hi everyone so let's solve this assignment 15 and if you have studied the integrator and differentiator circuit before then only I will recommend you to solve this assignment and uh if you have studied it deeply no matter from wherever you have studied if you have studied it deeply if you have studied it in good manner then only you should be attempting this assignment if you are just watching this assignment just for the uh practice purpose then uh you should avoid it right so the questions will be very good this is the first one if I had show I don't know if I had showed you the question or not so I will show you the question this is the second one this is the third this is the fourth and this will be the fifth so five questions are there we will solve it one by one so let's start with the first one this is a very basic question this question we have already solved these kind of question in the theory part only we have soled you need to draw the V waveform so what is the what will be the V waveform at equals to capacitor will be shorted right at equal to Z capacitor will be shorted there will be a negative feedback involved capacitor is shorted so negative feedback will be there this zero will be copied here and it will be carried out to the output zero will be copied here will be carried out to the output so V not at equal to 0 plus 0 Vol what will be your V at tal to Infinity V not value at tal to Infinity at tal to Infinity capacitor will be open cired at tal to Infinity this capacitor will be open cired still negative f is present because why it is present because resistor is there so negative feed is still present so this node will be zero only so if this is zero so simply this circuit would become what inverting amplifier so if this is inverting amplifier so your V note will be minus 5 Vol how will your V note graph go your V note graph will go something like this this is your this is your final output V Note versus T you needed to draw and that will be somewhat like this why we are making it because I know that output is not saturated here output is not saturated because saturation voltage is plusus 10 volt so output is not saturated here if output is not saturated then certainly your at Infinity it will go to minus5 in this man right from 0 to Minus 5 they could have asked anything in this question they could have asked any value at a particular Point as well so you can simply write the value of V not t as well you can write the equation of V not t that would be - 5 into 1 - e^ - t by 2 What will be the to will it be 2 RC r r by 2 C or RC it will be RC only because this is simply behaving as a current Source this is simply a current source and this is like what is the equivalent resistance across the capacitor that is R only so your this is your time constant that is RC this is the V not equation and this will be the wave for they could have asked you the IC equation as well what if they had asked you the IC equation so what is the current at t equal to 0 IC value at tal to 0 plus what is that current that is 5 by R right and what is the IC value at T to Infinity that is 0 ampere because Capac will be open cired that would be 0 amp right all the current will be flowing through this resistor as stud said this will be open cired so current would be what that current would be current would be 0o amp so what will be your ICT equation that was 5 by r e to the^ minus t by 2 to is RC into UT that's it this is your ICT this is your ICT this is your V not t well and good V not is there I is there anything they could have asked they could have asked the equation of this one as well what about VX VX will always remain at zero VX will always be 0 volt VX will always remain at 0 volt say yes or no why it will always remain at 0 volt because output is not sated so virtual short will always be valid so VX will always remain at 0 volt right so they could have asked anything in this particular question and you are able to solve it well and good let's move to the next problem so here something interesting will happen we will draw all the waveforms we will draw the v w as well we will draw the VX waveform as well but here in the question what I have asked you I have asked you the VX value at T = to 2 second VX value at T to 2 second so what do you think how you should approach this question so simple way we will write down the V note equation we will see where the V note will try to go so initially what will be the V note potential at equal to 0 plus since capacitor is shorted there is no charge on the capacitor negative feedback is available there capacit is shorted so negative feedback is available so this zero will be copied towards the output Z Vol and at Infinity where it will try to we note at Infinity this Capac will be open cired so it will become a simple inverting amplifier so at Infinity it will try to go to minus 15 volt right but Satur voltage is 10 volt only saturation voltage is 10 volt so output will be saturated output will be saturated what is the equation of vot the actual equation of V is this it is trying to go to minus5 from zero right with time constant being RC so what your vot was trying to do this is what your vot was trying to do V Note versus p and it started from from Zer it was trying to go to -15 it was trying to go to -15 but in between it will meet 10 minus 10 started from 0 volt was trying to go to-5 but in between it met minus 10 here it met minus 10 it met minus 10 after that output will be saturated after that output will be saturated right so at T = to T1 at T = to T1 we know T1 would be equal to minus 10 and output will saturate output saturated and if output is saturated what does that mean virtual short will not be valid after T greater than T1 virtual short will not be valid so after after that what will