 Okay let us start we want to do our major work in this course is to do some circuits which amplify okay so we are looking for amplifiers and as I said last time both technologies are possible using a BJT or using a MOSFET we will more concentrate on MOSFET but as pedagogy or as the word things went through bipolar came first so let us give him some credit to it and then go over to MOSFET so I will not actually show you a bipolar amplifier I will say okay if I change from MOSFET equivalent circuit to this equivalent circuit you can as well evaluated for bipolar but before that let me quickly see the biasing assembly in either cases so let us say I will do you have done it but I quickly want to go through for a BJT we already discussed earlier that if I have a circuit which is to act as an amplifier sorry sorry sorry I must somehow see that this transistor remains in active mode so if I draw a characteristics IC versus VCE which is nothing but V0 and if I write this equation VCC is equal to IC RC plus VCE so this is a straight line this is a straight line so if I draw this straight line on the same characteristics maybe somewhere something like this okay this value is of course VCC and VCC by RC this is called what load line so if I have fixed the transistor for a given capital IB that is the base current over which small IB is going to be imposed let us say this is my IB which is IB dash we may call then the device has fixed bias current of base which is IB dash which I can fix that is what I want to fix this is the characteristic which on which I want to operate for the transistor IB dash is the base current IB dash is the base current which is DC current I want to fix the DC current for the transistor which is on this characteristic and if I then connect RC on this then I have a load line like this which intersects this IC VC characteristics at this point and we know this point is called the operating point or the quiescent point and for this value the DC value of VCE we call VCQ and this current we call ICQ these are the DC currents DC current and voltages at which the device is going to operate for given load of RC is that clear this is called the operating point or the quiescent point this is called this is what we want to bias I want to fix for a given RC given IB dash so that I will have now these values so what is the typical criteria VCQ should have I can have a different load lines for example I may have a load line something like this or I may have a load line something like this what does it changes the intersection of load line with the transistor characteristics that means it can be here or it can be here sorry here or it can be somewhere else depends on RC I choose essentially what is going to change the value of VCQ as well as ICQ if the device has to remain in active mode what should be roughly where VC should be always it should not be close to 0 or small value because there the device will enter on this side saturation if I have a very small base current then the device be beyond may not be 0.6 volt and therefore direct itself will not turn on is that correct so I cannot have RC such a value that I have a characteristic somewhere like this then there is no base current available there is no transistor action so choice of RC must be such that this VCQ should be roughly between I should not say exactly but half roughly half of the VCC so if you keep around half the VCC it need not be exactly at VCC VCC by 2 but around VCC by 2 if you get your VCQ you are ensuring for any value of IB you are always you will always remain in active mode so choice of VCQ should be such that you are always in active mode and typical value should be such that it is roughly half and do not see if it is 12 volt or 10 volt supply is it 4 volt yeah 4 volt is still with well within the range of active mode okay so do not insist that it has to be 6 volt but it can be as close as possible to 50% VCC it will be ideal for you because it will guarantee your device to remain in active mode okay the reason we are not very much worried is too much because change in IB which I shown here is how much it will be small IB few millivolts so that variation in this IB value across this is so small that for the same RC value change in IC values will be also very small larger because there is a beta gain going on so it will be larger than change in IB but still it will be small enough but that also should remain within active mode of the transistor transfer should not come out of this and therefore once I said that I see I cannot have large input signal because it may then swing device to the saturation or cut off side okay. So since we normally will never use larger signals this course we may not actually work on large signal amplifier there are issues there are requirements but as of now let us a small signal we must therefore assume that by same logic we will show you later that it must transistor if it has to remain in the saturation which is similar characteristics it should be half the we please remember it is not half is a sangrass and number okay it can be 0.6 it can be 0.4 of EDD or 0.35 of EDD but it should be away from the linear side and it should be away from the cut off side as long as device remains in active mode your amplification is guaranteed okay. So these are the issues when I decide what value of RC I should choose or given an RC what VCC if I given where I am going to be to find the operating point this is what we are essentially saying when I say I want to know whether I am in active mode is that clear this is what essentially we say evaluate VCEQ and ICQ once this DC values are fixed we have done small signal analysis what do we do then we can calculate GMR the everything value required once I know my DC values I have my everything available on my platter