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Diffstat (limited to '3682/CH2/EX2.15/Ex2_15.sce')
| -rw-r--r-- | 3682/CH2/EX2.15/Ex2_15.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/3682/CH2/EX2.15/Ex2_15.sce b/3682/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..b337e3b30 --- /dev/null +++ b/3682/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,28 @@ +// Exa 2.15
+
+clc;
+clear;
+
+// Given data
+
+// Referring Fig. 2.26, we get
+Vcc=10; // Volts
+R1= 4.7*10^3; // k Ohm’s
+B=100; // Current gain(>>1)
+Vbe=0.75; // Volts
+
+// Solution
+
+disp(" Node ‘A‘ is at transistor Q1, Node ‘B’ is at transistor Q2 and Node ‘C’ is at transistor Q3.");
+
+printf('\n From Fig. 2.26 at node ‘A‘. I = Ic1 + Ib1 + I1‘ ...Eqn(1)');
+printf(' \n Also at node ‘B‘. I1‘ = Ic2 + Ib3.');
+printf('\n Putting value of I1‘ in eqn(1) we get I = (approximately) 2Ic. \n');
+
+I = (Vcc-Vbe)/R1; // By ohm‘s law
+
+printf('\n The calculated value of I = %.2f mA. \n' , I*1000);
+
+Ic3 = I/2;
+
+printf(' The collector current of Q3 is equal to the collector current of Q1 and Q2 due to mirror action. \n Therefore, the emitter current IE3 = Ic3 = Ic = I/2 = %.2f mA. \n ',Ic3*1000);
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