 Let us start what we have left last time. We are looking for amplifiers and as I said this course for the first time maybe I do not know. We will concentrate more on mass technology mass device based amplifiers, oscillators, opiants. Though as and when I need and as and when I feed I will give you something equivalence of bipolar as well. So not that bipolar is out I keep saying every time. So today we shall look into typically mass amplifiers. We have already solved one problem for mass but just to give you more details there can be three possibilities of an amplifier made in mass with mass devices. One is N-mass amplifier. So the word here is the transistor which gives you a GM transistor which gives you a GM is called the driver is that word correct the driver transistor is the one where input is received and which gives you GM time something at the output. So for example, take the first figure this transistor is called driver transistor. So if it is N-channel coin it is called N-mass amplifier. If it is a second part like it is a P-channel device that is called P-mass amplifier. If there are both P-channel and N-channel gates are connected this will be called CMOS. We have already done CMOS or I do not know whether you may be doing now in your digital course. So this is called CMOS amplifiers and then the first part of course is the driver. What kind of driver you have decides the name N-mass, P-mass or CMOS and there are other names also we give one is called based on loads. So you have an N-mass amplifier with first one is resistive load. You can see resistance is the load resistance. In the case of fourth the resistance is replaced by a transistor that is equivalent of that and this transistor acts like a resistor okay. So it is called and this transistor we shall see the latter. This is device is always in saturation and a saturation transistor resistance. We know why it is higher in saturation because the characteristics are very flat R0 is roughly resistance. So typically by changing the size of the resistor I can change the W by L of this transistor and hence the resistance. So the resistance of the load is varied by changing the size of the load transistors okay. If instead of N-channel device as a load I can also have a P-channel, this circle stand for sorry I should have made this is where I differentiate between N-channel and P-channel. A circle at the gate is a P-channel device. No circle at the gate is an N-channel device. So if you have a P-channel device and to make N-channel device this was my gate, this was my drain, this was my source. Can you tell me since mass transistors are identical from source to drain most cases there will be asymmetric in some case but otherwise source drain can be interchanged. So why did I call this as drain and not as source? Why this upper lead here was not called source because they are identical. I can call this drain and this source but I did not. In N-channel device which carriers make transport electrons is that correct? Since electrons come from source to drain the actual current will flow from where positive terminal to negative since the actual current is going down the electron current is going up that means source should be lower than the upper side because otherwise electrons cannot move up is that clear that is why they were given such names. So in same logic can you think what I did here which will be source here upper one because now holes also travel in the same directions as the normal current positive polarity down. So this is source this is drain and again gate is connected to drain so whenever gate is connected to drain whether in N-channel or P-channel transistor will enter saturation and as equivalent resistance can be evaluated by finding the W by E right the equation will do that and we know what is the resistance we are offering. So what exactly we did we replace this so called RD by equivalent resistors what is the advantage we get you are not done any technology you are not aware but just for the heck of it any idea why I am trying to replace resistances by transistors less area typical values of this resistance will be 40k okay 20 to 40k. Now if I substitute if I put an actual silicon it will take more than 20 transistor area for one resistor okay so to an integral circuit what is our importance as many transistors I can pack per unit area that is the density of the transistor and that is what I am increasing every year okay that is what Moore's law said pack pack so if I put resistor all that advantage I will lose so I am trying to particularly integrate circuits will normally never use RD but for our circuit course as much if I put RD does not really matter because it is external I can go to the breadboard and buy I will take some 40k resistors from the rack and put there okay we can test it but in real life when it goes to chip we never put any resistances on chip any day unless it is necessary in case of RF circuits. So is that correct there are two kinds of loads N channel saturation P channel there is third one which I forgot maybe I can show you here which is this load transistor is there is a line shown here means already transistor is on without VGS applied and it enhances as VGS increases where does it shut off at minus VT okay at minus VT so this is called depletion loads so there are three possibilities of course you can always say another one which is maybe more generic let us say only N channel loads and this I called VGG if VGG is equal to VDD then what is the status of this transistor VGG is equal to VDD that means I am connecting to power supply it is a saturation that is what we did if VGG is greater than VDD by 1 VT more than a VT then what will be status of this it is in linear mode it is in linear mode so what is the resistance in linear mode extremely large linear and R is very low so in some requirement if you want lower resistance what do we do we do not put it to VDD but put it to higher voltage and actually use linear circuit linear device there is that correct so that R can be further reduced in case as I say I am not saying where do you think R smaller will be required what are smaller means does current increase or decrease increase so if you have this is what digital people do if you have a capacity load here and this is to charge this what will be far is larger what is the time constant larger so switching frequency will be lower so if all is lower switching frequencies are higher so is the bandwidth connected to it is that correct to somewhere power and frequency are getting related did you get the point larger the current higher is the switch okay lower the currents that is low power lower speeds coming up is that correct so this is somewhere connected to us so our