 Okay good morning to everyone and so you have heard about our virtual laboratory efforts and we hope that you all of you will actually use that very fruitfully and pass it on to your students and the idea of course is that eventually the students must benefit and in that sense we are all partners and I hope you can take it forward. So my session is mostly about how to use simulation in teaching in particular teaching electronics and so we will be using a simulation platform called SQL which has been developed at IIT Bombay and so but if you are already used to some other simulation tool that is fine it is not so much the which simulator you use it is how you use it so that is really the idea. How many of you are teaching electronics courses all of you? How many of you have taught the topic of let us let us take a few topics and let us begin with the topic of resonance how many of you have taught resonance RLC resonance everybody has taught RLC resonance okay. So this is we will just use it as an example as to see how simulation can be used effectively to improve the understanding the students have of few topics. So let us take a very simple RLC circuit are you all you are using simulation tools like Peace Pies or Matlab and so on alright. So you are all familiar with the simulation packages and simulation is often used because things get too complicated for our mind but it can also be used extremely effectively for teaching purposes. So let us start with a simple RLC circuit okay. So let us just take this oldest source as 1 angle 0 alright. So let us say you have taught this topic you have taught the students about resonance resonance frequency quality bandwidth etc and now we are trying to reinforce their ideas with the help of some simulation tools okay ideally you could do this whole thing in the lab but it does not happen because it takes too much effort so it is very good to do that with simulation tools alright. So what how would you what is the first thing you will demonstrate with the simulation tool see that. So okay let us have some values for this L and C 1 million D 1 micropirate right okay. So what is the first thing you would like to demonstrate with this circuit frequency response and you would like to show them that the frequency the response peaks at the resonance frequency given by 1 over 2 pi square root L C can somebody calculate this for me anybody has a calculator these days your mobiles also have calculators can somebody calculate this number I am trying to do this in a sort of live mode because then you can do the same thing. So I will treat you for this for this purpose as a class of students and so when you teach the students will be doing this calculations for you. So can you somebody give me this number anybody has a mobile or a calculator okay approximately so 1 million D 1 micropirate 1 over 2 pi 10 is to square root of 10 is to 50 hertz it sounds too low to be there is a difference there is a problem with the order of magnitude the number is correct alright let us see what is square root of 10 is to 9 5 kilowatts that is right it should be about 5 kilowatts okay square root 10 is about 3 right 3 and 2 pi is about 6 so this turns out to be about 5 kilowatts alright okay so let us let us do this and actually show that the frequency is 5 kilowatts in the evening session you will I hope most of you will come for this session all this in the evening in that you will actually learn to use this circuit simulation package and you can do this example yourself. So let us quickly make up this circuit I will tell you in more detail how to use the similar right now I am just sort of doing it for you so essentially you need to pick up components place them on your canvas okay we have got R L and C in place what do we know what else do we need we need a voltage source okay and we need something called ground okay ground actually is required for any circuit simulator to work because that gives us the reference voltages that defines the reference that defines the voltages with respect to a particular node so let us say we have a ground as well then just make the connections simple alright if you can do the same if you are using piece by already you can do the same thing with piece by the simulation tool is really not an issue the issue is what you do with it alright okay what is the next step values okay so let me make this amplitude as 1 and frequency we are going to vary say it does not matter let me make this okay let me make it as 1 ohm right now and later we see if you can change that L 1 mic a milli handy C 1 micro farad alright so now our circuit is ready in this particular simulator you need to inform the simulator as to what you want to record okay so that is done by let me also name these components I named these as R L C and D S now let us include some output variables and what is the output variable you would like to look at is basically the response in this case is the current right so and the current is the same through all of them so we can look at the current through any of these things let me choose the register and there is an output AC variable called I 1 AC okay of the register so let me put that here so that is my output variable alright so now the description of this circuit is complete and now I can go back go to the so-called solve section now we have defined the circuit now we need to tell the simulator what it is that you want to do you want to do a transient analysis AC simulation DC simulation and so on that is done by define defining a solve block so in this case we were interested in AC simulation frequency domain response so