happen in the circuit that we will check out but till now we got to know that at tal to T1 output will be saturated so first let's find the value of T1 so what that value of T1 would be - 10 = -5 into 1 - e ^- T1 by RC value of RC is given RC is 1 second value of RC is 1 second okay so from here we will get the value of T1 here I need to put T1 because T = to T1 you are getting output value to be minus 10 so here I need to put T1 so 10 by 15 - one Ln 10 by 50 - 1 Ln some 1.09 D1 = to 1.0 9 SEC at 1.09 SEC at 1.09 SEC your output is saturated for T greater than T1 4 T greater than 1.09 second so will that output always be at minus 10 will that out will that output will always be at - 10 no we cannot say that we need to check the VX Behavior we need to check the behavior of VX where the VX node is trying to go what is the be aare of VX so that we need to check and then we will comment if your if your output will be there at minus 10 or it can change so first we need to determine the VX potential now so after that here virtual s not valid so VX will not be at zero now virtual sord not valid previously virtual short was valid here the virtual short was valid right till output was saturating so virtual short was valid so what would be the equation what would be the waveform of VX till T = to T1 I'm drawing from 0 to T1 from 0 to T1 from 0 to T1 I drawing it will be 0 only this is the time T1 T1 is 1.09 second till this time till this time VX is zero only after that what will happen to VX virtual short is not valid so what will happen let's check that out mob distracting this was your R this was your R this was your capacitor C output is saturated to minus 10 right output is saturated to minus 10 so T = to T1 would be the potential of capacitor C can you tell me 10 volt here right because your VX VX node like here it is 5 Volt or whatever voltage it is 15 I guess yeah and input you are having 15 volt so your VX node for T less than 1.09 second here it was 0 volt only so T = to 0 plus it will not immediately change no this potential will not immediately change at tal to 1.09 plus so here I can write like this would be zero only like the capacitor voltage is not going to change immediately first question is how can you say the capacitor voltage is 10 volt because this node was at zero this node was at zero and this node was at minus 10 that means it would have been charged to 10 volt only no that's why I'm saying it so initially your VX potential is at zero only initially your VX potential is at zero only at study state where it will try to go at study state at study State what will happen the capacitor will be open circuited capacitor will be open circuited no one can change this potential we are assuming that at study State V not is minus 10 only this is what we are assuming we are assuming at stady State VX is minus 10 only then we will see the behavior of VX and we will determine if the VX stays at minus 10 or not so this is our assumption that till study State VX will be minus 10 only this is our assumption why this is the Assumption because we don't know that this potential is changing or not because this potential can change with respect to the opam because this is the output of opam so this can change so first we are taking an assumption that your output will always be minus 10 then we will find the behavior of VX so what will be the behavior of VX at tal to Infinity so where it is trying to go at tal to Infinity it is trying to go to 2.5 volt say yes or no so here VX node is increasing VX node is increasing okay so the VX node is increasing from 0 to 2.5 so if VX node is at zero then your output is minus 10 if VX node is at 0.1 VX node is at 0.1 which is equals to V negative potential and what is your V positive potential that is zero so V negative potential is more than V positive potential that means output will be minus 10 only say yes or no if your V positive potential is 2.4 volt at that time as well your output will be minus 10 so always your output will be minus 10 only say yes or no so what we assumed that your output always remains atus 10 that is correct what did you understand here that first we got to know that our output is saturated but we are not sure if the output will remain at T atus 10 volt only or not because for T greater than Z the behavior may change but we assumed the output is at minus 10 volt only So based on that what will be the behavior of VX the behavior of VX is that it will rise from 0 to 2.5 then we checked if VX is rising from 0 to 2.5 will our output be always at minus 10 yeah what if we checked the behavior and we got to know that VX was going down from 0 to minus 2.5 then can you apply can you write this equations no because what will happen if your VX is going down from 0 to- 2.5 the very next moment where it will try to go minus 0.1 at that time only your output will change to 10 say yes or no we have seen these kind of circuits no in the theory part we have seen these kind of circuits where the VX the behavior of VX decides the vde potential the vde potential immediately changes say yes or no and if we V note potential immediately changes then again the behavior of VX will be different then we again find the uh equation of VX and we will see where the output goes and if the output is going something different then again we will change the vote potential and then we will see that VX will stay at zero only because one node is trying one node is trying to get it up and one one node is trying to get it down so to 0.1 to minus 0.1 0.1 - 0.1 it will stay at zero so this concept I have given multiple times in the theory part right but here this concept is not used here your output is staying at- 10 volt only well and good so output will stay at- 10 volt only so output will always stay at minus 10 and VX