Simple Switcher Topologies and Their Morphology
Sanjaya Maniktala

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Today's topic is Simple Switcher Topologies and their Morphology. Today's seminar will be given by Principal Applications Engineer Sanjaya Maniktala. Welcome, Sanjaya. Please go ahead with your seminar.

MANIKTALA:
(Slide 1) Thank you, Wanda. I'm Sanjaya and I'll be presenting the seminar. On the first slide we have the topic, which is Simple Switcher Topologies and their Morphology. I know what you feel, that there are two words with "ology" here, and that makes it sound very tough, but I hope it'll be clearer to you as I go along.

(2) I'll go to the second slide here, and that is my contact information. If at any time you have questions you can always send it directly while the seminar is in covering. But you can also send me questions at my email address, and I'll try to answer them in the next few days.

(3) Morphology is basically a study of structural form. What structural form are we studying? Let me define the scope of this presentation a little early. We are going to focus essentially on the three fundamental topologies of switching power conversion. They are the buck, the boost, and the buck-boost. We are going to focus more on the problems we face when we apply integrated switches. Integrated switches are by definition switches which have a control and a switch inside them. The problem with that is that there is less flexibility for the user, so he cannot necessarily use a switcher defined for a certain application in a different topology. We will talk a little bit about sepic, Cuk, and other such topologies, and I'll try to show that they are basically composite topologies and they should be viewed in that light.

(4) How did this study evolve? As an applications engineer, I take a lot of customer calls. One of the nagging questions we have which we feel we don't do full justice to is questions like, "Can I do an inverting regulator?" If you go to for example the schematic which is here, it is the LM2591HV. I have two schematics there. For our purposes they seem to be at first sight similar; you don't see much difference. But if you look very closely, you will see that the schematic on the top is the inverting regulator and the schematic on the bottom is the standard typical application circuit, which is the buck regulator. Very rarely in the accompanying application information we make it clear that the topology has changed completely when we do an inverting regulator. What are the problems with that? The customer says, "You have a 2591HV and you say it's a one-ampere device. That can only take to mean that you can get a load current of one ampere." But you have to keep in mind that that sort of statement is true only for the buck. If we go to the other topologies-the boost or the buck-boost-and a top schematic is the buck-boost, then there is no defined correlation between your output load current and the switch current anymore. Therefore this device, even if it is applied-and it can be applied to this inverting configuration-it cannot do a one-ampere load current anymore. We are also responsible for some of the confusion which has come up, and the schematic in that sense is confusing. By the way, that datasheet was done by me, so I'm responsible for that, too.

(5) Looking at some other competitors, you see even more garbled schematics. This is the same schematic, a positive-to-negative converter, but it looks very confusing and you don't see where the input power is coming and where is the output at first sight. So you have to struggle a little bit. Here's another one after this.

(6) This is a similar schematic, which is really confusing. You cannot tell what is going on-at least I cannot, and most customers find it hard to understand. One of the problems with confusing schematics is that when we apply a device in one topology to another topology, it is very helpful to try and keep in mind where the switch is internally. Since the switch is internal, we tend not to show that. But you should draw it out and you will see most clearly what is the power flow because a switching power converter is basically that. You have to understand where the power is coming in and how it is being delivered to the output.

(7) I'm going to think aloud for some time about the topologies and then you'll probably see a picture emerge from that. That's probably the way I also looked at it to start with until I saw a larger picture emerge, and that is the picture I'll try to present to you. Let's start at the very beginning. We have here a standard buck converter as shown in introductory textbooks, and I'll transition to a boost in four steps and show you the problems. In most introductory textbooks, they tend to show the two switches in the form I have shown it, as mechanical switches, basically. You have to keep in mind that if there were mechanical switches, it doesn't really matter which one you're controlling and which is following. The only condition is that they have to be interlocked and when one turns on, the other turns off. Hypothetically speaking, we could have an identical device on the top position and on the bottom position. We'll come to that. The problem is that switches are not perfect. What we normally do is we place a switch at the top position and the diode at the bottom position, and I marked it out. If you try reversing that, and accepting the fact that switches are not perfect, you will see that you will basically get a dead shot. You should keep in mind that switches are not perfect, and we'll see the problems with that, too.

(8) One of the things they talk about in textbooks is they say if you take a buck converter and you apply power to the output terminal, then on what were its input terminals you will get a boosted voltage. That's what I'm trying to show you, how we can do that. Here in this slide I'm applying power to the right side of the schematic now so that becomes the input, and I'm going to try and draw power from the left side. That looks fine, but it's not the way we normally like to see it in schematics drawn. We tend to view schematics as an input on the left side and as an output on the right side.

(9) Let me just flip the same schematic horizontally by 180 degrees.  You'll see that now it is looking very close to a boost converter. But there is a problem. You see if there were mechanical switches, this would have worked again. But as I told you, in the buck converter we put the switch in the upper position.  Here, if I don't swap the positions of the switch and diode, this will not work. That's an important lesson to keep in mind, that the structure of the switch is the determining factor also in how we can apply a switcher to different applications.