let's make that AC simulation alright in AC simulation one thing we are going to do is to vary the frequency right we are expecting a frequency cutoff frequency of 5 kHz so we should vary the frequency from let's say 10 Hz to 100 kHz so that we can see a peak so that is done by a statement called vary frequency if you have any questions please stop me at any time okay now you see that there is something called vary type and that says linear whenever you change frequency it should always be changed in logarithmic question otherwise it gets compressed otherwise all the points will get concentrated towards the high frequency okay so let me put and let me make this a number of the number of points fairly large 500 alright now we are going to change the frequency from 10 Hz to 100 kHz logarithmically with 500 points so that defines this particular what next we need to define an output file and that is done by the output block and we need to select some output variables now since our input our output variable is AC type of variable we can select phase or magnitude as output to be written to a file okay as you make the changes on the right hand side things will change also in the sentiment okay so with that our description is now complete and we can now run this program one more thing we can do is we can display these properties on the canvas just take a minute okay so that we know what is what so we have an R of 1 ohm we have an L of 1 milli and we have a C of 1 micro can you see all of can you all see can you all see okay alright better now okay or and these things are dynamic so suppose I change a property of this inductor that will reflect in the label okay so now we are ready to run the program if you run the program the if the program is successful you will see a message here at the bottom called program completed and the menu will automatically shift to this graphs the middle one okay because that is what the user will typically like to do so then select the output file select an x axis the middle one is the x axis the right one is the y axis y axis we will choose the magnitude of the current okay and then let's graph now this is not fitting in in your this is a laptop so the resolution is not is a slight problem this window should really be fitting into your into your center window but can you see the reasons right this is of course not the best way of doing it because whenever there is a frequency involved you should always use the lock scale so let's do that now there is a frequency response is it 5 kilo Hertz it is 5 kilo this is 1000 Hertz 10,000 Hertz and so let's expand that part so that you can see better all right now you can see better so this is about 5 kilo Hertz okay so this if you are teaching students and if you show such an example the students immediately get in their head that there is such a thing called peak and resonance and so on and soon after you cover the theory they immediately understand try to connect to what you're doing in class what else would you like to show them what happens if I change the value of R for example if I make R if I make R go from one right now it is one ohm suppose I make it two ohms what will happen to that what will happen to the curve we all know that the resonance frequency itself will not change one or else see that is independent of but the bandwidths or the sharpness or the quality will change so bandwidth is given by for this circuit it is given by R over L so if you increase the resistor the bandwidth will increase by a factor of 2 and the quality is given by the resonance frequency by the bandwidths so in effect the bandwidth will increase then the quality will decrease so in other words the sharpness will decrease in other words the circuit will become less selective so let's see if that happens I'll go back to the circuit editor okay so I have changed this R to 2 and now let me run the program again and let's replot okay so actually the data is there but it has got sort of over most of it is overlapping here so you can't really see so let's expand this part see what we get okay now it's clear that now can you see the quality is the sharpness has decreased all right so that is of course you do see a problem here that these problems the points actually not enough so we need to increase the number of points see right now you have a data point here and your data point here but you would really like to have some more points in between okay so let's increase that number and repeat this simulation okay now it's looking better okay so it's so this these kinds of things are just no they will depend on the problem how many number of points is enough and so on and of course you can get a visual sort of feedback from the plot that you see all right so that is just one part of resonance circuits okay and this is what we mostly teach in our classes right a lot of hidden things which we teach but they don't quite come out what else can you see in this circuit which will improve the understanding of the student can you check on the bandwidth we have the data we can see what the bandwidth is right so let's do that so what is the expected bandwidth and what is the bandwidth we are actually getting so with two ohms our bandwidth is r over l and in terms of hertz I think we will have to how much 2k this is two ohms l is 1 milli henry 2k will get divided by 2pi right so 2k by about about 300 hertz okay so 300 hertz means if you are plotting the frequency response like this okay this whole thing is 300 hertz right so we should have 5 kilo hertz minus about 150 hertz 5k minus 150 hertz and 5k plus 150 hertz so that is those are our 3db