 However some of the interesting and some of the non-linear kind of them I would like to discuss or at least show you. Many of these circuits are given in the books, particularly this Smith and Federer's book. So you need not have to really write down what I am saying because these are actually taken from the book itself. However there are tricks something which I might have added but not much. One of the use of a opamp is to convert what I should say V2I converter or I2V converter and this is a typical current to voltage converter. You have a current source and obviously we know in the opamp if it is in the negative feedback system and the V plus is grounded the current entering here from the current source must enter through RF because no current can enter opamp why we say so because input impedance is treated very very high roughly infinite in ideal cases infinite. Since that moves through this so one can write I2 which is 0-V0 by RF is equal to I1 which is nothing but IS so V0 is IS times RF and therefore one can say output voltage is proportional to the input current therefore it is called voltage to current converter. If you put a resistor here it will be a normal gain amplifier which can be output multiplied by something more can be actually voltage to voltage converters just to say you that why it is called voltage to current converters. This is very simple most of us uses indirectly or directly but there is something which is of an interest normally all amplifiers can be voltage to current converters but here is something which is interesting if you want voltage to current converter one of the thing which you are expecting please remember in this circuit it was I2V but if you are doing V2I the I will be decided by the load current how much load you put larger the load currents will be proportional to that actually but what is a good current source at the output that current should be independent of the load that is what we are looking so here is this modified circuit of normal V2I any opium is a V2I converter that is not a big issue but here is something a modified V2I converter which shows that the output current is independent of the load here is my load circuit here is my normal opium this is my R1R2 feedback input is given here now what I am doing from the output I have a resistor which is connected to the load and this is the current which I am interested in oil oil is my output current which I want to be proportional to V in is that correct what is my requirement oil should be proportional to V in and not functions of ZL itself okay is that point clear what are we looking for we do not want if I change ZL the oil should not change in normal sense it will and we do not want that to happen so I did it I took a feedback from this common point and return it to V plus and put a resistor down this is a very simple interesting circuit please remember this is a negative feedback and this is to some extent positive feedback going on we are trying to adjust the net feedback now okay we also know V minus is V plus okay and not equal to 0 because this potential will not be 0 now okay is that point this V plus will not be 0 now there is a divider going on so obviously V plus is not and that is equal to the potential VL is that point clear this VL is same as this and therefore V plus is actually V minus so V minus is V plus equal to VL and not 0 okay as we normally assume because there we always say V plus is grounded in this case V plus is not grounded okay so we can say how much is VL is essentially oil times ZL is the voltage drop across ZL oil times ZL is VL okay now what I am trying to say that is I should be independent of ZL that is what I am looking for I also know current entering here must enter to the feedback so I1 is I2 as no current enters opium okay in what cases current can enter opium if the RIs are not very high and the other resistances are small enough some current may enter opium proportionately but our assumption as of now is we are close to ideal opiants no current or practically 0 current actually okay this is an assumption in real life there is a modification has to be done because it may not be very very accurate in most cases but as I say difference potential will be order of less than a millibolt so our assumption of V minus equal to V plus which makes RI close to infinity is not very invalid assumptions okay please look at the circuit if you are drawn keep watching that I all that I did is I am now trying to get the current equation V in minus VL please remember V minus is VL so V in minus VL by R1 is VL minus V0 by R2 which is the current flowing from this is that I1 and I2 are equal so I1 is V in minus VL by R1 and I2 is VL minus V0 by R2 they are equal but I know VL is IL RL right now IL ZL so V in minus IL ZL by R1 is IL ZL minus V0 by R2 this is one equation of interest for example the second equation please note down all that I am doing is substituting VL equal to IL ZL is that correct substituting VL is IL ZL in my earlier equations substitute VL by IL ZL is that okay no great thing no great action going on just simple substitutions now I am interested in the current entering from feedback side through a resistor R3 and that current I declare it as I3 so I say I3 how much is I3 will be from here in the figure this potential minus this potential divided by R3 this potential minus this potential divided by simple ohm's law nothing great so I said I3 is V0 minus VL by R3 but I will again resubstitute VL by V0 minus IL ZL by R3 this may be my interest of second equation but if you see at this point at this node what are the this node and this node are same is that correct they are connected so what is the at this node what is the Christian law will say at this node currents current some some should be 0 so I said okay I3 is IL plus I4 I IL plus I4 is I3 okay but how much is I4 VL by R4 but VL is IL ZL by R4 so I