BB@ 6:35pm 04-08-2012
Sir what are the main differences between Windows XP & LINUX?
Replied on: 4:22pm 04-14-2012

I presume you are Biroja Bandhu of 3rd year in our college. If so, I had already discussed it in the class.

Nevertheless, remember that Windows (any version) is proprietary software belonging to Microsoft Corporation and as such the user, apart from being having to buy it, has no access to its source code.

On the other hand, Linux is open-source software that comes with a GNU public license. One may use it for free and also have full read and write access to its source code. One may even redistribute it (for free or otherwise) after modifying the source code, but has to make the modified source code available to the recipient as well.

There are many other differences in features which you may find using google.

Birojananda Bondhu@ 3:27am 03-17-2012
Sir, what would be the c-program for
Newton-Rapshon method
Eular Method
Simpson method
Replied on: 3:23pm 03-19-2012

program to implement numerical integration through Simpson's method
has already been taught to you in the last class:

Read in the limiting x values: x[1] and x[N] from the keyboard (N should be odd)
Calculate the step size
Create the complete x grid:
calculate the function to be integrated fx[1] to fx[N] at all the x points
f[i]=pow(x[i],3.0) + pow(x[i],2.0) + 1.0; // assuming the function to be x^3+x^2+1



program to implement Newton-Raphson's method to solve an algebric equation in x (say x^3-2=0)

Start with a trial solution for x (say xi)

for (i=1;i<=10000;i++)
fx=pow(xi,3.0) -2.0;


program to implement Euler's method to solve a 1st order differential equation
(say dydx=f(x,y)=x+y given y0(x0)=y(at x0)=0 for x0=0)

// use the relation: y1=y0+h*dydx0 (where y1=y(at x1) and y0=y(at x0) dydx0=dydx at x0
// h = spacing between successive x values, the samller resulting in more accurate solutions

Start with given values of x0,y0
Calculate dydx0
Calculate y1=y0+h*dydx0
Calculate dydx1=f(x1,y1) // x1=x0+h
Calculate y2=y1+h*dydx1
y1,y2,y3.... give us the required solutions at x1,x2,x3......

Soubheek Nath@ 11:32pm 02-21-2012
Sir, plz give a feedback about this qus. I've little bit a problem to answer it..

Year 2010 , sixth paper:

Discuss the varius operating modes of 8255. (8 marks)
Replied on: 12:14pm 02-23-2012

I have just posted a relevant note on my website:
www dot abhipod dot bravehost dot com
You can find the link on the "Home" page
or you may use the "Supplementary Notes" link in the "Teaching" page.

I will be discussing actual interfacing examples with the 8255A in the class and show you how one may
program the 8085 microprocessor accordingly.

Soubheek Nath@ 11:15pm 02-21-2012
How do i change the background colour of Pspice plot screen from black to white??
Replied on: 12:37pm 02-23-2012

As you have told me in the class, you have already found the answer..
to change the background colour to brightwhite in the pscipe.ini configuration file.

I may add a few words.

In the demo version, the configuration file could be pspiceev.ini instead of pspice.ini.

Moreover, you would have to locate the file in the windows directory (which is in C: drive by default) and not in C:\Program Files\...

BTW, even if the background colour is black under PROBE DISPLAY COLORS settings, it should create no problems while printing as the default setting for background colour is brightwhite under PROBE PRINTER COLOURS setting.

You may change them if you want using a plain text editor like notepad.

APoddar@ 3:28pm 12-04-2011
Contn. of reply to Krishna moorthy's Question..
Pl: see below: (I had to split the reply because of the constraint of packing in too much text in the reply form in
Replied on: 3:46pm 12-04-2011

In a Bipolar Junction Transistor (BJT), say an n-p-n transistor (emitter is of n-type, base is of p-type and collector is of n-type), biased in the active mode, the emitter-base junction (EBJn) is forward biased and the collector-base junction (CBJn) is reverse biased. Considering the fact that the concentration of majority carrier electrons in the n-type emitter is made much larger than that of the majority carrier holes in the p-type base by appropriate doping and making the base relatively much thinner, the forward bias of the EBJn ensures that the most of the emitter current IE is due to the majority carrier electrons diffusing across the EBJn and injecting into the base, where they become the minority carriers. Very few of these injected minority electrons recombine with holes in the base, giving rise to a small base current, while the remaining are diffused and swept across the reverse biased CBJn giving rise to a collector current IC which is almost equal to IE.
A negligible drift current does flow in the reverse direction across the CBJn since the reverse bias of the junction aids the drift of the thermally generated minority carriers across CBJn (holes in the n side and electrons in the p side).

Thus we see that the current in the above BJT is primarily due to electrons.

Using a similar reasoning we could say that in a p-n-p transistor, the current flow is primarily due to holes.

If you are a student of electronics or physics, consult any good text-book on basic electronics to have a more complete understanding.

krishna moorthy @ 1:59pm 12-01-2011
many books will say that current in bjt is both by electrones and holes.but what i understood is hole is only a vacant space and when electrone moves rthrough holes we feel that holes move in opposite derection.but current flow can cuse by only electronics. how the holes can cause flow of electronics?
Replied on: 3:43pm 12-04-2011

The motion of the holes in a semiconductor can be understood by picturising the motion of electrons in the valence band as they hop from one atom to another leaving behind vacant sites which in turn are being filled with other valence electrons themselves. The resulting current which is in an opposite direction to the motion of electrons (note a hole has a charge of + one electronic charge), however, is said to be due to hole motion only (as per convention).