points so that is where the response should turn out to be 1 hour square root 2 times I max 1 hour square root 2 is about 0.7 and I max in this case is what is I max I max is 1 divided by 2 2 ohms right so 1 this is 1 angle 0 volts so this will turn out to be about 0.7 by 2 0.35 okay so when the current becomes 0.35 ampere the frequency should be about 5 minus 5k minus 1 so 5 minus 0.3.85 4.85 kilo hertz and 5.15 kilo hertz okay so let us let's check this let's check whether this is actually happening and all these things are you know things that you can do in class live absolutely no problem and that gives us gives us a lot of confidence so I have expanded this frequency so what is the peak that we expect 0.35 ampere sorry not 0.35 0.5 ampere right so you can see this is 0.1, 0.2, 0.3, 0.4, 0.5 so it is indeed 0.5 that's one thing you can definitely point out and it is at 5 kilo hertz as we expect the frequency now our interest is the 1 over square root 2 point the 3db point and that is at 0.35 right so 0.35 so these are the points that we are interested in okay so we can expand this further and see what we get so 4.8, 4.9, 4.85 okay so that's roughly what we expected it to be and this on this side 0.35 so about 5.2 okay alright so of course we made a very rough calculation if you do it precisely then you will get better but certainly the order is correct okay what else would you like to see show your students yeah Xc and Xn that's right so first Xc and Xm we cannot actually show the impedance itself but we can show in a simulation setup we can show the phase of the current so what will the phase look like angle of y so we know that if omega is small 1 over j omega c dominates is large and if omega is large then j omega c j omega l dominates right so then the current in this case the first case the current is the phasor will go as 1 over 1 angle 0 by 1 over 1 by j omega c so that will have a phase of 90 degrees j omega c so this has a phase of 90 degrees and the other extreme in case 2 you will have 1 over j omega l and that is a phase of minus 90 degrees and the phase of course at resonance is exactly 0 right so that is something that we can show so let's look at the phase of the current okay so again let's make the x axis lock axis so that you can see it better alright so that is very clear low frequencies the phase is 90 degrees at resonance which is somewhere here the phase should be exactly 0 and at high frequencies the phase is minus 90 minus 50 minus 100 so minus 90 okay so that is something that you can show and of course as you change the resistance value this will become sharper less sharper alright what else would you like to show the students transient analysis yeah that is something you can show we will do that later what else but in the frequency domain itself there is so much more information that you can show that is sharpness that we already see okay what about these voltages vr vl this also actually has a lot of information okay so we know that as so our current is 1 angle 0 by r okay so what is going to happen to vr let's look at the phasor diagram so this is your imaginary axis real axis this is your source voltage right okay now at resonance of course the source voltage and this resistance vr is going to be the same and at any other point this is the vr is either going to lag or it is going to lead up to 90 degrees okay so now in this particular situation let's take this particular case this is your vr this is vs and in this case your vc is minus j by omega c okay vl will lead right vl will lead this and vc will lag right and the resultant of these two will essentially add up to add up with vr and then give you vs alright so this phasor diagram of course we cannot show with the simulation tool but we can look at how vr v is vc and vl change with frequency okay alright and at a given frequency you can the student will have access to the angles and magnitudes of all of these quantities so that he or she can sketch the phasor diagram okay so what about vc itself what will how will that change suppose I plot magnitude of vc as a function of frequency this is my frequency so how will this change as a frequency becomes very small magnitude is going to become 0 as a frequency becomes very small the magnitude is going to become vs right as the magnitude all of this if the if omega is small this 1 over j omega c is going to be very large and all of this will essentially just drop across the alright so we already have two data points at very low frequencies we have vc equal to vs at very high frequencies we have vc equal to 0 what about f0 that is the part that many students have not understood in my experience okay so now let us deal with that part what is what is what happens at f0 greater than vs how much greater quality factor that is right okay now this is these are things that I find the students do not quite understand but if you show them a plot like the one that we are going to see they will immediately remember this all their life so vc at resonance is equal to vl at resonance and it is equal to q times vs this is very well known result very easy to derive nice result and q is in our case what is q q is omega 0 by bandwidth and in our case the omega 0 is 1 over 2 pi 1 over square root let us not worry about this 2 pi 1 over square root we will see divided by r over l okay so this is going to be 1 over r square root l by c that is the quality factor and let us calculate it in our case 1 over r r is 2 ohms l is 1 milli and c is 1 micro so 1 over 2 1 milli by 1 micro square root 10 times 10 right so this is about square root 10 is about 3 right 3 by 2 1.5 