substitute again in this equation third which is V0 minus IL ZL by R3 is equal to IL plus I4 which is IL ZL by R4 say if I do this and collect terms now I am I am what is the interest I am looking I am looking for load current in terms of V in and what I am expecting at the end it should not be function of ZL if I get it I achieved what I am starting with is that point clear my ultimate aim is to get IL in terms in terms of V in but the term should not have any ZL term okay because if that happens and the source is output current is function of the load itself which no one will like because I do not know what I am going to connect later okay so I want a current output which is only function of input voltage so if I do this collect the terms as I did from these three equations I am going to get IL into R2 by R1 upon ZL plus R3 minus 1 minus I just collected IL terms equal to V in upon R2 by R1S now what is the condition I should have that IL should not be a function of ZL the upper equation shows there is ZL terms appearing is that correct ZL terms appear if I want IL to be only functions of resistances okay independent of ZL what is the condition I should meet that R2 by R1 R3 R2 upon R1 R3 what is the term I am talking which is functions of Z4 this term must be equal to then this terms will cancel is that clear if this R2 by R1 R3 is equal to 1 upon R4 these two terms will cancel is that correct so the condition that is IL will not be a function of ZL will be R2 by R1 R3 is equal to 1 upon R4 and if I then substitute here IL is 1 upon R4 into V in this R2 upon R1 R3 has been replaced by 1 upon R4 now this R2 upon R1 R3 has been replaced by 1 upon R4 so we say IL is V in by R4 is that correct so we are now got the output current which is independent of ZL and only a proportional to input voltage but under what condition this was derived that R1 R2 upon R1 R3 must be equal to 1 upon R4 if this condition is met the output current will be independent of any load value whichever as long as this identity holds this will always be valid is that correct so choice of resistances can make output current proportional to V in but independent of ZL this is very simple circuit and has been used almost extensively everywhere when you want current source to the next stage proportional to the input voltage in the opamp there is no real difference between them is that clear yeah that is what I say in opamp there is no nothing much because opamp allows all AC DC inputs so is that so this is a slightly different circuit which is standard V2I is always known to us just put a opamp and it is V2I converters okay but here we found out that that time will be what the load goes may have a function of that so we just wanted to find a condition in which output currents will be independent of the load itself this is something slightly additional circuit which we added there to make the point here is that clear so these circuits are also given in books but maybe on sites you can see the some websites but this is a standard circuit which is used extensively on the chip okay therefore I thought I should show you that how do we actually get current sources for the next stage okay is that correct what should be it called there is this is a current source proportional to what voltage so what is it should be called CS voltage control current source this is voltage control current source VCCS is that point clear the name is VCCS voltage control current source by similar logic we can create current control voltage sources current control current sources everything is possible by some small tricks here are there is that clear to you these are the tricks which will use whenever we need for the driving something what we need and what is available as an input we can convert corresponding it to be dry the next stage dry you understand what the word I keep using dry when I actually give that output of that stage to the input of the next called driving okay I drive it okay that point here there is of course we are not talking any technology here not many good students in this we have a process called diffusion in the case of making integral circuits when p type or n type impurities are introduced into the other kind of material and the process is called diffusion so what first we put some impurities on the surface and then we call drive in drive in okay impurities get in the second two circuits I will not spend much time but you they are very important circuits in our analysis of many circuits these are essentially why I am showing you this because I am going to go for filters and I need these kind of circuits there op-amp can be utilized to create a differentiator as well as an integrator can you think if I have a differentiator and I want to convert it to integrator what should be the without thinking much integration is opposite that of differentiation CDV by DT is I so we is in one of one C integral IDT this fact that I just have to do opposite of that makes integral integrator convert to differentiator and differentiator going to integrate so we will see both of them quickly if do not want this is my a capacitor in the series of the input through a feedback register or the current entering capacitor is CDV by DT now that DC cannot go madam once I put a series capacitance I am only talking of time dependent turns so CDV in by DT is the current through capacitor must pass through R which is 0-V0 by R and if I therefore write V0 then V0 is – RCDV in by DT is that correct please look at what is the current in R 0-V0 by R that is equal to CDV in by DT essentially saying V0 T is – RCDV in by DT so what I am saying output is proportional to differentiation differential of input is that correct this is differentiator if you look at from the transfer function side put it 1 upon CS here do all the analysis get a transfer function V0 by VNS which is only a CR if I plot this as the gain function there