(10) Here we finally have the boost stage and I had to swap the position of the switch and diode to be able to make it work. You can again realize that if there were mechanical switches, it didn't matter which was the so-called switch and the diode. But because the switches are not perfect, you need to do this; otherwise, it won't work.

(11) So I played around with the schematic and I generated a boost. Suppose I even manage to switch the positions of the switch and diode and I got it working. I want to show you in the next slide here that you must understand that-probably there's an error on the title: "The Buck and the Boost Compared." I want to show you that they're topologically different. Even if I did some jugglery and I managed to make it work, they're different topologies in their essence. For example, in the top block I have a buck regulator. I apply a duty cycle of 0.25. I put in an input of 12V and I get an output of 3V. That follows from the equation D=Vo/VIN.
I ask the question, suppose I manage to flip it around? So I start with an input of 3V. What boosted level do I get if I have the same duty cycle? No, I don't get back 12V; I only get back 4V. The underlying reason for this is that the buck is very different from in fact the two other topologies-that's the boost and the buck-boost-in the sense that the buck is the only converter which continues to deliver power to the output, both while the switch is on and when it is off. Whereas for a flyback or boost in a buck-boost topology, what we do is basically when the switch is on we store energy in the inductor, and only when the switch goes off does the current flow into the output. So they are inherently different. That doesn't mean that we cannot use an integrated switcher meant for one application in another application, but we have to understand the limitation arising.

(12) To summarize, a perfect switch does not exist. Diodes are passive switches and we don't have really any effective control over them. The only way we make them turn on and turn off is by causing the voltage across them to change. That we do by switching a bipolar, an active switch. For example, an active switch, we have FETs; we have bipolar transistors; they have a certain voltage rating. Taking an NPN transistor as an example, it has a 30V rating. But you have to understand that a 30V rating is from collector to emitter; it's not the other way around. In the other way around, it'll be a few volts, 6V to 10V; otherwise, you'll damage the device. Similarly, the FETs have an internal body diode, and they can continue to conduct even when we tend to cut them off if the voltage changes.

(13) What really is restricting us? We did see that the switches are not perfect. What else? Now here I have three schematics. The top one is what we call the standard buck converter. I'm assuming that the bottommost one is another one, which I'll talk about. But the middle one is the one I transited to, so that's the positive-to-positive boost. I told you that we needed to swap the positions of the switch and diode. Assuming I could do that here, what else is a problem? The problem is if you go to the topmost schematic, you see that there's a control block, which I am now showing. A control block supplies power to the internal circuitry, and that is driving the switch. It normally demands a DC rail. In a standard buck converter IC, integrated IC, the positive supply to the control comes from exactly the same position as the collector of this transistor, which is the upper rail. Now if I try to transplant this integrated switcher into a standard positive-to-positive boost topology, you can see the problem. The control is connected to the collector of the transistor. Now in this topology, that node is switching up and down. You're not going to get a DC rail for your control anymore. If I had alternatively connected the control or had the flexibility to attach the control to the left side, which is to the positive rail, I would have got it working. But that flexibility does not exist in a standard buck converter IC.
What else can I do with this buck converter IC? The lowermost schematic, I see that this is the so-called negative-to-negative boost configuration. Here I can actually get it to work because the control supply will come from the collector, and that is quiet now. That's not swinging. The bottom, which is the ground of the control, that is uncommitted. I can connect it to the lower rail and I can get it to work. So in fact it is true: The positive-to-positive buck can work; rather, an IC meant for the positive-to-positive buck can do the negative-to-negative boost. But it cannot do the positive-to-positive boost for the reasons as I mentioned.
You can see a pattern starting to emerge here. I have this yellow highlighting. First thing I'll tell you is I'm going to adopt a certain terminology which you must be aware of. That is every power supply, every converter has something called a "switching node." That switching node is the point where the switch, the diode, and the inductor connect. Irrespective of any topology or configuration, let us call that a "switching node." Look at the switching node. In the topmost schematic, you see that you had the diode position and the cathode of the diode is connected to the switching node. Look at the central schematic. You will see that it is the anode which is connecting to the switching node. We know that that topology could not work with this particular IC. The lowermost schematic, you see something else. You see this work. By some strange coincidence, the cathode is again connected to the switching node. In fact, this is the basis for the presentation here, that I discovered a pattern that, just by looking at the diode, you can actually predict what topology or what configuration a given IC can handle. To give you a preview of what I'm going to present, the bottom line is that an IC, which has been designed for a certain configuration as a switching node, it can work for any other topology which has a similar structure around the switching node. Now I'll move to a polling question here, and you can send in your responses.