will exponentially rise from 0 to 2.5 VX will exponentially rise from 0 to 2.5 what will be the vxt equation vxt equation would be 2.5 from 0 to 2.5 it will rise 1 - e to the power T minus T1 T1 would be 1 1.09 by RC 1 second why I why I subtracted this value because we are starting from 1.09 right so you need to find VX potential at 2 second so what that would be 1 minus 2 - 1.09 divided by 1 SEC this is the negative sign here + 2 + - A to the power -1 + - into 2.5 1.48 volt so is that the option available 1.5 volt is available so nearly you will get 1.48 volt 1.48 volt your VX value at 2 second is 1.48 volt in the examination if they are asking you good questions from integrator circuits they will ask these kind of problems right and you can draw the waveforms as well how will you draw the waveform of VX and V note this is very simple actually this question was not that tough next question will be more interesting okay so all these question I have framed by myself only I try to pick different values and and then I frame the questions right and V note potential also can make VX was at zero at some particular time and then it rises it will rise up to 2.5 and here it was it rises row to minus 10 only after that it saturates minus 10 it saturates now solve this question try doing it on your own have the patience have the faith you have studied enough and you can solve this question you have to draw VX and vform okay it is very easy it is not tough at all so let's again start with the same thing at T equals to0 this will be shorted if this is shorted that means virtual short is valid so this 10 will be copied here now this node is not zero this 10 will be copied here so V not at T = to 0 plus v not value at tal to 0 Plus that would be 10 volt and V note value at tal to Infinity what that would be at tal to Infinity where it will try to go atal to Infinity what will be the same again virtual short is still valid because resistance is there so virtual short is still valid then what will be your V potential you can find that vote potential how will you find it simple it is opam circuit only nothing else this is your resistance R this is your V potential this was your resistance R this was your resistance r here you are having 5 volt since virtual short is valid this node will be 10 volt only this is 5 volt this 10 will be copied here right so this node is 5 volt this node is 10 volt from here 5 by R current is flowing in this direction 5 by R is flowing in this direction then this should be 15 volt right yeah so V note at Infinity will be 15 volt from 10 volt it is trying to go to 15 so we'll write down the equation in this manner it is trying to go to 15 from 10 volt with time constant being r see this is what it is trying to do but will it be able to do it no it will not be able to do it because our saturation voltage is 12 volt so at St state it was trying to go to 15 but it will not be able to do it because our saturation voltage is saturation voltage is 10 volt no 12 volt right regation voltage is 12 volt 10 to 15 it was trying to go but at T at time T goes to T1 this is 12 Vol time T = to T1 at T = to T1 what happens output saturates at 12 volt let's find the value of T1 first 12 is equal to 15 minus E to the 15 - 5 into e ^ minus T1 by RC value will again be 1 second RC value I did not write it here it should be 1 second only okay should be given as 1 second rcal to 1 second but here they asked only the waveforms not the values so uh if they asked only the waveforms then certainly you do do not need to find the value of T1 because they are asking only the waveforms but to make it easy easy for you I am writing the values as well 3x 5 it would be 3x 5 Ln 0.5 D1 would be 0.5 second D1 will be 0.5 second this time is 0.5 second until that time your output is also sorry your VX node is also at 10 volt VX node this is your VX node no this is your VX node VX node till that time virtual short is valid so VX node will be at 10 volt only till that time your VX node will be at 10 volt only this node will be at 10 volt till T = to T1 which is5 second well and good now output is saturated to 12 volt output is saturated to 12 volt now so what will happen to the VX node that we need to check because we don't know if the output will stay at 12 volt or not this is the condition we need to check if the output stays there on not for T greater than 0.5 second what is the scene this is the resistance R now virtual short is not valid so you cannot say anything about it so this node is 5 volt right this is resistance r then there is one more resistance then there is a capacitor till what potential capacitor would have charged this is 12 volt right this is 12 volt output is saturated 12 volt and here you are having for T less than 0.5 second just at T = to 0.5 minus this was 10 volt so this would have charged to 2 volt in this direction would have charged to 2 volt in this direction right so your VX potential VX potential at T = to 0.5 plus what that would be that would be equals to 10 volt only at Infinity where it is trying to go at Infinity where the VX is trying to go at Infinity VX will try to go to this is resistance r no this distance R only so at Infinity your VX will try to go to 12 into r + 5 into R / 2 r that means 8.5 so your VX will try to go from 10 to 8.5 so let's get back to the Circuit this is your VX so your VX is trying to go from 10 to 8.5 let's just say if it goes to 9 volt 10 to 8.5 it is going so if it goes to 9 volt output have any problem or not because output is saturated at 12 volt if your VX node that means your V negative node is at 9 volt and V positive node is at 10 volt output is having an issue no so output will stay at 12 vol only understood if it is going to 8 6 volt because 10 to 8.58 is going so if VX node is at 8.6 volt