is a 0 going here at 1 upon RC and it will give 20 dB per decade rise is that correct so what why I showed you if there is a capacitor here at the input at that 0 point the gain will start 20 dB per decade this is something you should remember so wherever there is a capacitor or a differentiator circuit it is essentially in a transfer function Bode plot saying at that 0 gain will or transfer function magnitude will start rising by how much 20 dB per decade can you convert it to octave also what is it equivalent will be they keep writing many books 6 dB per octave calculate that they go to the base 10 to the base 8 convert it to logarithmic simple okay from going to one this to the other so is that point clear what is the point I am saying this is what we wanted but this essentially what I am trying to show that whenever I see a capacitor there in a transfer function of this kind I am as if trying to rise my magnitude by 20 dB per decade after 1 upon RC value okay this fact I am going to utilize in my filter design is that worth clear to you why I am showing you this ultimately my aim is not to use differentiator integrate just for the heck of it I am going to use them as can you think about what is let us say per say after somewhere this have occurred like this so what kind of filter I am looking for high pass so below that nothing is passing beyond that everything will pass so a capacitor series made I will do now if I put a series this is what you should understand in principle circuit is not very great because there is nothing very great happening in the circuit but what is principle behind is this oh if I do this oh then I am allowing other frequency what will I need otherwise I will also need low pass I also will need a low pass so what is a low pass will be initially it will pass everything at certain point of frequency it should not pass I had anything so if I want to invert this I should see integrator may be doing the same thing integrate because opposite of this will be a low pass and opposite of differential will be a integrator so an integrator essentially will help you to get a low pass system a series capacitance will give you differentiator will give you like to give you some kind of high pass applications can you now mix the two what can you get then I can get both band pass as well as band reject why because then I right now I put both frequency like the if I do this and I can reject it so a general principle of filter is essentially taken from a simple transfer function which is essentially capacitor related is that point why we were all the time worried about that where the poles are where the poles are because at the end many applications will require some frequencies to pass and some frequencies be stopped okay this is where the filters are most important in all applications okay so without going into much detail for an integrator if you see an integrator I have an input a resistor so what do I do in integrator exchange capacitor with a resistor wherever there was a resistor put a capacitor whether a capacitor put a resistor and it will become inverse okay V in by R is – CD V0 by DT so V0 okay only thing you should remember in the case of integrator this initial value has to be specified because capacitor may come pre-charged in most cases if I want VCO to be 0 what should I do let us say initial because once you do it it will get charged now for the next application I want VC to be 0 again what you have to what should I do is K against which to be a resistor that K a circuit like on the path they got Joe go discharge Karneko allow correct input is that correct we will see that whenever we want to discharge a capacitor a parallel path must be provided through a resistance which is switched why it is called switched one there because then during the process of integration I do not want that switch to be open this switch to be open when I am doing like this but when I do not want to do this I want to discharge the charge put on the capacitor through our resistor is that correct this is something we do always all integrators is that clear the charge otherwise what will happen the next time it will pile on that okay is that correct so if I want this term to go 0 every next time then I must actually discharge the capacitor before the next integration is going to be is that correct so there is some resistor with a switch is always connected across the capacitor to remove the charge from that is that correct this is a technique which is universal in all analog digital or whatever you say this is the standard technique of removal of charge pre-charge system it is got a transfer function one upon a CR so there is a pole this will go up to from here to here 20 dB per decade and this is essentially saying if initially it was constant at this point this will give me a low pass I can adjust these two values where from it should start okay so I can say I can create a low pass putting an integrator I can create a high pass using a differentiator will come to filters more detail but this is just a basic principle behind filters before we go to filters here another interesting circuit you have already done half way rectifiers in your first year course what is the way we feel rectify anything normally if you have a diode okay load here and a signal here what is the way say this is a half way rectifier we say so whenever this becomes positive signal the diode conducts okay diode conducts so all of it is transferred to RL okay whenever this becomes negative diode switch reverse bias and no output is seen is that correct we say rectified AC became DC one direction unidirection now the problem with any diode will require at least 0.65 or 0.6 or called gamma as we say voltage drop for turning it on okay cut in voltage is roughly 0.6 to 0.65 typically 0.65 so at least till that happens the device will not turn on is that clear now this fact means that