choice is not in our normal circuit we hardly care of power because we say a breadboard Pankha laga then get a coolant darling nothing can be done inside a chip like this so we have to be extremely worried about what power we actually dissipate per unit area on silicon chip and if you have seen my some graph microprocessor these days are consuming so huge a power it is like a rocket nozzle of a huge launcher rocket launchers so many thousand degrees of centigrade so one has to understand that why we are worried in power specifically on a chip is that clear but not in so much in like if you have seen a desktop computers you see it is a big box a huge fan is sitting backside cooling it all PCBs are cooled by huge fans is running at very high speed around 600 RPMs so what is the advantage there you have a space and you have a fan you have a power supply 220 volts no problems on a chip 1 1.2 volts supply or 2 1.5 volts supply of a battery inside that is that so you have no way to run a fan there that here and therefore on the this fact has to be understood because when you go ahead you should realize why sometimes we we are so catchy about power power power because there on that chip we are only worried about power as of now in our course I may not say so much but just to make you point why all these difference were shown because these are the kinds of amplifiers we may use in chip as well as on the breadboard that is called discreet and integrate discreet we may never use this kind why why do I use it I have a good register so this has to be understood that why I am showing you both all the time because in future you will be hardly working on a breadboard you will actually work on chips that correct so you must know what is the difference and I go on chip designs or chip circuits there is a just now I said different kinds of loads some of them I showed some others are also can be shown here this already can be replaced by a current source since you say odd is higher a current source the output resistance is higher or lower higher if it is a current source the output resistance is very high so I say if I put a current source here fixed current source which I can create from where a mirror you can see if I push a current here reference current this current I can adjust to any value is that clear this I can show you in a defam how do I actually bias but just take it equivalent current source can be transferred to an amplifier as a load by a simple current mirrors okay simple current mirrors and another thing which I just now said about the saturated loads how do I calculate resistance I connect gate and drain to VDD write the equations device if it is in saturation first assume in saturation let us say it is because VGS- VT is less than VDS VGS equal to VDD it means VGS- VT is always less than VDD if I use this delta ideas by delta VDS is essentially the one upon R of this this which is if I differentiate this so R is beta lambda VDD- VT is that clear so by what is beta contains beta- into W by L so changing the size I can change the resistance lambda is a technology parameter beta- mu C ox is a technology parameter VT is a device parameter technology parameter so all that you can do is change W by L so different sizes of transistors will be different resistances is that okay is that okay so this is how the loads are created in mass amplifiers last people is it clear loads can be created out of transistors themselves okay and what is the advantage I say the area of this will be 100th of the area of the safe for equivalent value of resistances okay and that is what we are really looking for how to calculate resistance just delta I by delta V and inverse of that is your R so nothing great happens just write equation differentiate and get your resistance is that okay of course this equation which I read you need not ready just want to show you that what is it you just I mean I just I need not have written all I would have directly written here but I thought you should know why how do I calculate this is only to show you methods this formula is not relevant this is the method which I am showing how to evaluate R's in a transistor okay okay as something you can see okay I thanks for suggesting if you leave lambda per se lambda is 0 for example this equation is quadrate I mean it is a parabolic equation okay now if you see a IV characteristics of a diode how does it look it is something like this and if I put a parabolic equation it is something like this so it closely replicates the diode characteristics it is actually opposite depends on the parabolicity it may be much higher or much lower than that but it is normal cases it would be almost as a following diode equivalence of that is why it was called diode connected is that clear please remember this this is only valid when Vgs- Vt is small enough okay if it is too high then it will go like this then it may not actually look like diode characters is that okay why diode connected okay okay there are other ways of defining the amplifiers which is basically circuit based and these are an amplifier can we have a common source terminal what do we mean by common between input and output whichever terminal is common we say it is a common that so if source is common between input side and a drain side output side then we say common source the other possibility it may be a common gate okay common gate is essentially what means the gate is common between input and output so where will be input at the source side and where will your output at the drain side gate is common between input and output and the last is common drain okay now this common drain is very important circuit for every one of us okay and it is also called source follower in BJT what it will be called equivalently source in BJT is what emitter so it is called emitter follower emitter follower is identical at least in nature has common source amplifier common drain amplifier which is called source follower will do some analysis for common source please read the Milma this over set us with book for bipolar circuits equivalent okay okay so here is the first amplifier which we did partly last time but repeat again you have a common source please remember now here is gate here is drain and here is a source okay now you have a source which is common to input and common to output therefore it is given a name common source amplifier now the only difference between this and the other one is this source has a source resistance RS source has a source resistance RS which is shunted by a source capacitance CS is that okay so at DC what will happen to this CC CS at DC means frequency 0 is the impedance lower or higher higher open circuit so if for a DC as if capacitance does not exist CS does