Questions/Answers about Switching Power-Supply Topology (part 1)

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This section serves to highlight and summarize the spectrum of key SMPS topology-related design issues that should be kept in mind when actually designing switching-mode power supply converters (or for a job interview in related field!).

Questions and Answers

Question---1: For a given input voltage, what output voltages can we get in principle, using only basic inductor-based topologies (buck, boost, and buck-boost)?

Answer---The buck is a step-down topology (V_o < V_in), the boost only steps-up (V_o > V_in), and the buck-boost can be used to either step-up or step-down (V_o < V_in, V_o > V_in). Note that here we are referring only to the magnitudes of the input and output voltages involved.

So we should keep in mind that the buck-boost also inverts the polarity of the input voltage.



Question---2: What is the difference between a topology and a configuration?

Answer---We know that, for example, a 'down-conversion' of 15 V input to a 5 V output is possible using a buck topology. But what we are referring to here is actually a "positive-to-positive" buck configuration, or simply, a "positive buck." If we want to convert -15Vto -5 V, we need a "negative-to-negative" buck configuration, or simply, a "negative buck." We see that a topology is itself fundamental (e.g. the buck) - but it can be implemented in more than one way, and these constitute its configurations.

Note that in the down-conversion of -15V to -5V, we use a buck (step-down) topology, even though mathematically speaking, -5 V is actually a higher voltage than -15 V! Therefore, only magnitudes are taken into account in deciding what the nature of a power conversion topology is.

Similarly, a conversion of say 15V to 30V would require a "positive boost, "whereas, -15V to -30V would need a "negative boost." These are the two configurations of a boost topology.

For a buck-boost, we need to always mentally keep track of the fact that it inverts the polarity (see next question).

Question---3: What is an "inverting" configuration?

Answer---The buck-boost is a little different. Although it has the great advantage of being able to up-convert or down-convert on demand, it also always ends up inverting the sign of the output with respect to the input. That is why it's often simply referred to as an "inverting topology." So for example, a "positive-to-negative" buck-boost would be required if we want to convert 15 V to -5 V or say to -30 V. Similarly, a "negative-to-positive" buck-boost would be able to handle -15 V to 5 V or to 30 V. Note that a buck-boost cannot do 15 V to 5 V for example, nor can it do -15Vto -5 V. The convenience of up- or down-conversion (on demand) is thus achieved only at the expense of a polarity inversion - the conventional (inductor-based) buck-boost topology is useful only if we either desire, or are willing to accept, this inversion.

Question---4: Why is it that only the buck-boost gives an inverted output? Or conversely, why can't the buck-boost ever not-invert?

Answer---In all topologies, there is a voltage reversal across the inductor when the switch turns OFF. So the voltage at one end of the inductor "flips" with respect to its other end.

Further, when the switch turns OFF, the voltage present at the swinging end of the inductor (i.e. the switching node) always gets "passed on" to the output, because the diode is then conducting. But in the case of the buck-boost, the "quiet end" of its inductor is connected to the ground reference (no other topology has this). Therefore, the voltage reversal that takes place at its other end (swinging end) is also a voltage reversal with respect to ground.

And since this is the voltage that ultimately gets transmitted to the output (which is also referenced to ground), it's virtually "seen" at the output.

Of course the output rail continues to stay inverted even when the switch turns ON, because the diode then stops conducting, and there is an output capacitor present, that holds the output voltage steady at the level it acquired during the switch off-time.

Question---5: Why do we always get only up-conversion from a boost converter?

Answer---Inductor voltage reversal during a switch transition occurs in all dc-dc switching topologies - it just does not necessarily lead to an output reversal. But in fact, inductor voltage reversal is responsible for the fact that in a buck, the input voltage is always stepped-down whereas in a boost, it's stepped-up. It all depends on where the "quiet" end of the inductor connects to. In the boost, the "quiet" end connects to the input rail (in the buck, to the output rail). Therefore, since the swinging end of the boost inductor is connected to ground during the switch on-time, it then flips with respect to the input rail during the switch off-time, gets connected to the output through the conducting diode, and thereby we get a boosted output voltage.

Question---6: What is really "ground" for a dc-dc converter?

Answer---In a dc-dc converter there are two input rails and two output rails. But one of these rails is common to both the input and output. This rail is the (power) "ground". The input and output voltages are measured with respect to this reference rail, and that gives them their respective magnitudes and polarity.

Question---7: What is "ground" for the control IC?

Answer---The reference rail, around which most of the internal circuitry of the IC is built, is its local (IC) ground. This rail comes out of the package as the ground pin(s) of the IC.