but I want that means for certain value you cannot actually switch it on a here it will start so in fact the this is a weakness of this I want to make a more precision I say as close to 0 can I get it or maybe few micro volts 50 micro volts instead of 600 millivolts I want 50 micro volts the number was which one though these will be therefore called what precision rectifiers is that correct why precision I want this which over we are occurring at close to 0 or at least in micro volts okay not in millivolts this is what we are looking for and therefore this circuit is essentially called a rectifier I will again as I say I will not dwell too much in detail I just to the principle behind these circuits are given in books so it is not anything great I am talking I am only trying to explain what is the basic system we are working at there is a diode in series to the output of a op-amp which is what is this connection will be called follower so we are using a follower out in between the load and there is a diode and we expect v0 will be then switching on off at very close to 0 or as close to 0 as is possible now think of it diode has a voltage drop of 0.6 volt okay for cutting what is the typical gain of an op-amp I mean this defam or op-amp 10 to power 4 or 10 to power 5 so if I see input if output is 0.6 volt what will be input v-v plus is my vd how much will be that 0.6 divided by 10 to power 5 how much is this this is 60 micro volt okay so in 60 micro volt difference the diode will conduct is that correct or not conduct 60 micro volt okay 6 that depends on the gain I use okay all right so is that point clear that very small change in v in differential I can make output go high or low I mean after the rectifier because diode will be switched off at less than 0.6 is that correct anything less than that will switch off in earlier case it could not because it is a 0.6 voltage drop was required to on and off the switch okay this fact that I can now switch off or switch on even at a very low differential is something very great because then the output will be immediately 0 going to cut off or going to full signal passing by just small change is that correct so is that what precision is clear to you why I call precision because small change at the input can rectify now because the gain is very high the diode drop of 0.6 is equivalent to in 6 micro volt at the input which means small change here will actually create 60 micro 600 millivolt at the output which will turn it on or turn it off this is why we say it is precision rectifier so the theory behind is a diode current is ISQVF by NKT eta KT how much is eta in silicon devices if I plot IV as a push characteristics actually diode come in so not 0 sorry a point 6 a power QV by how much eta KT of the beach make QV by KT of the or you have to take QV by 2 KT of the physics guy is a silicon germanium say related nahi a principle of devices diffusion kaha generation accommodations kaha consa current cup dominate kartha hai uska proportionality hai I just told you that some way before you go look at it the normal slope starts with 2 KT 1 upon 2 KT then becomes one upon KT kind then again become 2 KT kind which turns start dominating in the currents which phenomenons therefore it is proportion to different way this is just to show you why it is precision the word was people there is they are not explained in some books okay so I thought at least you should know why these are called precision rectifier so a precision guy rate diode lagai amy diode hai to precision precision consa precision amplifier jo lagai a usse a rahe opam se a rahe this fact you must understand opam is doing great business for you indirectly it is not actually participating because current to Bahar we are thin ikal nahi par yeh itne se karan wo game khel gaya that is why it should be important okay here yeh di aap kisi chis ka logartham nikal nahi chata hai ache a computers aap logon ekabhi padha hai ke naam there used to be initially when we started education in your time 30 40 years ago we used to have a computer those are digital computers have not come okay we only knew a some a you know when we say logic and get bataya to home expert manager there were other kinds of computing systems which were there those analog computers we have worked on that analog computers okay so analog computers massive differential integrator ii lagte hain but we could do lot of games using analog computing us make abhi kabhi aisa demand war ki logartham nikal nahi hai so consa circuit usame use war tha yeh wo hai it is normal circuit instead of a resistor here I have put a diode here is that correct I have put a diode here what is the input karan we in my arm is that correct but that same current must go through this which is the I2 which must be the diode current because please remember diode does not listen anything it only passes as far as diode theory I just calculated ID current in the rectifier ID I have did you see that current ID IS e to the power QF by KT is the ID current is that diode current now please take it if ID is 0 how much is V0 0 no connection that means if ID 0 that means there must not be any current here why there should not be current here because no current can enter open if there is no current here there cannot be current here if initially there is no capacitance output is 0 if there is no power current passing through this okay so if ID we can say that means you can say last voltage but if capacitor it will be finally discharge through internal open okay so if this is 0 we in has to be 0 then there is no current here but if diode conducts what does that mean this potential must be larger than V0. If that is so one can say if the this is a voltage of VF VF plus V0 must be 0 because this is 0 is that correct which means V0 is minus VF and VF KT by Q log I minus log IS a VF so V0 is eta KT by Q log V in minus log R minus log IS or to