The Principles of Switching Power Conversion--Understanding the Inductor

Capacitors/Inductors and Voltage/Current

In power conversion, we may have noticed that we always talk rather instinctively of voltage rails. That is why we also have dc-dc voltage converters forming the subject of this book.

But why not current rails, or current converters For example? We should realize that the world we live in, keenly interact with, and are thus comfortable with, is ultimately one of voltage, not current. So For example, every electrical gadget or appliance we use runs off a specified voltage source, the currents drawn from which being largely ours to determine. So For example, we may have 110 V-ac or 115 V-ac 'mains input' available in many countries. Many other places may have 220 V-ac or 240 V-ac. So if For example, an electric room heater is connected to the 'mains outlet,' it would draw a very large current (~10-20 amperes), but the line voltage itself would hardly change in the process. Similarly, a clock radio would typically draw only a few hundred milliamperes of current, the line voltage again remaining fixed. That is by definition a voltage source. On the other hand, imagine for a moment that we had a 20 A current source outlet available in our wall. By definition, this would try to push out 20 A, come what may - even adjusting the voltage if necessary to bring that about. So, even if don't connect any appliance to it, it would still attempt to arc over, just to keep 20 A flowing. No wonder we hate current sources!

We may have also observed that capacitors have a rather more direct relationship with voltage, rather than current. So C = Q/V, where C is the capacitance, Q is the charge on either plate of the capacitor, and V is the voltage across it. This gives capacitors a somewhat imperceptible, but natural association with our more "comfortable" world of voltages.

It's perhaps no wonder we tend to understand their behavior so readily.

Unfortunately, capacitors are not the only power-handling component in a switching power supply! Let us now take a closer look at the main circuit blocks and components of a typical off-line power supply as shown in Fgr. 1. Knowing what we now know about capacitors and their natural relationship to voltage, we are not surprised to find there are capacitors present at both the input and output ends of the supply. But we also find an inductor (or 'choke') - in fact a rather bulky one at that too! We will learn that this behaves like a current source, and therefore, quite naturally, we don't relate too well to it! However, to gain mastery in the field of power conversion, we need to understand both the key components involved in the process: capacitors and inductors.

Coming in from a more seemingly natural world of voltages and capacitances, it may require a certain degree of mental re-adjustment to understand inductors well enough. Sure, most power supply engineers, novice or experienced, are able to faithfully reproduce the buck converter duty cycle equation For example (i.e. the relationship between input and output voltage). Perhaps they can even derive it too on a good day! But scratch the surface, and we can surprisingly often find a noticeable lack of "feel" for inductors. We would do well to recognize this early on and remedy it. With that intention, we are going to start at the very basics... .

The Inductor and Capacitor Charging/Discharging Circuits

Let's start by a simple question, one that is sometimes asked of a prospective power supply hire (read "nervous interviewee"). This is shown in Fgr. 3.

Note that here we are using a mechanical switch for the sake of simplicity, thus also assuming it has none of the parasitics we talked about earlier. At time t = 0, we close the switch (ON), and thus apply the dc voltage supply (VIN) across the capacitor (C) through the small series limiting resistor (R). What happens?

Fgr. 3: Basic Charging/Discharging Circuits for Capacitor and Inductor

Most people get this right. The capacitor voltage increases according to the well-known exponential curve VIN ×(1-e-t/t), with a 'time constant' of t = RC. The capacitor current, on the other hand, starts from a high initial value of VIN/R and then decays exponentially according to (VIN/R) × e-t/t. Yes, if we wait "a very long time," the capacitor would get charged up almost fully to the applied voltage VIN, and the current would correspondingly fall (almost) to zero. Let us now open the switch (OFF), though not necessarily having waited a very long time. In doing so we are essentially attempting to force the current to zero (that is what a series switch is always supposed to do). What happens? The capacitor remains charged to whatever voltage it had already reached, and its current goes down immediately to zero (if not already there).

Now let us repeat the same experiment, but with the capacitor replaced by an inductor (L), as also shown in Fgr. 3. Interviewees usually get the "charging" part (switch-closed phase) of this question right too. They are quick to point out that the current in the inductor behaves just as the voltage across the capacitor did during the charging phase. And the voltage across the inductor decays exponentially, just as the capacitor current did. They also seem to know that the time constant here is t = L/R, not RC.

This is actually quite encouraging, as it seems we have, after all, heard of the 'duality principle.' In simple terms this principle says that a capacitor can be considered as an inverse (or 'mirror') of an inductor, because the voltage-current equations of the two devices can be transformed into one another by exchanging the voltage and current terms. So, in essence, capacitors are analogous to inductors, and voltage to current.