if this node is at 8.6 volt output is having any problem no because V positive node is again higher so output will be always at 12 volt so the assumption that we made that output will be at 12 volt only so that is correct that output will always be at 12 volt so here your V note will always be at 12 volt and this will be the equation VX will follow so equation of VX would be from 10 to it is trying to go to from 10 to it is trying to go to 8.5 so 8.5 plus 1.5 e to the^ minus t by not t by RC T minus 0.5 by RC RC is RC is 1 SEC okay and here I can add U T minus 0.5 okay and what about V note V note T would always be at 12 volt for U T minus 0.5 after that it will always be at 12 volt so we can draw the final waveforms you can draw these WS so it always remains at 12 volt and what about this one from 10 to it is going to 8.5 these kind of things in the examination will certainly be asked in the future they are not going to ask you the standard circuits now they will frame very good questions here that is the speciality our this is the speciality of our channel that we will deal whatever that is going to come in the examination not whatever that has come whatever that has come that is certainly be in a very short manner or in a very short time but whatever that can come that is also important well and good let's move towards the next question we are slowly solving our all the questions slowly slowly we are moving let's solve this question this is also very very good question here the concept of network analysis will also be used and I'm going to use that directly okay a lot of people will be getting confused here that how to find the vde potential at tal to 0 plus okay yeah so the main question is you need to find V not potential at tal to 0 plus so what will be the value of V note at tal to 0 plus that is my question at tals to Infinity anyone can can tell because this will be open cired this will also be open cired so V not value at tal to Infinity will be minus 5 volt that anyone can tell but at T to 0 plus volt will be the same that you need to tell me see what is the answer answer will be minus 2.5 how so at T = to 0 Plus at T = to 0+ what is is the frequency content here at equal to 0 plus if you have studied network analysis if you studied the transient part you can tell me what is the frequency content atal to 0 plus Omega equal to Infinity if you studied the Trent part from prep Fusion you know it at equal to 0 Plus at equal to 0 there is sudden jump from zero to directly it is jumping to 5 Vol that means there is infinite frequency content now if there is infinite frequency content then how can you replace your par combination of RC this is R this is C this is giving you some finite impedence finite impedance and at Omega go to Infinity this will give you very low impedence very low impedence right because frequency content is very high so it will give you very low impedence so very high impedence let's just say some not very high some finite impedence that is some 50 and very low importance it would be some one let just say 1 micro so what will be the parall combination of this 50 into 1 micro divided by 50+ 1 micro that would nearly be this will be nearly 50 only 5050 will get cancel it will nearly be 1 micro only so if frequency is infinity so parallel combination of RC can be replaced with simple capacitor parallel combination of RC can be replaced with simple Capac so at tal to Z how you can replace the circuit at T = to 0 plus how you can replace this circuit at T = 0+ I'm making this circuit at T = to 0+ this will be open CED right this will also be open circuited because parall combination RC will be replaced with only capacitor so this will be the circuit that's it this you can remove this also you can remove like this I mean in this manner I can make the circuit better this is how it will look right negative feedback is there vir will still be valid this node will still be zero you can apply the how what will be the output voltage output voltage will simply be 0 minus V 0 minus V divided by 1 upon CS or you can write cs plus 0 - V divided by 1 upon 2 CS divided by 1 upon 2 CS right that is the impedence impedence provided by the capacitor that would be 1 by 2 CS or you can write into 2 CS that would be equals to Z so V not by V you will get min - 1 by2 Cs and CS will go there so V by V you will get minus 1 by2 VN is 5 Vol no VN is 5 Vol V is equal to 5 Vol V not would simply be equals to - 2.5 volt at T = to 0 plus this is all happening at T = to 0 plus v note value at T = to 0 plus is - 25 volt what will be the value at tal to Infinity at T = to Infinity V not value at T goes to Infinity at T goes to Infinity capacitor will be open Circ so V not value will be minus 5 did your output saturate no output did not saturate fromus 2.52 it is going to minus5 - 2.5 to Minus 5 so output did not saturate so what will be your V waveform you can write the equation as well V note equation I can write here what will be the V note equation it is trying to go to minus5 from initial value of 2.5 to the^ minus t by what will be the time constant yeah that is very interesting what will be the time constant what do you think what will be the time constant I guess it should be 2 RC only because first tell me it is a first order circuit or second order circuit yeah that was interesting question it is a first order or second order tell me if it is a first order or second order circuit you can find the transformation it will be first order circuit only and intuitively how you can tell intuitively how you can tell it is first order or second order directly I can see in the trans function certainly I'm going to get first order only let's check that out what is your V not s by VNS the 2 C parallel with r what it give you C parall with r what it