say minus eta KT by Q log V in by IS so V0 is proportional to logarithm of V in with a constant this is the constant part okay so this is essentially called logarithmic amplifier if this is 0 no current if there is no current that is diode is switched off no current can enter here either okay. However if the current has to pass this potential has to be 0.65 or this should be at least higher than this that means this potential plus this potential must be this this plus this some sum of the two that is V0 plus VF must be 0 because this potential is 0 at this node is that correct which means V0 must be minus VF when the diode conducts correct when it is not conducting this potential has to be smaller than that only then it is not conducting essentially if it is not conducting there is no potential here if there is no potential or there should not be any current coming from so there is no 0.65 drop it can create so diode is switched off okay rectify canals go irrespective the expression which I got whenever diode will conduct it will do logarithmic situation it has to conduct is that correct cut in a lot of VF and I cut in which is essentially given in terms of the values given to me but it is 0.65 any you say jada hona chai is that clear then only it will conduct so we say I say a anti-log amplifier kele kya karna chai ya exponential anti-log is say exponential kya karna chai ya resistor hai vaha diode laga ye jaha diode emai resistor laga ye so it will become a anti-log amplifiers exponential me aapko karke nahi dikhaunga sif circuit dikha dito so is it clear to you there is something mirror is going on mirror is slightly different from copy okay. This is very interesting analog circuit ac labor is interesting here chara side or say other function a lot then there is nothing great I am so I am just repeating opamp basic theory current enters here goes through here if the diode is conducting this will happen if not conducting it will be any way switched off okay. So yaha par mene nikala hai v0 is minus IR exponential QVF by KT so v0 is proportional to QVF by this essentially exponential of the drop BF is a function of all I whatever you are looking okay. So ye isko kya volte exponential or anti-log amplifiers so you have a log amplifiers anti-log amplifiers you have precision this I will ask you now you try maybe you look in the book there are few more circuits in analog you did in your second year or maybe doing now there is something called clamp there is something called limiter use diodes there a precision clamp and precision limiter kya se banayenge is this you try yourself to circuits all the circuit kaha se liya main hai jod normal diodes se hum bana sakta hai yaha karke dikhaya jaar hai opamp kisa better hai. I want to now create a precision clamp what do you mean the precision clamp exact value par bo voltage vaha se banna nahi chai okay or precision limiter when a current should not exceed beyond a given value at any cost okay precision on a chase so doh circuit hai jo ke diode ke saar pehle hum ne design kya hua hai unko opamp ke saar design kya hi and see that they are better than the without opamp circuits okay ye se rectifier dikhaya vahisa wo doh bhi hai rectifier dikhaya uske ye se a clamps or limiters behind ye unka bhi properties using opamp should be better than without using opamp actually do you understand this simple logic it deho actually opamp does nothing all that we are doing is only this circuit is that clear so why opamp then because otherwise it is not this virtual ground create hote hain physically ground kya to divider ho jayega virtual ground ne lai maa karan andar chala jayega yaha karan nahi jaar hai je jo facility aap create kar rahe hai that is essentially because of an opamp is that correct so opamp property directly is not appearing in anywhere here do you see anywhere opamp ka gm nahi aaraya avi nahi aaraya bhi nahi kuch bhi nahi le nahi aam doh but we remove opamp and everything will be lost is that clear so please do not feel that opamp does not play any role opamp actually is playing all the all the roles indirectly okay and that is what you have utilized those properties okay. Okay, nothing is happening in the triangle anyway, okay, now that you have this is a very trivial query. The output stage actually sinks it is that correct so the current path is like this is that clear so this fact has to be understood that opamp looks as if it is open circuited but it is not it is essentially providing a path internally is that clear these are trivials but this makes opamp theory more and more interesting that how opamp is doing all its machines okay. Okay so there is an amplification sorry there is an multiplier which can be analog what do you want to multiply I want to multiply two voltages V1 and V2 okay so I say okay I go back to my simple maths I say log of AB is log A plus log B log of AB so AB is my product is a product of A and B is anti log of log A plus log B is that correct anti log of log A plus log B. I should create one logarithmic amplifier to create log A similarly I should create logarithmic amplifier for B log B then I must create some of the two because log A plus log B take summer I create log A I create log B I sum them by summer and then I take anti log of that okay so I get AB as multiplication a theory analog multipliers is that clear so given two voltages to multiply first create logarithmic of first voltage logarithmic of second voltage then add the how do you sum them summer is the easiest circuit which you already done A key input per do know though through input log out resisted then you sum them and once you get some them put the anti log amplifier to create the multiplier this is my V1 the first block is giving me log V1 this is log V1 this is log V2 what is this summer though input a V1 kind put lia log V2 kind put lia this goes some Kia a anti log amplifier lagaya anti-logger system and