From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- sample_notebooks/DeepTrambadia/sc201_1.ipynb | 85 ++++++++++++++++++++++++++++ 1 file changed, 85 insertions(+) create mode 100644 sample_notebooks/DeepTrambadia/sc201_1.ipynb (limited to 'sample_notebooks/DeepTrambadia/sc201_1.ipynb') diff --git a/sample_notebooks/DeepTrambadia/sc201_1.ipynb b/sample_notebooks/DeepTrambadia/sc201_1.ipynb new file mode 100644 index 00000000..aa4f487b --- /dev/null +++ b/sample_notebooks/DeepTrambadia/sc201_1.ipynb @@ -0,0 +1,85 @@ + + + + +

IPython QtConsole 3.2.0

Python 2.7.10 |Anaconda 2.3.0 (64-bit)| (default, May 28 2015, 16:44:52) [MSC v.1500 64 bit (AMD64)]

Type "copyright", "credits" or "license" for more information.

IPython 3.2.0 -- An enhanced Interactive Python.

Anaconda is brought to you by Continuum Analytics.

Please check out: http://continuum.io/thanks and https://anaconda.org

? -> Introduction and overview of IPython's features.

%quickref -> Quick reference.

help -> Python's own help system.

object? -> Details about 'object', use 'object??' for extra details.

%guiref -> A brief reference about the graphical user interface.

In [1]: import math

...:

...: #(a)

...: #initialisation of variables

...:

...: E=10 #E in V

...: R=1 #R in Kohm

...:

...: #Calculations

...:

...: Id=E/R #Eq.(2.2)

...: Vd=E

...: print "The current Ic is= %fmA "%(Id),";Vd=0V"

...: print "The diode voltage is= %fV"%(Vd),";Id=0A"

...: print "The resulting load line appears in Fig. 2.4. The intersection between the load line and the characteristic curve defines the Q-point as"

...: print "The level of VD is certainly an estimate, and the accuracy of ID is limited by the chosenscale. A higher degree of accuracy would require a plot that would be much large and perhaps unwieldy"

...:

...: #(B)

...: print "(B)"

...: Ir=9.25 #Ir in mA

...: Vdq=0.78 #Vdq in v

...: Vr=Ir*R

...: print "Vr = Ir*R = Idq*R %d="%(Vr),"or"

...: Vr = E-Vdq

...: print "Vr = E-Vdq = %f" %(Vr)

...: print "The difference in results is due to the accuracy with which the graph can be read. Ideally,the results obtained either way should be the same."

...:

...: #Graph solution to example 2.1

...:

...: import numpy as np

...: import matplotlib.pyplot as plt

...:

...: Vd = np.linspace(0.0,10.0)

...: Id = np.linspace(0.0,10.0)

...: Id= -Vd + 10

...: plt.plot(Vd, Id)

...: Vd = [0,0,0.1,0.1,0.2,0.2,0.3,0.3,0.3,0.3,0.4,0.5,0.6,0.7]

...: Id = [0,0,0,0,0,0,0,0,0.1,0.1,0.3,0.7,2.0,10.0]

...:

...: plt.plot(Vd, Id,'yo-')

...:

...: plt.xlabel('Voltage (v)')

...: plt.ylabel('current (mA)')

...: plt.title('About as simple as it gets, folks')

...: plt.grid(True)

...: plt.savefig("test.png")

...:

...: plt.show()

...:

...: print "example 2.2:"

...: print "repeat the example 2.1 for R =2"

...:

The current Ic is= 10.000000mA ;Vd=0V

The diode voltage is= 10.000000V ;Id=0A

The resulting load line appears in Fig. 2.4. The intersection between the load line and the characteristic curve defines the Q-point as

The level of VD is certainly an estimate, and the accuracy of ID is limited by the chosenscale. A higher degree of accuracy would require a plot that would be much large and perhaps unwieldy

(B)

Vr = Ir*R = Idq*R 9= or

Vr = E-Vdq = 9.220000

The difference in results is due to the accuracy with which the graph can be read. Ideally,the results obtained either way should be the same.

example 2.2:

repeat the example 2.1 for R =2

In [2]:

\ No newline at end of file -- cgit