not exist is that okay one upon Omega C if Omega is 0 the impedance offered is infinite this CS will offer infinite impedance against RS means open circuit to RS so RS is the only resistance so we say a DC capacitances are this capacitance is open but there are two more capacitances I have put here their values also should be such that at DC values they should act like an open circuit what does that mean that the DC voltages here which I am going to apply should not get connected to power supply they should be blocked to power supply they may modulate other side that is that correct this DC value here should not get directly connected to input source so what should we do there put a capacitance which is called series capacitance called coupling capacitance what does it decouple in fact it should be called actually decoupling capacitance what it decouples the AC source from DC source is that clear AC source from DC source similarly from the output this is VC VDD I want to remove this I do not want to couple this VDD which is the power supply DC power supply to the AC output which is V0 okay so is that point clear to you at this point what will be the signal DC plus AC and I do not want DC to be output I want to only see AC amplifiers okay so what should I do I should block DC this easy to is again the same blocking capacitor or a decoupling capacitor between power supply and the AC signal IC outputs so what should be their value should be 1 upon Omega C I want to be very very high oh sorry 1 upon Omega C for DC it has to be it will always be open circuit but for a lower frequencies or moderate frequencies what should you act whenever for AC signal it should get in so what should be the impedance offered by these practically 0 practically 0 so what should be the values of them because Omega are not going to be very high we are you going to use smaller frequencies as of now so C should be very high so typically some micro farad point 47 micro farads or a micro farads could be the decoupling capacitors or this is that some experiment you may be doing you can think why we are putting a larger value of C in the series connections same way for this at all the AC frequencies it should it should not it should act as open circuit so what should be the value here please remember I repeat for DC it should be open but for AC it should actually we are like a short circuit to RS I do not want RS to be part of my AC circuit so what should be the value of CS please think of it 1 upon Omega C okay I want at a given frequency of operation this should act like a short circuit C should be again high enough because then only it will act like a and which capacitors normally you put there have you seen there say electrolyte large value capacitors which we put across this device is that clear so this sees purpose is at AC frequencies it will short what RS but for DC frequencies it itself will open up and only RS will appear in the DC network is that correct this is how a amplifier biasing will be done now bias is done as usual like a fixed bias as shown here RG1 RG2 to VDD and to the gate and equivalent resistance is how much perth from the venance please remember here it is very easy no gate current divider is RG2 upon RG1 plus RG2 times VDD that is your feminist bias what you are want VGS and how much is resistant will offer the parallel combination on that value I have kept here as RG is that correct RG is nothing but RG1 parallel RG2 is that correct so that is that resistance is not shorted or opened by capacity said it is bias it must sit here if it if RG does not exist what will not exist the DC bias for this will not exist is that correct DC bias will not exist this is the most important part in the actual circuit when I draw so having shown you a common source amplifier what do I what is the my ultimate aim I must lie I will like to find V0 divided by okay I think I we are making mistake so we call it RS dash here because you know this RS is a source resistance so we start calling IR series or our source or because our source is sitting here so that name is should not get confused okay is that okay maybe they call it in the book I think if I am not wrong they are calling R signal okay R signal generator okay R signal is what they are talking about okay fine if I have this circuit then I can draw equivalent of that at the input side you have a input source V in you have a series resistance of the signal source RS dash you have the RG larger resistance of bias network RG and they have a gate here is that okay this is if you are just seen the circuit you can see this is the input side okay from gate okay this is my source from gate to source there is no connection as far as AC is concerned there is no connection between gate to source please look at the circuit again if you just a minute between gate to source there is an insulator so there is no connection between which is in contrast to what bipolar if this while your base current how much current would probably be flowing here beta plus 1 IB so there is a much easier case in the mass because there is nothing actually comes out from the other side okay so ideal device in some sense now for this other side at the output equivalent of that side you can see what is the output model we have said whatever is the VGS signal appearing here at the gate with reference to source which is grounded GM times VGS is the current source GM times VGS is the current source shunted by what are 0 which is the output resistance of the transistor okay output resistance of the transistor so this is actually equivalent transistor source GM VGS parallel R0 what is this already RD is the drain resistance which we have put there and what is this additional RL I am showing you here this is external load which I do not know this may be coming from external side so that is my actual load which may be RL okay so if I show you now it is R0 parallel RD parallel RL now we also need to find apart from gains two more branches what is the problem I said in connecting the two circuits in bipolar the output impedance and the input impedance are sometimes equal or bad okay one is much lower than the other so it is called loading effect now I want to see how much is the R in and how much is the R out of my circuit so that can I cascade what do in my cascade the output of the first stage is given to the input of the next stage is that kind is called cascade so can I cascade another word which I am going to do after this is cascode what could it be different from what is the word I use earlier also I once said cascode amplifiers have one advantage what does what is that disadvantage it gives over the normal amplifiers I said you something not we are not talk bandwidth so you are we will