Usually, this is connected on the PCB directly to the power ground. However, there are exceptions, particularly when an IC meant primarily for a certain topology (or configuration) is rather unconventionally configured to behave as another topology altogether (or just a different configuration). Then the IC ground may in fact differ from the power ground.

Question---8: What is "system" ground?

Answer---This is the reference rail for the entire system. So in fact, all on-board dc-dc converters present in the system usually need to have their respective (power) grounds tied firmly to this system ground. The system ground in turn usually connects to the metal enclosure, and from there on to the "earth (safety) ground" (i.e. into the mains wiring).

Question---9: Why are negative-to-negative dc-dc configurations rarely used?

Answer---The voltages applied to and /or received from on-board dc-dc converters are referenced by the rest of the system to the common shared system ground. By modern convention, all voltages are usually expected to be positive with respect to the system ground. Therefore, all on-board dc-dc converters also need to comply with the same convention. And that makes them necessarily positive-to-positive converters.

Question---10: Why are inverting dc-dc converters rarely used?

Answer---We usually cannot afford to let any given on-board converter attempt to "redefine" the ground in the middle of a system. However, inverting regulators can on occasion be used, especially if the converter happens to be a "front-end" converter. In this case, since the system effectively starts at the output terminals of this converter, we maybe able to "define" the ground at this point. In that case, the relative polarity between the input and output of the converter may become a "don't care" situation.

Question---11: Can a buck regulator be used to convert a 15 V input to 14.5 V output?

Answer---Maybe, maybe not! Technically, this is a step-down conversion, since V_o < V_in.

Therefore, in principle, a buck regulator should have worked. However in practice there are some limitations regarding how close we can set the output of a converter in relation to the input.

Even if the switch of a buck regulator is turned fully ON (say in an all-out effort to produce the required output), there will still be some remaining forward-drop across the switch, VSW, and this would effectively subtract from the applied input V_in. Note that in this fully ON state, the switcher is basically functioning just like an LDO, and so the concerns expressed in Section 1 regarding the minimum achievable headroom of an LDO apply to the switcher too, in this state. As an example - if the switch drop VSW is 1 V, then we certainly can't get anything higher than 14 V output from an input of 15 V.

The second consideration is that even if, for simplicity, we assume zero forward voltage drops across both the switch and the diode, we still may not be able to deliver the required output voltage - because of maximum duty cycle limitations. So for example, in our case, what we need is a (theoretical) duty cycle of V_o/V_in = 14.5 V/15 V = 0.97, that's , 97%.

However, most buck ICs in the market are not designed to guarantee such a high duty cycle.

They usually come with an internally set maximum duty cycle limit ('DMAX'), typically around 90 to 95%. And if that's so, D = 97% would be clearly out of their capability.

A good power supply designer also always pays heed to the tolerance or spread of the published characteristics of a device. This spread is usually expressed as a range with a specified "min" (minimum), a "max" (maximum), and a "typ" (typical, or nominal). E.g., suppose a particular IC has a published maximum duty cycle range of 94 to 98%, we cannot guarantee that all production devices would be able to deliver a regulated 14.5 V - simply because not all of them are guaranteed to be able to provide a duty cycle of 97%. Some parts may manage 97%, but a few others won't go much beyond a duty cycle maximum of 94%. So what we need to do is to select an IC with the published 'min' of its tolerance range greater than the desired duty cycle. E.g., a buck IC with a published DMAX range of 97.5 to 99% may work in our current application.

Why did we say "may" above? If we include the forward drops of the switch and diode in our calculation, we actually get a higher duty cycle than the 97% we got using the "ideal" equation D = V_o/V_in. The latter equation implicitly assumes VSW = VD = 0 (besides ignoring other key parasitics like the inductor's DCR). So the actual measured duty cycle in any application may well be a couple of percentage points higher than the ideal value.

In general we should remember that whenever we get too close to the operating limits of a control IC, we can't afford to ignore key parasitics. We must also account for temperature variations, because temperature may affect efficiency, and thereby the required duty cycle.

Question---12: What role does temperature play in determining the duty cycle?

Answer---As mentioned in Section 1, it's generally hard to predict the overall effect of temperature on a power-supply's efficiency, and thereby on its duty cycle variation with respect to temperature. Some loss terms increase with temperature and some decrease.

However to be conservative, we should at a minimum account for the increase in the forward drop of the mosfet switch. For low voltage mosfets (rated ~30 V), the increase in RDS (on-resistance) in going from room temperature to "hot" is typically 30 to 50%. So we typically multiply the published room temperature on-resistance by 1.4. For high-voltage mosfets, as used in off-line power supplies, the increase is about 80 to 100%. So we typically multiply the room temperature on-resistance by 1.8.

