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diff --git a/1223/CH1/EX1.6/Ex1_6.sce b/1223/CH1/EX1.6/Ex1_6.sce
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+clear;
+clc;
+//Example 1.6
+Is=10^-13;//(A)reverse saturation current
+V_PS=5;//(V)applied voltage
+R=2;//(KOhm)Resistance in circuit
+V_T=0.026;//(Volt)terminal voltage
+//V_PS=Is*R*(exp(V_D/V_T)-1)+V_D
+//5=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D
+//let right side of equation be x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D
+V_D=0.6;//(V)
+x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D
+//so the equation is not balanced
+V_D=0.65;//(V)
+x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D
+//again equation is not balanced .solution for V_D is between 0.6V and 0.65V
+V_D=0.619;//(V)
+x=(10^-13)*(2000)*(exp(V_D/V_T)-1)+V_D
+//essentially equal to the value of the left side of the equation i.e 5V
+printf('\ndiode voltage=%f V',V_D)
+I_D=(V_PS-V_D)/R;//(A)diode current
+printf('\nthe diode current=%f mA',I_D)