But wait! Why are we even interested in this exotic-sounding new principle? Don't we have enough on our hands already? Well, it so happens, that by using the duality principle we can often derive a lot of clues about any L-based circuit from a C-based circuit, and vice versa - right off the bat - without having to plunge headlong into a web of hopelessly non-intuitive equations. So in fact, we would do well to try and use the duality principle to our advantage if possible.

With the duality principle in mind, let us attempt to open the switch in the inductor circuit and try to predict the outcome. What happens? No! Unfortunately, things don't remain almost "unchanged" as they did for a capacitor. In fact, the behavior of the inductor during the off-phase is really no replica of the off-phase of the capacitor circuit.

So does that mean we need to jettison our precious duality principle altogether? Actually we don't. The problem here is that the two circuits in Fgr. 3, despite being deceptively similar, are really not duals of each other. And for that reason, we really can't use them to derive any clues either. A little later, we will construct proper dual circuits. But for now we may have already started to suspect that we really don't understand inductors as well as we thought, nor in fact the duality principle we were perhaps counting on to do so.

The Law of Conservation of Energy

If a nervous interviewee hazards the guess that the current in the inductor simply "goes to zero immediately" on opening the switch, a gentle reminder of what we all learnt in high school is probably due. The stored energy in a capacitor is CV2/2, and so there is really no problem opening the switch in the capacitor circuit - the capacitor just continues to hold its stored energy (and voltage). But in an inductor, the stored energy is LI 2/2. Therefore, if we speculate that the current in the inductor circuit is at some finite value before the switch is opened and zero immediately afterward, the question arises: to where did all the stored inductor energy suddenly disappear? Hint: we have all heard of the law of conservation of energy - energy can change its form, but it just cannot be wished away! Yes, sometimes a particularly intrepid interviewee will suggest that the inductor current "decays exponentially to zero" on opening the switch. So the question arises - where is the current in the inductor flowing to and from? We always need a closed galvanic path for current to flow (from Kirchhoff's laws)!

But, wait! Do we even fully understand the charging phase of the inductor well enough? Now this is getting really troubling! Let's find out for ourselves!

The Charging Phase and the Concept of Induced Voltage

From an intuitive viewpoint, most engineers are quite comfortable with the mental picture they have acquired over time of a capacitor being charged - the accumulated charge keeps trying to repel any charge trying to climb aboard the capacitor plates, till finally a balance is reached and the incoming charge (current) gets reduced to near-zero. This picture is also intuitively reassuring, because at the back of our minds, we realize it corresponds closely with our understanding of real-life situations - like that of an over-crowded bus during rush hour, where the number of commuters that manage to get on board at a stop depends on the capacity of the bus (double-decker or otherwise), and also on the sheer desperation of the commuters (the applied voltage).

But coming to the inductor charging circuit (i.e. switch closed), we can't seem to connect this too readily to any of our immediate real-life experiences. Our basic question here is - why does the charging current in the inductor circuit actually increase with time. Or equivalently, what prevents the current from being high to start with? We know there is no mutually repelling "charge" here, as in the case of the capacitor. So why? We can also ask an even more basic question - why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument - because we know Ohm's law (V = I × R) all too well. But an inductor has (almost) no resistance -it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to "hold-off" any voltage across it? In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor - we are not very clear! Further, if what we have learnt in school is true - that electric field by definition is the voltage gradient dV/dx ("x" being the distance), we are now faced with having to explain a mysterious electric field somewhere inside the inductor! Where did that come from? It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of 'induced voltage.' This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor - it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed, as in Fgr. 3) is the 'induced voltage.' Note: We also observe that the analogy between a capacitor/inductor and voltage/current, as invoked by the duality principle, doesn't stop right there! For example, it was considered equally puzzling at some point in history, how at all any current was apparently managing to flow through a capacitor - when the applied voltage across it was changed. Keeping in mind that a capacitor is basically two metal plates with an interposing (non-conducting) insulator, it seemed contrary to the very understanding of what an "insulator" was supposed to be. This phenomenon was ultimately explained in terms of a 'displacement current,' that flows (or rather seems to flow) through the plates of the capacitor, when the voltage changes. In fact, this current is completely analogous to the concept of 'induced voltage'- introduced much later to explain the fact that a voltage was being observed across an inductor, when the current through it was changing.

So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase in Fgr. 3, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps in to try to keep the current down to its initial value (zero). So we apply 'Kirchhoff's voltage law' to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm's law). As time progresses, we can think intuitively in terms of the applied voltage "winning." This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff's voltage law). That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.

Why does the applied voltage "win"? For a moment, let's suppose it didn't. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other - and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.

We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday's/Lenz's law). And this "allows" for the additional drop appearing across the resistor, as per Kirchhoff's voltage law! But we still don't know how the induced voltage behaves when the switch turns OFF! To unravel this part of the puzzle, we actually need some more analysis.