not remember it immediately we said gain bandwidth product of an amplifier is constant okay this is called figure of if you increase gain bandwidth goes on we will see the expression that it cascode what it does it breaks this this limit okay that is exactly is what is it doing the gain can be enhanced without losing bandwidth and that is exactly what all these years we were looking for I do not want to lose my gain at the cost of increasing the band we are retaining bandwidth at higher gains this is exactly what cascode does okay so we will go to cascode amplifier later but let us right now think that this word should not be confused any day okay so what is Vgs value if you look at this is a I hope that you have done basic circuit course reasonably good this is simple mesh equation this voltage equal to this plus this into V in the potential divider so it is Rg upon Rg plus RS dash time V in is your Vgs and which is the case when it is equal to V in when Rg is much larger than RS dash you can say okay all that input signal is at the available at the as Vgs but this condition must be met is that correct this condition must be met if not you should use whatever value Rg RS dash as you use there it is something like selling since there will be something in denominator larger than numerator it will be less than Vgs will be less than V in but it may be 0.9999 V in so why should I worry about 0.59 I say V in okay so the idea is to know where to when these are all things which I am trying to show you in real circuits what do I assume in this course since you are doing analysis please write everything okay I am not giving you values so in analysis you should show what it is but in real life I do not use that I say okay this is past okay this has to be understood why we quickly do things many much earlier than others not because we cannot solve this we just know how anyway that same okay okay at the output side what is the current only current source is GM Vgs but the direction of current is want from drain towards source is that correct GM Vgs but the output is always shown as plus minus so what should be the sign should be given opposite because GM Vgs will actually go from like this but voltages are measured like this so we say V0 is minus this minus is very important what does it mean in circuits minus means invert or a phase of 180 degree phase of 180 degree so if your input signal is in one phase the output is always 180 degree out of phase from the input this is the most important characteristics of a transistor that the output is 180 degree out of phase from the input which is always 180 degree okay but if you manipulate some external circuit this 180 plus or minus you can do additionally and that is exactly what we do in our real life to change this 180 plus minus something to get the phase I want but transistor per say will always give you 180 it cannot change it to 179 or 185 this is fixed 180 minus sign minus is coming from where J square is that correct J is 90 J square is 180 is that correct this J vector J is 90 so J square is 180 J into J is that clear this is coming from very simple circuit theory so this is R0 all these resistances look to be in parallel R0 Rd all everyone is in same across this so GM times R0 Rd RL into V in is the output V0 by V in is the voltage gain which is minus GM times R0 parallel Rd plus however this was under an assumption that Vgs is V in is that correct it was an under assumption that Vg is V in otherwise what would have come here Vgs and then what should I replace with Rg multiply Rg upon Rg plus RS to make it same value as the gain is that correct if I would have said this is not V in but then Vgs then I would have replaced this Vgs by Rg upon Rg plus RS times V in and then another factor I would have multiplied here is Rg upon Rg plus RS dash if V in is not Vgs is that okay this is as simple as that okay so now you got the gate MOS gain so what it depends on R0 is not in your hand to some extent you are it is in your hand what is it how does it how R0 gets controlled by no lambda is fixed okay so ID 1 upon lambda ID is IDS is R0 so the ideas which is ideas is coming from where the bias network ideas DC bias current is decided by the bias network Vgs minus VT now that value is giving you capital IDS is that correct DC value and that may decide what the GM 2 pi 2 beta ideas is GM so GM actually decides how it is under root please remember GM is proportional to root of ideas R0 is inversely proportional to ideas so by adjusting ideas I can say some way GM times R0 can be modified however RD are there are externals okay so that normally what will happen RD and RL will be smaller much smaller compared to R0 how much will be R0 typically mega ohms how much will be RD RL few hundred tens of kilo ohms in normal case RD parallel RL will be smaller than R0 so you can even neglect R0 is that correct neglect R0 but otherwise as a theory you can retain it so by changing the bias current what does bias current means in essentially biasing point your operating points if you vary your gains will also correspondingly is that okay that is what exactly we are trying to show is that okay everyone has seen okay there are two more parameters of interest as I say for input impedance and an output impedance okay before we show it I will give detail little how to calculate for any network I think you must have done it if you are given a network you can always find the output impedance and an input input how do I calculate these impedances short the outputs okay and from the input side apply an input source with the input resistance RS of you can even make 0 whatever current and comes out of this BX by RX is input input by same logic short circuit all voltage sources at the input side is that clear open all current sources independent current so please remember this is independent actual sources short the voltage source open the current source apply VX at the output find the current entering from the that node VX by X is R0 so the method is identical to the circuit theory which we are done sometimes by observation we will use it otherwise you can always do this for any larger circuit is that correct this time this is so obvious so I did not do analysis so I just want to show you that this is what I am going normally RL is not considered to be part of output resistance why it is external I do not know what people will give me okay so normally R0 is taken across RD even some people believe actually you should measure intrinsic R0 also that is without even the drain resistance is that correct so it will be always R0 in most cases so okay if I do that then one can see from the circuit sorry there is no actual as seen from here this is the only resistance it is seen