to evaluate the small signal values and therefore small signal gains and small signal whatever we want to is that clear so the idea behind fixing this is very crucial and therefore it is called quiescent point fixing or BJ amplifier biasing is very crucial for our small signal analysis that clear the small signal characteristics are governed by the DC or bias DC points and therefore they should be within your control so that you can fix the other values as per your requirement of the small signal gains or impedances or whatever you are doing or bandits is that okay so having told the importance of biasing let us quickly look for an example there are few methods in which BJT biasing is done one of the simplest method you can see something like this is to actually connect a resistance RV and of course there will be RC okay this may go to actually VCC sorry this is RC sorry this is your VCC point okay VCC point and this is your VBE let us take a case VCC is say 12 volt I have chosen some value so beta is 100 for the sake of it and what is VB on we shall always assume in the bipolar 0.7 why what is the criteria we chose that that is the value I choose diode will be sufficiently forward base emitter junction is sufficiently forward biased such that some emitter collector currents are available to us okay 0.6 just turn on 0.75 saturation so we must remain in between 0.65 to 0.75 which is 0.7 okay please actually if you want to evaluate may not be 0.7 in real calculate if you do measurements you may find 0.69 points on 1.685 do not worry too much evaluation is for assuming that it is close to 0.7 okay so that I assume 0.7 volt okay now let us say we want this device to have VCEQ equal to 6 volt and ICQ of 1 million this is the operating point I want to have that okay so that I is that clear this is 6 volt is roughly half of the VCC therefore it will we roughly know it we are always in active mode okay then how much is VCE VCE no by formula VCC minus ICRC VCC minus drop across the load by a current IC is the remainder VC so VCE is equal to this minus drop across RC is that okay this voltage minus drop across RC which is ICRC must be the remainder is VC is that correct and VCQ is what we have been given but this is the equation we should we can also say from this side let us say current in this is IB tell me what is the mesh you are getting from VCC to IB to this to the ground so anyone can give me another equation VCC equal to I be into RB plus BB on so I have one equation here I have second equation here is that clear if I substitute the values given to me and how much is IB ICQ given to me is 1 milliamp therefore IBQ is 1 milliamp by 100 so it is 10 micro amp okay okay so what is the value we are trying to find from this I want to know what is the value of RB and what is the value of RC 2 equations to unknowns we should be able to evaluate both values is that equation 2 equations to and everything else is known to us VCQ is known to us VCC is known to us IC is known to us RC we want to know here this is known to us this we are we know now RV we do not know but this we know 2 equations which are to unknowns RC and RB 2 equations to unknowns always solvable one find IB value from here beta times that substitute here and evaluate for IB there or vice versa IC by something you substitute here and get the value of RB so I did this analysis is that clear 2 equations to unknown should be easily solvable and I get the value of RC which is typically 6 kilo ohms and RB I got is 1 mega ohm or 1.1 mega ohm kind of this is a single resistor biasing okay now there is some problem in this which you read in the book why this is not really preferred by now one must understand what is the word I am saying why it is not preferred is that point clear why why why it not be preferred by seeing or why it is something go can go wrong with this circuit you can see IC and IV are related through which term beta is that correct even actually VV on both these terms are temperature dependent VV on may not be that strongly dependent but IC this beta is very strongly dependent beta is also function of IC the level of IC also decides beta so if there is a variation in beta what will happen from this what will what will shift accordingly if there is a variation in beta values for the same RC RB values you have chosen what will be moving out the VCQ and ICQ may not be same as what we actually designed for is that clear so this is external environmental dependent the operating point is not always fixed is that clear is that clear as beta varies IC IV will vary accordingly and you have evaluated for a given value of beta RC RB values but now they are fixed okay but beta very so VCQ and ICQ will actually not be same as what they were earlier this means the device characteristics or device parameter has interfered in your bi-C is that correct and we will not like this to happen this is why I said such a this is if I say if X variation happens here in denominator there is another same variation occurs it cancels some if you say ratioed circuits okay water vary there it may also vary here also beta is not coming into my final picture of bi-C then I say okay I have a very stable bias points okay not that any system has that kind of 100% stable points but closer to this or better than this is what we are looking for never say that it is always fixed will never be fixed but how much we can tolerate how much we should be allowed to tolerate that the beta value which is IC by IB small IC by be there for the signal we are assuming that should remain constant within that range is something is not changing I damn care what else it changes as long as the slope there does not change very greatly I say fair enough okay so I should always say an engineering that variation is going to come how much you should be