gives you r parall with 1 by CS what it gives you R upon RCS + 1 so what this will give you what this will give you R upon 2 R CS + 1 and what this will give you R upon RCS + one right so what will be the transform function we not as by vs that would simply be minus RF by this importance divided by this importance that would be minus of RCs + 1 divided by 2 RCS + 1 that would be the transformation so it is a first order C only because there is only one pole and what is the time constant time constant would be 1 by sorry time constant would be 2 RC so here I should write first order first order with time constant of 2 RC well and good first order with time constant of 2 RC time constant of 2 RC but output did not even saturate output did not even saturate here so you can draw the V not waveform now how will your vform Go - 2.5 2 - 5 - 2.52 - 5 Vol well and good everything is clear let's move TOS our last question so you can try solving it on your own how will you solve it V set potential is very high it is given that we vet potential is very high so output will not saturate at T equals to Z initially virtual short is valid or not although here there is z amp current inductor will be open circulated but still capacitor is providing zero impedance so there is still a negative present right there is still a negative feedback present so negative feedback is present atal to 0 Plus at T to 0 plus negative feedback is there that means what does that mean virtual sword is valid if virtual short is valid then what will happen what do you think this node is zero what is this input this is a impulsive input impulsive in there this node has become zero now this node is zero and here you are having pulse pulse input that is 5 delta T 5 delta T the voltage across the inductor is impulsive now say yes or no what is the voltage across the inductor that is impulsive that is 5 delta T in this direction what is the voltage at tal to 0 plus I'm talking about at tal to 0 Plus that is f delta T what is the current equation current equation of inductor current equation I this I'm calling I that would be 1 by L starting from 0 0 plus you can say 0 minus you can say till some time T from 0us to time T 1 by L integration VT dot DT right 1 by L vt. DT 1 by L and 0 minus or 0 plus you can also say from there it is starting till time T what is the potential across it that is f delta T 5 Delta t u f d DT what will be the integration of this that would be F UT right integration of this will be step so 5 by L UT this will be the integration say yes or no this is your ilt so your ilt is 5 by L UT I is 5 by L UT what happened is that initially there is a virtual short concept valid because there is a negative feedback so the voltage across the inductor is fixed now and that is impulsive if there is impulsive voltage that means there will be step current in case of capacitor if there is in inite or impulsive current that means there will be step voltage the concept is across the inductor the current cannot change instantaneously across the inductor what was the same your i0 minus what was your il0 minus i0 minus was 0 ere right so what you would expect at IL 0 plus I 0 plus what you would be expecting 0 ampere right but that is not correct here this is not correct why so because there is impulsive voltage across it if there is impulsive voltage across the inductor then the current will be step the current is immediately changing so at tal to 0 plus what is the current the current is 5 by L the current is 5 by L so this particular circuitry is behaving as a current Source this is acting as a current source which is driving 5 by L current say yes or no so I can replace the circuit how I can replace the circuit now so I will just copy it I will remove this part this is simply acting as a current source that has this direction okay it is simply acting as a current source that is 5 by L UT so it is driving 5 by L current initially at tal to 0 plus what will be your V note at tal to 0 plus what will be the V note V note will be 0 volt at tal to Infinity what will be V not why it is 0 volt because capacitor is shorted complete current will be flowing through the capacitor because capacitor is shorted right and at tal to Infinity what will be V capacitor will be open circuited but nothing is happening with this current this current is constant because this is coming from the inductor this current is constant so at tal to Infinity what will happen the capacitor will be open circuited so complete current will be flowing through this if complete current is flowing through this then what will be your V note V note will simply be minus 5 by L into R say yes or no initially this is simple RC circuit with respect to current input step current input so initially complete current is flowing through the capacitor but there is no charge on the capacitor so V not is zero but at equal to Infinity V not will be minus 5 by L into R minus 5 by L into R so this is the same right with time constant being RC so you can write down the V note T equation that would be minus 5 R by L 1 - a ^ minus t by RC into UT this will be your V waveform from 0 till minus 5 L by r always it remain saturated there time con being RC since they have given saturation voltage are very high so that means output is not being saturated output is not saturating then this will be the waveform we note is starting from zero to this well and good so now the concepts are crystal clear to you you must have solved a lot of problems is there any other one no these are the final problems that we have solved so in that multiple concepts are use the video are short but you must have understood a lot of new things here okay then if you found it helpful you can share it with your friends okay so let's meet you in the next video thank you