(14) Let me summarize what we learned from here. Yes, we learned that by applying power at the output of the buck converter we do get a boosted output voltage, but we needed to swap the designations of switch and diode because we don't have perfect switches. For an integrated switcher, we also saw that we have a control block, and the control block needs to get power. There's another issue, and that is the drive signal. The drive signal, the control is driving the switch, and it has to be able to turn it on and off. That also we must be able to continue doing when we try to take a switcher meant for a certain application and try to use it in another. There's another problem, and that is about the feedback. The feedback is luckily a surmountable problem. You can fix a certain circuitry in a way that the IC will get the correct information and react accordingly. The feedback then is usually available, and there's lots of things you can do to get it to work. That's not going to be a stumbling block.

(15) How about the boost? Again there's a title mistake. This should be "Boost and Buck-Boost Compared." I'm not going to go too deep into this, but one thing I want you to realize is that the boost and the buck-boost are almost exactly the same topology. In a sense, you can consider them as different fundamental topologies; but in another sense, they're exactly the same. The bold lines show you a connection by which you get the boost. You have your output capacitor and the negative terminal of that output capacitor is going to the lower rail, and you get a boosted output. If you disconnect this lower terminal of the capacitor, the negative terminal of the capacitor, and connect it to the upper rail, you get a buck-boost topology. If you work out from the duty cycle equations what is the output voltage, as I have an example here which I hope you can see. I have an input of 12V, the duty cycle is 60%, and I use the standard equations for a boost and a buck-boost to see what is the output voltage. Then for the boost case I get 30V as the output, and for the buck-boost I get 18V. You realize that 18V plus 12V, which is the input, is exactly equal to 30V.
What have I done? Basically what I have done is that the output voltage-rather, the voltage not across the cap but at its output upper terminal. In both cases it remains the same if you measure it with respect to the lower rail. The difference between the boost and the buck-boost is essentially that our reference point changes. What we call the ground and the boost, which is the bold three horizontal lines. In the buck-boost we had it dotted at three horizontal lines, and that is what we call the "ground." When we say that there's a certain output voltage, we are measuring it with respect to different points. But as a matter of fact, the voltage on the positive terminal of the capacitor remains the same with respect to the lower input rail. They are very topologically similar. We will also realize that there is no difference in fact between an IC which was meant for a boost application or a buck-boost application. They are actually interchangeable. There is a difference in the feedback, but that, as I told you, we can circumvent and we can overcome.

(16) Here's another slide, which shows you an important difference between the buck and the rest of the topologies, which is the buck-boost and the boost. As I told you, the buck continues to deliver current to the output, even when the switch is on. For the buck, we can show that the average inductor current is equal to the load current. However, for both the boost and the buck-boost, it is the average diode current which is equal to the output load current. If you work backwards, what you find is that it's only for the buck-boost and the boost that the correlation between the switch current and the load current is not so simple as it is for the buck. For the buck, when you say that it's delivering five amperes, the switch is also seeing five amperes, +/- 20% ramp portion. But for the buck-boost and the boost, the inductor current rises as you increase your duty cycle, and it is Io/(1-D). That is the reason why a buck IC cannot work for the same load current in a different application. For example, a buck IC; suppose you have a one-ampere buck IC. Its current limit is set to about say 1.3 amperes or 1.5 amperes to account for the little ramp which you have. Since the load current is equal to the switch current, you can get one ampere of load current. But when you apply that IC to a buck-boost topology, your switch current goes up to Io/(1-D).  For each case you actually have to work out the duty cycle to know how much load current you can get because the switch current limit is fixed. The switch current limit is always internal and its purpose is to predict a switch.

(17) While seeing the pattern emerge from all this, I realized that the only easy way to understand which IC can do which application is to think in terms of something called the "LSD cell." This is not as colorful as it sounds. It stands for Inductor, Switch, and Diode. We'll see, as I mentioned earlier, that any IC which is primarily designed for a certain application which uses a given type of LSD cell, if you start thinking in terms of LSD cells, then you can go to any other topology or configuration and look at the LSD cell in that. If the LSD cell is the same as in the original application, that IC can most probably be used. Of course you'll have to account for the fact that you'll have to correctly calculate what your voltage rating is, how much load current you can get; but yes, it can work.

(18) This is a little bit of the terminology I have. The compatible topologies, they have the same LSD cell, and then there are incompatible topologies which are different LSD cells.  You can ask the question that does that mean that if the LSD cell is different I can never use that IC? It turns out that for one particular IC, which is the boost or the buck-boost IC, you can so-call enforce an opposite cell. We'll see that there's a trick involved in doing that. But on the face of it, that is not very obvious.

(19) The bottom line is a switcher is a switcher is a switcher. You have to think of it as something which basically is a switch which is turning on and off. You have a diode and you have an inductor and you need to start thinking in terms of LSD cells.