able to tolerate is that your expertise okay how good you can get to is that point clear so this is something very trivial but has to be understood as a analog design or sorry analog circuit people okay so other method which is most popular fixed bias methods is this is that okay okay call this RB1 call this RB2 there are two ways you can solve this circuit you can solve maybe one third way which is approximately what are the two ways I will say one circuit one way of doing is around this you apply to the left of this line what you can apply equivalently you can create an equivalent heaven in source and the resistance for this this is a VCC and there are two resistances is that clear this is a VCC and there are two resistances so I can actually make a feminine source and feminine series resistance to that voltage source the other method which you may solve later I will use them in let us say this is IB1 and let us say this is IB2 but this current is the actual base current so what is the relationship we have IB1 is IB2 plus this I agree with you the way it is at the we are applying at this node the feminine's point at this node so a loop as if is only on this side is that correct the mesh is essentially on this side I agree with you but during whatever I put it when I calculate the currents that value will be taken care through the RE and other VB values so I will find what equivalent voltage I am going to get at actual VB value I am going to I will evaluate that VB finally anyway I am I will first get a V feminine and from there I will get VB value which is what exactly I am getting okay so as long as I get that value that is that is what I am trying to say different methods will be small variation so one of this is if what is the current in IB1 will be VCC by VB divided by IB1 is that correct this voltage minus this voltage divided by RB1 is IB1 VB upon RB2 is IB2 sorry VB upon RB2 is IB2 so I know these two current in terms of VCC and resistances but I do not know IB right now okay I do not know IB but I have another equation if I look at the outside what is the other equation on the output side VCC is ICRC plus VCE plus how much IE but I IB and ICs are related various ways one of course is this which you can use if you wish anytime or I is equal to beta plus one times IB and IC is beta times you can use any of these expressions okay IC is beta times IB I is beta plus one or we can also say I is IB plus IC now one can see from here what if there is a current flowing here okay is V will be now normally if sources emitter is grounded there is no voltage drop we said this is grounded but now there is a resistance here current is flowing so this VE value will be how much IE RE I RE so VE how much is this voltage then will be IE RE plus VB on is that correct and if I am evaluating a VB from this side as well and I equate them is that clear so I can I will be able to get the exact value of IB I am evaluating at that point for a given VB I am fixing from where I RE plus VB on is VB so I now two equations to evaluate VB equate them and I can then be able to get the IB value and therefore once I know IB IC is known I is known I know everything else is that clear so the tricks of the trade is one method is use this technique I use this equation and substitute IB in terms of IC or I is okay or convert everyone into IB which are ready and solve for equations is that correct but there is a simplified method I can give you which is called alternate method instead of even doing this if you say your beta is very high how high say 100 plus 200 150 100 or plus at least 100 or above how much typical IC you can get let us say VCC is 5 volt how much IC you expect by highest of IC VCC by RC so let us say RC is 1 kilo ohm or 5 kilo whichever let us say 1 kilo ohm so 5 by 1 kilo means how much current I can flow 5 million so the maximum current I may flow for a 5 volt supply with RC of 1 case 5 million that divided by 100 how much it will be 10 microns if IB 1 and IB 2 are at least one order or 2 order means how much 10 times is one order 100 times is 2 orders if IB 1 and IB 2 are at least in order or above 2 orders more than IB I can neglect IB is that clear I can neglect IB and solve so every time you may not do long calculations if your beta is given large enough but only thing in this course do not do right now saying that I forgot and left IB do not like show me solve automatically see that value is small enough but for a quick calculation people say how much so I can tell you how much because IB will be smaller so this current IB 1 will be then equal to IB 2 and solve that is the divider okay that is the simplest divider so I know VB without much thinking is that correct this is what when you do in the lab when you go to the lab and let us say you are biasing you need not worry too much about IB that time just choose R1 R2 for a given VB you are looking for is that correct may not be exact but that is not needed because in numbers 10.001 is relevant irrelevant okay then is as much as 10.001 but 10.1 or 10.15 may not be different I mean they should not be said they are no different there is a equivalence change and therefore when to leave some things has to be verified is that so I gave you some idea if beta are higher IB 1 equal to IB to just register ratio is good enough for you is that correct this is what actual circuit designers do and the they are on the board they just connect them and they say oh it is coming why we are allowed to do this because we are not interested in VCQ or VBQ or ICQ exactly what do we want to confirm that device remains in active mode if it is 5.5 or sorry 2.5 or 2.6 or 2.4 how I damn care for as long as I am within that so even with this calculations if I get value of V CQ ICQ good enough to remain in active mode I have no compuctions I do