From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

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 1757/CH14/EX14.10/EX14_10.sce | 70 +++++++++++++++++++++++++++++++++++++++++++
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 create mode 100755 1757/CH14/EX14.10/EX14_10.sce

(limited to '1757/CH14/EX14.10')

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+
+// Example14.10  // Design a video amplifier of IC 1550 circuit
+clc;
+clear;
+close;
+Vcc = 12 ; // V
+Av = -10 ;
+Vagc = 0 ; // at bandwidth of 20 MHz
+hfe = 50 ; // forward emitter parameter
+rbb = 25 ;  // ohm  // base resistor
+Cs = 1*10^-12 ;  // F  // source capacitor
+Cl = 1*10^-12 ;  // F  // load capacitor
+Ie1 = 1*10^-3 ; // A // emitter current of Q1
+f = 1000*10^6 ; // Hz
+Vt = 52*10^-3 ;
+Vt1 = 0.026 ;
+
+// When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3
+// i.e Ic1=Ie1=Ie3
+Ie3 = 1*10^-3 ; // A // emitter current of Q3
+Ic1 = 1*10^-3 ;  // A  // collector current of the transistor Q1
+
+// it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite
+re2 = %inf ;
+
+// emitter resistor of Q3 
+re3 = (Vt/Ie1);
+disp('The emitter resistor of Q3 is = '+string(re3)+' ohm ( at temperature 25 degree celsius) ');
+
+// the trans conductance of transistor is
+gm = (Ie1/Vt1);
+disp('The trans conductance of transistor is = '+string(gm*1000)+' mA/V ');  // Round Off Error
+
+// the base emitter resistor rbe
+rbe = (hfe/gm);
+disp('The base emitter resistor rbe is = '+string(rbe/1000)+' K ohm ');  // Round Off Error
+
+// the emitter capacitor Ce 
+Ce = (gm/(2*%pi*f));
+disp('The emitter capacitor Ce = '+string(Ce)+' F ');  // Round Off Error
+
+// the voltage gain of video amplifier is
+// Av = (Vo/Vin) ;
+// Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) 
+ // At Avgc = 0 i.e s=0 in the above Av equation
+alpha3 = 1 ;
+s = 0 ;
+// Rl = -((alpha3*gm)/(rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av))); 
+
+// After solving above equation for Rl We get Rl Equation as
+Rl = 10/(37.8*10^-3);
+disp('The value of resistance RL is = '+string(Rl)+' ohm ');
+
+// there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency 
+Rl = 675 ;
+// fa = 1/(2*%pi*Rl*(Cs+Cl));
+// after putting value of Rl ,Cs and Cl we get
+fa = 1/(2*3.14*264.55*1*10^-12);
+disp('The pole frequency fa is = '+string(fa*10^-3/1000)+' M Hz '); // Round Off Error
+
+
+//fb = 1/(2*%pi*Ce*((rbb*rbe)/(rbb+rbe)));
+// after putting value of Ce rbb and rbe we get
+fb = 1/(2*%pi*6.05*10^-12*24.5);
+disp('The pole frequency fb is = '+string(fb*10^-3/1000)+' M Hz ');
+
+fc = 1/(2*%pi*Cs*re3);
+disp('The pole frequency fc is = '+string(fc*10^-3/1000)+' M Hz ');
+
+disp(' Hence fa is a dominant pole frequency ');
-- 
cgit