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-rwxr-xr-x182/CH5/EX5.1/example5_1.sce15
-rwxr-xr-x182/CH5/EX5.1/example5_1.txt3
-rwxr-xr-x182/CH5/EX5.2/5_2.txt3
-rwxr-xr-x182/CH5/EX5.2/Example5_2.sce21
-rwxr-xr-x182/CH5/EX5.3/5_3.txt4
-rwxr-xr-x182/CH5/EX5.3/example5_3.sce22
-rwxr-xr-x182/CH5/EX5.4/5_4.txt1
-rwxr-xr-x182/CH5/EX5.4/example5_4.sce12
-rwxr-xr-x182/CH5/EX5.5/5_5.txt2
-rwxr-xr-x182/CH5/EX5.5/example5_5.sce15
-rwxr-xr-x182/CH5/EX5.6/5_6.txt1
-rwxr-xr-x182/CH5/EX5.6/exampke5_6.sce17
-rwxr-xr-x182/CH5/EX5.7/5_7.txt2
-rwxr-xr-x182/CH5/EX5.7/example5_7.sce11
14 files changed, 129 insertions, 0 deletions
diff --git a/182/CH5/EX5.1/example5_1.sce b/182/CH5/EX5.1/example5_1.sce
new file mode 100755
index 000000000..ba9eda331
--- /dev/null
+++ b/182/CH5/EX5.1/example5_1.sce
@@ -0,0 +1,15 @@
+// to find the high and low output voltage values
+// example 5-1 in page 120
+clc;
+//Given data
+Vcc=5;// DC source in volts
+Io=1e-3;// output current= 1mA
+R1=1e+3;//R1=1K-ohm
+Vi=0;//lowest input voltage
+Vd=0.7;// silicon-diode drop in volts
+//calculation
+printf("High output voltage=%d V\n",Vcc-(Io*R1));
+printf("low output voltage=%.1f V\n",Vi+Vd);
+//result
+//High output voltage=4 V
+//low output voltage=0.7 V \ No newline at end of file
diff --git a/182/CH5/EX5.1/example5_1.txt b/182/CH5/EX5.1/example5_1.txt
new file mode 100755
index 000000000..1ce44d1bb
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+++ b/182/CH5/EX5.1/example5_1.txt
@@ -0,0 +1,3 @@
+
+1.Highest out put voltage=4 V
+lowest output voltage=0.700000 V \ No newline at end of file
diff --git a/182/CH5/EX5.2/5_2.txt b/182/CH5/EX5.2/5_2.txt
new file mode 100755
index 000000000..a70319809
--- /dev/null
+++ b/182/CH5/EX5.2/5_2.txt
@@ -0,0 +1,3 @@
+
+2.Vb1=-1.657143 V
+Vc1=4.395973 V
diff --git a/182/CH5/EX5.2/Example5_2.sce b/182/CH5/EX5.2/Example5_2.sce
new file mode 100755
index 000000000..8987b11e1
--- /dev/null
+++ b/182/CH5/EX5.2/Example5_2.sce
@@ -0,0 +1,21 @@
+// To find the collector and base voltages
+// example 5-2 in page 121
+clc;
+// Given data
+R1=15e+3; R2=27e+3; Rc1=2.7e+3; R11=R1; R21=R2; //resistance values in Ohm where R11=R1' and R21=R2'
+Vc2=0.2;// collector voltage of on transistor in volt
+Vce=Vc2;//collector-emitter saturation voltage in volt
+Vbb=-5;//dc power supply in volt
+Vcc=5;//dc power supply in volt
+//calculations
+Vr1r2=Vc2-Vbb;//voltage across Ri and R2 in volt
+Vr1=(R1/(R1+R2))*Vr1r2;// voltage across R1 resistor in volt
+Vb1=Vc2-Vr1;// base voltage
+printf("Vb1=%.1f V\n",Vb1);
+//with Q1 off
+Vrc1=(Rc1/(Rc1+R11+R21))*(Vcc-Vbb);
+Vc1=Vcc-Vrc1;// collector voltage in volt
+printf("Vc1=%.1f V",Vc1);
+//result
+//Vb1=-1.657143 V
+//Vc1=4.395973 V \ No newline at end of file
diff --git a/182/CH5/EX5.3/5_3.txt b/182/CH5/EX5.3/5_3.txt
new file mode 100755
index 000000000..d2e9dede0
--- /dev/null
+++ b/182/CH5/EX5.3/5_3.txt
@@ -0,0 +1,4 @@
+3case 1,
+Total current=0.460000 A
+case 2,
+Total current=0.006900 A
diff --git a/182/CH5/EX5.3/example5_3.sce b/182/CH5/EX5.3/example5_3.sce
new file mode 100755
index 000000000..d998d4f5d
--- /dev/null
+++ b/182/CH5/EX5.3/example5_3.sce
@@ -0,0 +1,22 @@
+
+//example 5-3 in page 124
+clc;
+// Given data
+// 3(1/2) digit display
+If1=20e-3;//forward current per segment of led=20 mA
+If2=300e-6;//forward current per segment of lcd
+//calculations
+for n=1:2
+ if n==1
+ I=If1;
+ else I=If2;
+ end
+ It=3*7*I+2*I;// each digit has 7 segments and there are three digits with a half digit that has 2 segments
+ printf("case %d,\n Total current=%.0f mA\n",n,It*1000);
