hey guys. In the last video, we looked at Peter and seated voltages. In this video, we will see different ways to combine these two voltages to generate a temperature-independent bandgap voltage. So, in the last video, we derived this Pete hat and seat at voltage equations. So, here we have two aims. One is to make these two slopes equal, and next is to combine these two voltages to form the band gap reference voltage. So, we notice that in the Peter equation, we have this parameter n, which is basically the ratio of two devices or the ratio of two currents. So, one simple solution may be to choose n in such a way that these two slopes become equal. So, let's see what value of n will be required. So, if we put these values and do the calculation, we get this value, which is more than 12 billion. Obviously, it is impractical to have a device ratio of 1 to 12 billion, so we can discard this option to be very impractical. Our next option is to somehow amplify the pitot voltage. Let's

choose the value of n to be a more reasonable value of say 8. So, in order to match the slopes of Peter and seat at, we need to amplify PT by around 11. And this amplification needs to be very accurate so that we don't introduce errors and inaccuracies. The most popular way to do it is to change the pitot voltage into a pitot current and then pass that current into a bigger resistor to create a bigger pitot voltage. So, the whole process looks something like this. First, we somehow drop the pitot voltage across the resistor R1. That creates a pitot current. And then, that pitot current is flown into a second lesser R2. And we get an amplified pitot voltage. Notice that this is a ratio metric amplification. We need to say that the gain of the amplification depends on the ratio of two similar devices. And that is why it is a very accurate way of achieving the required gain. So, let's see how we can build this kind of circuit. So, we start

with two diodes d1 and d2, which have a ratio of 1 to n. It's implied that d2 is a parallel connection of identical diodes. Now, if conditions are right, then P, that voltage, would appear across these two terminals. So, let's put our first resistor now. For the moment, we assumed that the voltages v1 and v3 are equal and also that these two branches have the same current. Then, the voltage across R1 is our P tat voltage. So, in order for this circuit to work, we need to ensure that v1 and v3 are equal and these two branches have the same current. One way to achieve these two conditions is to use feedback. So, here we have got a high-gain differential amplifier and two identical PMOS transistors, which act as a controlled current source. The purpose of this high-gain amplifier is to make sure that the voltages v1 and v3 are equal, and the purpose of these PMOS transistors acting as control current sources is to make sure that the current

in these two branches are equal. Now let us try to figure out which terminal goes to the positive terminal of the amplifier and which goes to the negative. Now there is a simple way to look at it and there is a complicated way to look at it. In this video, I will take the simple day, which ignores some subtleties but works just as fine to figure out the polarity of the amplifier. Recall that the voltage across the diode is rather insensitive to the current into it. That means small changes in the current do not change its voltage a lot. Now let us assume for some reason current in these two branches drops. In order to recover this drop in the current, this feedback needs to increase the gate-to-source voltages of these P Moss's. Now voltage V even doesn't change much with this drop in the current because it is the voltage across the diode. The same goes to v2, but voltage across r1 does change because of this change in current. So overall v3 reduces, so we see that the polarity of v3 and the output of

amplifier are same, and that means v3 should go to the positive terminal of this amplifier. Now, there is a complicated argument to arrive at the same conclusion that uses that concept of negative and positive feedback around the loop, but that is for another video. So, once we set up this to properly, we get repeated current in these branches. So, now we can push this beat-add current into a register to get our amplified Peter voltage. So, we have achieved our first objective of getting a pitot voltage. As for the second objective, which is adding Peter and seat-at voltages, remember that we can add two voltages simply by putting them on the top of each other. Now, our seated voltage is simply the voltage across our diode. So, if we add a diode in this branch, it should do the job. And finally, here we are, our bandgap reference voltage. So, what is the value of this voltage? So, VF is the sum of VC tat and VP tat, and here are the equations of the seat-at and VP tat. Now

