initial commit / add all books

8 files changed, 193 insertions, 0 deletions

diff --git a/2492/CH2/EX2.1/ex2_1.sce b/2492/CH2/EX2.1/ex2_1.sce
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index 000000000..45774bc43
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+++ b/2492/CH2/EX2.1/ex2_1.sce
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+// Exa 2.1

+format('v',9)

+clc;

+clear;

+close;

+// Given data

+n_i = 1.5 * 10^10;// in /cc

+p = n_i;// in /cc

+n = n_i;// in /cc

+miu_n = 1400;// in cm^2/V-s

+miu_p = 450;//in cm^2/V-s

+q = 1.6 * 10^-19;// in C

+E = 20;// in V/cm

+a= 5;// cross section area of Si bar in cm^2

+sigma_n = n*q*miu_n;// in mho/cm

+sigma_p = n*q*miu_p;// in mho/cm

+// Electron current density 

+Jn = sigma_n*E;// in A

+Jn= Jn*10^6;// in µA/cm^2

+disp(Jn,"Electron current density in µA/cm^2 is");

+// The hole current density 

+Jp = sigma_p*E;// in A/cm^2

+Jp= Jp*10^6;// in µA/cm^2

+disp(Jp,"The hole current density in µA/cm^2 is");

+//The total current in the bar 

+total = (Jn+Jp)*a;// µA/cm^2

+disp(total,"The total current in the bar in µA/cm^2 is");

+format('e',8)

+// The resistivity of the bar 

+rho = 1/(n_i*q*(miu_p+miu_n)*10^2);// in ohm-cm

+disp(rho,"The resistivity of the bar in ohm-cm is");

diff --git a/2492/CH2/EX2.2/ex2_2.sce b/2492/CH2/EX2.2/ex2_2.sce
new file mode 100755
index 000000000..849b9adbc
--- /dev/null
+++ b/2492/CH2/EX2.2/ex2_2.sce
@@ -0,0 +1,26 @@
+// Exa 2.2

+format('v',5)

+clc;

+clear;

+close;

+// Given data

+V_F = 20;// in V

+Vin = V_F;// in V

+V_BE = 0.7;// in V

+R1 = 500;//resistance in ohm

+R2 = 10;// resistance in ohm

+// Peak current though the diode

+Ifpeak = (V_F-V_BE)/(R1+R2);// in A

+Ifpeak = Ifpeak * 10^3;// in mA

+disp(Ifpeak,"The peak current through the diode in mA is");

+R_L = 500;// in ohm

+// Peak output voltage

+Vpeakout = Ifpeak*10^-3*R_L;// in V

+disp(Vpeakout,"The peak out voltage in V is");

+// For ideal diode

+Ifpeak = V_F/R_L;// in A

+Ifpeak = Ifpeak * 10^3;// in mA

+// The peak output voltage for ideal diode 

+Vpeakout= Ifpeak*10^-3*R_L;// in V

+disp(Ifpeak,"The peak current for ideal diode in mA is");

+disp(Vpeakout,"The peak output voltage for ideal diode in V is");

diff --git a/2492/CH2/EX2.3/ex2_3.sce b/2492/CH2/EX2.3/ex2_3.sce
new file mode 100755
index 000000000..66b3e9ed7
--- /dev/null
+++ b/2492/CH2/EX2.3/ex2_3.sce
@@ -0,0 +1,25 @@
+// Exa 2.3

+format('v',6)

+clc;

+clear;

+close;

+// Given data

+t = 27;// in °C

+t= t+273;// in °K

+q= 1.6*10^-19;// electron charge in C

+v = 200*10^-3;// in V

+kt_by_q= 0.026;// in V

+Io= 3*10^-7;// in A

+// For large reverse bias I= Io*(%e^(q*v/(k*t)))

+I= Io*(%e^(v/kt_by_q)-1);// in A

+I= round(I*10^6);// in µA

+disp(I,"The current flowing through the diode in µA is");

+Idaso = Io*2^7;// in A

+// r_ac = dv/di = 1/( Io*(1/kTdividedq)*(%e^(v/(kTdividedq))) );

+r_ac = 1/( Io*(1/kt_by_q)*(%e^(v/(kt_by_q))) );

+disp(r_ac,"The ac resistance in ohm is : ");

+kt_by_q = 0.032;// in V

+I1 = Idaso * ((%e^(v/(kt_by_q))) - 1);// in A

+I1 = I1 * 10^3;// in mA

+delI = I1-(I*10^-3);// in mA

+disp(delI,"The increase in forward current in mA is");

diff --git a/2492/CH2/EX2.4/ex2_4.sce b/2492/CH2/EX2.4/ex2_4.sce
new file mode 100755
index 000000000..792a9e905
--- /dev/null
+++ b/2492/CH2/EX2.4/ex2_4.sce
@@ -0,0 +1,20 @@
+// Exa 2.4

+format('e',9)

+clc;

+clear;

+close;

+// Given data

+V = 30;//applied forward voltage in V

+R_L = 3;//load resistance in  k ohm

+R_L = R_L * 10^3;// in ohm

+Imax = V/R_L;// maximum diode current in A

+Imax = Imax * 10^3;// in mA

+slope = -1/R_L;// in mho

+plot([V,0],[0,Imax]);

+xlabel("V_F in volts");

+ylabel("I_F in mA");

+title("DC load line");

+disp("DC load line shown in figure")