The motion of free electrons in the conduction band constitutes a separate current component due to electron motion only.

In a pure or intrinsic semiconductor, the number of electrons in the conduction band equals the number of holes in the valency band. As such the hole and electron current contributions are equal and both contributions must be taken into account.

In an impure or extrinsic semiconductor like an n-type semiconductor, the no: of free electrons in the conduction band is much higher than that of holes in the valence band (free electrons are the majority carriers and holes are the minority carriers), as a result of which we may consider the current to flow due to these free electrons only. In an impure p-type semiconductor, on the other hand, the no: of holes in the valence band is much higher than that of free electrons in the conduction band (holes are called the majority carriers and free electrons the minority carriers), as a result of which we may consider the current to flow due to these holes only.

umada@ 11:40pm 11-15-2011
can you teach me physics? at least basics of physics.
Replied on: 1:07am 11-20-2011

Hey Umada,

Thanks for visiting my site. Well, I am not sure what your question exactly meant. If you have any particular question on a topic in physics, do post it over here, I sure will try to give an answer.

The following sites might be of help. Do take a look!

The Physics Classroom:


Goutam Das@ 12:41am 09-14-2011
Sir, plesae refer me some good books for semiconductor physics.
Replied on: 12:59pm 09-18-2011

Books on Semiconductor Physics of Streetman, Sze, Neamen should be helpful..
Fundamentals have been covered nicely in Millman Halkius's Electronic Devices & circuits book as well.

Goutam Das@ 12:05am 09-14-2011
Sir, I have found an application of BJT in cc configuration that it is used in impedance matching purpose. But I couldnot get any information about impedance matching.
Last day we covered 'hybrid parameter model' portion of BJT.
Tomorrow class at 5.40: RC coupled circuit, Darlington configuration, frequency response of BJT, Cut off frequencies
Replied on: 12:46pm 09-18-2011

Impedance matching refers to the transfer of maximum power from the source to a load..Look up the Max: Power Transfer Theorem in any book on Electrical Circuits.

The CC amplifier or emitter follower, having unity voltage gain, a high current gain of beta, a high input impedance and a low output impedance, can be used to match the output of an amplifier having large output impedance with a low impedance load like a speaker, requiring large load currents.

The high input impedance of the CC amplifier also prevents the amplifier, whose output it is connected to, from being loaded.

The Darlington pair configuration can also be used as an emitter-follower to match amplifiers and loads with higher degree of mismatch..

Don't worry..We will cover these topics in college shortly.

Goutam Das@ 1:15am 09-12-2011
Thank you sir. I have collected this book.
Goutam Das@ 12:27am 09-06-2011
a question was given in GATE 2001 from BJT, The current gain of a BJT is i) gro ii) gmr iii) r ?
Replied on: 11:49am 09-09-2011

Hello Gautam,
I am not too sure about the answer! g and r usually represent conductance and resistance respectively which obviously are not dimensionless as is the current gain..Did you go through the whole question?
I presume that the list has something to do with the g-parameters in a g-parameter representation of a 2-port BJT network.
I1(V1,I2) = g11*V1 + g12*I2
V2(V1,I2) = g21*V1+g22*I2
g12 = I1/I2 with V1=0 (g12 represents the reverse current transfer ratio with input port short circuited)
Similarly g11 is the input conductance with output open-circuited, g21 is the voltage gain with output open-circuited and g22 is the output resistance with input short circuited.
The nomenclature for the parameters could be different.
Hope this helps.
You may refer to Sedra/Smith's Microelectronic Circuits Book Appendix B.

By the way, do not get too bogged down with GATE questions, rather concentrate on your B.Sc. course for the time being..You would be able to appear in GATE only once you have completed (or about to complete) your M.Sc.

Goutam Das@ 11:22am 08-29-2011
Sir, my first year exam is over, could I know about our next session classes for second year?
Replied on: 7:34pm 09-01-2011

I had already told you in the class day before yesterday..
Till all of you take provisional admission to the college, you are welcome to attend my classes informally from 5pm onwards.
I plan to take the IInd year classes on Monday, Tuesday and Saturday and the IIIrd year classes on Tuesday, Wednesday and Friday.
IInd year practical classes will be taken by P.Pal,on Wednesday.
I am working on the routine which will start being effective once you take provisional admission. Do contact the college office to know the dates for the admission.
Pass on the information to those who have just appeared in the Part II exams as well.

hello@ 11:00pm 08-24-2011
Could you post a few links to websites of electronics or IT companies who would be interested in recruiting M.Sc. Electronics passouts?
Replied on: 7:27pm 09-01-2011

Vivek Singh@ 12:26am 08-06-2011
What are the prospects after studying electronics honours?
Replied on: 12:32am 08-06-2011

You should go for a post graduate degree in radio physics or electronics or information technology. This will help you get a good job in the IT field. You may also pursue research and do a PhD. This will help you build a good career in the academic field.

Visit my career guidance link in my Teaching page for related information.

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