The Effect of the Series Resistance on the Time Constant

Fgr. 4: Inductor Current during Charging Phase for Different R (in ohms), for an Applied Input of 10 V

Fgr. 5: Inductor Voltage during Charging Phase for Different R (in ohms), for an Applied Input of 10 V

Let us ask - what are the final levels at the end of the charging phase in Fgr. 3 - that is, of the current in the inductor and the voltage across the capacitor. This requires us to focus on the exact role being played by R. Intuitively we expect that for the capacitor circuit, increasing the R will increase the charging time constant t. This is borne out by the equation t = RC too, and is what happens in reality too. But for the inductor charging circuit, we are again up against another seemingly counter-intuitive behavior - increasing R actually decreases the charging time constant. That is in fact indicated by t = L/R too.

Let us attempt to explain all this. Looking at Fgr. 4 which shows the inductor charging current, we can see that the R = 1 ? current curve does indeed rise faster than the R = 2 ? curve (as intuitively expected). But the final value of the R = 1 ? curve is twice as high.

Since by definition, the time constant is "the time to get to 63% of the final value," therefore the R = 1 ? curve has a larger time constant, despite the fact that it did rise much faster from the get-go. So that explains the inductor current waveforms.

But looking at the inductor voltage waveforms in Fgr. 5, we see there is still some explaining to do. Note that for a decaying exponential curve, the time constant is defined as the time it takes to get to 37% of the initial value. So in this case we see that though the initial values of all the curves are the same, yet For example, the R = 1 ? curve has a slower decay (larger time constant) than the R = 2 ? curve! There is actually no mystery involved here, since we already know what the current is doing during this time (Fgr. 4), and therefore the voltage curves follow automatically from Kirchhoff's laws.

The conclusion is that if, in general, we ever make the mistake of looking only at an inductor voltage waveform, we may find ourselves continually baffled by an inductor! For an inductor, we should always try to see what the current in it’s trying to do. That is why, as we just found out, the voltage during the off-time is determined entirely by the current. The voltage just follows the dictates of the current, not the other way around. In fact, in Section 5, we will see how this particular behavioral aspect of an inductor determines the exact shape of the voltage and current waveforms during a switch transition, and thereby determines the crossover (transition) loss too.

The Inductor Charging Circuit with R = 0, and the "Inductor Equation"

What happens if R is made to decrease to zero?

From Fgr. 5 we can correctly guess that the only reason that the voltage across the inductor during the on-time changes at all from its initial value VIN is the presence of R! So if R is 0, we can expect that the voltage across the inductor never changes during the on-time! The induced voltage must then be equal to the applied dc voltage. That is not strange at all - if we look at it from the point of view of Kirchhoff's voltage law, there is no voltage drop present across the resistor - simply because there is no resistor! So in this case, all the applied voltage appears across the inductor. And we know it can "hold-off" this applied voltage, provided the current in it’s changing. Alternatively, if any voltage is present across an inductor, the current through it must be changing! So now, as suggested by the low-R curves of Fgr. 4 and Fgr. 5, we expect that the inductor current will keep ramping up with a constant slope during the on-time. Eventually, it will reach an infinite value (in theory). In fact, this can be mathematically proven to ourselves by differentiating the inductor charging current equation with respect to time, and then putting R = 0 as follows [...]

So we see that when the inductor is connected directly across a voltage source VIN, the slope of the line representing the inductor current is constant, and equal to VIN/L (the current rising constantly).

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Current Surge Voltage Arc Current Source Voltage Source Switch Closed Switch Open Switch Closed Switch Open

Fgr. 6: Mirror Circuits for Understanding Inductor Discharge

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Note that in the above derivation, the voltage across the inductor happened to be equal to VIN, because R was 0. But in general, if we call "V" the voltage actually present across the inductor (at any given moment), I being the current through it, we get the general "inductor equation":

dI/dt = V/L (inductor equation)

This equation applies to an ideal inductor (R = 0), in any circuit, under any condition.

For example, it not only applies to the "charging" phase of the inductor, but also its "discharging" phase! Note: When working with the inductor equation, for simplicity, we usually plug in only the magnitudes of all the quantities involved (though we do mentally keep track of what is really happening - i.e. current rising or falling).

The Duality Principle

We now know how the voltage and current (rather its rate of change), are mutually related in an inductor, during both the charging and discharging phases. Let us use this information, along with a more complete statement of the duality principle, to finally understand what really happens when we try to interrupt the current in an inductor.

The principle of duality concerns the transformation between two apparently different circuits, which have similar properties when current and voltage are interchanged. Duality transformations are applicable to planar circuits only, and involve a topological conversion: capacitor and inductor interchange, resistance and conductance interchange, and voltage source and current source interchange.