+end
+//result
+// case 1,
+//Total current=0.460000 A
+//case 2,
+//Total current=0.006900 A
+ \ No newline at end of file
diff --git a/182/CH5/EX5.4/5_4.txt b/182/CH5/EX5.4/5_4.txt
new file mode 100755
index 000000000..80b791c41
--- /dev/null
+++ b/182/CH5/EX5.4/5_4.txt
@@ -0,0 +1 @@
+4.output frequency=244 hertz \ No newline at end of file
diff --git a/182/CH5/EX5.4/example5_4.sce b/182/CH5/EX5.4/example5_4.sce
new file mode 100755
index 000000000..a2d1fcccd
--- /dev/null
+++ b/182/CH5/EX5.4/example5_4.sce
@@ -0,0 +1,12 @@
+// to find the out put frequency in fig 5-10
+// example 5-4 in page 130
+clc;
+//Given data
+To=1e-6;//oscillator time period=1 micro-second
+N=16;// modulus number of the counter = 16
+n=3;// number of counters
+//calculations
+T=To*(N^n);//out put time period
+printf("output frequency=%d hertz",1/T);//output frequency
+//result
+//output frequency=244 hertz \ No newline at end of file
diff --git a/182/CH5/EX5.5/5_5.txt b/182/CH5/EX5.5/5_5.txt
new file mode 100755
index 000000000..4ed1f43de
--- /dev/null
+++ b/182/CH5/EX5.5/5_5.txt
@@ -0,0 +1,2 @@
+5.number of pulses counted for Vi=0.750000 V are 75000
+number of pulses counted for Vi=0.900000 V are 90000
diff --git a/182/CH5/EX5.5/example5_5.sce b/182/CH5/EX5.5/example5_5.sce
new file mode 100755
index 000000000..f5403b3ff
--- /dev/null
+++ b/182/CH5/EX5.5/example5_5.sce
@@ -0,0 +1,15 @@
+//To find the number of pulses counted
+//example5-5 in page 131
+clc;
+//data given
+Vr=1.25;//peak voltage of ramp in volts
+tr=125e-3;//time period of the ramp=1.25 ms
+T=1/(1e+6);// frequency =1 Mhz and time period of the clock pulses is 1/f
+for Vi=0.75:(0.9-0.75):0.9,// analog input voltages for which clock pulses has to b found
+ t1=(tr/Vr)*Vi;//time period of the comparator high out put
+ N=t1/T;// pulses counted
+ printf("number of pulses counted for Vi=%.2f V are %d\n",Vi,N/100);
+end
+//result
+//number of pulses counted for Vi=0.750000 V are 750
+//number of pulses counted for Vi=0.900000 V are 900 \ No newline at end of file
diff --git a/182/CH5/EX5.6/5_6.txt b/182/CH5/EX5.6/5_6.txt
new file mode 100755
index 000000000..abc2f6952
--- /dev/null
+++ b/182/CH5/EX5.6/5_6.txt
@@ -0,0 +1 @@
+6.for less 1percent error,use n=7 \ No newline at end of file
diff --git a/182/CH5/EX5.6/exampke5_6.sce b/182/CH5/EX5.6/exampke5_6.sce
new file mode 100755
index 000000000..f0231db72
--- /dev/null
+++ b/182/CH5/EX5.6/exampke5_6.sce
@@ -0,0 +1,17 @@
+
+// example 5-6 in page 133
+clc;
+//Given data
+//error should be less thsn 1%
+// for less than 1% error count>=100
+n=6;
+N=0;
+while(N<100)
+N=(2^n)-1;//count value
+if(N<100)
+n=n+1;//increment n and check weather N has exceeded 100
+end
+end
+printf("for less 1percent error,use n=%d\n",n);
+//end
+// for less 1percent error,use n=7 \ No newline at end of file
diff --git a/182/CH5/EX5.7/5_7.txt b/182/CH5/EX5.7/5_7.txt
new file mode 100755
index 000000000..21ac8386d
--- /dev/null
+++ b/182/CH5/EX5.7/5_7.txt
@@ -0,0 +1,2 @@
+
+7.out put voltage=6.250000 V \ No newline at end of file
diff --git a/182/CH5/EX5.7/example5_7.sce b/182/CH5/EX5.7/example5_7.sce
new file mode 100755
index 000000000..fed413092
--- /dev/null
+++ b/182/CH5/EX5.7/example5_7.sce
@@ -0,0 +1,11 @@
+
+// example 5-7 in page 135
+clc;
+//Given data
+D=8; C=0; B=2; A=0; //corresponding analog input voltages for the digital input 1-0-1-0
+Vi=10;//input voltage in volts
+//calculation
+Vo=(D+C+B+A)*Vi/16;// output voltage
+printf("out put voltage=%.2f V",Vo);
+//result
+//out put voltage=6.25 V \ No newline at end of file