+disp(slope,"The slope of the line in mho is");

+

+// Note: There is calculation error to find the value of slope because -1/3kΩ = -3.33*10^-4 mho,  not = -3.33*10^-3 mho

diff --git a/2492/CH2/EX2.5/ex2_5.sce b/2492/CH2/EX2.5/ex2_5.sce
new file mode 100755
index 000000000..db2bb8eb6
--- /dev/null
+++ b/2492/CH2/EX2.5/ex2_5.sce
@@ -0,0 +1,23 @@
+// Exa 2.5

+format('v',6)

+clc;

+clear;

+close;

+// Given data

+Vp= '[4+0.2*sin(omega*t)]';// in V

+Ip= '[4+0.3*sin(omega*t)]';// in mA

+//The instantaneous power dissipated, P= Vp*Ip = [4+0.2*sind(wt)]*[4+0.3*sind(wt)] 

+// and putting, sin^2(omega*t)= 1/2-1/2*cos(2*omega*t)

+P= '[16.03+2*sin(omega*t)-0.03*cos(2*omega*t)]*10^-3';// in W

+disp(P,"The instantaneous power dissipated in the diode in W is : ")

+// Pmax occurs when omega*t=90, so

+omega_t= 90;// in °

+Pmax= [16.03+2*sind(omega_t)-0.03*cosd(2*omega_t)]*10^-3;// in W

+disp(Pmax,"The maximum value of instantaneous power dissipated in W is : ")

+// Pmin occurs when omega*t=-90, so

+omega_t= -90;// in °

+Pmin= [16.03+2*sind(omega_t)-0.03*cosd(2*omega_t)]*10^-3;// in W

+disp(Pmin,"The minimum value of instantaneous power dissipated in W is : ");

+// The average power dissipated

+Pav=(Pmax+Pmin)/2;// in W

+disp(Pav,"The average power dissipated in the diode in W is : ")

diff --git a/2492/CH2/EX2.6/ex2_6.sce b/2492/CH2/EX2.6/ex2_6.sce
new file mode 100755
index 000000000..549de380c
--- /dev/null
+++ b/2492/CH2/EX2.6/ex2_6.sce
@@ -0,0 +1,20 @@
+// Exa 2.6

+format('v',9)

+clc;

+clear;

+close;

+// Given data

+R_L = 50;// in ohm

+V = 10;// in V

+R = 5;// in ohm

+V_D = (V*R)/(R_L+R);// in V

+R_D = (R_L*R)/(R_L+R);// in ohm

+I_D = V_D/R_D;// in A

+I_D = I_D * 10^3;// in mA

+plot([V_D,0],[0,I_D])

+xlabel("V_D in volts");

+ylabel("I_D in mA");

+title("DC load line");

+disp("DC load line shown in figure")

+slope = -1/R_D;

+disp(slope,"The slope of the dc load line is : ");

diff --git a/2492/CH2/EX2.7/ex2_7.sce b/2492/CH2/EX2.7/ex2_7.sce
new file mode 100755
index 000000000..aba6b1ea5
--- /dev/null
+++ b/2492/CH2/EX2.7/ex2_7.sce
@@ -0,0 +1,26 @@
+// Exa 2.7

+format('e',8)

+clc;

+clear;

+close;

+// Given data

+V_T= 26*10^-3;// in V

+T = 300;// in K

+V = 0.25;// in V

+I=  10 * 10^-3;// in A

+// I = I_S*( (%e^(V/(n*kTdividedq)))-1 ) = I_S*( (%e^(V/V_T))-1 );

+I_S= I/(%e^(V/V_T)-1);// in A

+disp(I_S,"The reverse saturate current in amp. is : ")

+format('v',6)

+// For 1 mA current

+I = 1;// in mA

+I = I*10^-3;// in A

+V = (1/38.46)*log(I/I_S);// in V

+disp(V,"The bias voltage needed for 1 mA in V is");

+// For 100 mA current

+I = 100;// in mA

+I = I * 10^-3;// in A

+V = (1/38.46)*log(I/I_S);// in V

+disp(V,"The bias voltage needed for 100 mA in V is");

+

+// Note: Answer in the book is not accurate.

diff --git a/2492/CH2/EX2.8/ex2_8.sce b/2492/CH2/EX2.8/ex2_8.sce
new file mode 100755
index 000000000..e6eabc950
--- /dev/null
+++ b/2492/CH2/EX2.8/ex2_8.sce
@@ -0,0 +1,22 @@
+// Exa 2.8

+format('v',6)

+clc;

+clear;

+close;

+// Given data

+V_T= 25*10^-3;// in V

+// I = Io*( (%e^(V/V_T))-1 );

+I = 1;// in mA

+I = I * 10^-3;// in A

+V = 0.15;// in V

+Io = I/( (%e^((1/V_T)*V))-1 );// in A

+Io = Io * 10^6;// in µA

+disp(Io,"The reverse saturation current at room temperature in µA is");

+// Io doubles for every 10°C-rise in temperature. Thus at 40°C

+Io_new= 4*Io;//new value of reverse saturation current in µA

+disp(Io_new,"The reverse saturation current at 40 °C in µA is");

+I = 100;// in mA

+I = I * 10^-3;// in A

+// I = Io*( (%e^((1/V_T)*V))-1 );

+V = (1/40)*log( I/(Io*10^-6) );// in V

+disp(V,"The forward bias voltage in V is");