so whatever current entering V by is only RG whatever is voltage here divided by R as RG is the current so this divided by that current is same as RG so input resistance is RG input resistance is RG very high how much it will be any mega ohms because RG1 and RG2 also are in mega ohms so because of them we will always have RG very very high is that correct RG very very high so input resistance of a mass amplifier is normally decided by the bias network and typically is the order of few mega ohms is that correct typically why I am saying typically because value you choose RG and RG2 I do not know if they smaller then it will be smaller as well okay what will be the output resistance seen again from here the same logic short here then the VGS is only coming from here so what is the resistance will be offered here you open this RD parallel R0 forget about RL if you see a resistance coming from the outside and if I put a voltage source and find current V by parallel this is the current this I am opening okay this I am opening so if I put a V0 and two parts that means parallel so V0 by is the parallel combination of R0 parallel RD so the output resistance is R0 parallel RD typically R0 is of the order of mega ohms RD is of the order of few kilo ohms tens of 20 kilo ohms so what is the actual output resistance can be talked as RD so output resistance of a mass amplifier in a common source technique is just the drain resistance which I apply and that drain resistance decide what apart from it also please remember drain resistance also decide the operating point is that correct VDD by RD this is giving the maximum current the load line is decided by the RD and the RD is essentially the output resistance if you want accuracy you say RD parallel R0 yes so what is is that clear since I did not do actual calculations I probably messed up but otherwise you can think that by observation that is correct so if RG is not RS dash I repeat RG upon RG is this is the gain of a mass common source amplifier is that okay this is most important in the case of is this amplifier will be have a larger gain or a smaller gain okay the three amplifiers I told you which are the three I said common source common drain and common gate what do you believe which will have highest voltage gains gate is common to source and drain whatever is source will pass to drain that is what source drain current is same now so it is a unity roughly okay it depends on the load values at the input and output I same both sides is that correct so it will be dependent on the values of RD and RS you are actually missed whereas there is no GM factor there is no booster there in the case of source follower what does that follower means what so follows what if it is a source follower what does that mean the output voltage follows input voltage as it is that means great gain it unity so common source amplifiers is the only amplifier which will you reasonable amount of voltage gains okay is that point clear so they had to understand but then why the other two are so much talked about they must have something additional with them at the cost of what gains okay I do not have gain but I must be giving you something better which may still require you to use otherwise I will not study common gate and common drain why should I if I have no gain no additional advantage so why should I use so there we will like to see if I change over from common to other do I achieve something else at the cost of something else is that clear to you this is what we are looking for other two amplifiers there is another common source amplifier which is very popular okay another before we quit here this equation you can see the R0 is equal to how much 1 upon ? ID is that correct ? is what parameter we call technology this is decided by the technology GM under root to beta dash what is beta dash Mu C ox again technology W bile for external circuit like I say no control so even that is fixed by someone else all that I can vary the ideas is that correct so if there is a change in this so-called technology parameters because of what environment let us say basically temperature even humidity changes but at least basically temperature if those value will change what will happen to these gain they will also change with the particularly with the temperature is that correct so if you have a design an amplifier in a system which should have a gain of say 100 or gain of 40 it may either go down depend on the temperature which side it goes or may go up either way and then the next circuit will see different input voltage than what you actually plant okay and they itself they also will get corresponding changes so at the end of the day cascading circuit may not show the actual gain it may be either much smaller or much higher in which it may saturate and it may not even amplify is that correct so I am clearly worried that the gain function should not be a very strong function of technology if that happens then I say I am safe because then I know what gain I am using is that correct but let us see whether this the next circuit does that and at what cost now if you see a what is the difference between the last circuit and this circuit no everything else is same except the source resistance is not bypassed by CS is that clear source resistance is now available to you both in DC as well as in AC is that correct is it okay source resistance is not bypassed by any capacitance so this is called source amplifier with source degeneration okay now can you see RS for an AC equivalent circuit we will show draw it this will be now part of the input as well as part of the output earlier IRS was getting bypassed so no worries now you have a problem that RS is common to both sides okay and let us say if I draw an equivalent circuit of this where does this RS appear and why should it then say the gain will be independent roughly independent of technology parameters is that clear why I am worried about technology parameters because the temperature is a strong which will increase do what you okay even if you put a fan why fans cannot cool very quickly all that which law does not allow that to happen Newton's law of cooling is proportional to the difference of temperature nothing much I can accelerate it by removing the heat around but the actual cooling will go by law of cooling is that kind proportion to the difference you are looking is that clear do you know still recollect your law of Newton's law of cooling that their rate of cooling is proportion to the difference of the temperatures okay so initially it will faster and then it starts cooling as you come closer to room temperature 20 or degree to 23 degree you may take 2 hours because the difference is very small now okay so the worries with