so about 15 right so actually if you look at this plot this is if this is vs this is going to be much larger than vs at f0 so if you plot it versus you will get something like this suddenly it shoots up at and then it becomes at some point it becomes 0 okay alright this is very very important point people do not realize that there are such huge voltages at resonance across capacitor and the inductor let us look at this so to do this we need to add some more variables to our circuit okay and that is done by using some more additional elements okay let me just name this node okay let me tell you what I am doing I have got these voltmeters okay each one of these is actually connected to this point but the connections are not being shown otherwise it will look very messy so they are being assigned the same names so that the connections are there internally but not shown explicitly in the circuit diagram okay so and also let me add the variables that those are those go with this so one of these voltmeters is actually connected here this is connected here this is connected here so now we need to add output variables which is a voltage of this voltage of that and voltage of that and that will give us vrvcl vl and vc oh no this actually this will not do it this one does not have AC part so let me what I will do I will just open an existing circuit so that it saves us some time this is the one I wanted so this circuit is ready made and it is already made up let me change the values suitably like we have okay this measures the resistor voltage that measures the inductance voltage and that measures the capacitor voltage and all these variables are already defined as the output variable you can see that here vrac is defined as the output voltage of this element and so on let us run this okay so now I am plotting vr okay let us plot actually all of this together see what we get or let us just plot the inductance first or what we are looking at capacitor right capacitor okay now it should be very clear okay so this is one at low frequencies equal to vs then at resonance which is somewhere here becomes 16 times we said 15 but that was approximate so it becomes about what we expected and then finally when the voltage becomes when the frequency becomes large it will go to so this is some this is a figure which is extremely informative and it is something that can stick in the students mind very well all right what happens if I change the resistance to 5 ohms okay what will happen to that figure so it was 2 ohms now I made it 5 ohms what will happen to that figure okay this point will obviously not change at very low frequencies this point will obviously not change at resonance now the quality is better or worse resistance has increased the quality has decreased and therefore you will not get quite as high as 15 or 16 we will get some so if you do the all these things in class there is a very good chance that the students will sort of remember it you know without studying the topic further it is there is an excellent chance of that happening so you can see that the sharpness has decreased and the magnitude has come down okay what about vl vl should do just the opposite vl if the frequency is very small then vl should be small 0 frequency is large then the entire voltage is going to come across the resistance is going to be 1 and in between at resonance it will go through a peak so it should do like this something like that so let us just take it out that is exactly what happens so omega very small j omega l is small the voltage is 0 finally it will become 1 at omega increases considerably and at resonance it will go through q times the vs okay all right what about phase okay suppose I want to plot the phase of vl for example angle of vl what will that do very high frequencies the angle should be what happens at very high frequency very low frequencies this is like a short circuit and the capacitance actually will dominate the capacitance impedance is minus j omega c it will be much larger than r so the phase of this current will be let us work it out I cannot do all of this in my head so I will be one angle 0 approximately minus j by omega c right so it will be minus it will be plus j omega c that is i and your vl is the current times the impedance so what is the phase is minus 180 or 180 depending on how you look at it okay so let us check this so in short we are expecting the phase to go from plus minus 180 to 0 at high frequencies okay okay so that is exactly what happens 180 here and finally it is 0 whether 180 is it 180 or minus 180 how did how should we choose it is very simple at resonance what is the phase of this j omega l at resonance what is the phase of i r the current 0 j omega l times i that is plus 90 degrees okay so at resonance which is at 5 k somewhere here the phase will be exactly 90 degrees all these things you can bring out in class if you are showing an example like this okay so this is not really about teaching resonance as such but how to use a simulation tool to improve the students understanding so let us take some other example not related to resonance but something else let us say diodes you would have all done examples like this in your courses okay so there is a simple diode circuit two diodes this is my input voltage and this is my output voltage let us take some numbers for these let us say this is two volts let us say this is 1 k 0.5 k okay and let us say that this diode voltage diodes have a drop of 0.7 volts when they conduct like we typically do all right so the student is asked what would you ask the student to do in this case we will ask you to ask