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+// Exa 2.18
+clc;
+clear;
+close;
+format('v',6)
+// Given data
+V_CC= 12;// in V
+bita_min= 30;
+R1= 15;// in k ohm
+R2= 100;// in k ohm
+R_C= 2.2;// in kohm
+V_BB= -12;// in V
+V_BE= 0.7;// in V
+// Part (i)
+Vi= 12;// in V
+V_BEsat= 0.8;// in V
+V_CEsat= 0.2;// in V
+// Applying KVL to B-E circuit, Vi= I1*R1+V_BEsat or
+I1= (Vi-V_BEsat)/R1;// in mA
+// Applying KVL to -12 V supply,
+I2= (V_BEsat-V_BB)/R2;// in mA
+// Applying KVL to input loop,
+I_B= I1-I2;// in mA
+// Applying KVL to output loop, V_CC= I_C*R_C+V_CEsat or
+I_C= (V_CC-V_CEsat)/R_C;// in mA
+I_Bmin= I_C/bita_min;// in mA
+if I_B>I_Bmin then
+    disp("Part (a) :")
+    disp("As the value of I_B ("+string(I_B)+" mA) is greater than the value of I_Bmin ("+string(I_Bmin)+" mA)")
+    disp("Hence the transistor is in saturation region")
+end
+Vo= V_CC-I_C*R_C;// in V
+disp(Vo,"The output voltage in volts is : ")
+
+// Part (b)
+I2= (V_CC+V_BE)/R2;// in mA
+// and I1= (V_CC-V_BE)/R1;// in mA          (i)
+I_B= I_Bmin;// in mA
+I1= I2+I_Bmin;// in mA
+// Now from eq(i)
+R1= (V_CC-V_BE)/I1;// in k ohm
+disp("Part (b)")
+disp(R1,"The value of R1 in k ohm is : ")
+
+// Part (c)
+I_C= 0;// in mA
+Vo= V_CC-I_C*R_C;// in V
+disp("Part (c) : Transistor is in cutoff")
+disp(Vo,"The value of Vo in volts is : ")
+
+// Note: There is some difference between coding output and answer of the book. This is why because in the book the calculate value of I_C is 5.36 mA/ 30 = 0.178 mA while accurate value is 0.179 mA
+ 
+
+
+
+
+