1 files changed, 36 insertions, 0 deletions
diff --git a/3682/CH2/EX2.12/Ex2_12.sce b/3682/CH2/EX2.12/Ex2_12.sce
new file mode 100644
index 000000000..322ec5837
--- /dev/null
+++ b/3682/CH2/EX2.12/Ex2_12.sce
@@ -0,0 +1,36 @@
+// Exa 2.12
+
+clc;
+clear;
+
+// Given data
+
+//Referring fig No. 2.21, we get
+
+Vcc=9; // Volts
+Vbe=0.7; // Volts
+R1=30*10^3; // Ω
+Re=1.94; // Ω
+B=125;  // current gain
+VT = 25*10^-3; // Volts
+
+// Solution
+
+Iref= (Vcc-Vbe)/R1;
+printf(' The value of Iref = %.3f mA. \n ',Iref*1000);
+// Also at Node A.- Iref=Ic+3*Ib. i.e Ic = Iref*(B/(B+2))
+// Assuming IB3 of widlar source negligible.
+// Therefore putting back value of Iref we get values of Ic1
+Ic=Iref*(B/(B+3));
+Ic_mA = Ic*1000;  // in mA
+
+printf('\n  The value of Ic1 = Ic2 = %.3f mA. \n ',Ic*10^3);
+// Calculating Ic3 using eqn 2.74 ;
+
+// Re = (VT/(Ic3*(1+1/B)))*ln(ic_mA/Ic3);
+// Re - (VT/(Ic3*(1+1/B)))*ln(ic_mA/Ic3) = 0;
+
+deff('y =  f(x)', 'y = (Re-(VT*log(Ic_mA/x))/(x*(1+1/B)))'); // here x = Ic3
+[x,v,info]= fsolve(0.01,f);
+
+printf(' \n  By trial and error method, we get Ic3 = %.4f mA.\n',x);