9 files changed, 78 insertions, 0 deletions
diff --git a/991/CH20/EX20.1/Example20_1.sce b/991/CH20/EX20.1/Example20_1.sce
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+//Example 20.1.
+clc
+format(6)
+disp("    CMRR = Ad / Acm = 10^5")
+acm=(10^5)/(10^5)
+disp(acm,"    Therefore, the common-mode gain, Acm = Ad / CMRR =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.2/Example20_2.sce b/991/CH20/EX20.2/Example20_2.sce
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index 000000000..3ce1cd6b3
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+++ b/991/CH20/EX20.2/Example20_2.sce
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+//Example 20.2.
+clc
+format(6)
+sr=20/(4) // in V/us
+disp("    The slew rate,        SR = dVo / dt")
+disp(sr,"    SR(in V/us) =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.3/Example20_3.sce b/991/CH20/EX20.3/Example20_3.sce
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index 000000000..ee99ec349
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+++ b/991/CH20/EX20.3/Example20_3.sce
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+//Example 20.3.
+clc
+format(5)
+disp("The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,")
+disp("      SR = 2*pi*f*Vm / 10^6 = 0.5 V/us")
+vm=50*(20*10^-3) // in volts
+disp(vm,"The maximum output voltage,  Vm(V) = A*Vid =")
+disp("The maximum frequency of the input for which undistorted output is obtained is given by,")
+f=(0.5*10^6)/(2*%pi*1) // in kHz
+x1=f*10^-3
+disp(x1,"     fmax = SR*10^6 / 2*pi*Vm =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.4/Example20_4.sce b/991/CH20/EX20.4/Example20_4.sce
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index 000000000..1734e724c
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+++ b/991/CH20/EX20.4/Example20_4.sce
@@ -0,0 +1,11 @@
+//Example 20.4.
+clc
+format(5)
+disp("The 741C has typical slew rate of 0.5 V/us. Using Eq.(20.8), the slew rate is,")
+disp("      SR = 2*pi*f*Vm / 10^6 = 0.5 V/us")
+vm=(0.5*10^6)/(2*%pi*(40*10^3)) // in volts
+disp("    = 3.98 V peak-to-peak",vm,"The maximum output voltage,  Vm(V peak-to-peak) = SR*10^6 / 2*pi*f =")
+disp("The maximum peak-to-peak input voltage for undistorted output is,")
+vid=3.98/10 // in volts
+format(6)
+disp(vid,"      Vid(V peak-to-peak) = Vm/A =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.5/Example20_5.sce b/991/CH20/EX20.5/Example20_5.sce
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index 000000000..87056908b
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+++ b/991/CH20/EX20.5/Example20_5.sce
@@ -0,0 +1,5 @@
+//Example 20.5. refer fig.20.10.
+clc
+format(6)
+af=-10/1
+disp(af,"  The closed-loop voltage gain Af = -RF / R1 =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.6/Example20_6.sce b/991/CH20/EX20.6/Example20_6.sce
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index 000000000..93cf57e13
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+++ b/991/CH20/EX20.6/Example20_6.sce
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+//Example 20.6. refer fig.20.11.
+clc
+format(6)
+af=1+(10/1)
+disp(af,"  The closed-loop voltage gain, AF = 1 + RF/R1 =")
+beta=1/(1+10)
+disp(beta,"  The feedback factor,  beta = R1 / R1+RF =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.7/Example20_7.sce b/991/CH20/EX20.7/Example20_7.sce
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index 000000000..52c468008
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+++ b/991/CH20/EX20.7/Example20_7.sce
@@ -0,0 +1,6 @@
+//Example 20.7. refer fig. 20.16.
+clc
+format(6)
+v=-(2+3+4) // in volts
+disp("The output voltage is given by,")
+disp(v,"    Vo(V) = -Rf/R * (V1+V2+...+Vn) =")
+\ No newline at end of file
diff --git a/991/CH20/EX20.8/Example20_8.sce b/991/CH20/EX20.8/Example20_8.sce
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index 000000000..e69bee383
--- /dev/null
+++ b/991/CH20/EX20.8/Example20_8.sce
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+//Example 20.8.
+clc
+format(5)
+disp("1.  Given: fL = 1 kHz")
+disp("2.  Since R and C values are not given, let assume C = 0.01 uF")
+r=1/(2*%pi*(10^3)*(0.01*10^-6))
+x1=r*10^-3  // in k-ohm
+disp(x1,"3.  Therefore, R(k-ohm) = 1 / 2*pi*fL*C =")
+disp("4.  Given pass band gain A = 1 + Rf/Ri = 2 i.e. the value of Rf = Ri")
+disp("Let Rf = Ri = 10 k-ohm. The high pass circuit values are shown in Fig.20.31")
+\ No newline at end of file
diff --git a/991/CH20/EX20.9/Example20_9.sce b/991/CH20/EX20.9/Example20_9.sce
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index 000000000..7a2d9aaeb
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+++ b/991/CH20/EX20.9/Example20_9.sce
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+//Example 20.9. refer fig.20.35(a).
+clc
+format(6)
+disp("(a) From Eq.(20.32), the time period, T = 2RC ln(R1+2R2 / R1)")
+disp("          T = 2RC ln(116*10^3 + 2*100*10^3/116*10^3)")
+disp("          T = 2RC ln(316*10^3/116*10^3)")
+disp("          T = 2RC        (since ln(316*10^3/116*10^3) = 1)")
+disp("Given     f = 1 kHz,  T = 1/f = 1 ms")
+disp("That is,   2RC = 1*10^-3 sec")
+disp("Therefore, the time constant RC = 0.5*10^-3 sec")
+r=(0.5)/0.01 // in k-ohm
+disp(r,"(b) With      C = 0.01 uF,  R(k-ohm) = 0.5*10^-3/0.01*10^-6 =")
+disp("(c) Maximum value of differential input voltage is")
+x=2*14*(100/(100+116))
+disp(x,"          2*Vsat*(R2 / R1+R2) = ")
+disp("Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V")
+\ No newline at end of file