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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);
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