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+// Exa 2.16
+
+clc;
+clear;
+
+// Given data
+
+// A level shifter as shown in fig. 2.31
+// Assuming Ideal silicon transistors  
+Vbe = 0.7; // Volts
+// B(current gain) has very large values
+Vcc = 15; // Volts
+Rc = 10*10^3; // Ω
+Re = 5000; // Ω
+
+// Solution
+
+printf(' From fig. 2.31 we get that, transistors Q1 and Q2 form a current mirror.\n');
+printf('  so Ic1 = Ic2 = I and that can be found by Ohm‘s law as ');
+I = (Vcc - Vbe)/Rc; // Ω
+printf(' I = Ic2 = %.2f mA. \n', I*1000 );
+printf('  Now, the difference V1-V2 can be found using KVL as ');
+dV = Vbe + I * Re;  // KVL between end points
+printf(' %.2f V. \n',dV);