(20) I have been talking about ground. It's also helpful when you're studying all these different permutations and combinations to realize that the so-called designation of the ground is something we thought of. There are several types of ground. The power ground is what I'm showing in this schematic. The story goes like this, that in any power converter you have two input rails and you have two output rails. One of these rails is always common, and that is the rail you designate as a ground. There is nothing sacred about it. You must rather think in terms of what is the high level and what's the low level rather than always think in terms of where's a ground. That will help you in applying the IC to different applications.

(21) Here I have what is called a "negative-to-negative configuration." It's because the common rail is the upper rail, the dotted line, and therefore I have what is called a "negative to negative," because the input as measured with respect to that ground is negative now, and same with the output.

(22) This is the negative-to-positive configuration, and I just straighten it out. You can see in the lower schematic what it normally looks like.

(23) This is the positive-to-negative configuration.

(24) As I mentioned, there's something called the "IC ground," and that need not be the same as the power ground.

(25) To be able to talk in a more generic sense, what I also did was I said I'll define something called the "N-switch" and the "P-switch." The N-switch is basically an N-FET or an NPN because you know they behave similarly. You have to take the voltage high to turn them on if they are N-channels or an NPN.  If you take it low, you turn it off. So there are similarities, and just for convenience I'm going to start referring to them as N-switches and P-switches more often.

(26) This is how you turn them off, so they're similar; the bipolar and the FET are similar. There is a difference, and I'll point it out a little later.

(27) This is the LSD cell. Now if you look at the top one you see, first I start with two mechanical switches and then I break it up into what are real switches. The top I call a "positive LSD cell" simply because it's my designation because the cathode is looking at the switching node. The other cell on the right of it is what I called the "negative LSD cell." It's got the anode connected to the switching node. You can see that this is the only way you can possibly make it work. You have a high rail, and when it passes current then you have to let the diode free you. So these are the only two possible positions.
Now turning from mechanical switches to rail switches, I have shown FETs here. I don't know whether you can see clearly, but A is an N-channel set; B is a P-channel set; D is again an N-channel set; and C is a P-channel set. This is the way you have to look it up.

(28) In terms of this, I had the following terminology. I have what I call the Type A cell; that's an N+ cell. It is an N-switch; it could be an N-channel FET or an NPN bipolar. The cathode is the switching node or the LSD node, as I'm calling it here, too. If you look at this chart, you'll see that for an N-switch you have only two possible cell choices; that's the Type A and the Type B. For the P-channel you have only two; that's the Type C and the Type D.

(29) Here's a simple lookup table; you can refer to it.

(30) What are configurations? What's the difference between a topology and a configuration? You must understand that when we say "I have a step-up or a step-down regulator," what I'm referring to is basically the magnitudes of the input and output voltages. We ignore the signs. Those are the topologies. We have the buck, we have the boost, and the buck-boost. But as you are seeing now, we could have something called the "positive-to-positive buck" or a "negative-to-negative buck." These qualifiers are what I'm going to call the "configuration," all the topologies.

(31) The buck-boost will take a given voltage and change it to either a smaller voltage or a larger voltage. That's the quality of the buck-boost. You can show that in the process the sign is always inverted. If I start with a positive voltage, I will get a negative voltage. If I start with a negative voltage, I will get a positive voltage.

(32) Here as a reference chart I have all the possibilities of the buck. Of course I've shown it with a FET, but the similar thing holds true for an NPN or a PNP bipolar transistor. So you have the positive-to-positive buck, you have the negative-to-negative buck, and you have two implementations of each. One is with an N-channel device and one with a P-channel device.

(33) Here are the boost configurations. You must notice something here, which I should have pointed out in the last slide. In each schematic I have given the cell type here: A, B, C, or D on the lower corner of these schematics. I have also shown you the drive levels. You see the small waveform, the lowest schematic. You have the drive levels shown here. The gray portions are the place where the switch turns on, basically. I've also shown you the input rails and the ground. You will notice a pattern here. You will notice that there is a problem because, for example, take A here. In A you will see that to turn, these are the signals you have to apply to the gate of the FET. You have to apply a voltage which is higher or outside the input rail. Same goes true for C. You will notice that A and C always have the requirement that you have to apply a drive signal which is outside the input rail, whereas B and D you don't have any such requirement. What happens is that for A and C, how do you get a voltage which is outside the input rail? You need to typically apply a bootstrap circuit, which bumps up the voltage.

(34) These are the buck-boost configurations, and again you will see the same pattern. A and C need to be driven outside the input rail, and that poses a problem. Yes, if instead of the FET you have a bipolar transistor, you can still make it work. But we will pause now for another polling question here. You can send in your responses, and I will continue to the next slide. I am trying to get the slide to-yeah, I did.

