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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+//Average power dissipated by the gate
+//Given Data
+Vbb = 1.15;//in V
+Vee=5.20;//in V
+Vbe5=0.7;//in V
+R=1.18;//in K
+r=300;//in ohm
+Vbecutin=0.5;//in V
+
+//If all inputs are low then we assume that Q1,Q2 and Q3 are cutoff and Q4 is conducting
+Ve=-Vbb-Vbe5;//Voltage at Common Emitter in V
+//Current I in 1.18K Resistor
+I = (Ve+Vee)/R;//in mA
+I1=I;
+//Output Voltage at Y
+vy = -(r*I/1000)-Vbe5;//I is in mA so 1000 is multiplied
+Vbe = vy-Ve;
+if(Vbe<Vbecutin)
+ v=0.7;//voltage across Q5 in V
+ rQ5 = 1.5;//in K
+ i = (Vee-v)/rQ5;
+ v = 0.75;//from the graph in V
+ Ve = -v-Vbe5;
+ Vbe4=-Vbb-Ve;
+end
+
+vo = -vy-v;
+
+Vb4 = Vbb;
+Vc4 = -(I*r)/1000;//in V
+Vcb4 = Vc4+Vb4;
+Vb1 = v;
+Vc1 = vy+Vbe5;
+Vcb1 = Vc1 + Vb1;
+
+Vbe1 = Vbe5;
+Ve = -(Vb1+Vbe1);
+I = (Ve + Vee)/R;
+I2=I;
+
+I =(I1+I2)/2;
+disp('mA',I,'I=');
+I2 = (Vee-v)/rQ5;
+I3 = (Vee+vy)/rQ5;
+I = I + I2 + I3;
+P = Vee*I;
+disp('mW',P,'Power dissipated = ');
+
+//end \ No newline at end of file