We can thus spot our "mistakes" in Fgr. 3. First, we were using an input voltage source applied to both circuits - whereas we should have used a current source for the "other" circuit. Second, we used a series switch in both the circuits. We note that the primary function of a series switch is only to interrupt the flow of current - not to change the voltage (though that may happen as a result). So if we really want to create proper mirror (dual) circuits, then forcing the current to zero in the inductor is the dual of forcing the voltage across the capacitor to zero. And to implement that, we obviously need to place a switch in parallel to the capacitor (not in series with it). With these changes in mind, we have finally created true dual circuits as shown in Fgr. 6 (both are actually equally impractical in reality!).

The "Capacitor Equation"

To analyze what happens in Fgr. 6 we must first learn the "capacitor equation" - analogous to the "inductor equation" derived previously. If the duality principle is correct, both the following two equations must be valid:

V = L (dI/dt) (inductor equation)

I = C (dV/dt) (capacitor equation)

Further, if we are dealing with "straight-line segments" (constant V for an inductor and constant I for a capacitor), we can write the above equations in terms of the corresponding increments or decrements during the given time segment.

V = L

(inductor equation for constant voltage)

(capacitor equation for constant current)

It’s interesting to observe that the duality principle is actually helping us understand how the capacitor behaves when being charged (or discharged) by a current source! We can guess that the voltage across the capacitor will then ramp up in a straight line - to near infinite values - just as the inductor current does with an applied voltage source. And in both cases, the final values reached (of the voltage across the capacitor and the current through the inductor) are dictated only by various parasitics that we have not considered here - mainly the ESR of the capacitor and the DCR of the inductor respectively.

The Inductor Discharge Phase

We now analyze the mirror circuits of Fgr. 6 in more detail.

We know intuitively (and also from the capacitor equation) what happens to a capacitor when we attempt to suddenly discharge it (by means of the parallel switch). Therefore, we can now easily guess what happens when we suddenly try to "discharge" the inductor (i.e. force its current to zero by means of the series switch).

We know that if a "short" is applied across any capacitor terminals, we get an extremely high current surge for a brief moment - during which time the capacitor discharges, and the voltage across it ramps down steeply to zero. So we can correctly infer that if we try to interrupt the current through an inductor, we will get a very high voltage across it - with the current simultaneously ramping down steeply to zero. So the mystery of the inductor "discharge" phase is solved - with the help of the duality principle! But we still don't know exactly what the actual magnitude of the voltage spike appearing across the switch/inductor is. That is simple - as we said previously, during the off-time, the voltage will take on any value to force current continuity. So a brief arc will appear across the contacts as we try to pull them apart (see Fgr. 6). If the contacts are separated by a greater distance, the voltage will increase automatically to maintain the spark. And during this time, the current will ramp down steeply. The arcing will last for as long as there is any remaining inductor stored energy - that is, till the current completely ramps down to zero. The rate of fall of current is simply V/L, from the inductor equation. So eventually, all the stored energy in the inductor is completely dissipated in the resulting flash of heat and light, and the current returns to zero simultaneously. At this moment, the induced voltage collapses to zero too, its purpose complete. This is in fact the basic principle behind the automotive spark plug, and the camera flash too (occurring in a more controlled fashion).

But wait - we have stated above that the rate of fall of current in the inductor circuit was "V/L." What is V? V is the voltage across the inductor, not the voltage across the contacts.

In the following sections, we will learn that the voltage across an inductor (almost always) reverses when we try to interrupt its current. If that is true, then by Kirchhoff's voltage law, since the algebraic sum of all the voltage drops in any closed circuit must add up to zero, the voltage across the contacts will be equal to the sum of the magnitudes of the induced voltage and the applied dc rail -- however, the sign of the voltage across the contacts (i.e. its direction) will necessarily be opposite to the other voltages (see the gray triangles in the lower schematic of Fgr. 6 ). Therefore, we conclude that the magnitude of the voltage spike across the inductor is equal to the magnitude of the voltage across the contacts, minus the magnitude of the input dc voltage.

Finally, we know everything about the puzzling inductor discharge phase!

Flyback Energy and Freewheeling Current

The energy that "must get out" of the inductor when we try to open the switch is called the 'flyback' energy. The current that continues to force its way through is called the 'freewheeling' current. Note that this not only sounds, but in fact is, very similar to another real-world situation - that of a mechanical spinning wheel, or a 'flywheel.' In fact, understanding the flywheel can help greatly in gaining an intuitive insight into the behavior of an inductor.

Just as the inductor has stored energy related to the current flowing through it, the flywheel stores energy related to its spinning action. And neither of these energy terms can be wished away in an instant. In the case of the flywheel, we can apply "brakes" to dissipate its rotational energy (as heat in the brake linings) - and we know this will produce a progressive reduction in the spinning. Further, if the brakes are applied more emphatically, the time that will elapse till the spinning stops entirely gets proportionately decreased. That is very similar to an inductor - with the induced voltage (during the off-time) playing the part of the "brakes" and the current being akin to the spinning. So, the induced voltage causes a progressive reduction in the current. If we have a higher induced voltage, this will cause a steeper fall in the current. In fact, that is also indicated by the inductor equation V = L dI/dt!