him to do several things one plot plot v o versus v i that is always the step number one and then once we once the student has got this then we will we can ask him or her to plot no the output voltage if the input is sinusoidal or triangle then what happens and so on but the the key really is this v o versus v i all right so d1 conducts if v in is greater than 2.7 d2 conducts if v in is less than 0.7 minus 0.7 so we mark these points let us say this is over 1 volt 1 minus 1 volt plus 1 2 and so on okay so we mark these points 2.7 let me extend this 2.7 is somewhere here minus 0.7 all right in between neither this diode conducts nor this diode conducts and of course there is no voltage drop across this resistor and in v o will be equal to v in and so that is the first thing that we can so there you between these two points we will have a straight line which goes to the origin all right beyond this what happens then when d1 starts conducting you then write this current i d1 is equal to v in minus 2.7 by r0 plus r1 etc right and then you calculate this v o all right so what it will do is essentially it will cause a change in slope at this point so the slope was 1 earlier and at this point the slope will change similarly similar things will happen here and the slope will change all right so that is roughly what the students should see if you actually simulate this circuit and or if you go to the lab and hook up this circuit what is the slope r1 divided by r1 plus r0 right so how much this slope r2 divided by so it is 0.5 by 2.5 right 1 by 5 okay so if we look at the craft carefully we should be able to see all of these details so let us do this I will use a very simple diode model here which has got a large off resistance and a small on resistance so let us make a make the on resistance 1 mili ohm and off resistance let us say 10 mega ohms very small very large okay and then we will copy this into another diode okay now we need resistors okay and also we need a voltage source okay and at a DC voltage source and of course I need a ground okay now in this for this source the input source we can actually use a AC source because then we will be able to see all that result as well so let us call that AC alright now we can quickly do the wiring now these are things you can either assign as a homework to the students or you can just do it in class because really it does not take much time and if there is a problem with time you can always make up these yourself and just run the examples in the print in front of the students in class so this was 2k okay let me also write these values here so for our reference so let me call this D1 all this D2 and these things have got a parameter called the on voltage on so we will put that as 0.7 volts now this AC source alright one question is this is so this is our V in and this is our V out so we need something to measure this output voltage naming the nodes is not really important or not really required but often like to do it because what else okay we need to set this voltage at 2 volts that is it yeah and this input voltage we can let this go from minus 5 volts to plus 5 volts so an amplitude of 5 volts and a frequency we can frequency really does not matter because we are only going to look at it as a sort of low frequency DC circuit so let me make the amplitude 5 and let me make the frequency that is a 50 hertz alright so let me make up alright what all do we want to see that is the next thing we need to worry about what would you like to see definitely we would like to see this input voltage output voltage what else would you like to see simulation the good thing is you can see a lot of things lab it is not always possible to see away so what is from the students perspective or to from the point of his understanding what would be a good thing to see in this circuit apart from diode conduction so the student would like to see when exactly this diode starts conducting when exactly this starts conducting because that is the basis of our calculation so it is good to see this diode current so let us add all these variables one is the input voltage here P1 output voltage diode current okay and let me name these things okay so that is it alright so now the next step is what you want to do with this circuit okay and what we want to do with this circuit is now this is a time varying source so we should only do a transient simulation not a DC simulation okay so let us go back to solve blocks add a solve block and change that to transient okay as soon as you change it to transient the you will need to be you will need to fill up these details like what is the starting time what is the ending time what is the delta t t and let us simulate for just two cycles which is 40 milliseconds because the frequency is 50 hertz delta the constants a constant let's say a small number like 0.01 m 0.01 millisecond and in transient simulation you need to choose a method you will learn more about this when you read the manual and so on so let me choose the backward Euler method okay we have a very simple linear problem not linear problem but with a very simple problem with a simple diode model so there should not be a problem with convergence and so on then we have an output block and output block we will include the variables vs v out id1 and id2 that should do it okay so the program has completed and I hope it things are things have worked so first thing we would like to see is what is the output what is the output voltage as a function of that is the input voltage that is the output voltage let us see what that looks like okay so that is indeed what you would expect