I is that point clear so if you have given a function now you can understand how we were doing in analog computations of arithmetic you can see how we were doing it we were trying to use opams everywhere to create arithmetic is that correct arithmetic they were constant one kia ja sakta sorry hain here parik register batana bool kia meh constants ko to hamehsha one kia ja sakta by choice of values is that okay char block do log do log bananika ek sum karnaika aur ek anti-log karnaika the product aga apka ishe bhi a chai multiplier analog ka hain was simple hain ek opam se kia ja sakta hain 4 quadrant multiplier bool kia plus minus multiplication ho sakta hain okay since we are looking for ultimately what is the importance of my opam for what I am looking for I am looking for what we call there is a word in electrical engineering you are also doing a course in that name is called signal processing is that correct the word is signal processing what does signal processing means there is an input signal on which you do some machine and get some output signal okay now what machine we do is a different game okay which transforms Nikol sakta hain which time to frequency convertors have kia ja sakta but there is something processing is done one of the signal processing requirement is filtering that means certain part of the signal I want to pass and certain I do not want is that correct so this is first signal conditioning the scope will tell us miss one of the major component of signal conditioning is filter is that correct now if you want to learn filters you must know a little bit of bode plots because only then you know how the filters can be created what else you will require in signal conditioning other than filters your go last part of now course full analog kia with a in world is as you are learned by now world is digital like or not like 90% of the chips is used as a system is used are digital but all nature signals are always analog so what do you do so you must convert our analog signal into digital that is also another signal conditioning or signal processing I am convert an analog signal into a digital signal and for some reason I want to display like gharikak as I tick tick tick I want to display that like in computer also if you see many laptops of this they show a watch like running analog they must be having a digital to analog converters is that correct so the last part of this course will be analog to digital converters and digital to analog converters which ones will be maximum used a to D because you need not require every time display which is analog you need only this so all the processing first thing is there are many other processing part but at least the two of them which are most crucial for us one is filtering the other is a to D converters so we were looking for frequency response which is essentially for any amplifier or any system is interesting if you have an op-amp which is an open loop case then the open loop gain is a 0 upon 1 plus j by omega 0 this is a transfer function for a normal open loop amplifiers op-amp a pole h omega 0 per gain is falling 20 log V2 by V1 after omega 0 and at 0 the gain is DC gain which is a 0 this is a transfer function of open loop op-amp and therefore omega 0 is the dominant pole yeah I give pole only to be acceptable assuming there is a dominant pole far away from others what does that mean what is the cry that what it will satisfy automatically if I have a dominant poles far away from other poles and zeros what does that actually guarantee me if the a pole on a pole again so come up what will be the phase margin then positive system will be stable enough is that correct system will be stable enough okay so right now assumption is there is a dominant pole away from other poles only we are looking for the first pole okay okay at please remember this is an interesting fact you can see omega 0 to omega t jane me a 0 to 1 0 pe ana chaiya kyo aisa aisa theory kyo aisa because every decade you move on the frequency how much DB it goes 20 DB or 20 log a 0 essentially is what you are going down is that correct so the is kajo ratio hoga omega t this triangle case after the cut away 0 is that correct it let us say 10 times this is 10 times is that correct 20 log per decade we have 20 DB head we are 0 is that clear so what does it trying to tell you it says that if you have a dominant pole approximation valid system is stable omega 0 into a 0 is the unity gain frequency which is essentially saying because unity gain means you have your a but what you need to gain bandwidth is constant please look at it what I said you have a 0 omega 0 gain bandwidth is that clear again again 1 into you say gain bandwidth product is constant for any amplifier so omega 0 is the bandwidth gain is a 0 which is constant band no no the bandwidth is the point at which the gain start following that is the first pole is the bandwidth is that correct so a 0 into omega 0 is the gain bandwidth product is constant even here it should be constant so here if omega t is this the unity gain must be achieved what is this frequency gain bandwidth constant mean product of gain into frequency at that point must be same everywhere gain fall hoga tabhi frequency birthday yeah per what is omega t how do I define this is one so what is the multiplier 1 into omega is gain into frequency but that means this frequency into one must be same as anywhere which wherever you go that is how omega t was derived in fact if you see gm by c was derived on that principle please go back and read what we said please read it this equation is always valid if dominant pole a prognosis valid omega times is does karka dekhona up coming up a example we believe a kidna be frequency ratio koro what I can a dekhona again UC staff say fall hoga na single pole a dhya dhan record double pole a to go valid mehura is he live only bola it is a dominant pole approximation okay