Hi, welcome to Unlock Layout and Design and today we are going to discuss a very popular topic which is called as the Bandgap Reference or the BGR which is found in almost all ICs. So in this video we will discuss about what is the need for reference, then we will write down the BGR block diagram and then we will see how to generate CTAT and PTAT. What does this mean? We will explain that one. Then we will look into the actual circuit of BGR.
Then we will speak about what is known as the trimming. And then we will discuss about startup circuit which is present in almost all bandwidth preferences. And we discuss one more architecture known as a sub 1 volt architecture. Then we will discuss about matching concepts in layout. And then finally we will discuss why the name BGR is coined.
So first of all let's try to understand what is the need for a reference voltage. Okay so if we know that integrated circuits like they should work in all temperature zones like it could be a desert region where there is like extreme high temperature or it could be a snowfall region or a polar region where there is like freezing temperatures like 20 degrees below zero. So such kind of regions also it should work.
what i need is i need is a reference voltage independent of temperature so that i can compare it with another voltage that is dependent on temperature and then i can display and do all the calculation based on that for example if 27 degrees is represented as 1 volt then and this doesn't change with any temperature right 27 degrees also it is 1 volt, 120 degree also it is 1 volt and minus 40 degree also it is 1 volt. I have a 1 volt reference which will not change with temperature. Then I can generate one more voltage which will be proportional to temperature. Then I can compare with this 1 volt and I can say that now the temperature is say some 80 degrees or 45 degrees okay or minus 20 degrees.
All those things. So basically I need this reference. This need not be always temperature only.
For example I have 1 volt which I use which will not change at all with anything and I will make this 1 volt represent the speed of the vehicle. So when I have 1 volt the speed is 1000 rpm or say some 100 km per hour. So now If the speed of the vehicle changes, if it goes up or comes down, I can definitely compare it with this 1 volt and make a decision or display that temperature.
If it is 2000 RPM, then maybe it will read it as 2 volt, then I can say that the speed is 2000 RPM. I can display this one. Or if it is like 0.5 volt, then I can say that it is like 500 RPM. Okay, so on and so forth.
So now let's try to understand how should the reference be on a chip. On a chip, how should this reference voltage be? So this reference, whatever we are speaking, it should be independent of three things, which is called as PVT variation. What do you mean by P, V and T? P means it is process.
So we know that we have the ICs which are fabricated on different processes. So it could be like CMOS. 130 nanometer or 65 nanometer or finfet etc or by cmos etc okay and this is like voltage or the supply voltage okay so if they are run by battery definitely that the voltage will not be constant it will be changing or with uh battery getting drained it will reduce so definitely with respect to this supply voltage so typically if it is like 3.3 volt it should run with 3.63 and also like 2.97 volt okay so 10 percent if i increase or decrease it should still be working and it should be independent okay so it should give the same voltage irrespective of my process process will have like a slow fast and typical okay and also different processes like this okay and also like temperature. Normally the ICs will work from minus 40 degree to 125 degree centigrade.
During this entire temperature range it should give a constant voltage. So this is about PVT. So it should be independent of PVT. So first let's try to see how to make the reference independent of temperature. So the idea to generate a reference is, so it's very difficult for us to get something which is constant with respect to temperature.
Everything will vary with respect to temperature. So what we do is we will identify two quantities, one quantity which is always proportional to the temperature. Suppose temperature increases say from 0 to 100 degrees, this quantity also increases. That quantity is called as p tat. p tat means...
proportional to absolute temperature. So, I will identify one voltage which is proportional to the absolute temperature. So, I will call that as p tat. So, then I will also find out one more quantity which is called as c tat which is complementary to absolute temperature. That means as and when temperature increases, this voltage decreases.
This is what you can see here. Here the temperature increases, this voltage increases. then here the when the temperature increases this voltage decreases okay so now what i do is i will add up these two so this is my p tag this is my c tag i will add both of them then what do i get i will get something like constant right so that's what is shown here so that's what is my reference which is independent of temperature now temperature is changing from say 0 to 100 degrees but centigrade but the voltage is constant because we are adding two different type of voltages which have opposite temperature coefficients so this is the general idea how we generate a band gap reference so how do we generate the bgr on chip okay so we need to identify c tat and p tat as we discussed now so we know that vbe okay so if i have a bgt like this okay, is the emitter base collector.