us is that ambient but temperature you may remove but the actual cooling will be slow is that correct now that means there will be a variation of gains irrespective whether you like you do not like but I will like I would not like that so I should true something which this circuit probably does the circuit analysis almost identical the only difference do you see something difference in equivalent I had drawn now this RS I had drawn from source to the ground but remember this already and RL are not going to source now is that clear where are they going in the actual circuit already going to the ground RL going to the ground but not going to the source because source is not grounded source has an RS going to the ground so equivalent have you seen this please look at this circuit this RD is grounded this RL is grounded this is of course your V0 and this is your source which is not grounded what is it so I should not connect RD parallel R0 parallel RL now is that point clear the difference what is going to get is this that RD RL R parallel but not parallel to R0 is that correct because they are to grounded R0 is only going to source because that is how I define R0 for the sake of it maybe we can say lambda is very small first case we may say RG is much larger than RS sorry this call RS dash and we say R0 first simplicity this is your V in with reference to the ground okay please remember this V in or this value is not same as VGS please look at the circuit again this V in because this is grounded but this is not grounded is that correct is that clear V in is grounded but this terminal is not going to the source is not going to ground but to the source so I now write a Kirchhoff law okay I start from here V in and I calculate the voltages how it falls so first is V in is equal to VGS this is V in so VGS plus what is the current flowing in GM RS GM times VGS is the current flowing in the RS is that okay this is VGS plus this drop I want to know what is the V into the ground this voltage plus this voltage VGS plus IRS whatever flowing through this this is my V in is that okay how much is VGS I will find out so this is VGS plus GM times VGS into RS is equal to V in is that okay I repeat this voltage plus this voltage is the ground to this voltage is that okay divider this plus this is this so GM VGS into RS plus VGS which is equal to VGS time 1 plus GM RS so how much is VGS now VGS is V in upon 1 plus GM RS is your VGS is that okay just solve this equation so VGS is equal to V in upon 1 plus GM RS is that correct if RS is 0 what is the condition we get V in is VGS that is what we did if RS is 0 now I say RS present so all that it has reduced something this is called degeneration the word is degeneration input is degenerated now by 1 plus GM RS term okay compared to earlier but what is the advantage of doing all this you must be doing something out of it if this is open circuit at same current passes through RS so GM VGS into RS is the drop across RS plus VGS must be equal to V in solve it and you get this expression okay simple circuit analysis how much is V 0 from here from our circuit analysis this current is actually putting GM VGS is the only current source which is also passing through RD or if R0 is infinite again the idea is R0 infinite so I write GM VGS RD parallel RS is V 0 then I replace V 0 by how much V in upon 1 plus GM RS so I get AV is equal to V 0 by V in minus GM upon 1 plus GM RS times RD parallel or L okay I repeat if RS GM RS is much larger than 1 what is GM RS is larger it may be GM is typical of the order of how much 10 to power minus 3 or kind RS may be 10 kilo ohms so it will be around 10 okay so you may say if GM RS is larger than 1 GM upon 1 plus GM RS is 1 and then you get sorry then GM RS GM cancels then I get RD parallel divided by RS can you now see the voltage gain does not have a term which is technology related is that okay the voltage gain of a common source amplifier with source not bypassed or degenerated source source this then we get a gain which is ratio of load to the source resistance is that correct none of these two terms are anyway technology dependent what is the criteria only to do this GM RS should be larger than 1 or RS should be larger than 1 upon GM that is what exactly how to make a RS value so RS is should be greater than 1 upon GM GM is decided by the currents you are already passing in the DC currents so you find out the choice of so now it said you must understand there is a problem in calculate and a bias network you need an RS value and to get RS greater than GM also you are another any inequality going on there solve together such that both conditions are simultaneously satisfied is that clear what I said in calculation of a bias point operating point RS is appearing there also but the condition I am now saying and GM is you are using so condition I am saying use GM 1 upon GM should be smaller than RS okay so you make some choice of RS first get that bias point using this and reach it whether that satisfies your proper by saturation or active mode conditions and then also see that this value is equally satisfied is that correct if that does not happen then you may have a problem of instability this does not happen normally we know how much to keep so that value 10 times is good enough for all cases okay so what is the advantage of this again RDs will be order of RD RL will be order of few tens of kilo ohms RS will be of the same order so what will be typical gains will be 2345 typically the gains will be 2345 even 1 if they are equal okay so what does that mean we have achieved something at what cost at the cost of voltage gain I say okay I am I am stabilizing it okay I am getting now an amplifier which is temperature independent okay why it is temperature independent because even these are temperature dependent term R has a positive temperature coefficient but since they are in the same assuming same material carbon resistor the temperature coefficient of R and numerator denominator may almost similar and therefore they are independent of temperatures is that clear so what is the advantage I got if I want a very stable gain but not very large gains then I should use common source with degenerated source resistance this is how so at very large cost I paid I would have got a gain of 100 and now I am getting 5 okay but I am now saying that is exactly what circuit people want okay they want to know is it stable is it fixed okay is that okay so something so wherever the first stage of amplifier or wherever you feel that it is more prone to ambience there you first use a low gain amplifier which is common source degenerated degenerated source then ahead you may use normal amplifier for next boost which probably is away from it so