not worry on this is that clear so when you go on the I hope you have started your lab and some excellent have been performed when you do amplify design hopefully you will then you must someone ask you or TA may not be knowing but tell TA this is very easy said you also learn from me okay quickly if I do them in SQL and we THV R sorry RTH at the base for R1 R2 this I am calculating for this RB1 R2 now if I do this can you suggest how much will be VTH blow away if the dividery and RB2 divided by RB1 plus RB2 into VCC and how much is RTH short the source okay short the source these two if you are short this RB1 RB2 are in parallel so if I now substitute this equivalent circuit is that point clear what I only applied for the biasing network a Thevenin source and Thevenin equivalent of a circuit is that clear last people is that clear simplest this is only representation of bias network as Thevenin source plus Thevenin's resistance if I do so or maybe we will draw the other circuit right here so that RTH VTH and then from here you have a standard resistance RE RC this value is known to us so this current is IB okay I will give some values and solve for it now in circuit I know this if I am given RB1 RB2 I am given RE and RC what values I will be given they were designed either the values of resistances will be given then I can find the operating point if I am given an operating point then I will not be given RC and RB and I will have to evaluate by design inverse operation for the same is that correct what is the difference between analysis and design I repeatedly saying if the values are given when you solve for final operating point we say analyze the circuit and got this operation but in real life what is fix important for you not the values of resistances or capacitance we are interested in operating point we know this is the point I want to operate if I want to operate at this point what should be the value this while call it as design is that point the inverse of that is designed if you go down from top to bottom you say you analyze okay this course is mostly analysis sometimes I will show you the design aspect of the circuit okay so I saw this maybe if you are noted I will just give you some values I have chosen okay these values are not again very sankro sank but taken from a book though I this is a unsolved problem and I do not know one of the book old notes I checked so but these are unknown I mean unsolved problems so I solved for my own sake whether I still do it or can do it or not hopefully I am able to do it every time I choose a problem okay sorry so I calculate Vth which is 12.2 upon 56 plus 12.2 into 10 volt and that comes out to be 1.79 what is the formula given RB 1 upon RB 2 upon RB 1 plus RB 2 time Vcc is the feminine source so that is 1.79 RTH parallel combination of RB 1 and RB 2 that I evaluated as 10 K so here is my circuit finally which you should look 10 K feminine source Vth is equal to 1.79 volt okay then this is my IBQ this is my transistor this is my 400 ohms this is my 2 K and this is my VcQ so this is the equivalent circuit equivalent still not done very much because we are not interested in small signal so we need not expand this transistor area transistor this into equivalent small signal is that point clear why I did not draw the small signal of transistor I am only looking for DC values is that correct if I am looking for AC values and what should I do first I should break this into RBV dash everything with which yesterday we did and then solve the circuit is that okay in this we are not done that because we are not interested in GM or this value right now so we are only looking for DC this value is VB on sorry which is 0.7 volt I be please remember IBQ dash into 10 K plus I into 400 is that okay so I would say Vth Q is 1.79 equal to IBQ into RTH plus VB on plus I into R rather IEQ now how much is IEQ I said in terms of IB beta plus 1 beta 0 plus 1 is beta 0 is same as beta for all practical word possess yes otherwise I must specify beta DC independent of AC beta okay now using this equation I have evaluated IBQ so substitute IEQ by IBQ here IBQ here RTH I know this I know this I know this I know so I calculate IBQ as 21.6 micro amps correspondingly ICQ is beta times 100 times 21.6 micro amp which is 2.16 million and if you wish I can also calculate IEQ multiplied by 101 instead of 100 this is 2.18 just to get an idea V now let us calculate firstly VB or VCEQ how much is VCEQ Bolo by equation Bolo VCC minus ICQ into RC plus IEQ into RE RE IEQ ICQ I evaluated RE RC are known VCC is known so this value I get 4.8 volt so what is the operating point for this how do I specify operating point ICQ is 2.16 million and VCEQ is 4.8 volts what is VCC value 10 volts okay or how much I said 10 volts are we close to half so obviously these values must be concocted by me so that I come close to VCC by 2 okay or the inverse way I did I find where should be I close to that okay so is that correct the choice of RC RE RB 1 RB 2 should be such that you come close to 50% of the VCC please remember this is not exact 50% less than 50% it can become 5.2 also or 5.4 also or 4 also just 4 even 3.5 is good enough less because VC will be linear when it will be less than 0.6 volts anything less than 0.6 or 0.7 volt is now in saturation device is getting saturated is that correct so as long as you are less than diode drop you are not in saturation is that correct so you can be safe as long as it is more than 1 volt you are still in active mode but there the swings are very small because you are very at the smaller characteristic ends okay so you should go little ahead so around 3 4 5 6 is good enough to large RC what will it to larger this what will it create problem let us say I choose 8 volts what will it means no no just think of it in that characteristics