If I connect it like this, okay, this is called as a connecting like a diode, this VBE voltage, okay, this VBE voltage is actually a CTAT voltage. What do you mean by CTAT? Complementary to absolute temperature.
That means if temperature increases, this voltage will reduce, okay. That is what is shown here, okay. So, let us see here. So, this is what is shown here. So, with temperature, I am going to give a constant current source like this.
So, current is constant. So, that time, this is my VBE, the difference between base and emitter is called VBE, that will reduce with respect to temperature. But how quickly or how fastly will it reduce?
It has got a rate of reduction in voltage as minus 2 millivolt per degree centigrade. Means, if the voltage is say 750 millivolt at 0 degrees. So when it is like 1 degree, then it will be reduced by 2 millivolt means it will be 748. So then 2 degree it will be 746. So like this it will keep on reducing. So this is nothing but the CTAT just got minus 2 millivolt per degree centigrade.
Okay, then there is something known as VT. Vt which is known as the thermal voltage. This is one quantity which is actually p tat. That means to say whenever the temperature increases this voltage increases. So that is what is shown here in the below diagram.
So this one is increasing with respect to temperature. So Vt is one quantity which is the thermal voltage which will increase with respect to temperature but unfortunately this will increase only at a rate of 0.085 millivolt per degree centigrade whereas this is decreasing at minus 2 millivolt per degree centigrade so now obviously these slopes are not same so this guy is like reducing very quickly you But this guy is increasing very slowly. So we have to do something to bring up, bring the slope like this.
So we have to multiply this with some number. That number is what is shown as this m. We can calculate that. So that is what is this number m and then I will add so that both of them have like the same slope. This is minus 2 millivolt per degree centigrade and this one is plus 2 millivolt per degree centigrade after multiplication.
Then I will get a temperature independent reference which is a constant. So that is what is shown here. V out is VBE which is a CTAT plus VT which is a PTAT multiplied by m. so then i will get my bgr okay so as we discussed how do we generate c tat so just i have a current source which is connected between vdd and this bgt here bgt is like diode connected we call this as diode connected whenever collector and base they are shorted okay so vpe is this voltage And this has a CTAT behavior.
It will be something like this at minus 2 millivolt per degree centigrade. Okay. So let's try to understand P-TAT generation circuit.
Okay. So I have an arrangement like this wherein I have IC here and IC same IC here. Here I have only one transistor and here I have N transistors.
Okay, so they are all connected in parallel. So whenever I have a current Ic here, this is VBE1. Okay, so this VBE is given by this equation.
VBE is equal to VTln Ic by Is. What is Ic? Ic is nothing but the collector current and what is Is?
Is is nothing but the saturation current. So I will write VBE1 is equal to VTln IC, I will just call it as IC by IS. So in this second case what will happen?
So what happens to the current in each transistor? The IC will get divided into the N transistors. So what happens to the current in each transistor? It will be IC by N. See they are all connected in parallel.
So each current will be Ic by N. So I will write the equation VBE2 is equal to VT LN Ic by N divided by Is. So now I will take the difference of VBE1 and VBE2. I will write VBE1 minus VBE2 is equal to, so this will be VTLN, this quantity minus this quantity, that will be division, m by n. So if this is m and this is n, or P and this is Q, P by Q.
So that will be VT ln IC by IS divided by IC by IS into n, which will be nothing but VT ln n. And these things will get cancelled and only n will come. So if I see the equation VBE1 minus VBE2, is VT ln n.
I will just write only this equation now. So VBE1 minus VBE2. So this is my VBE1. Okay, this node, this node is shorted.
So it is nothing but the same voltage here. And this one is VBE2. VBE1 minus VBE2 is nothing but VT. L N N. So this N will be a number like 8. So L N N will be a constant and V T we understood that is the p tat quantity.
It is a p tat quantity. So now we got when if I subtract V BE1 and V BE2 if I make this arrangement. If I subtract VBE1 and VBE2, I will get a V tag. So difference in VBE itself will give me a beta so this is important so vbe as such only vbe is c tat and vbe1 minus vbe2 is beta hope this is clear so finally we look into the pgr circuit so we know that vbe1 minus vbe2 is p tat and that is nothing but vt ln n okay natural logarithm of n which is the multiple so here i have a circuit wherein i have used a op-amp okay i have used a op amp so this circuit this one is this voltage is nothing but this vbe1 only okay so i will write this as vbe1 and because of virtual ground what do you mean by virtual ground virtual ground means whatever voltage is there here the same voltage will come here when the gain is high So we will assume that the optimum gain is high.
