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 2-1 in page 77
+
+clear; clc; close;
+
+// Given data
+rho1=1.5; // Resistivity of p-side of Ge diode in ohm-cm
+rho2=1; // Resistivity of n-side of Ge diode in ohm-cm
+e=1.6*10^-19; // Charge on an electron in C
+mu_p=1800; // Mobility of holes
+mu_n=3800; // Mobility of electrons
+
+// Calculation
+N_A=1/(rho1*e*mu_p);
+N_D=1/(rho2*e*mu_n);
+printf("(a) rho = 2 ohm-cm\n");
+printf("N_A=%0.2e /cm^3\n",N_A);
+printf("N_D=%0.2e /cm^3\n",N_D);
+printf("The height of the potential energy barrier is:\n");
+V_0=0.026*log((N_A*N_D)/(2.5*10^13)^2);
+printf("V_0=%0.3f eV\n\n",V_0);
+printf("(b)For silicon:\n");
+N_A1=1/(rho1*e*500);
+N_D1=1/(2*e*1300);
+printf("N_A=%0.2e /cm^3\n",N_A1);
+printf("N_D=%0.2e /cm^3\n",N_D1);
+V_01=0.026*log((N_A1*N_D1)/(1.5*10^10)^2);
+printf("The height of the potential energy barrier is:\n");
+printf("V_0=%0.3f eV",V_01);
+
+// Result
+// (a) For Ge, V_0 = 0.226 eV 
+// (b) For Si, V_0 = 0.655 eV
diff --git a/181/CH2/EX2.1/example2_1.txt b/181/CH2/EX2.1/example2_1.txt
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+(a) rho = 2 ohm-cm
+N_A=2.31e+015 /cm^3
+N_D=1.64e+015 /cm^3
+The height of the potential energy barrier is:
+V_0=0.227 eV
+
+(b)For silicon:
+N_A=8.33e+015 /cm^3
+N_D=2.40e+015 /cm^3
+The height of the potential energy barrier is:
+V_0=0.656 eV 
+\ No newline at end of file