RC should be very small so that the current because now you are looking for this point at very much higher value so where it will actually hit very high base currents is that correct at those currents the maximum current which device can supply from the power supply is not available to you so RC will require say 10 ohms okay so the power dissipation will be so high that your circuit may not function is that correct so too much VC is also not good too small a VC also not good so halfway is what you should look is that as a decision maker you should decide where you should be around around is half okay so this finishes bipolar this similar things can be done by other this take IB value also and solve again leave IB and solve again and you will find that in all cases these value will be very close I must tell you if I leave IB and solve for this I may get 4.8 you may get 4.7 or 5 4.9 not great changes will happen so what is the method I am suggesting if you do rigorously also you will get close to same result if you do on rigorously also you will get the same effort then you should do rigorously initially because you are not aware how bad or how would the system given to you but after an experience you know this is good enough okay so this we start with our first goal of the service which is MOSFET amplifier so before looking into other MOSFETs amplifiers which are different kinds first look for its biasing I repeat I keep saying by DC first why because DC decides the AC values phone signal values of personally look for DC values one of the circuit shown here as a biasing you have a okay the difference between bipolar and this the voltages are called VDD VDS VGS kind of thing or VSB if not stated VSB is 0 is that point clear if not stated VSB is what is VSB the substrate bias or bulk bias is if not stated so essentially if I am not stating you anything this is what I am doing is that clear if stated then you will have to evaluate things at least what things you will have to evaluate for such cases if VSB exists the capacitances those are direct functions are substrate bias yesterday we did that isn't it otherwise for this biasing case we need not worry because these are DC cases capacitance do not play any role so we just look for DC biasing how much is the gate voltage from that this this is VGS plus IDS times RS drop across this plus this drop across RS plus VGS is the gate voltage is that correct or you never say given a gate voltage first it will subtract VGS and the remainder will allow us PS divided by RS as your ideas is that correct is that equation clear however how much since this is a DC case I am solving how much is ideas devices in assumption is devices in you want to make it in saturation so we start yesterday someone raised an issue correct but we wanted it so we assumed it okay and see that we do get that so we ideas is beta by 2 VGS minus VT what is the assumption I am doing in all DC cases lambda equal to 0 is small signal also I may use lambda 0 but where I will not use lambda 0 yesterday I said which case I will not use or 0 because that makes it infinite okay so only for that I will keep lambda otherwise on analytical solutions I may not use lambda often unless lambda is high 0.1 then I cannot neglect it if it is 0.002 or 0.001 I said damn is that clear to you so sometimes you may say sir all that big things you talked about lambda and all that and now you are making every time 0 so this is analytical solution otherwise it becomes nonlinear situation and difficult to analytically solve okay so this is beta by 2 Vov square VGS minus VT is over voltage or excess voltage overdrive so VGS is VG minus IDS RS or VG minus beta by 2 Vov square into RS is that correct so if I fix VG and I know how much Vov I normally will operate for a given value of RS what I am fixing VGS in a way I am fixing VGS otherwise also from where if I fix my Vov what I am fixing VGS because VGS minus VT is Vov if I say Vov is 300 200 mV or 400 mV plus VT is the VGS is that correct so VGS can be fixed from where a variety is the way by actually right now I should not use Vov you should write VGS minus VT take VGS on the other side because it will be two terms VGS is that clear this VGS minus VT will appear from here this is VGS a quadratic equation will appear in VGS and you will have to solve is that clear in real life right now I would say okay the basic idea is I can get VGS of my choice by making a choice of RS and making a choice of VT and this is always if I am given a VT I am given beta's then I know for a given RS what will be my VGS is that correct this RS will use this letter and very important is called D generation resistance RS is called D generation this is a very important role both in biasing as well as in the amplifier designs RS RS is very very important parameter that is source resistance which is extra this is not that RSS RS value which is source resistance of the device this is externally put resistance which is RS this is external put drain resistance RD now why I am showing you this the way I am now will start biasing is this okay I have an amplifier which is a common source is that point here why it is called common source source is going to be grounded or at least going down and source is also on the two loops which two loops I am trying meshes two meshes I am talking source is on this side and is on this side source is available to you on so it is a common source between output and input is that clear that is why it is given a name common source source is common to output and common source is also common to input and therefore it is called common source is that clear so if I am in common source amplifier this will look biasing will be something of this kind okay do you see it is similar to what we did just now for bipolar you can see