So whatever voltage is there at this negative terminal, it will come at the positive terminal as well or both these terminals will have the same voltage. So here also I will get VBE1 and we know this voltage we know already that is nothing but VBE2, this voltage. So now that is VBE2. So, what is the voltage across R1?
Voltage across R1 is nothing but VBE1. This side it is VBE1, this side it is VBE2. VBE minus VBE2 which is nothing but VTLN. So, this voltage is the VTAT voltage.
And this VBE2 is? C-TAT voltage. I have a mechanism here.
I am adding C-TAT and P-TAT. So now will I get a band gap? We will address this issue. We will see whether we will get a band gap.
So we understood the voltage across this one is P tag and the voltage across this is C tag. If we add these two are we going to get a band gap? No.
See because our vve has got minus 2 millivolt per degree centigrade and vt has got a temperature coefficient of plus 0.087 millivolt per degree centigrade so how much should i multiply 0.087 to get the same plus 2 millivolt per degree centigrade So if I multiply this Vt by 22.9 or roughly say 23 or I can take it as 24 whatever number. Basically I have to multiply by 23. If I multiply by 23 I will get plus 2 milli volt per degree centigrade. Then I can add that to Vbe which has got negative temperature coefficient.
then i will get a bg band gear reference okay okay so let's try to understand the bgr circuit so we have this one is vp tat and this one is c tat so i have to add these two to get VBGR. So, at the moment, what is VBGR? So, it is VP tat, VC tat, which is nothing but VB2 plus VP tat.
We saw VB1 minus VB2 is VT LN N. And we also saw that VT should be multiplied by 23. So, normally, what is this LN N? If LN N is 23, then I can just add. So, if LN N I should be 23. What should be this number?
How much multiplier I should have? So this multiplier should be around 10 lakh. Okay. So or even more. Even after that I will not get 23. Okay.
This will be a very big number and entire chip will be filled with that. So normally how much we use is around 8. Okay. So this one will be around 8. 8 multipliers and this will be 1 multiplier. So we will have 1 BJT here.
and 8 BZ is here. So now ln 8, so this will be ln 8 which will be around 2, okay. So basically it is VBE2 plus VT into 2. So now I need actually 23. So roughly I have to multiply it by another 10, okay. So this 2 I got, I have to get 23 or I have to multiply by 11, 11 or 12, okay. So, if I multiply by 11 or 12, I will get the required equation.
So, then I can add it with VBE2 and then I will get my VBG. So, how do I do that? So, now, this is the final circuit of BGR.
What I have done is I have put one extra resistor here. What is the current that is flowing here? If we know the voltage across this, that divided by this resistance is the current okay and the same current is flowing here also the same current is flowing here also so what is the current here we know vbe minus vbe vbe1 minus vbe2 are which is nothing but vt ln n is the current that divided by r1 is the is the voltage vt ln n is the voltage and that divided by r1 is the current that is flowing here okay the same current is flowing here also Okay, so that current into R2 will be nothing but the voltage across this, this voltage.
So we have like V p tat 1, V p tat 2, I will call it as V p tat 2 and V be 2. Okay, so now this into this is nothing but R2. This current into R2 will give me the voltage. here so i have to add this voltage along with this and here what i will get vbg so let's write that equation okay vbg so that is nothing but this current the same current is flowing here and this vb2 so vbg is equal to vb2 plus vt lnn plus Vt ln n divided by R1 into R2. So, if I rewrite the equation, it is nothing but Vb2 plus Vt ln n into 1 plus R2 by R1.
So, this is now 2. And this together should be around 11 out of 12. So if I take R2 to be 10K and R1 to be 1K, this one will be 10. 10 plus 1 will be 11. So if this one is 10K and 1K, then what is that I will get? I will get a V band gap which is Vc tat plus Vp tat. so this entire thing is vp tat so quickly we can understand that vpe is a c tat for that i will add a vp tat which is nothing but a difference in vpe that is vp tat but that is not enough so i will add extra then i will get my vpg Okay, so this is about this circuit is not very easy to understand. I have tried my level best to break it down into pieces and make everybody understand.
