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

---
 608/CH18/EX18.01/18_01.sce | 12 ++++++++++++
 608/CH18/EX18.02/18_02.sce | 11 +++++++++++
 608/CH18/EX18.03/18_03.sce | 13 +++++++++++++
 608/CH18/EX18.04/18_04.sce | 14 ++++++++++++++
 608/CH18/EX18.05/18_05.sce | 15 +++++++++++++++
 608/CH18/EX18.06/18_06.sce | 14 ++++++++++++++
 608/CH18/EX18.07/18_07.sce | 14 ++++++++++++++
 608/CH18/EX18.08/18_08.sce | 16 ++++++++++++++++
 608/CH18/EX18.10/18_10.sce | 13 +++++++++++++
 608/CH18/EX18.11/18_11.sce | 31 +++++++++++++++++++++++++++++++
 10 files changed, 153 insertions(+)
 create mode 100755 608/CH18/EX18.01/18_01.sce
 create mode 100755 608/CH18/EX18.02/18_02.sce
 create mode 100755 608/CH18/EX18.03/18_03.sce
 create mode 100755 608/CH18/EX18.04/18_04.sce
 create mode 100755 608/CH18/EX18.05/18_05.sce
 create mode 100755 608/CH18/EX18.06/18_06.sce
 create mode 100755 608/CH18/EX18.07/18_07.sce
 create mode 100755 608/CH18/EX18.08/18_08.sce
 create mode 100755 608/CH18/EX18.10/18_10.sce
 create mode 100755 608/CH18/EX18.11/18_11.sce

(limited to '608/CH18')