that it is not that much temperature dependent is that clear so this particularly happens in instrumentation or measurement systems the sensor part is in an environment which may be bad okay so there when I boost the signal I must have a constant gain signal is that correct once I come out of that then I can amplify because I have come out of the system is that clear so the whenever there is an instrumentation sensor measurements the first stage should be this kind because it should give you almost stable outputs is that correct and then as you get the signal you have a 1 microvolt signal and you say now you have a 5 microvolt fixed signal the next stage may be a normal common source amplifier is that okay so I am not really worried about large boost but I want to get out of this temperature variation and have a good signal for me this is what essentially so 2 amplifiers we saw one is common source the other is common source with source degeneration here is some quick problems I will not solve it fully I just want to show you how do it typically you are given a value of this from this hour sorry said that means books so you can see the values Rg1 Rg2 have been given like this VDD is 5 volt okay in analog circuit in my first slide I show it is a dual rail supply what does that mean I said you may have a power supply at the top VDD side plus and at the source side minus but the actual voltage will be VDD-VSS and since it is minus actually it sums so if a 5 volt supply I want to use total so what should we VDD VSS if they are to be equal to end 2.5 plus minus 2.5 down is that it has some advantage maybe in the quiz I will ask you why I should have 5 to 0 is not as preferred at 2.5 to minus 2.5 there must be some advantage no difference is still 5 so why I want 2.5 minus 2.5 there must be something more to it look at it okay right now the value is chosen in from said us meet of these RD parallel RL is given to you 7k source degenerated resistance is half K source resistance signal source resistance is 0 beta n is 2 millivolt per lambda is taken 0 threshold is 0.8 okay what does this beta n means it includes W by L it includes W by L beta n dash W by L together I have given you a value of 2 milliampere volt square. So what is the RG value in our equivalent circuit RG1 parallel RG2 so whatever a parallel combination from there the DC analysis if I do which I did VGSQ is given 1.5 volt VDSQ is 6.24 this is I have calculated you can I am directly in your DC points VGSQ is 1.5 VDSQ is 6.25 ID current flowing is given to us how much somewhere given to us 0.5 okay if not given assume it now okay 0.5 million then GM is equal to 2 beta n IDSQ which is is that okay if these values are given for given RG1 IDSQ I am given as 0.5 milliamps then GM can be calculated as 1.4 milliampere volt is that correct please remember you must write correct units in circuits please do not say marks have been directed for circuit only units oh it is correct sir sir to have but correct name I make this but marks may not so please remember every place your units must be correct okay R0 is given lambda 0 to infinite so what is the gain for this amplifier GM upon 1 plus GM RS RD parallel RL substitute all values the gain is 5.76 how much gain I told you it will be around 45 2345 so how much gain I got 5.76 or around 6 you can say if you use 1 plus GM RS is GM RS then RD by RS so look at the value now since GM RS is not very large compared to the other ones very large compared to one without if I use this expression then I get gain of 14 whereas if I use exactly what is the why I am saying you this essentially means this identity is not correct in the value given to you here is that clear if GM RS is not much larger than one you may land in such situation you may get only you will get much larger gains which actually does not exist okay so in calculations do not a priori believe that GM RS is larger than one and remove it okay so I intentionally chose this value in the book such that I will show you that I keep saying it is RD by RS RD by RS but do not believe me every time because the value is given will decide whether but if you do not do this and only do this it will take automatically correct value irrespective whether it is lower or smaller whatever it is is that correct so this example was just to make a warning signal to you that do not go by just my statements because unless you verify this okay since we are as I say equivalent common emitter source amplifier common emitter amplifier okay all this the amplifier shown last time also what is the what things I left in all equivalent circuits capacitances why I left it I say I am working at low frequencies and all of them are open or short as far as the requirements I am not considering them but where will I consider them the high bandwidths I say up to where this will work at that time I will use full this and that is our next frequency response of an amplifier is our next topic so I am avoiding that just here to show that because that anyway is going to come here also CC 1 CC 2 C are so chosen that it acts like a short circuit for AC short circuit for AC short circuit for AC otherwise for DC they are all open circuit I repeat CC 1 CC 2 and C are so chosen that they act like a short circuit for AC signals and open circuits for DC signals to you see this is identical to mass there is if you replace this transistor by a mass transistor how does it differ there is some difference is happening what is the difference happening this terminal there was how much gave to source open is that correct now there is a R pie is that clear to you the first thing you are now seeing that base is getting connected to emitter through our pie which in the case of gate is that clear this is the major difference you are getting from bipolar to us are you must to buy for now you can see sorry I forgot this I had still draw RE for AC so I what did I do I actually put a short across the show RE shorted this he will bypass by past DC at the time but AC it will short the RE itself in the circuit layer this is our pie this is GM VBE drop across our pie this is R0 this is your RC this is your external load as what shown earlier what is this RBB da RBB shown here what is this resistance RBB the bias network RB1 parallel RB2 RBB is RB1 parallel RB2 same fixed bias we are done okay bias network RB1 is a parallel network RBB since in this case RE is bypass this is essentially grounded so one can say this whole is one connection everywhere which one oh sorry I am very sorry I told you know actually it is reveal so I told you that actual there is a RE small resistance then emitter resistance is obvious so emitter is not exactly the lead output there is a small RE resistance there parasitic so we