However, we have also learned something more fundamental about the behavior of an inductor, as described next.

Current Must Be Continuous, Its Slope Need Not Be

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SPIKE! Current Voltage dI/dt (or Voltage) dV/dt (or Current) Possible and Acceptable Possible but Unacceptable Impossible Possible and Acceptable Possible but Unacceptable Impossible INDUCTOR CAPACITOR

Fgr. 7: Inductor Current Must Be Continuous, But Its Slope Need Not Be. Capacitor Voltage Must Be Continuous, But Its Slope Need Not Be.

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The key word in the previous section was progressive. From a completely mathematical/geometrical point of view now, we should understand that any curve representing inductor current cannot be discontinuous (no sudden jumps allowed) - because that will in effect cause energy to be discontinuous, which we know is impossible. But we can certainly cause the slope of the current (i.e. its dI/dt) to have "jumps." So we can, For example, change the slope of current (dI/dt) in an instant - from one representing a rising ramp (increasing stored energy), to one representing a falling ramp (opposite sign, i.e. decreasing energy). However, the current itself must always be continuous. This is shown in Fgr. 7, under the choices marked "possible."

Note that there are two options in the figure that are "possible." Both are so, simply because they don’t violate any known physical laws. However, one of these choices is considered "unacceptable," because of the huge spike - which we know can damage the switch. The other choice, marked "acceptable," is in fact what really happens in any switching converter topology, as we will soon see.

The Voltage Reversal Phenomenon

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VIN I(t) I(t) VIN - +

Switch Closed, Switch Open

Fgr. 8: How the Voltage Reverses on Attempted Interruption of Inductor Current

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We mentioned there is a voltage reversal across the switch, when we try to interrupt its current. Let us try to understand this better now.

An intuitive (but not necessarily rigorous) way to visualize it’s shown in Fgr. 8. Here we note that when the switch is closed (upper schematic), the current is shown leaving the positive terminal of the applied dc voltage source - that being the normal convention for describing the direction of current flow. During this on-time, the upper end of the inductor gets set to a higher voltage than its lower end. Subsequently, when the switch opens, the input dc source gets disconnected from the inductor. But we have just learnt that the current demands to keep flowing (at least for a while) - in the same direction as previously flowing.

So during the switch off-time, we can mentally visualize the inductor as becoming a sort of "voltage source," forcing the current to keep flowing. For that reason, we have placed an imaginary (gray) voltage source (battery symbol) across the inductor in the lower half of the figure - its polarity in accordance with the convention that current must leave by the positive terminal of any voltage source. Thus we can see that this causes the lower end of the inductor to now be at a higher voltage than its upper end. Clearly, voltage reversal has occurred - simply by the need to maintain current continuity.

The phenomena of voltage reversal can be traced back to the fact that induced voltage always opposes any change (in current). However, in fact, voltage reversal does not always occur. For example, voltage reversal does not occur during the initial startup ('power-up') phase of a boost converter. That is because the primary requirement is only that the inductor current somehow needs to keep flowing - voltage takes a backseat.

So hypothetically, if a circuit is wired in a certain way, and the conditions are "right," it’s certainly possible that voltage reversal won't occur, so long as current continuity can still be maintained.

However, we must be clear that if and when a converter reaches a 'steady state,' voltage reversal will necessarily occur at every switch transition.

For that we now have to understand what a "steady state" is:

A Steady State in Power Conversion, and the Different Operating Modes

A steady state is, as the name indicates - stable. So it’s in essence the opposite of a runaway or unstable condition. But we can easily visualize that we will in fact get an unstable condition if at the end of every cycle, we don't return to the current we started the cycle with - because then, every successive cycle, we will accumulate a net increase or decrease of current, and the situation will keep changing forever (in theory).

From V = Lfli/?t, it’s clear that if the current is ramping up for a positive (i.e. applied) voltage, the current must ramp down if the voltage reverses. So the following equations must apply (magnitudes only) [...]

Here the subscript "ON" refers to the switch being closed, and "OFF" refers to the switch being open. VON and VOFF are the respective voltages across the inductor during the durations ?tON and ?tOFF. Note that very often, ?tON is written simply as "tON," the switch on-time. And similarly, ?tOFF is simply "tOFF," the switch off-time.