(35) For most of the remainder of the presentation I will focus on the most common ICs, which involve N-switches. Sure, there are ICs coming out with B-switches, and notably the buck with a B-switch we'll take it up towards the end. But to get a better idea of the pattern here, let's stick to N-switches. The purpose of this is to tell you that the B-switch configurations can be automatically generated from all the N-switch configurations by something called "inversion." I'm trying to show you here what inversion involves is from the top to bottom. You change 5V to -5V, zero remains zero. Change the direction of the diode, flip the polarities of the capacitors, and you basically get it to invert. Even the drive voltage, 2V becomes -2V. You can take any schematic which has been drawn with an NPN transistor and invert it in this manner, and you get the corresponding schematic for the B switch. Notice that the designations of the collector and the drain and the gate don't change, so don't try to flip those.

(36) Here I give you a live example of how we do it. What I'm doing is I'm starting with the positive-to-positive buck and I'm trying to give you the reflected or the inverted topology, which is the negative-to-negative buck using a P-switch. I invert it and I do all the things I told you in the previous slide. Then I flip it around to make more sense and I get exactly what I had earlier for the negative-to-negative buck with a P-switch. You'll notice that the cell configuration has changed from A to C.

(37) Having understood topologies/configurations, we need to highlight the fact that it's important to keep in mind how the IC itself is constructed because that certainly plays a key part in what applications it can handle, besides of course the cell structures. When you do a survey of most of our common ICs, you find that they fall into two basic categories.

(36) I'm calling this a "Type 1" IC; what it is is a boost or a buck-boost. The problem is that sometimes we market an IC as a boost IC, sometimes as a buck-boost IC, and we may be guilty of not making it clear that they are basically the same. A boost application IC can be used for a buck-boost application right off the bat. Look at the ICs here. You'll see that this is the way a so-called boost or a buck-boost IC is constructed with an N-switch. On the left I have a FET; on the right I have a bipolar transistor. You'll notice the bold lines; you should. You can see that the control block in the driver stage, the lower rail is connected to the emitter or source. You will also see that the supply to the control block is virtually uncommitted. It is not connected to the transistor. We'll see that this affords us an additional degree of flexibility when we try to use this IC for a different application. You will notice one more thing here is that you may need a bootstrap, as I mentioned earlier, which has not been shown here. For a FET, depending upon the fact that the drive is inside or outside the rail, you might need a bootstrap.

(39) Now this is a Type 2 IC, which I am calling a "Type 2 IC." These are standard buck ICs. You will see the difference here. The difference is that the collector or the drain is connected to the upper rail of the control block. You will see that this restricts the ability of the device significantly to handle different applications.

(40) Summarizing, we have Type 1, which is our so-called boost or buck-boost IC; and we have Type 2. If in doubt, look at the positive supply to the control block. Is it uncommitted or is it connected internally? If it is uncommitted, in all likelihood it's going to be a Type 1 IC; and if not, in all likelihood it's going to be a Type 2 IC.

(41) Now coming back a little to the point. NPN-switches are generally easier to drive. We're comparing the case where we had an NPN-switch as compared to say an N-channel set. You know that for example in a Darlington, we can take the base of the transistor and connect it to the collector and it turns on. We don't really need a bootstrap. But for a FET, as you know, we need to take the gate, which is a control terminal, several volts above the source. If the transistor turns on, that could well take it outside the input rail.
What is the disadvantage of the NPN? You can get it to turn on without a bootstrap, but it will be a little more dissipative than a FET because you need the CE drop across the transistor basically to pump enough base current and to keep it on. Though you don't need a bootstrap for the NPN, you will have a problem because it is going to be a little more dissipative and the drop across the switch is going to be more. The only way to reduce the drop is to use a FET, and that forces you on many occasions, depending upon the topology and configuration, to use a bootstrap.

(42) Doing a little survey of ICs here, I picked some parts randomly from our portfolio. We have an LM1575/2575. This is our first generation buck IC. It's a 52 kHz switcher and it uses an NPN transistor. You can see that the collector of the transistor is basically connected to the control block. This is a buck IC and this is a Type 2 IC by our definition.

(43) Here we have the LM2590HV. This is again a one-ampere part. It's a full-feature version of the 2591. You will see that it uses again NPN transistors.
You will also notice that the ground, though not shown, is uncommitted. It's also probably not very clear that the way the power to the control block is coming. It is clear in this particular one. You can see it coming from the pin on the right which is going to the collector of the transistor, so this is a Type 2 IC.

(44) Summarizing, for a Type 2 IC with an NPN-switch there is going to be a certain drop across the switch. It's about 1.4V typically, and that drop is needed to keep the switch on.

(45) Here's another-this is a third generation part: LM2670, the N-FET, because it is going to be used in a buck converter IC. You saw from our previous diagrams that that would require the rails to be outside the input rail. You will need a bootstrap for this. What are the advantages and disadvantages of a bootstrap? The bootstrap means that you cannot have a controller with 100% duty cycle because every cycle you need to switch off to be able to refresh the bootstrap capacitor.