diff --git a/608/CH18/EX18.01/18_01.sce b/608/CH18/EX18.01/18_01.sce
new file mode 100755
index 000000000..40e5ece49
--- /dev/null
+++ b/608/CH18/EX18.01/18_01.sce
@@ -0,0 +1,12 @@
+//Problem 18.01: A differential amplifier has an open-loop voltage gain of 120. The input signals are 2.45 V and 2.35 V. Calculate the output voltage of the amplifier
+
+//initializing the variables:
+Vi2 = 2.45; // in Volts
+Vi1 = 2.35; // in Volts
+A0 = 120; // open-loop voltage gain
+
+//calculation:
+Vo = A0*(Vi2 - Vi1)
+
+printf("\n\n Result \n\n")
+printf("\n the output voltage is %.0f V",Vo)
\ No newline at end of file
diff --git a/608/CH18/EX18.02/18_02.sce b/608/CH18/EX18.02/18_02.sce
new file mode 100755
index 000000000..8064ecc4b
--- /dev/null
+++ b/608/CH18/EX18.02/18_02.sce
@@ -0,0 +1,11 @@
+//Problem 18.02: Determine the common-mode gain of an op amp that has a differential voltage gain of 150E3 and a CMRR of 90 dB.
+
+//initializing the variables:
+Vg = 150E3; // differential voltage gain 
+CMRR = 90; // in dB
+
+//calculation:
+CMG = Vg/(10^(CMRR/20))
+
+printf("\n\n Result \n\n")
+printf("\n common-mode gain is %.2f",CMG)
\ No newline at end of file
diff --git a/608/CH18/EX18.03/18_03.sce b/608/CH18/EX18.03/18_03.sce
new file mode 100755
index 000000000..9aadaa6a5
--- /dev/null
+++ b/608/CH18/EX18.03/18_03.sce
@@ -0,0 +1,13 @@
+//Problem 18.03: A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and the CMRR.
+
+//initializing the variables:
+Vg = 120; // differential voltage gain 
+Vi = 3; // in Volts
+Vo = 0.024; // in Volts
+
+//calculation:
+CMG = Vo/Vi
+CMRR = 20*log10(Vg/CMG)
+
+printf("\n\n Result \n\n")
+printf("\n common-mode gain is %.3f and CMRR is %.2f dB",CMG, CMRR)
\ No newline at end of file
diff --git a/608/CH18/EX18.04/18_04.sce b/608/CH18/EX18.04/18_04.sce
new file mode 100755
index 000000000..d0adb3b57
--- /dev/null
+++ b/608/CH18/EX18.04/18_04.sce
@@ -0,0 +1,14 @@
+//Problem 18.04: In the inverting amplifier of Figure 18.5, Ri = 1 kohm and Rf = 2 kohm. Determine the output voltage when the input voltage is: (a)+0.4 V (b) -1.2 V
+
+//initializing the variables:
+Rf = 2000; // in ohms
+Ri = 1000; // in ohms
+Vi1 = 0.4; // in Volts
+Vi2 = -1.2; // in Volts
+
+//calculation:
+Vo1 = -1*Rf*Vi1/Ri
+Vo2 = -1*Rf*Vi2/Ri
+
+printf("\n\n Result \n\n")
+printf("\n output voltage when the input voltage is 0.4V is %.1f V and when the input voltage is -1.2V is %.1f V",Vo1, Vo2)
\ No newline at end of file
diff --git a/608/CH18/EX18.05/18_05.sce b/608/CH18/EX18.05/18_05.sce
new file mode 100755
index 000000000..1df26bb38
--- /dev/null
+++ b/608/CH18/EX18.05/18_05.sce
@@ -0,0 +1,15 @@
+//Problem 18.05: The op amp shown in Figure 18.6 has an input bias current of 100 nA at 20°C. Calculate (a) the voltage gain, and (b) the output offset voltage due to the input bias current. (c) How can the effect of input bias current be minimised?
+
+//initializing the variables:
+Ii = 100E-9; // in Amperes
+T = 20; // in °C
+Rf = 1E6; // in ohms
+Ri = 10000; // in ohms
+
+//calculation:
+A = -1*Rf/Ri
+Vos = Ii*Ri*Rf/(Ri+Rf)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the voltage gain is %.0f",A)
+printf("\n (b)output offset voltage is %.5f V",Vos)
\ No newline at end of file
diff --git a/608/CH18/EX18.06/18_06.sce b/608/CH18/EX18.06/18_06.sce
new file mode 100755
index 000000000..c09c5c0b0
--- /dev/null
+++ b/608/CH18/EX18.06/18_06.sce
@@ -0,0 +1,14 @@
+//Problem 18.06: Design an inverting amplifier to have a voltage gain of 40 dB, a closed-loop bandwidth of 5 kHz and an input resistance of 10 kohm.
+
+//initializing the variables:
+Vg = 40; // in dB
+bf = 5000; // in Hz
+Ri = 10000; // in ohms
+
+//calculation:
+A = 10^(Vg/20)
+Rf = A*Ri
+f = A*bf
+
+printf("\n\n Result \n\n")
+printf("\n the voltage gain is %.0f, Rf = %.0f ohm and frequency = %.0f Hz",A, Rf, f)
\ No newline at end of file
diff --git a/608/CH18/EX18.07/18_07.sce b/608/CH18/EX18.07/18_07.sce
new file mode 100755
index 000000000..0fe86458a
--- /dev/null
+++ b/608/CH18/EX18.07/18_07.sce
@@ -0,0 +1,14 @@
+//Problem 18.07: For the op amp shown in Figure 18.8, R1 = 4.7 kohm and R2 = 10 kohm. If the input voltage is- 0.4 V, determine (a) the voltage gain (b) the output voltage
+
+//initializing the variables:
+Vi = -0.4; // in Volts
+R1 = 4700; // in ohms
+R2 = 10000; // in ohms
+
+//calculation:
+A = 1 + (R2/R1)
+Vo = A*Vi
+
+printf("\n\n Result \n\n")
+printf("\n(a) the voltage gain is %.2f",A)
+printf("\n(b) output voltageis %.2f V",Vo)
\ No newline at end of file
diff --git a/608/CH18/EX18.08/18_08.sce b/608/CH18/EX18.08/18_08.sce
new file mode 100755
index 000000000..526809b08
--- /dev/null
+++ b/608/CH18/EX18.08/18_08.sce
@@ -0,0 +1,16 @@
+//Problem 18.08: For the summing op amp shown in Figure 18.11, determine the output voltage, Vo
+
+//initializing the variables:
+V1 = 0.5; // in Volts
+V2 = 0.8; // in Volts
+V3 = 1.2; // in Volts
+R1 = 10000; // in ohms
+R2 = 20000; // in ohms
+R3 = 30000; // in ohms
+Rf = 50000; // in ohms
+
+//calculation:
+Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)
+
+printf("\n\n Result \n\n")
+printf("\n output voltageis %.1f V",Vo)
\ No newline at end of file
diff --git a/608/CH18/EX18.10/18_10.sce b/608/CH18/EX18.10/18_10.sce
new file mode 100755
index 000000000..709985d98
--- /dev/null
+++ b/608/CH18/EX18.10/18_10.sce
@@ -0,0 +1,13 @@
+//Problem 18.10: A steady voltage of -0.75V is applied to an op amp integrator having component values of R = 200 kohm and C = 2.5 μF. Assuming that the initial capacitor charge is zero, determine the value of the output voltage 100 ms after application of the input.
+
+//initializing the variables:
+Vs = -0.75; // in Volts
+R = 200000; // in ohms
+C = 2.5E-6; // in Farads
+t = 0.1; // in secs
+
+//calculation:
+Vo = (-1/(C*R))*integrate('-0.75', 't', 0, 0.1)
+
+printf("\n\n Result \n\n")
+printf("\n output voltage is %.2f V",Vo)
\ No newline at end of file
diff --git a/608/CH18/EX18.11/18_11.sce b/608/CH18/EX18.11/18_11.sce
new file mode 100755
index 000000000..e07676410
--- /dev/null
+++ b/608/CH18/EX18.11/18_11.sce
@@ -0,0 +1,31 @@
+//Problem 18.11: In the differential amplifier shown in Figure 18.16, R1 = 10 kohm, R2 = 10 kohm, R3 = 100 kohm and Rf = 100 kohm. Determine the output voltage Vo if: 
+//(a) V1 = 5 mV and V2 = 0 
+//(b) V1 = 0 and V2 = 5mV 
+//(c) V1 = 50 mV and V2 = 25mV 
+//(d) V1 = 25 mV and V2 = 50mV
+
+//initializing the variables:
+V1a = 0.005; // in Volts
+V2a = 0; // in Volts
+V1b = 0; // in Volts
+V2b = 0.005; // in Volts
+V1c = 0.05; // in Volts
+V2c = 0.025; // in Volts
+V1d = 0.025; // in Volts
+V2d = 0.05; // in Volts
+R1 = 10000; // in ohms
+R2 = 10000; // in ohms
+R3 = 100000; // in ohms
+Rf = 100000; // in ohms
+
+//calculation:
+Vo1 = -1*Rf*V1a/R1
+Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b
+Vo3 = -1*Rf*(V1c-V2c)/R1
+Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)
+
+printf("\n\n Result \n\n")
+printf("\n (a)output voltage is %.3f V",Vo1)
+printf("\n (b)output voltage is %.3f V",Vo2)
+printf("\n (c)output voltage is %.3f V",Vo3)
+printf("\n (d)output voltage is %.3f V",Vo4)
\ No newline at end of file
-- 
cgit