let's check this out so this is minus 0.7 right this is 0 minus 1 so minus 1 minus 0.7 2 3 2.7 and that is a slope of 1 that you can get out from the axis and then this slope what is this slope it should be 1 by 5 is it 1 by 5 1 by 3 so in other words if I go 3 here okay the axis are different let me expand it okay so in other words if I go from 1 by 3 right okay if I go from 3 to 3.3 I should go by 0.1 there right so is that happening 3 to 3.3 and from here to here is 0.1 1 by 3 okay similarly the other slope should be 1 by 5 that is the other slope so if I go from here to here that's I should go here from there to there by 1 so this is 0.5 and this is well let's it should be 1 by 1.5 you can verify that so you can show all these things in class alright so that is not all we would also like to see the diode currents that's very important let me plot the diode current ID 1 versus VS and that is what we would expect that the diode 1 starts conducting at 2.7 and otherwise is 0 diode 2 will start conducting at minus 0.7 and then it is then it conducts at that point 0 minus 1 so this is minus 0.7 so all of these things if the students actually sees these in the class definitely as a as a good impact on the student and things stay in the mind because the picture is always worth a lot alright what else would we can we see now since we have this since we have a time varying source we can actually see the time varying output so this is an input AC source so we should be able to see some clipping kind of thing so let's look at things with respect to time so that's the this is when the slope is 1 and then the input and output should be exactly equal and this is where the slope is 1 by 3 and 1 by 5 diode currents that's the diode currents so this is where diode 2 is conducting and that is where diode 1 is conducting and you can of course look at these things together with the input voltage and so on to make sense of it okay what else do you teach in basic electronics Thevenin circuits let's look at a Thevenin circuit example all right let's look at Thevenin requirement problem I see many familiar faces in the audience I have seen you before I have seen you before I have seen some other people okay so we want to reduce this circuit in to this form let's take some number 10 volts 2 ohms 6 ohms okay a very simple circuit actually you can show many more complicated circuits but the point is how do you use a simulation tool to show this Thevenin equivalent resistance any idea you use p-spires before right so normally the simulation packages will not give you the Thevenin and Thevenin voltage or resistance directly so how do you go about it okay so the answer is very simple all right so so this circuit and this circuit are the same and now suppose I connect a voltage source here okay an external voltage source called v0 right and I look at this current i0 and suppose I plot v0 versus i0 what will that give me that will give me a straight line let's see let's see what the equation is very simple v0 is v Thevenin plus i0 times rth if I have my signs are correct yeah okay i0 itself is v0 minus vth no wait wait wait wait that is not what I want I want v0 versus i0 so this is it right so v0 versus i0 is a straight line v0 versus i0 x intercept equal will be equal to y intercept will be v y intercept will be vth yeah when i0 is 0 okay and when v0 is 0 minus vth by rth right okay somewhere here vth by rth okay so I should get a straight line which looks like this okay let me do better than that let me plot it against this current let's say i prime so this is vth minus i prime times rth okay so in this case I will get o versus i prime so now the only thing or the only difference is the direction of the current is defined like that so now the y intercept is the same x intercept is positive did I make a mistake u equal to x right so now the same straight line except it is a mirror image of that one alright so in other words we can get from this figure v Thevenin immediately just look at this number and what is actually this thing called is nothing but the short circuit current or the Norton right Norton equivalent okay so in our case in this example if I plot this if I make a plot like this what should I get for vth in this example 2 by 8 2.5 so I should get a vth of 2.5 and I should get a Norton current or a short circuit current of 10 by 6 and from these of course I can get the resistance 10 by 6 is 1.67 ampere and of course the Thevenin resistance in this case it is just 2 parallel 6 which is 1.5 ohms and that should be also equal to this ratio 2 by 8 into 10 divided by 10 by 6 is the same thing okay so several ways of looking at it so but the key is if you are using a simulator then you can produce a plot like this and from this this plot has got all the information that you need about the Thevenin circuit so let us do this simple problem okay so that is the circuit we are looking at this is 6 ohms and this is 10 volts okay and the output variables that we would like to see is the current through this source but I want to define it like this so let me just put an ammeter there okay and we are going to look at this voltage as a function of this this current so let me put those as that output variable this voltage here and this current here okay so very simple circuit so let me go to the solve blocks these are DC simulation and we are going to vary a parameter what parameter are we going to vary in this we are going to vary the input voltage so we can get many points okay so let me just call this some name PS so element is VS parameter is VDC values let us say we want to go from what should we choose