Now suppose we are able to create a circuit in which the amount the current ramps up by in the on-time (fliON) is exactly equal to the amount the current ramps down by during the off-time (fliOFF). If that happens, we would have reached a steady state. Now we could repeat the same sequence an innumerable amount of times, and get the same result each and every time. In other words, every "switching cycle" would then be an exact replica of the previous cycle. Further, we could also perhaps get our circuit to deliver a steady stream of (identical) energy packets continuously to an output capacitor and load. If we could do that, by definition, we would have created a power converter! Achieving a steady state is luckily not as hard as it may sound. Nature automatically tries to help every natural process move towards a stable state (without "user intervention"). So in our case, all we need to do on our part is to provide a circuit that allows these conditions to develop naturally (over several cycles). And if we have created the right conditions, a steady state will ultimately result. Further, this would be self-sustaining thereafter. Such a circuit would then be called a switching "topology"!

Conversely, any valid topology must be able to reach a state described by the following key equation fliON = fliOFF = fli. If it can't get this to happen, it’s not a topology. Therefore, this simple current increment/decrement equation forms the litmus test for validating any new switching topology.

Fgr. 9: Different Operating Modes of Switching Regulators

Note that the inductor equation, and thereby the definition of 'steady state,' refers only to the increase/decrease in current - it says nothing about the actual (absolute) value of the current at the start (and end) of every cycle. So there are in fact several possibilities. We could have a steady state in which the current returns to zero every cycle, and this is called a 'discontinuous conduction mode' (DCM). However, if the current stays pegged at some non-zero value throughout, we will have 'continuous conduction mode' (CCM). The latter mode is the most common mode of operation encountered in power conversion. In Fgr. 9 we have graphically shown these operating modes (all in steady state). We also have some other modes that we will talk about very soon. Note that in the figure, the "square" waveform is the voltage across the inductor, and the slowly ramping waveform is the inductor current. Let us make some related observations:

a) We see that the voltage across the inductor always reverses at every switching event (as expected in steady state).

b) We note that since the inductor equation relates voltage to the slope of the current, not to the actual current, therefore, for a given VON and VOFF, several current waveforms are possible (all having the same dI/dt for corresponding segments).

Each of these possibilities has a name - CCM, DCM, BCM (boundary conduction mode, also called critical conduction mode), and so on. Which of these operating modes actually occurs depends on the specific circuit (i.e. the topology) and also the application conditions (how much output power we are demanding and what the input and output voltages are).

c) The inductor voltages, VON and VOFF shown in the figure, are related to the application conditions VIN and/or VO. Their exact relationship will become known a little later, and we will also learn that it depends on the specific topology.

d) A key question is - what is the exact relationship between the average inductor current and the load current? We will soon see that that too depends on the specific topology. However, in all cases, the average inductor current ("I_AVG"or"IL") is proportional to the load current ("IO"). So if For example IO is 2 A and IAVG is 10 A, then if IO is decreased to 1 A, IAVG will fall to 5 A. Therefore on decreasing the load current, we can get IAVG to decrease, as indicated in Fgr. 9.

e) Typically, we transit automatically from CCM to DCM, just by reducing the load current of the converter. But note that we will necessarily have to pass through BCM along the way.

f) "BCM" is just that-a 'boundary conduction mode'- situated exactly between CCM and DCM. It’s therefore a purely philosophical question to ask whether BCM should be viewed as CCM or DCM (at their respective extremes) - it really doesn't matter.

g) Note that in all the cases shown in Fgr. 9, with the exception of DCM, the average inductor current IAVG is just the geometrical center of the ramp part of the current waveform. In DCM however, we have an additional interval in which there is no current passing for a while. So, to find the average value of the inductor current, a rather more detailed calculation is required. In fact that is the primary reason why DCM equations turn out looking so complicated - to the point that many engineers seem to rather instinctively ignore DCM altogether, despite some advantages of operating a converter in DCM instead of CCM.

Note: Expectedly, all DCM equations lead to exactly the same numerical results as the CCM equations - when the converter is in BCM. Practically speaking, we can freely pick and choose whether to use the CCM equations, or the more formidable looking DCM equations, for evaluating a converter in BCM. Of course, there is no reason why we would ever want to struggle through complicated equations, when we can use much simpler equations to get the same results!

h) What really is the average inductor current "IAVG" as shown in Fgr. 9? A nice way to understand this parameter is through the "car analogy." Suppose we press the gas pedal of a car. The car responds by increasing its speed. In an analogous fashion, when we apply a voltage across an inductor (the on-time voltage "VON"), the current ramps up. Subsequently, suppose we press on the brakes of the car. The car will then respond by decreasing its speed. Similarly, when the applied voltage is removed from the inductor, voltage reversal occurs, and an induced voltage (the "brakes") appears across the inductor, "VOFF." Since it’s in the opposite direction as VON, it causes the current to ramp down. So now, if we press the gas pedal (VON), followed by the brakes (VOFF), in quick succession, and with the right timing, we could still make the car continue to move forward despite the constant lurching. It would then have a certain average speed - depending on the ratio of the gas pedal duration and the subsequent braking duration. In power conversion, this "lurching" is analogous to the 'current ripple' fli = fliON = fliOFF. And quite similarly, we have an 'average inductor current' IAVG too, as shown in Fgr. 9.