Ill. 1: Positive-to-Positive Step-up/Step-down Converters: SEPIC CONVERTER; NON-ISOLATED TRANSFORMER-BASED BUCK-BOOST CONVERTER

Question---13: How can we convert an unregulated input of 15 V to a regulated output of 15 V?

Answer---The term "unregulated" implies that the stated value just happens to be the 'typical' (usually center) of a certain range, which may or may not yet have been defined. So an "unregulated input of 15 V" could well mean say 10 V to 20 V, or 5 V to 25 V, or 12 V to 18 V, and so on.

Of course, ultimately, we do need to know what this input range really is. But it should already be apparent that for a 15 V to 15 V conversion, if the input falls at the lower end of its range, we would need to up-convert, and if the input is at its upper end, we would need to down-convert. Therefore, we must choose a topology capable of performing both step-up and step-down conversions on demand.

How about the buck-boost? Unfortunately, the standard inductor-based buck-boost also gives us an inverted output, which we really don't want here. What we need is a non-inverting step-up/step-down topology. Looking around, a suitable candidate for this is the 'SEPIC' (single ended primary inductance converter) topology. See Ill. 1. It is best visualized as composite topology - a boost stage followed by a buck cell. Though this "boost-buck" combination needs only one switch, it requires an additional inductor, and also entails significantly more design complexity. We may therefore wish to consider a derivative (or variant) of the conventional buck-boost topology, but with the inductor replaced by a transformer. In effect, what we are doing is - we are first separating (isolating) the input from the output, and then reconnecting the windings of the transformer in an appropriate manner so as to correct for the inversion. Thus we get a non-inverting or 'non-isolated transformer-based buck-boost'- sometimes simply called a 'flyback' topology.

Question---14: It is much easier to find "off-the-shelf" inductors. So why is a transformer-based buck-boost even worth considering?

Answer---It is true that most designers prefer the convenience of off-the-shelf components, rather than custom-designed components (like transformers). However, high-power off-the shelf inductors often come with two identical windings wound in parallel (on the same core), (though that may not be immediately apparent just by looking at the datasheet). Further, the ends of these two windings are sometimes completely separated from each other (no galvanic connection between the windings). The reason for this may be that from a production standpoint, it doesn't make sense to try and solder too many copper strands on to a single pin/termination. So the intention here is that the two windings will be eventually connected to each other on the PCB itself. But sometimes, the intention of leaving separate windings on an inductor is to allow flexibility for the two windings to be connected to each other either in series, or in parallel, as desired. So for example, if we place the winding in series, that would reduce the current rating of the inductor, but we would get a much higher inductance. If in parallel, the inductance would come down, but the current rating would increase. However, in low-voltage applications, where safety isolation isn't a concern, we can also exploit this inductor structure and use it as a 1:1 transformer. And that would be very helpful in correcting the polarity inversion of the buck-boost. In other words, an off-the-shelf inductor could now serve as a 1:1 transformer!

Question---15: In an inductor with split windings (1:1), how exactly does its current rating and its inductance change as we go from a parallel configuration to a series configuration?

Answer---Suppose each winding has 10 turns and a DCR (dc resistance) of 1 ?. So if it's used in parallel configuration, we still have 10 turns, but the effective DCR is 1 ? in parallel with 1 ?, i.e. 0.5 ?. When a series configuration is used, we get 2 ? and 20 turns. We also know that inductance depends on the square of the number of turns. So that goes up four times.

What about the current rating? This is largely determined by the amount of heat dissipation the inductor can tolerate. But its thermal resistance (in deg C/W) isn't determined by the winding configuration, rather by the exposed area of the inductor, and other physical characteristics. Therefore, whether in series or in parallel configuration, we have to maintain the same total I^2R loss. E.g., suppose we call the current rating in parallel as "IP", and in series "IS," then as per the DCR in our above numerical example, we get IP ^2 × 0.5 = IS ^ 2 × 2 so IP = 2 × IS Therefore, in going from a parallel to a series configuration, the inductance will quadruple and the current rating will halve.

What happens to the B-field? Don't we have to consider the possibility of saturation here? Well, B is proportional to LI/N (see Section 2, "DC-DC Converter Design and Magnetics").

So if inductance quadruples, I halves, and N doubles, the B-field is unchanged!

Question---16: Is there any difference between the terms "buck-boost" and "flyback"?

Answer---The answer to that may well depend on whom you ask! These terms are often used interchangeably in the industry. However, generally, most people prefer to call the conventional inductor-based version a (true) "buck-boost," whereas its transformer-based version, isolated or non-isolated, is called a "flyback."