there it was RB1 RB2 here I made it RG1 RB2 it is RS there it was RE this was RC there it is RD as far as match is concerned I do not see anything different from what we did but what is the added advantage we got here there in the bipolar this current and this current we also have this current IV is IG available to us in this no because MOSFETS do not have DC currents so then can you tell me how much is VG RG2 upon RG2 plus RG1 times VDD is this voltage divider potential divider is it okay since there is no current here that is what the case which case I am now talking I will be 0 case small I will be case is same as this case so this VG is nothing but RG2 upon RG1 to RG2 times VDD is my VG now you see other equations RD let us say this potential is VD so VDD minus VD by IDS is RD then RS is how much BS divided by IDS now here also one more advantage IE and ICs are not same it is small change but there is an IB in between whereas in the case of MOSFET the current in source to drain is same okay so IDS is same for both side as well as drain side so VS upon IDS is RS VD minus VD upon IDS is RD what is the actual transistor current beta N by 2 VGS minus VT square or beta N by 2 VOV square this is the drain current IDS current for the transistor I know this if I know this and if I am not given this I must be given VD or VS if I am given this I know the value of RD and RS or given value of RD and RS I will be able to get VD and okay so let us do some calculations an example given from said Ryan Smith and me okay here is a problem please this is very relevant for us because this is what we are going to use often we are an example of same transistor biasing MOSFET which is N channel MOSFET unless stated otherwise all MOSFET I am going to use NMOS otherwise I will specifically say it is PMOS otherwise all transistors are used in this are NMOS beta N so even if I do not put N word there assume beta N okay given to me is beta N what is beta N actually in numbers beta N dash into W by L of the transistor so right now sizes have been taken Mu C ox is known and the total product of all that Mu C ox W by L is 1 milliamp per volts let us say the bias current I am looking for is half million bias current I am looking for is half million assume right now lambda 0 VDD is 15 volt okay VT is 1 volt okay too high value I chose but does not matter because this is given in the books I chose the same value RD is 10 kilo ohms and RS is 1 kilo ohm okay sorry it is not 10 it is a 1 kilo ohm no sorry both are 10 sorry sorry sorry I am very sorry let it be both 10 because I get 10 in Yemen so how much is VDS VDD-IDS RD plus RS in this case we do not have to IRS or ICR because IC and I here same so it is IDS times RD plus RS so I get VDS equal to 5 volt VDS equal to 5 volt now for saturation VGA VDS must be greater than VGS-VT so VDS plus VT should be greater than VGS so VGS should be less than 6 volt is that correct for device to remain in saturation VGS must be less than 6 volts so let us see now that R1 R2 given to us whether that gives me VG which is less than 6 is a criteria clear to you given this current biasing for you I evaluated VDS which is my what is my operating point half milliamp and 5 volt are my DC operating point VDS Q and IDS Q is half milliamp and 5 volt okay now for this I want to know what is that RG1 RG2 is that provide VG VGS which should be less than 6 volts is that clear so let us do that calculations how much is VG how much is VG VG is VGS plus IDS RS so I substitute VGS plus 0.5 milliamp into 10 K so VGS plus 5 is VG we also know IDS from this characteristics beta n by 2 VGS-VT so I evaluate VGS equal to 2 volt is that correct VGS equal to 2 volt VG is 7 volt please remember how much is VG 7 volt is that correct this is VGS is 2 volt plus 5 volt so the how much is the voltage at the gate terminal I am looking DC 5 plus to 7 volt how much is my VGS 2 volt is it less than 6 volt it is less than 6 volts so whatever value you are looking for we have already got now VGS is less than 6 volt which means transistor is in such but it is saying otherwise you assume saturation and you are getting yes but as long as both side agree it is fair enough it is like a loop this changes that that changes this if stability occurs the values are okay is that correct it is something like this feedback system I change that you change me but if you stabilize these are valid on both sides is that clear so assumptions are not very absurd though initially it looks or you are assumed and you say proved obviously okay so for given what is VG value from the network we could find please remember what is VG we said if you see this expression how much is VG RG 2 upon RG 1 plus RG 2 times VDD divider is that clear it is a divider so if I use this expression I get VG is equal to RG 2 RG 1 into VDD so 7 is RG 2 plus listen to 15 I evaluate a relationship 8 RG 2 is 7 RG 1 is that clear 8 RG 2 is equal to 7 RG 1 so I assume one of the values of RG 1 or RG 2 and therefore the other is known to me using RG 1 and RG 2 values what value I am going to get VG is always equal to 7 volt for this VG 7 volt how much is VGS 2 volt for which my value guarantee that VG should be less than 6 volt is also met for which my currents VGS- VT is going to be half million and for this my VDQ will be 5 volts so I achieved the biasing by making a choice of RG 1 and RG 2 if given values of already and RS is that clear to you so this is called design this is a bias network design is that design work here I evaluated the value of RG 1 and RG 2 for a given as if I want this I want idea of half million I specified you okay why I am interested in IDQ what is it going to decide in mass transistor and a small signal GM under root 2 beta ideas is going to be my GM so if I fix my ideas I know how much GM I am looking GM decides