1 files changed, 79 insertions, 0 deletions
diff --git a/1691/CH3/EX3.10/e3_10.sce b/1691/CH3/EX3.10/e3_10.sce
new file mode 100755
index 000000000..765d8e221
--- /dev/null
+++ b/1691/CH3/EX3.10/e3_10.sce
@@ -0,0 +1,79 @@
+//example3.10
+clc
+disp("The circuit of self biased binary is shown in the fig. 3.80")
+disp("Assume Q1 is OFF and Q2 is ON. As Q2 is in saturation,")
+disp("V_CE(sat)=V_CE2=0.4V")
+disp("V_BE(sat)=V_BE2=0.8V")
+disp("a) Calculation for the stable state currents and voltages")
+disp("Draw equilvalent circuit from base of Q1 to collector of Q2.")
+disp("Another equivalent circuit from collector of Q1 to base of Q2 is shown in the fig.3.81")
+disp("To calculate the various voltages, is necessary to calculate the currents I_c2, I_B2 for ON transistor Q2. The currents I_C1=I_B1=0 mA as Q1 is OFF.")
+disp("To obtain I_c2, I_B2 let us obtain. Thevenins equivalent once across collector and ground and other across base and ground, for ON transistor Q2.")
+disp("Consider Thevenins equivalent across collector of Q2 and ground as shown in the fig. 3.83(a) while Thevenins equivalent across base of Q2 and ground as shown in the fig. 3.83(b).")
+disp("Referring fig 3.83(a) we can write,")
+i=(20*45)/49.7
+format(7)
+disp(i,"V_oc(in volts)=I(R1+R2)=(V_cc*(R1+R2))/(R1+R2+R_c)=")
+r=(45*4.7)/(45+4.7)
+format(6)
+disp(r,"R_th(in k ohms) = (R1+R2)parallel to R_c with V_cc -N short  =")
+disp("Referring fig 3.83(b),")
+v=(20*15)/(30+15+4.7)
+format(6)
+disp(v,"V_OC(in V)=I*R2=(V_CC * R2)/(R1+R2+R_c)=")
+t=(15*34.7)/(15+34.7)
+format(7)
+disp(t,"And,  R_th(in k ohm)=R2 parallel to(R1+R2)=")
+disp("Applying KVL to base-emitter loop,")
+disp("-I_B2(10.473)-0.8-0.39(I_B2+I_C2)+6.036=0")
+disp("0.863(I_B2)+0.39(I_C2)=5.236")
+disp("I_B2+0.0359(I_C2)=0.482   Now multiply by 0.39,")
+disp("0.39(I_b2)+0.014(I_C2)=0.1879     ..(1)")
+disp("Applying KVL to collector emitter loop,")
+disp("(-I_C2)(4.255)-0.4-0.39(I_B2+I_C2)+18.108=0")
+disp("-0.39(I_B2)-4.645(I_c2)=-17.708    ..(2)")
+disp("Adding equations (1) and (2) we get,")
+disp("-4.631(I_C2)=-17.5201")
+c=(-17.5201)/(-4.631)
+format(6)
+disp(c,"I_C2(in mA)=")
+b=(-17.708+((4.645)*(3.783)))/(-0.39)
+disp(b,"and,      I_B2(in mA)=")
+disp("From this, the various voltages can be obtained as,")
+v=((0.346+3.783))*(0.390)
+format(5)
+disp(v,"V_EN(in V)=(I_B2+I_C2)*R_E =")
+n=0.4+1.61
+disp(n,"V_CN2(in V)=(V_CE2+V_EN)=")
+b=0.8+1.61
+disp(b,"V_BN2(in V)=(V_BE2+V_EN)=")
+w=(2.01*15)/45
+disp(w,"V_BN1(in V)=(V_CN2*R2)/(R1+R2)=")
+v=0.67-1.61
+format(5)
+disp(v,"V_BE1(in V)=(V_BN1-V_EN)=0.61-1.61=")
+disp("For cut-off, V_BE1 is 0V given, but actually it is still less i.e. -0.94 V. This ensures that Q1 is still OFF.")
+a=(((20*30)/(4.7+30))+((2.41*4.7)/(4.7+30)))
+format(7)
+disp(a,"V_CN1(in V)=")
+disp("b) To find (h_fe)_min")
+disp("For the ON transistor Q2")
+disp("I_C2=3.783mA,   I_B2=0.346mA")
+h=3.783/0.346
+format(3)
+disp(h,"Therefore, (h_fe)_min = (I_C2)/(I_B2)=")
+disp("Calculation of (I_CBO)_max")
+disp("To calculate (I_CBO)_max consider the circuit shown in the fig 3.85")
+disp("Obtain the Thevenins equivalent across terminal A and ground.The Thevenin voltage is V_A=V_B1=0.67 V")
+disp("Looking into terminals A and ground,")
+r=(34.7*15)/(34.7+15)
+format(7)
+disp(r,"R_th(in kohms)=(R_1+R_c)parallel to R2 =")
+disp("Hence Thevenin equivalent is: To find I_CBO_max, ")
+disp("V_BE(cut-off)=0V and V_EN =1.61 V ...Calculated earlier")
+disp("As V_BE= 0, base must be also at same potential as emitter with respect to ground.")
+disp("V_B1=V_EN=1.61 V   for(I_CBO)_max,")
+o=(1.61-0.67)/(10.472*10^3)
+format(11)
+disp(o,"I_CBO_max(in A)=(V_B1-V_TH)/(R_TH)=")
+disp("This is the maximum I_CBO")