However, we do understand that in a power converter, the output capacitor eventually absorbs (or smoothens) this "lurching," and thus manages to deliver a steady dc current to the load as desired.

i) Some control ICs manage to maintain the converter in BCM mode under all application conditions. Examples of these are certain types of 'hysteretic controllers' and self-oscillating types called 'ringing choke converters' (RCCs). However, we know that the current ramps down at a rate V/L. And since V depends on the input/output voltages, the time to get to zero current depends on the specific application conditions. Therefore, in any BCM implementation, we always lose the advantage of fixed switching frequency operation.

j) Most conventional topologies are nowadays labeled 'non-synchronous'- to distinguish them from more recent 'synchronous' topologies. In the former, a diode is always present (the catch diode), that prevents the inductor current from reversing direction, any time during the switching cycle. That is why, on reducing output power and/or increasing input voltage, we automatically transit from CCM to DCM.

However, in synchronous topologies, the catch diode is either supplanted or completely replaced by a 'low-drop' mosfet across it. So whenever the diode is supposed to conduct, we force this extra mosfet into conduction for that duration.

Since the drop across this mosfet is much lower than across a diode, not only do we manage to significantly reduce the conduction loss in the freewheeling path, but we can also now allow reverse inductor current - that is, current moving instantaneously away from the load. However note that the average inductor current could still be positive - see Fgr. 9. Further, with negative currents now being "allowed," we no longer get DCM on reducing output power, but rather enter FPWM/FCCM as described in the figure.

Note: It’s fortunate that almost all the standard CCM design equations (for non-synchronous topologies) apply equally to FCCM. So from the viewpoint of a harried designer, one of the "advantages" of using synchronous topologies is that the complicated DCM equations are a thing of the past! Though there are some new complications and nuances of synchronous topologies that we need to understand eventually.

The Volt-seconds Law, Inductor Rest, and Converter Duty Cycle

There is another way to describe a steady state, by bringing in the inductor equation V = L fli/?t.

We know that during a steady state fliON = fliOFF = fli. So what we are also saying is that In steady state, the product of the voltage applied across the inductor, multiplied by the duration we apply it for (i.e. the on-time), must be equal to the voltage that appears across the inductor during the off-time, multiplied by the duration that lasts for.

Therefore we get:

VON × tON = VOFF × tOFF

The product of the voltage, and the time for which it appears across the inductor, is called the 'volt-seconds' across the inductor. So equivalently, what we are also saying is that If we have an inductor in a steady state, the volt-seconds present across it during the on-time (i.e. current ramp-up phase) must be exactly equal in magnitude, though opposite in sign, to the volt-seconds present across it during the off-time (i.e. during the current ramp-down phase).

That also means that if we plot the inductor voltage versus time, the area under the voltage curve during the on-time must be equal to the area under the voltage curve during the off-time.

But we also know that voltage reversal always occurs in steady state. So clearly, these two areas must have the opposite sign. See the vertically and horizontally hatched segments in Fgr. 9.

Therefore, we can also say that the net area under the voltage curve of an inductor must be equal to zero (in any switching cycle under steady state operation).

Note that since the typical times involved in modern switching power conversion are so small, "volt-seconds" turns out to be a very small number. Therefore, to make numbers more manageable, we usually prefer to talk in terms of 'Et' or the 'volt-µseconds.' Et is clearly just the voltage applied across the inductor multiplied by the time in microseconds (not seconds).

Further, we know that typical inductance values used in power conversion are also better expressed in terms of "µH" (micro-henries), not H. So from V = L dI/dt we can write:

fliON = VON × tON

… (steady state, L in µH) Note: If in any given equation Et and L appear together, it should be generally assumed that L is in µH.

Similarly, if we are using volt-seconds, that would usually imply L is in H (unless otherwise indicated).

Another term often used in power, one that tells us that we have managed to return to the same inductor current (and energy) that we started off with, is called inductor 'reset.' Reset occurs at the very moment when the equality fliOFF = fliON is established. Of course, we could also have a non-repetitive (or 'single-shot') event, where the current starts at zero and then returns to zero - and that too would be inductor "reset." The corollary is that in a repetitive switching scenario (steady state), an inductor must be able to reset every cycle. Reversing the argument - any circuit configuration that makes inductor reset an impossibility, is not a viable switching topology.