this relation is valid for dominant pole approximations if that is so then omega t is unity gain bandwidth or gain bandwidth product GBW assumptions other poles occur at much higher than omega greater than omega t and system is stable this is my assumption which I have started a whole open loop now let us say I have an amplifier which has a closed loop system with a feedback of R2 and series of R1 input is at R1 input at me which is this kind non inverting kind same thing can be done for inverting do not worry too much about it is the closed loop amplifier ka gain kesa hoga by feedback theory this is which kind of feedback is this output V0 hack us ka part feed hora other V0 ko 0 kiya toh kya hoga no feedback so voltage is in shunt to that that means the shunt feedback connect kaha par hora evo a node par toh hoga kya connect hoga vishant hai karant hai so yeh konsa amplifier hai shunt shunt amplifier is that clear shunt shunt amplifier is that correct okay so many aapko us din bola tha ki you actually use an opamp this non inverting kind or inverting kind and see that shunt shunt amplifier jo ham ne normally derived kiya and use feedback and verify whether our theory what we said is valid okay yeh maya abhi assumption kar kachar no aap verify karo okay so the closed loop gain for this shunt shunt amplifier is aol upon 1 plus beta times aol so if I gave feedback factor beta if I derive open loop gain aol then I can get closed loop gain which is how much is closed loop gain for this 1 plus R2 by R1 so I must get using this as well 1 upon R2 by R1 and which is that value will be that will be a DC gain but I am not interested only now in DC gain I am interested in frequency dependence is that clear here aol is not a0 it is aol which is a frequency dependent terms so now I use this theory from the figure I see the feedback factor is given by R1 upon R1 to R2 or something like this now I write closed loop frequency dependent gain is aol upon 1 plus aol times beta I collect the terms normally this a is only aol open loop a0 a0 is a open loop is that term correct to me a is much greater than 1 plus R2 by R1 is the term open loop gain of an open is very large compared to 1 plus R2 by R1 how much is aol aon or open loop will be 10 to power 4 or 10 to power 5 so our assumption at if that is so I can reduce some terms out of that if this is valid I will get closed loop gain at DC value is a0 upon 1 0 a2 upon this which is roughly equal to R1 plus R2 so is that point clear acl0 is 1 plus R2 by R1 this is the value you got it under what condition you got this open loop gain is much larger in deriving that open normal characteristics also we kept saying that that the oil is very large the oil is very large so only the current goes through input to the feedback network this is the condition which we are satisfying this please look at it what did I say I substitute a0 upon 1 plus a0 upon 1 plus R2 by R1 if a0 open loop gain of an opium is very very large DC open loop gain is very very large this term a0 upon this can be approximated by 1 plus R2 by is that divided by a0 so 1 by a0 is 0 okay so is that clear the closed loop DC gain if I am the Nikola yeah 0 frequency gain Nikola so I am the A term Nikoli thing 1 plus R2 by R1 Nikola than I wonder what conditions we actually got it a0 is much larger than this value is that correct so is now I am clear that using a feedback when I make the same assumption again I get the same gain values which I have got it so is that now clear that the feedback amplifier theory is more universal and the normal solutions which we get we assume that automatically and solve for it because we say opium is ideal input resistance is higher gains are very high this assumption is actually coming from this expressions is that correct that a0 is much larger therefore this is the value which will get is that clear there that is why I say that anytime you do feedback it is much more general case you do whatever otherwise is a specific case okay so this example which I want to show you is how do I got that time by assumption is now provable that my assumption is this which I can still get the same value okay now if I substitute this back into my closed loop this then this is equal to 1 ACL 0 upon 1 plus j omega upon this okay so what is the new bandwidth now what we substitute j1 plus R2 R1 is made up so j term I am a person is the bandwidth Kiwi a omega not into a0 by ACL 0 is the new bandwidth term aga okay now dominant pole I have to pass is that correct let us see what we are done if you have a open loop this was your amplifier Bode plot is that correct if I have a closed loop the ACL 0 is the DC gain okay and it has a pole of omega CL 0 which is here omega 0 time a0 by ACL 0 which is this quantity or you have pole show again however according to me I say onwards it will still follow the same path to omega t is that point clear what I said I could have said I saw be hosakta hiya I saw be hosakta hiya I saw be hosakta hiya isn't it but I say no from beyond this value it will always follow the same path what does that mean the closed loop gain komega t should be same as omega t of open loop is that point clear what is the proof I had to do if it follows like this means omega t of closed loop must be omega t of open loop then only it will follow the same path from the same point so now we will prove that both omega t are actually equal is that correct if they are equal but the gain is that correct at the cost what I improved my bandwidth from omega 0 to omega CL 0 and unity gain bandwidth remains same now we must prove that unity gain bandwidth of a closed loop system is same as unity gain bandwidth product for unity bandwidth product for closed loop if we prove it my all assumptions that all that happens or bandwidth but can you think now if something else this can also occur in the itna niche aia the bandwidth is now so you can go anywhere