gave a name emitter dash inside to separate it from actual emitter need outside I repeat what I said this was a emitter dash and this was actually in the lead outside since this terminal was given earlier books by Shockley the person who actually get equivalent he called it emitter dash so base to emitter dash is VB dash it can be called V pie it can be called any this voltage across this is called VB dash because this terminal was given a name E dash emitter dash why because there is a RE small capital RS or what is RES I said small to the small resistance associated the emitter region itself in actual right now if not given all paracetics are 0 RC 0 this is 0 RS is 0 is that okay if given then use that huh so this is ground now because emitter is bypass now see will bypass the for AC so this is ground right now is that okay this point is ground now I already said that low frequency this is 0 this is 0 this is infinite please remember as CJ E plus C this should be infinite I want to open that across any capacitance the impedance should be very high okay series they should be short and in parallel they should be open is that okay so these are the conditions if I use the solutions now are in is being by II which is you can see from equivalent circuit are in is taken this resistance and in this resistance in series R pipe RB parallel RI RB is RB 1 parallel this how much is RI can you see from here between this how much is the resistance in R pie R pie so actual RI is only R pie is that correct but that will be parallel by RB be so the actual resistance is RB parallel R pie and if RB is larger than what will be RB parallel R pie if RB is larger than R pie then it is only R pie is that correct so actual value you may see in your calculation whether it is coming close to our buy but in problem solving do not make any conclusions just parallel and see what value is that correct okay so how much is the voltage across the R pie this value VB dash is R in upon R in plus oh this for now RS is okay because there is no source there this is signal source resistance so R in upon please take it this value please remember if this is R in R in upon R in plus RS time V in is your VB dash is that okay this voltage is this this plus this into V in as simple divider so that is the expression I use so typically it can be written as R in upon R in RS V in or V in RB parallel R pie RS RB parallel R pie write everything and if it is less than R pie upon R that is V in if RB is also larger so what is the purpose of doing this why I want to know this VB dash why I want this voltage because at the output the current source is GM times that value so I want to know what is the value which I am receiving there okay is that okay is that okay so I calculate VB dash as just now I said and now so what is VB dash it is just the divider between R pie and the source resistance yeah may harbour I say a lot of expression is I am giving you the real values quick check how much is so you can see from how much is the actual value of this we keep giving you all hints is because when you solve you know many time long expression forget some term somewhere here there and then your value becomes very odd so you should be able to know whether this value which you are getting is reasonable because it should be close to that value accuracy may be little more 1.001 is close to one so you are you getting close to one or you get 5 then why are you doing all that there is some bigger mistake you did that correct so yet these numbers which I am showing you is for this purpose roughly you should know where you are okay okay last part with pay finish now so now the G output side you are a current source of course this is also grounded because emitter is grounded GM VB dash is the current source R0 RC RL are the three resistances in parallel output is V0 I0 is the current here entering like this since I0 is entering like this for the parallel combinations are not here it should be shown here so I0 is minus GM VB dash so what is V0 what is V0 I0 times parallel combination of the three resistance is that okay three resistance IR is the output okay I0 R0 parallel RC however I0 is minus GM VB dash is minus GM R0 parallel RC plus RL VB dash so one gain at the VB dash itself you can say call AB0 dash V0 upon VB dash is minus GM R0 RC parallel RL but I am interested in what gain not with VB dash what gain I am interested in V0 by VN so I write AV is equal to V0 by VN just multiply by R pi upon R pi plus RS VN is VB dash so V0 by GM R pi upon R pi plus RS R0 parallel RC parallel what is GM R pi in bipolar's beta let us call this whole resistance parallel combination is RL dash then this gain is minus beta RL dash upon R pi plus RS more specific RII plus RS okay is this gain technology dependent yes beta is a temperature dependent term okay beta is so if I do not bypass RE what will be gain now as we did last assuming that that resistance takes care of 1 plus GM is higher than GM 1 then what will be the value will get the load resistance divided by the source I mean the emitter resistance is that correct let us say R0 is very high which will be so RC parallel RL divided by RE will be the gain in case emitter resistance is not bypassed and again it will be independent of temperature or external environment is that correct that will be called what degenerated emitter resistor okay no bypass but the gain will fall from how much huge number 100s to 5 6 8 10 kind of here beta is how much 100 plus if the ratio is higher or even equal the gain is please remember even if load is equal to RI plus RS beta being 100 the gain you are getting is 100 if I use by unbiased I get 100 but if I put bypass unbiased sorry emitter resistor it may be just RC RL divided parallel divided by RE that is it but it is independent of beta is that okay this is common emitter amplifier with emitter degenerated emitters okay earlier get a degenerated source okay what is degenerate word reduce okay you reduced it okay the effect was correspondingly reduced on the gain okay and therefore you degenerate by advantage you see run okay the final value I am sorry R0 is R out you see from here please remember external resistances are not you have taken care in calculation or output resist why because I do not know really okay R0 parallel RCs are the only two resistances I have which is the output resistance is that clear is that okay so three parameters you must tell me every time you do an amplifier analysis which are the three the voltage gain the input resistance and the output all three parameters must be evaluated before we say amplifier design is our analysis okay is that okay now so today we will stop here next time we will continue with it little more on mass other two masses