We have designed the pitot voltage such that its slope exactly cancels the seat at voltage slope, and we are left with V BG, which is approximately the silicon bandgap voltage. This voltage is approximately 1.2 volts. Now, we can generate this reference voltage in a slightly different way. So, notice that diodes are already available in these two branches. So, can we do away with this third branch? Indeed, we can by putting this R2 in these two branches instead. So, here we have the same reference voltage with a slightly lesser number of components. In fact, there are many more ways to rearrange this circuit. Now, all these configurations have their own pros and cons, which we will discuss in another video. Now, bandgap circuits are not only used to generate reference voltages but also the bias currents. We can reuse PT at current flowing in these branches as bias current. Now, this bandgap architecture is one of the most widely used bandgap architectures, but there are some disadvantages. The first is design complexity because of the amplifiers for some simple applications designing and

compensating amplifier may look like a lot of work. Second is the value of a rough voltage which is restricted to the gap voltage which is around 1.2 volts. If you need another voltage say 1 volt then you need to put another amplifier to do the job. Third limitation comes from the requirement of mineral supply in order to get the correct voltage we called as these 2p mosses act as current source. That means they need to operate in saturation region. So if VF is 1.2 volt and we assume around 200 to 300 millivolt across radius of these P mosses then we need at least 1 point 4 to 1.5 volts as supply. And if we add cascode to this stage then they come becomes even higher. Another disadvantage is variation of PTH current over temperature and process corners. If it is used as bias current current increases by 70% from minus 40 degrees Celsius temperature to 125 degree Celsius temperature. And if we also add the

process version of racer on the top of it. Then the ratio of the maximum current to minimum current can be more than 2. Now, no one architecture can address all these issues, but there are alternate architectures which can address some of these. Now, if design complexity and the area increased by the use of amplifier is your main concern, then there is another architecture which can be considered. Recall that in order for this circuit to work, the voltage across and current into these two branches should be equal. One way to ensure these two conditions is to put a current conveyor on the top of it. So, let's see how this works. The current mirror formed by primus P1 and P2 ensures that current in these two branches are equal. Coming to n masses, these have same gate voltages and same current flowing through it. As a result, they have same gate to source voltages, making V1 and V2 equal. Obviously, in order for this to work, sizes of P1 and P2 are equal. Similarly, sizes of N moanin and 2 are

equal so this current converter circuit ensures that Arvind has repeated voltage across it and hence repeated current in these two branches. And we already know the next step if we have the pitot current available. Now this circuit solves our complexity and area problem, but other problems are still there, maybe even worse. So let's now look at another architecture which solves almost every problem. So far we have considered adding seated and repeated voltages to generate our reference voltage, but that is not the only option. We can also think of adding PT and see that current to generate the reference voltage. In fact, we already have the pitot current; we just need to find a way to generate the seated current. Now we already have the seat at voltage in form of V1 and V2. So if we think very simply, if we put a resistor across these voltages, this resistor should have seat at current. But is it that simple? Yes indeed it is. So let's start again. So if you got a diode which is biased by a

constant current. We have already seen that the voltage across this diode is set at. Now, if we put a resistor in parallel to this diode, then the voltage across this resistor would also be set at. And if R doesn't change with the temperature, then the current flowing into it will also be a set current. Okay, now let's put everything together. So, we start with our family structure to generate the pitot current, and we put another resistor across this structure. So, here, the branch containing R1 will have PT at current, and the branch containing R2 will have set at current. And now we put all the infrastructure in place to make voltages and currents equal. And if we choose the value of R1 and R2 just right, we can make the total current temperature-independent. Here, Z stands for 0. Now, to generate the reference voltage, we simply dump this current into another resistor. So, as you can see, we don't need to put a diode anymore to generate the reference voltage, and this is because current itself is now

temperature independent now. By choosing different values of resistance or different values of currents, we can generate any desired reference voltage. This bandgap architecture is known as compensated current bandgap architecture or banbao architect. Bamba is the author of the famous paper which described this architecture. As you can see, we can now generate any reference voltage, removing one of the drawbacks of the conventional architecture. At the same time, this architecture requires lower supply voltages. Total supply, the comment is the voltage across the diode plus the voltage across this MOS transistor. These two add up to as low as 1 volt. The spread of resultant bias current has also reduced considerably. The bias current spread now only has the spread due to the process version of the register, which can be as low as plus minus 20%. The only remaining drawback is the design complexity and the area increased due to this amplifier. But this is not bad at all given the advantages this architecture provides. As a matter of fact, as supply voltages continue to reduce, this has become the default bandgap architecture. Ok, so in this video

we have looked at the three most commonly used bandgap architectures. We have overlooked many details and design considerations. In this video, I will cover some of these in more detail in the future videos. In the very next video, I will consider another very important aspect of bandgap circuits, and that is the startup. But for now, please post your comments below, and thanks for watching.

#24 Band-gap Reference Circuits - Architectures

#24 Band-gap Reference Circuits - Architectures

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