what the gains so essentially this ideas choice is the gain requirements is that correct the bias point which I fixed is essentially I am looking at some gains which I have someone is asking from me to design is that clear so inverse problems you got it why I redo once by this way because someone will tell I want a gain of 100 a gain of 100 or 50 or 50 then I will have to do reverse calculation and see how much GM I must provide you okay how much power I will dissipate so I should not go beyond this what age so I will fix some values and start calculating back till I get some numbers of everyone okay it will meet all your space this is what design is about okay so I did this okay before we quit another biasing which is very important for normally we may not do this in the case of discreet yes okay you have a point because normally will come in the ratio okay typically it should be RG 1 plus parallel RG 2 should be as high as possible okay the choice of RG 1 and RG 2 should be such that their parallel combination is sufficiently much higher in value okay typically one mega ounce is that clear the reason why I want that value to be higher will show you in a small signal case because that source resistant I do not want to use okay if that is smaller than that RS will hurt me here but if it comes I will use it I do not care if it is smaller but it should not be very low because it will consume then power at this gate in which I do not know is that correct please remember the power is not only on the transistor side but from power supply to the ground through RG 1 RG 2 also is going so if I choose smaller values it may still give me a ratio of what I am looking for and therefore VG may be still correct but the power dissipation may increase because it is VCC upon RG 1 plus RG 2 sorry VCC square upon RG 1 plus RG 2 so I am worried about the power also which I did not say it but essentially that decides how much RG should be kept if I keep too high in RG firstly you cannot get accuracy on those RG no no one makes those those resistances okay so you have a problem you should be somewhere in between which is available possible and something which is good enough for power is that correct and should be large enough compared to the source instant okay some can be engineering a exact kuch nahi choice score okay there is another way of biasing a transistor a mass transistor this let us say this is the time is called current biasing or current source biasing the word which is shown here is this but what I am really looking for is current source biasing yesterday or some days ago I have solved a circuit for you amplifier which I showed you this kind of biasing do you recollect this amplifier first day I showed you a mass amplifier and I say it is biased by fixed ideas okay this is called current source biasing what is it called that is the DC current in the transistor is fixed fixed okay that is why it is called current source what is the advantage of current source what is the resistance it will offer equivalent of infinity is care cross resistance in finite why I am worried this is that some people initially are 0 they were asking what is the output resistance of mass transistor are 0 if I see an equivalent there will be R0 here and the resistance of this which I call RR current source will be shunted across this R is that clear how do I calculate this is for AC what is the power supply value we give what is the status we give to VCC value for AC ground so if this terminal goes to ground this terminal goes to ground so these two registers are in parallel so do you expect that if this RCS is less than R0 then this R0 has no value because this will going to decide R is that correct output resistance of amplifier I want very high let us say but this may shine okay so what should be this value should be higher than R0 at least R0 that is R0 by 2 at least then I will get is that correct so the current source which I am going to use should have higher resistance of itself rather the system as an auto in sequence is that correct now this how do I get is one of the method which I am suggesting is that clear to you okay let us look at it this is called current mirror what do you mean by mirror if I have something here you say mirror and I put something here I see a image optics can am so if something in this arm I know a current this is my actual device which I want to use on an amplifier M2 is the transistor which I am going to use as a amplifier I want to fix this current ideas to for this fixed okay but I want to see that this current is controllable because bias point must be controllable so I see I have another current source I put somewhere here which reflects into this I can change this and then correspondingly this can be changed okay but once I fix this this current is also fixed this is called current mirror is that point clear this is called current mirror now let us see what how it mirrors let us say I have a power supply a resistance are and two MOS transistor M1 and M2 and how are how I am how am I connecting them the two gates are connected is that clear man is that clear two gates are connected but this common point of the two transistor gate is connected to the drain of M1 is that correct is connected to the drain of M1 however not only if I connect then nothing will happen so this is my actual amplifying transistor and this is my mirror part there I connect like this now since the gates are common source is common which is right now connect remember in many times in the transistor theory VSS may not be 0 it may be minus also okay but right now we assume 0 but it can be minus so VDD-VSS will actually adds VDD may be 2.5 volt VSS may be minus 2.5 volts so what is actual voltage I am flying 5 volts is that clear right now I assume VSS is 0 okay so how much is IDS 