(46) Here we have a summary of the differences. A FET-based IC would have usually a smaller drop, but it may need a bootstrap. That depends not on the IC but on the configuration or topology. For that you have to go back to the previous slides where I gave you all the topologies and configurations to see whether the drive signal is within the rails or outside the rails. For the bipolar you'll have a higher drop, but no bootstrap is needed and it is less complex.

(47) I have a question here is that this is a boost IC. We sell it as a boost IC, the LM2577. We don't make it very clear even on the front page whether it can do a flyback or a buck-boost application, but in fact it can. You know by now that the two topologies are similar and there's no reason why you cannot.

(48) If we look at the internal diagram, there's a problem here. You don't even see where the control-well, you don't see where the ground of the control is connected. Assuming that the upper side of the control is uncommitted, the ground must be connected somewhere, and it happens to be connected to the ground of the IC, and this makes it a Type 1 IC. I'm just teaching you how to recognize the IC, to think in terms of the type of IC and then subsequently the cell structure, and then you will get the picture here.

(49) As I mentioned, the LM2577, somewhere inside the datasheet we do show a flyback application, even though on the front page we didn't make it very clear.

(50) Here's a unique device in our portfolio. This is an LM1578/2578. The output transistor is actually uncommitted. This doesn't fall into the category of a Type 1 or Type 2. What it basically does is you can use it either way. You can use it as a Type 1 or a Type 2. It's a very interesting product, and I think we need to exploit the potential of this a little more.

(51) Here we intend to use it as a positive-to-positive boost or a positive-to-positive buck, even though they're entirely different. The other thing while looking at the ICs, as I told you once, you should try and look at what the position of the switch is and what pins it connects to inside the IC. As I showed you, very confused.

(52) The other confusing thing is that you should be careful that the labels we give to it are sometimes misleading. It varies even within a company, depending upon which engineer worked on it. Sometimes you might have the switching node then being called a switch; sometimes the output. Sometimes you could have the bootstrap being called the boot, so watch out for that. It's better to draw it out and see where's the switch.

(53) We are going to talk about flyback to refer to the buck-boost, but as one where we have primary to secondary isolation.

(54) Here's a brief summary of the zeta, Cuk, sepic, and boost-buck. This is extracted from an article on the National Edge under the title, "Slave Converters." You'll find it archived there. It was originally in the EDN magazine. You can go into it detailed; there's very little math here, but what I try to show here is that all these four topologies are actually almost identical. They consist of a boost stage followed by a buck cell. The only difference is that-in three of these, at least-we have knocked off a switch. This knocking-off of the switch was referred to originally by Cuk as the so-called "topological reduction of a switch." But even he recognized that there were composite topologies. You have to also understand that on the left side you can call it a boost, you can call it a buck-boost; it doesn't really matter. The interesting thing about all these topologies is that they share the same input/output transfer function. For each of them, Vo/VIN = D(1-D). It's no coincidence because they are actually all virtually the same. There are differences in the way the current is routed, and for that you actually have to work out the current paths. But from the voltage point of view, there's almost no difference. The grayed-out rails are rails which you could have implemented. One of the striking coincidences about this is that if you're regulating the output, you'll find that these grayed-out rails also give you regulated output if you want to use them. For more information on these composite topologies, I suggest you go through that article, "Slave Converters."

(55) Here's the logic behind the hidden applications. The Type 1 IC, which we are calling, is the boost application. It's the positive-to-positive boost. If you go back to the previous configurations, you will see that this is a Type B cell.

(56) The bottom line is that this is comfortable with any configuration, any topology where you have a Type B cell. A Type 1 IC cell, if you go back to our previous diagrams-and unfortunately I don't have time to again go through them. But if you refer back you will see that these are the applications that are possible simply on the basis of the LSD cell being the same.

(57) Now I come to actual implementations. This is the Type 1 IC, which is the boost/buck-boost IC, and this is its normal intended application. I have also given all the parameters on the top, which you need to verify that it is okay for a given application or not. Another thing you must keep in mind while designing is that in all cases when you make a buck topology, you are better off trying to design the inductor and verify the current limit at the maximum input voltage. Whereas for the boost and the buck-boost, because the inductor current goes up as the duty cycle increases, you have to do the design of the inductor and see the current limit at the lower input voltage.

(58) Here, a different implementation; you can see all the equations. I have given the Vswmax (Vswitch,max equation). In this case, for example, you must make sure that your switch pin-by the way, the switch pin has a different rating than the control pin in all our datasheets. The switch pin must have a rating which is greater than the maximum input voltage plus the output voltage because the output voltage is going to make it go up even more than the input voltage because it's a buck-boost. But you also need to have the IC start up, and that you will do at the minimum input voltage because that's where you have the minimum voltage and you still want it to start up. This is, by the way, a differential stage because you have to look at the two top terminals and you have to convey the information to the IC, whereas the IC is referenced to a different point. This is the -V end rail, which I have labeled here. To bring that information and convey it to the IC, you need this sort of a dirty, cheap differential stage.