So this is my VVG, this voltage, which is nothing but VV2, this one which is C-tac plus this plus this, which will be plus 1 plus. R 2 by R 1 into V T L and N. I can also write it as in terms of V V E 1. This is V V E 1. This voltage and this voltage.
voltage is same we know that one. So, this voltage plus this voltage that is nothing but V b e 1 plus 1 plus R 2 by R 1 into not 1 plus sorry. So, this is our V b g I can write this V b g s this V b e 1. plus this voltage.
This voltage and this voltage because of virtual ground it is the same. So, V b g is equal to V b e 1 plus V T L N N into R to the R 1. So, now, this one is combination of C tat and P tat. This is P tat and this is the multiplication factor.
So, I told you like around 23 multiplication is required. So, this is the multiplication factor. So, now, but this made it temperature independent. But then there will be process variations. So how to overcome that?
So for that what we do is we will do a process known as trimming. So because of process variation there will be small variation in VBE1, VBE2 all these things and VBG will be different. So normally when I add all these things it will come to 1.2 volt. All put together it will be VBG generally will be around 1.2 volt. So, whenever I have different process, so this 1.2 may become 1.21 or 1.18 or it can become 1.22, something like that.
But if I want a constant 1.2 volt, then what can we trim? We can trim this resistor or this resistor. This one, trimming this resistor is better because it is a higher value resistor. so what we will do is we will be able to add or subtract some resistance to this r2 so that this gain is adjusted and then we get the proper 1.2 volt again so this is the trimming in your bgr so if you see this is the vbg i have made this entire circuit diagram horizontal so here i have this resistor r1 r so all put together i have r1 in many this one r2 r3 r4 r5 r6 like this okay so now if i want to trim this is the major resistor which is there and all these things are there if i want to reduce the resistance what i will do is i will just close this switch if i reduce so this this could be some small amount suppose say this is 10k so this will be say some 10 ohms say 20 ohms 30 ohms 100 ohms 150 ohms like that okay so if you want to reduce this resistance what will i do i will just close this switch if i want to reduce 150 ohms if i close this switch this will be a short circuit and then current will not flow here but it will flow here but the resistance is zero okay so by closing these switches this will be like n-mass switch or a p-mass switch or a transmission gate by doing that we will do what is known as trimming So this is the method by which we add and subtract some resistance.
So it could be other way also. Now already I have some resistance. This switch is there which is already closed.
So that time this resistance is not there. If I want to add resistance I will just open it. If I open the resistance, resistance will get added. So now I have this resistor, I have open switch.
So if I close the switch, the resistance will get subtracted. But if already it is closed and if I open, it will get added. If I close the switch, it will get subtracted.
So much resistance is subtracted. So this is how we do trimming which is always used. So every engineer will come across this trimming in BGR. Another important circuit is a startup circuit. So in a BGR we have a op-amp.
This op-amp is stable for two currents. One is the normal rated current, how much ever like 10 microamp or 20 microamp whatever is the current or it is also stable with zero current. So now what happens is, so when it is zero, the body of reference voltage will not come at all. It will also become zero.
So we don't want that. We want the normal 1.2 volt reference. Here this is a separate circuit. We will discuss about this little later. So, but then we want the traditional 1.2 volt to come up.
So, for that what we do is we will have a startup circuit like this. What the startup circuit will do is it will pump in some current to the op amp when the power supply goes up from 0 to VDD. When it goes from 0 to VDD, what it does is it will pump in some current to this op amp and then the negative feedback loop.
will bring in the same current say some 10 micro or 20 micro whatever current is there it will come and my v band gap will come over so after that this startup circuit will get disconnected there will be some logic but it is very important to understand that there is a something known as the startup circuit which will be there during the startup so the main function of the startup current is to provide some current into this op-amp so that the op-amp will come and settle to the rated band gap voltage So we understood that the band gap voltage will be typically around 1.2 volt. So but then many a times the supply voltage itself will be like 1 volt. So if I get a band gap voltage of 1.2 volt it won't work.
So for that we have a different architecture called sub 1 volt band gap reference. So this is the circuit we will discuss this in a separate video. I just wanted to introduce that one here.
One other important point is we used a NPN transistor like this. So in all our discussions we used a NPN transistor like this. But in the circuit we used a NPN transistor. Here also we had one more. So N number of transistors.