1 flowing in this circuit VDD please look at it VDD-the please remember there is no gate current so whatever current is flowing in R is same current flowing in transistor there is no gate current it is not a transistor biasing base current is not they are also will neglect in case of bipolar mirrors we neglect that but assuming right now most answer loan so there is no great current so this current is 0 so whatever current is coming from here is going down so this current is VDD-VSS-please remember how much is this voltage VDS but since this point is connected to the gate so it is VGS 1 is that point clear VGS 1 is also VGS 2 and VDS 1 is equal to VGS 1 the way I have connected is that clear please remember this point is the VGS 1 but this point and this point are same so VDS 1 is same as VGS 1 I am shorting on this side is that correct so VDS 1 is equal to VGS 1 that VGS 1 is also equal to VGS 2 because they are also connecting the gates is that clear so that is also equal to VDS 1 okay but I do not know VDS 2 right now I know only VDS 1 is equal to VGS 1 equal to VGS 2 this is guaranteed by the by connecting like this since VDS 1 is same as VGS 1 sorry I am sorry since VDS 1 is now equal to VGS 1 it is larger than VGS minus VT so transistor is in saturation so you are one worry all the time whether transistor in saturation is met okay is that clear transistor is in saturation this mirror is most important in all integrated circuits okay so what is the current in the transistor beta n1 dash W by L into 1 VGS minus VT square assuming lambda 0 which is what I started with what is the current in the second transistor beta n2 dash W by L to size of the second transistor ratio of size W by L VGS 2 minus VT n square is that correct simple mass currents are given for both transistors what is the conditions we are looking into beta n dash is same for all n channel transistors is that correct VNC ox is not different transistor so beta n dash is same so these are same W by L may not be same M1 may have a different W by L and M2 may have different W by size may not be same how well VGS 1 is equal to VGS 2 is that clear VGS 1 they are actually connecting gates so VGS 1 is equal to VGS 2 this is same VT are same this is same so that if I take a ratio of IDS 2 by IDS 1 what is the IDS 2 current yes transistor in a technology will have same thresholds is that clear unless substrate bias is provided for a specific transistor its VT cannot be modified is that clear all N channel transistor will have unless stated otherwise we will have same VT okay or at least VT 0 will be same is that clear all P channels may have a different VT but will also have same because since you said it yeah there are new circuits in digital where we are using multiple VTs but in analog no one and them so far so this is common these are equal these are equal so the ratio of IDS 2 by IDS 1 is how much equal to W by L of 2 divided by W by L of 1 is that clear to you W by ratio of sizes so let us say I want IDS 2 same as IDS 1 so what should be the size of M2 the main transistor same as M1 okay or so I must know at least M1 size okay I must know M1 size then I know my M2 size for the ratio I am looking for if I want why it is called mirror the word given was if IDS 2 is same as IDS 1 then same current which I have a reference is now given to the biasing current is that why it is called mirror the current in my M1 which is called reference current which I am fixing is same as what is in the M2 are in proportional to that why is thrice ratio of the bias is that clear man this ratio maybe 234 so if it is 4 then 4 times the reference current I am biasing at the main transistor why I want this game to be done because I may choose smaller reference current for different circuit transistors amplifying parts I may actually bias them at different bias currents is that correct let us say differential amplifier we will see next time or next time whenever there the bias current will be different and a common source amplifier may have a different source current or bias current so we must have one generation of current source which multiples of that can be given to different parts of the circuit so this gives you allows me to that is that correct however I know IDS 1 which I call my reference which is VDD-VGS 1-VGS which is equal to beta nW1 by this since I know all the values here I will be able to evaluate the reference given a reference current I will be able to evaluate VGS and therefore I know what exactly is the values I am choosing R for that once I fix my R I know my reference current if I know my reference current I know proportionately the output current or the main transistor current which is being biased is that point clear so why it is called mirrored because whatever in the reference is mirrored to okay another interesting circuit which I did show may be interest for something I can do something like this I have another sir transistor here okay and extend this here and what will be current in IDS 3 no W by L 3 by W by L 1 is that correct so now I that is what I say if I choose a choice different the proportional current in different arms I can keep changing is that clear this is multiple areas can be given current from same source and this is a good current source because this I have sorted that this R I have adjusted such that this acts like a good current source okay that we will see next time how to make this as a good current source there will be because this currents which I am using are ensuring same VGS values reference is the first one the other one will be because biasing is done by current now not by VGS current is proportional VGS- VT so I fixed the current I get into VGS which means saturation is that correct that is the way I am doing now okay this is done in where integrated circuits