(59) Here is a negative-to-negative buck using the buck-boost IC. This is the way it is drawn. You still need a differential stage. All the equations are provided. Another thing you have to try and keep in mind is that every controller may have a duty cycle max limit, and you should be able to manage. Your application should not demand a duty cycle higher than the capability of the device, so in each case I have also provided that. This is just to control this. You can go to it to verify if your application is suitable for the device.

(60) Here's a negative-to-negative boost. You did need a differential stage. All you needed was a simple voltage divider.

(61) Now this is the trick. As I told you earlier, the Type 1 IC, which is the buck-boost IC or the positive-to-positive boost IC, this can do the opposite cell. It is primarily intended for a Type B cell, but it can do a Type A cell. That is because of the flexibility afforded by the fact that the supply to the control is not connected and is uncommitted. This is how you do it. You basically float the IC itself on the inductor on the switching node. Now some ICs may not like it, so you should actually build it to see whether this is going to work in principle it would. But some ICs may be uncomfortable with the noise that's generated, so for such cases you should try it out.
How do they solve the problem of the control getting a DC supply? Well, they rectify it. Because the control supply is uncommitted, you rectify it and you therefore need a decoupling capacitor also to the IC, which I've shown here. What you need to do to the output is you have a divider and you send it back to the input.

(62) Here is a Type A LSD cell again being implemented by a boost IC. It's a positive-to-positive buck, and it can be done. Here, notice that not only I had to rectify the supply to the control, but I also had to do the same for the output voltage because the output voltage with respect to the IC ground is again swinging because the IC ground itself is swinging. How is it going to sense the output voltage? You need to rectify that with respect to the IC ground and then put a voltage divider and bring that signal back to the IC.
This is a standard trick. What do you lose from here? Whenever you make a switcher which is the boost IC, do an application which is not commensurate with its original LSD cell design, then basically you have to do this particular trick. You lose something along the way because you're rectifying the output. You have a diode drop and all the variations possible with that, so the regulation is never going to be very good. But in some cases you might be able to live with that, so you can certainly try it out.

(63) Here are the summary of the applications possible for a positive-to-positive boost IC. The Type B, the yellowed portions are its natural cell choice, which is the original cell it was designed for, so it can cover any topology which has that. Those are the bold arrows here. But it can also cover the opposite cell. You remember that for the N-switch you have only two possibilities, the A or the B. So the opposite choice, what I talk about here is the A cell here. It can do the opposite cell, but it does it with that little trick of rectification which I talked about, so the regulation is not going to be very good.

(64) Now we turn our attention to the buck IC, which is actually the simpler of the two because it's restricted. It's positive control.  The pin to the control on the upper rail is connected to the collector or the drain of the transistor, and that makes it unable to handle any of the C-cells which it was not originally intended for. Basically this one can do only those cells which it was intended for. That cell, if you look at the positive-to-positive buck, it is a Type A cell so it can handle any configurations where you have a Type A cell. These are just a summary of that.

(65) Here are the actual implementations. These are all straightforward. The equations are given to you. You would be able to go to them as a checklist and you'll be able to see whether you are satisfying the condition.

(66) Here again we have a cell; it's a positive-to-negative buck-boost.

(67) I'm going to go through this a little faster because I'm running short of time here.

(68) We have enforceable choices for Type 1 IC, but for Type 2 IC there are no enforceable choices.

(69) This is a summary of their applications. As you can see, it was meant for a Type A application and it can only do a Type A application.

(70) I did briefly cover transformer-based applications, but I don't have much time to go to them. It's just as a reference, that in case tomorrow you want to build a transformer-based one, you can do it.

(71) (72) This is a differential sensing technique. This is an alternative to the cheap and dirty single transistor sense, which I talked about. This will give you a more accurate result. The equations are given on the right side. You are to be conscious of things like offset voltage of availability and where you may need to power the op amp externally.

(73) Here's a more accurate one. The equation for this is normally not given in any book, so I derived it. It's a higher gain stage because it has a divider on the output.

(74) Here, just in case you couldn't read some of those, they are given out here again.

(75) Very briefly I'll touch what we learned so far about the cell structure we did with an N-switch. Just briefly I'll extend it to a very common application coming up nowadays, and that is the buck IC, which uses a P-switch. In the lowermost section, I have that. Just on the basis of the query which I generated in terms of the LSD cells, I can predict which one will work and which will not work. AOK stands for yes, absolutely okay. OK is okay. POK, possibly okay; it may involve that you need to build it because the ground may be swinging, as I mentioned earlier, so try it out. NOK is not okay; it just won't work.

(76) Here I gave some design examples of the flyback regulator.

(77) It's easy enough, the math, and you can just go through the checklist and see whether this will work or not.

(78) Here's another example; I again go through some examples.