When we switch repetitively at a switching frequency "f," the 'time period' (T) is equal to 1/f. We can also define the 'duty cycle' (D) of a power converter as the ratio of the on-time of the switch to the time period. So:

(duty cycle definition)

Note that we can also write this as:

(duty cycle definition)

At this point we should be very clear how we are defining "tOFF" in particular. While applying the volt-seconds law, we had implicitly assumed that tOFF was the time for which the induced voltage VOFF lasts, not necessarily the time for which the switch is OFF (i.e. T - tON). In DCM they are not the same (see Fgr. 9)! Only in CCM do we get [...] and therefore D = tON tON + tOFF (duty cycle in CCM).

If working in DCM, we should stick to the more general definition of duty cycle given initially.

Using and Protecting Semiconductor Switches

We realize that all topologies exist only because they can achieve a steady state. In an "experimental" topology in which we can't make fliON = fliOFF happen, the inductor may see a net increase of current every cycle, and this can eventually escalate to a very large, almost uncontrolled value of current in just a few cycles. The name given to this progressive ramping-up (or down) of current (or inductor energy), one that is ultimately limited only by parasitics like the ESR and DCR, is called 'staircasing.' The switch will also turn ON into the same current, and can thus be destroyed - that is if the induced voltage spike hasn't already done so (which can happen, if the situation is anything similar to the "unacceptable" plots shown in Fgr. 7!).

Note: The very use of the inductor equation V = L dI/dt actually implies we are ignoring its parasitic resistance, DCR. The inductor equation is an idealization, applying only to a "perfect" inductor. That is why we had to put R = 0 when we derived it previously.

In an actual power supply, the "mechanical switch" is replaced with a modern semiconductor device (like the mosfet) - largely because then the switching action can be implemented reliably and also at a very high repetition rate. But semiconductor devices have certain electrical ratings that we need to be well aware of.

Every semiconductor device has an 'absolute maximum voltage rating' that, unlike any typical mechanical relay, cannot be exceeded even momentarily- without possibly causing its immediate destruction. So most mosfets don’t allow any "latitude" whatsoever, in terms of their voltage ratings.

Note: There are some 'avalanche-rated' mosfets available, which can internally 'clamp' the excess voltage appearing across them to some extent. In doing so, they are basically dissipating the excess energy associated with the voltage spike, within their internal clamp. Therefore, they can survive a certain amount of excess voltage (and energy), but only for a short duration (since the device heats up quickly).

There is also a maximum semiconductor device 'current rating,' but that is usually more long-term in nature, dictated by the comparatively slower process of internal heat build-up inside the device. So hypothetically speaking, we could perhaps exceed the current rating somewhat, though only for a short time. Of course we don't want to run a device constantly in this excess-current condition. However, under 'abnormal conditions,' like an "overload" on the output of the converter (or the extreme case of a shorted output), we may judiciously allow for a certain amount of "abuse" with regard to the current rating - but certainly not with the voltage rating! In a practical implementation, we have to design the converter, select the switch, and then lay it all out on a printed circuit board (PCB) with great care - to ensure in particular that there is no voltage spike that can "kill" the switch (nor any other semiconductor devices present on the board). Occasionally, we may therefore need to add an external 'snubber' or 'clamp' across the switch, so as to truncate any remnant spikes to within the voltage ratings of the switch.

To protect the switch (and converter) from excess currents, a 'current limit' is usually required. In this case, the current in the inductor, or in the switch, is sensed, and then compared against a set threshold. If and when that is momentarily exceeded, the control circuitry forces the switch to turn OFF immediately for the remainder of the switching cycle, so as to protect itself. In the next cycle, no "memory" is usually retained of what may have happened in the preceding cycle. Therefore, every switching cycle is started "afresh," with the current being continuously monitored to ensure it’s at a "safe" level. If not, protective action is again initiated, and can be repeated every cycle for several cycles if necessary, until the "overcurrent" condition ceases.

Note: One of the best-known examples of the perils of "previous-cycle memory" in implementing current limit occurs in the popular "Simple Switcher" family of parts (at www.national.com). In the "third generation" LM267x family, the control circuit rather surprisingly reduces the duty cycle to about 45% for several cycles after any single current limit event. It then tries to progressively allow the duty cycle to increase over several successive cycles back to its required value. But this causes severe output 'foldback' and consequent inability to regulate up to full rated load, particularly in applications that require a duty cycle greater than 50%. This condition is further exacerbated with large output capacitances, because the higher currents required to charge the output capacitor after the removal of an abnormal condition (e.g. output short), can lead to another current limit event (and consequent foldback for several cycles again) - before the duty cycle has been able to return to its desired value. In effect the converter goes into a continuous "motor-boating" condition on removal of the output short, and so the output never recovers. This is rather obliquely "revealed" only deep within the product datasheets (liability cover?).

With the introduction to power conversion now complete, we turn our attention to how switching topologies develop naturally out of the behavior of an inductor.