Then we had this resistor. Then the op amp. So this was the circuit. But then we will generally not have a NPN transistor. NPN transistor because that will come in a bi CMOS process, bipolar CMOS process, but normally we will only have a CMOS process.
So, what we do in that case is we will just use this kind of a parasitic PNP transistor. It is also called as substrate PNP. So, here this N-well, we will use a N-well, this is the P-substrate N-well. and this p plus all put together will act like a pnp transistor so how will that be if i connect it this way instead of connecting it like this earlier okay earlier i had it like this now i will just have it like this so then i have a pnp which is a substrate transistor which will be used in all bgrs okay So here what is important is this one PNP, this P is always connected to substrate and it is always 0 volt.
You cannot connect this to any other voltage. Okay, that is important. So let us try to understand the layout matching in BGR.
So as I told you there will be 1 is to 8. So there will be like one transistor, one BJT in one branch. And then we will have like n is equal to 8 in other branch. So now overall I have like 9 transistors. So how will I match them?
So the one transistor I will put it in the center. This is that one transistor. This is the BJT.
So in this one is the center one is the emitter. This one is the base and outermost one is the collector. So, here what I do is I will put the centromost one is the first transistor and I have 8 transistors which need to be matched with this one transistor. I have 8 transistors. I will place them.
all around 1, 2, 3, 4, 5, 6, 7 and 8 transistor. They will cover this one transistor and they are matched. But then this is not enough.
This is a reference block. So what we do is we will place dummies all around like this. So when I place dummies, this is how it looks.
So if you see this portion in the center is the actual 1 plus 8 which is like 9 transistors. These things, these 9 transistors go and sit here. And what you have on the periphery, these are all like dummy transistors.
All these are the dummy transistors. So when we place like this and route it this way the matching is appropriate and mind you matching is very very important in a BGR because this is a reference. So if you do not get the reference properly it will affect the entire chip.
So, what are the other things that we have to match? So, I told you this is like 1 and n, this is the P and P, substrate P and P here and here, so which need to be matched. What is that I have to match? I have to match this end, this resistor.
I have used one more resistor here so that there won't be any VDS mismatch for this. For that reason I will use this resistor. So now I have to match this with these two. So I told you if this is 10k this will be generally 100k. This will also be 100k.
These need to be matched. So if I don't match these two properly what will happen? See I told you what is the VBG? VBG is VBE1 plus VTLN, VTLNn into, VTLNn into R2 by R1. So if these two are not matched properly, then the gain will change.
So, instead of suppose say this is like now the gain is 10, 100K divided by 10K the gain is 10. So, now if I do not match this properly from 10 it may deviate to 11 or it may come to 9 and then the VBG will not be accurate. So, that is the reason why we have to match this properly and also we have to match this with respect to this. And these will be like bigger resistors.
Otherwise, the current consumption will be high. So this will have a lot of multipliers. So 10K and 100K, there will be a lot of multipliers and this will use like a lot of area.
So you have to match this one properly. These three resistors have to be matched properly. So this one I have shown how to match. And I have shown you these three resistors how to match. So other matching that we need to take care is this.
current mirror here. So this one is the current source whatever we saw earlier right we had this symbol. So it is a current source. These transistors act like a current source and they have to be same we had assumed that this is also IC this is also IC.
So if these are not matched then this current and this current will be different. So for that reason we have to match these two also properly. So inside this op-amp it will be generally like a two stage op-amp.
which we will discuss separately and in that we have different matching for the circuitry inside so in band gap what is that we have to match we have to match this one we have to match these resistors and we have to match these current mirror transistors so finally we will also understand why the name band gap reference so we know in any material we have what is known as the so this is the conduction band and this is the valence band in between there is a band gap gap in the band okay so this uh gap in the band for silicon is around 1.2 volt it is not volt it is electron volt okay electron volt is the measurement of the band gap energy of silicon. So between conduction band and valence band there is a distance and that is called as band gap and that for silicon is around 1.2 electron volt not volt. This is very similar to our band gap reference which will generate 1.2 volt not electron volt.
So because of this similarity they call it as band gap reference BGR which is called as BGR. that's all so thanks for watching the video so hopefully you understood the bandgap working and it's not a easy topic to understand so if you have understood and liked the video please hit the like button and don't forget to share and subscribe thank you