(79) It's a checklist, and you'll verify that the IC can work for a given application. As I told you, even though the LSD cells may be commensurate or enforceable, you still do have to check the current limits, the duty cycle max, the wattage rating, etc.

(80) Very briefly, layout practices. These are the critical things when you build it because to make it work you need to pay attention to the fact that the topologies are different in some respect. Unless you pay attention to the layout, you will not get it. For example, in the buck IC the input capacitor position is critical, but the output capacitor is not critical. But in both the buck-boost and the boost, you have to make sure that those trace lengths are small; otherwise, it will not work properly.

(81) Then there's some finer nuances when you this morphology. One of the things you should look at is the right half plane zero phenomena. The right half plane is explained intuitively in the following way. The buck topology doesn't have it, but the boost and the buck-boost do have this right half plane zero. It can be traced back to the fact that both for the boost and the buck-boost, you are building an energy into the inductor only when the switch is on, and you'd cycle energy to the output only when it's off. For example, if there's a sudden additional demand on the output, the converter responds by trying to increase the duty cycle. But that gives less time for the energy to flow into the output, so the voltage actually dips further, and that is the intuitive explanation for the right half plane zero. So there are some tricks for conquering the right half plane zero.

(82) Given here, the top one is the standard one we have in our data book.

(83) The other one is another trick, which we haven't really given in any of the datasheets but you can certainly try it out. This you can work out from the intuitive explanations. If the duty cycle is dipping in response to a sudden load step, then how to not let it dip too much. So in that sense you will be able to conquer the right half plane zero. That is here; this has been outlined here. This little thing is going to actually prop up the duty cycle a little, so the right half plane zero will be less prominent.

(84) These are the references which you have also on the lower side of the screen, I guess.

(85) Here's a thank you. I need to thank a lot of people who periodically review my literature, who have provided support for this event and others. I'll briefly go through them. At NSC Santa Clara I'd like to thank Wanda Garrett, Maurice Eaglin, Paul Greenland, Jon Cronk, Tanya Quach, Mike He for a lot of support along the way. At NSC in Colorado, Mark Hartman periodically reviews my literature. In our Rhode Island facility we have Leo Sheftelevich. In India I'd like to thank Dr. G. T. Murthy. And that's all. I'll hand it over to Wanda for the rest of it. Thank you very, very much.

MODERATOR:
Thank you, Sanjaya. You've given us a lot of food for thought. Rest assured, folks, that this material is available in the archives in about three hours. You can go through it at your leisure and review some of the details that Sanjaya was not able to spend time on this morning.
We have a survey that we're going to be sending you here shortly. Please do fill out the survey form.  In addition, if you have questions that you would like to send in to Sanjaya, please do that at this time.

Here's just a thank you from Hans. Hans says, "Thank you for this seminar, Dr. Maniktala. I am a student trying to build a 500-watt switch and power supply, and the seminar has been helpful."

MANIKTALA:
Thank you very much, Hans. Feel free to write to me and we can touch base if you have any further questions.
Aaron has also asked whether we have anything for audio amplifier power supply. I guess what he means is Class D. We do have some activity, but Wanda will probably answer that question.

MODERATOR:
Right. For a Class D amplifier for audio amplifiers, I have been in the lab several times when a number of years ago we were trying to use regular switching regulators intended for power supplies for audio applications. My goodness, it was awfully noisy at some times. At this time there are some really good Class D amplifiers intended for the audio amplifier application. I would direct you toward the LM4651 and the LM4652 for a dedicated Class D audio amplifier.

MANIKTALA:
I have a question from Ramesh. How to design the constant current mode sepic converter. This is a tricky one, but I'll suggest, since this seminar basically focused more on the fundamental topology, I would suggest you try to go through the article "Slave Converters" and you'll see that the sepic may not be as difficult as you thought it was. As far as the only tricky thing I know about that is the boost stability issue, and we ourselves don't have any great support on that yet. I'll try to see if I can find some information for you on that, but otherwise do try to look at the sepic as a composite topology of the boost and the buck, and it'll probably be very clear to you after that.

MODERATOR:
One of the tools that we do have available to everyone is our online design environment called the WEBENCH. We have a Power WEBENCH which includes many buck regulators, flyback regulators, boost converters. There are a couple of sepic converters that are included there. I believe we're working on an inverting buck-boost, although I'm not quite sure that it's ready yet. But this online design environment takes you all the way from your design requirements to a bill of materials and a simulatable design. If you wish to explore some of these topologies, that would be a good tool to use.
I believe we are at the end of our time at this point. I wish to again thank Sanjaya for taking us through this very interesting perspective on switching regulators. Thank you, everyone, for joining us for this seminar, Simple Switcher Topologies and their Morphology, brought to you by National Semiconductor and Yahoo! Broadcasts. Please do remember to fill out and submit your survey form. Thank you for attending, and good day.
(End of Presentation.)