and you find if I reduce my gain I will improve my bandwidth is that point clear if I improve my decrease my gain I will improve my bandwidth or vice versa if I reduce my bandwidth I can improve my gains in what case we said gain can still be boosted without loss of this cost code what is the penalty we paid there additional power we have to give because all transistors must get saturation videos of code in a better saturated current sink okay I got the power dissipate cost there but what is the advantage we got we broke this line okay okay so I have done this proof that omega t is same for both open and closed loop ACL at omega let us say omega t dash is for closed loop and omega t is for the open loop so ACL at omega co-omega t is how much unity is it unity gain and a unity ACL 0 upon omega t dash upon omega 0 to this last expression may may not have you could say omega t k jugga omega k jugga omega t dash please remember this expression may may omega t t dash likhra hai aur kuch nahi kar rahe aur yaha bhi omega t dash ki value rafte ACL 0 ACL omega dash itna hai unity that is the definition I am saying so that is equal to one is this function if square term is much larger than this bracketed term is larger than one which will be because this term is we can think like this then ACL 0 will be omega t dash upon omega 0 a 0 upon a one chhodhaya if this term is larger than one then I knew one square terms ko under root kiya to ye ka ye hi regeya so I get omega t dash upon omega 0 into a 0 by is that clear is ko upar karo fitan so you will get this expression so what is omega t dash now ACL 0 into omega 0 into a 0 by ACL ACL at omega t dash is one substitute that equal to omega t dash then I say if the condition is such that this term is much larger than one then I square under root goes then I say ACL 0 is this is ko either leg also ACL 0 is this and if I say that omega t dash is omega 0 a 0 by ACL 0 into ACL 0 ACL 0 cancels which is a 0 into omega 0 which is your omega so your omega t dash so under what condition this was valid that the omega dash t by omega 0 a 0 ACL 0 is much larger than this if it is guaranteed then the omega t dash will be always equal to omega t is that correct so our assumption in our figure and which always normally because ACL will always be lower than that this condition will be always met so onwards then the gain will fall with the same point 20 dB per decade down to omega t is that correct so our if I have an amplifier I can now think close loop karne se aur kuch ne hoga uska omega t same rahega uska close loop gain nicha jayga bandwidth proportionately a 0 by ACL 0 se is that clear that is what essentially feedback amplifiers do okay is that clear this is the strongest point of feedback that I have now control over the fall of gains as per my what I am looking for is that clear yes of case assumption they came in a dominant pole bo nahi hai toh isko bo complicated aata hai nahi kar sakta yeh bhi nahi hai but the per feedback laga nahi se naturally is pole split ho jaata hai so dominant pole feedback me kareeb kareeb guaranteed hai kareeb kareeb I will say in some cases it does not some other day so what is the advantage of feedback it always splits the poles uska kaya advantage of a dominant defect or jay say dominant pole Mila gain bandwidth product co-op fix kareeb fit sakta gain come coro feedback comes as soon as bandwidth has been this is the theory of opamp frequency response is that correct so close loop gain R2 or both I also explained to you that under what conditions this expression where is that that the gain is valid equal to 1 plus R2 by R1 only and only if the open loop game is very very large which all opamps why is here is always very large the defam followed by gain stage gives very large gains is that correct so is that why we are using opamp because this is guaranteed for everything by using a that point here this is an essential feature of opamp circuits is the open loop opamp go kabolengi defam because it is just a differential amplifier because if there is no feedback then it then it will saturate because as soon as signal V0 exceeds VDD it will saturate to VDD or to nonlinearity will start building amplifier will not remain amplifier is that point here this is the important factor which you all should consider when you work on opamps why opamps why analog people do not want to use single ended amplifiers okay or cascade amplifiers but they want to use pans is that clear opamp is therefore extraordinarily great device which can do almost everything what you are really looking for is that clear and I will still m chip it is controllable whatever values you want you can achieve those values very very accurately is that clear we will quit this but it introduce karkajangi so next time we are going to start this area for which we are waiting signal conditioning of paila part all these circuits so far shown here what was there they were independent functions now I say multiple functions can be used multiplier was multiple functions otherwise we will start looking use of now opamp circuits okay first thing will use them as filters okay is that clear these are four kinds of filters we want to use just for this the low pass filter cut off FH both the why it is called FH while it is the highest frequency up to which gain or transfer function is 1 okay this is the lowest below which it is 0 rest is high okay if I have a band pass I have between some frequency f1 f2 again is the trans function is 1 and if in between there is function in 0 rest it is 1 then we say it is band reject okay band reject so we will use this all four can you think that actually only two should be sufficient to create other two you score score the manipulate Kia to get you know manipulate is that correct overlap Koro or Bahar Koro you can always get band pass and band reject so basic two filters concept low pass or high pass see you then