this range V Thevenin is 2.5 so definitely this V Thevenin should be more than let us see well let us just go from some number to some number and if we do not get enough of this then we will change that output block so that should be you know solve this so our x axis should be I0 okay y axis should be V0 okay this happens sometimes when I make mistakes so we have made a mistake somewhere the mistake is actually I should be varying the output the voltage of this source and not this source okay this is a constant so that I made that mistake and we do have this so this is our x axis these are y axis right and let me just blow up this part so that so the short circuit current and the open circuit voltage which is the same as the Thevenin voltage this is the same as Vth by Rth let us read out these values so what is the open circuit voltage this is our 00 2. I think we should get 2.67 is it you should get 2.5 it should be exactly 2.5 so this 2.5 and the short circuit current is should be 1.67 1.5 1.6 1.7 so 1.67 if you want to plot this in more I mean read it more accurately you can always blow up this part now you can see that this is 1.667 okay okay so we are we are getting exactly what we wanted 1.6 what is the next thing you can teach the students next thing is so we are we will get from this Thevenin resistance equal to 1.5 and 7 voltage equal to 2.5 now we can ask the student that you connect instead of this whole thing we can connect a resistance and if I connect a resistance of let us say 2.5 ohms what should this current be so that is the problem that we can pose to the students so the answer is very simple this circuit is the same as this circuit is the same as 2.5 volts here R7 in 1.5 and then this is 2.5 all right so this current then should be should be 2.5 by 4 625 should be 0.625 so this is an exercise that the student can check out so we will just quickly do this what do we do this just remove this source remove this ammeter and connector resistor here as simple as that okay what is the value of this resistance 2.5 ohms okay and let us add this output variable the current of this resistor I will figure it out later now we are running out of time but anyway that is what we can show the students you can figure out the current then that will be good okay so I just wanted to give you a little tour of the various examples that are already there you can you are welcome to use these examples and they are in several directives there is a directive called BJT circuits digital circuits diode circuits e1 e1 0 1 is the course that we teach here there are lot of circuits there electronics general some more circuits op-amp circuits for electronics power systems I guess is not relevant here solar cells so if you are teaching any course in op-amp there are lots of circuits in op-amp as well so you know what this is what is this circuit yeah this is an integrator and this is a Schmitt trigger it is a square wave generator it is square wave or triangle wave generator so let us look at some of these reasons okay if you look at this output you should see a triangle wave if you look at this output you should see a square so that is what we will see and why is it not symmetric I mean why is it why is this not symmetric around 0 it is because there is an offset voltage okay there is an offset voltage here if you make this 0 then it will also go so all these things are very easy to show in class with the help of simulation what is this some of you would be teaching this this is a simple current mirror and this is a reference current and that reference current gets mirrored on this is a very simple one and it is not a very good one because it has got a variation in the output current because of the voltage here so we plot this current as a function of this voltage it will not be quite constant so that is so let us just plot that okay so that is what so this is the reference current on the left hand side and this is the mirrored current so you can see that the mirrored current is not quite constant there are things you can do to improve this and you can make a cascode connection and so on so I will just show you quickly that as well okay so this is the so called cascode connection and if you look if you do the same thing in this one if I plot if I plot this current as a function of this voltage here then things should be it should be much more constant so let me plot all all four transistor currents and you can see that there the variation is not even visible on this here of course things go wrong because things the transistors will enter saturation but that anyway you are not supposed to operate in this region okay so the so this advantages you can bring out with the help of simulation otherwise you do all this theory and then students are not sure whether they understand or not and then they have not seen any pictures so they they go back not knowing exactly how it works there are some digital circuits you are familiar with this circuit with an up down counter simple up down binary counter there is an up there is a mode input which goes to all of these circuits and then there is mode bar input and depending on whether mode is 1 or 0 you count up or down and so on these are things which are difficult to draw on the board so instead of drawing on the board you can just show show a simulation result so that is what it looks like there is the there is the clock q0 q1 q2 q3 okay I think our time is up so we will stop here in the evening we have an extra session it is outside your schedule in which you are welcome to learn to use the simulator and then try out some circuits okay thank you