Question---17: When and why do we need isolation? And how do we go about achieving it?

Answer---We must recognize that a (transformer-based) flyback topology may or may not provide us with isolation. Isolation is certainly a natural advantage accruing from the use of a transformer. But to preserve isolation, we must ensure that all the circuitry connected to the switch side of the transformer ('primary side') is kept completely independent from all the circuitry sitting on the output side ('secondary side'). See Ill. 1 in Section 1.

So for example, if in our attempt to correct the polarity inversion of a buck-boost we make a connection between the primary and secondary windings of the transformer, we lose isolation.

Further, to maintain isolation, besides making no galvanic connection between the power stages on either side of the transformer, we must not make any signal-level inter-connections either. So we must carry the feedback signal (or any fault information) from the output side to the IC, via one or more 'optocouplers.' The optocoupler manages to preserve primary-to-secondary voltage isolation, but allows signal-level information to pass through.

It works by first converting the secondary-side signal into radiation by means of an "led" (light emitting diode), beaming it over to the primary side onto a photo-transistor, and thereby converting the signal back into electrical impulses (all this happening within the package of the device itself ).

In high-voltage applications (e.g. off-line power supplies), it may in fact be required by law, to provide electrical isolation between a hazardous input voltage level and user-accessible ("safe") output terminals of the power supply. Therefore there is a "primary ground" at the input side of the transformer, and a separate "secondary ground" on the output side.

The latter is then to the ground of the system, and usually also to the earthed metal enclosure.

Question---18: In an off-line power supply, are the primary and secondary sides really completely isolated?

Answer---It is interesting to note that safety regulations specify a certain physical spacing that must be maintained between the primary and secondary sides - in terms of the RMS of the voltage differential between them. The question arises - how do we define a voltage difference between the two sides of a transformer that are supposedly separate? What is the reference level to compare their voltages? In fact they do share a connection! As mentioned, the secondary side ground is usually the system ground, and it connects to the metal enclosure and /or to the ground wire of the mains supply ("earth" or "safety ground"). But further down the mains distribution network, the safety ground wire is connected to the "neutral" wire of the supply. And we know that this neutral wire comes back into the primary side. So in effect, we have established a connection between the primary and secondary sides. It does not cause the user any problem, because he or she is also connected to earth. Therefore, the earth potential forms the reference level to establish the voltage difference across the safety transformer, and to thereby fix the primary-to-secondary spacing, and also the breakdown rating of any primary-to-secondary insulation.

Note that in some portable equipment, only a two-wire ac cord is used to connect it to the mains supply. But the spacing requirement is still virtually unchanged, since a user can touch accessible parts on the secondary side and complete the connection through the earth ground.

Question---19: From the standpoint of an actual power supply design procedure, what is the most fundamental difference between the three topologies that must be kept in our minds?

Answer---In a buck, the average inductor current ("IL") is equal to the load current ("IO"), that's , IL = IO. But in a boost and a buck-boost, this average current is equal to IO/(1 - D).

Therefore, in the latter two topologies, the inductor current is a function of D (duty cycle) - and therefore indirectly a function of the input voltage too (for a given output).

Question---20: In the three basic topologies, how does the duty cycle change with respect to input voltage?

Answer---For all topologies, a high D corresponds to a low input voltage, and a low D to a high input.

Question---21: What do we mean by the "peak current" of a dc-dc converter?

Answer---In any dc-dc converter, the terms "peak inductor current," "peak switch current," and "peak diode current" are all the same - referred to simply as the peak current 'IPK' (of the converter).

Question---22: What are the key parameters of an off-the-shelf inductor that we need to consider?

Answer---The inductance of an inductor (along with the switching frequency and duty cycle) determines the peak current, whereas the average inductor current is determined by the topology itself (and the specific application conditions - the duty cycle and load current).

For a given application, if we decrease the inductance, the inductor current waveform becomes more "peaky," increasing the peak currents in the switch and diode too (also in the capacitors). Therefore, a typical converter design should start by first estimating the optimum inductance so as to avoid saturating the inductor. That is the most basic concern in designing/picking an inductor.

However, inductance by itself doesn't fully describe an inductor. In theory, by choosing a very thin wire gauge for example, we may be able to achieve almost any inductance on a given core, just by winding the appropriate number of turns. But the current that the inductor will be able to handle without saturating is still in question, because it's not just the current, but the product of the current and the number of turns ('ampere-turns') that determine the magnetic field present in the inductor core - which in turn determines whether the inductor is saturating or not. Therefore, we need to look out for an inductor with the right inductance and also the required energy handling capability, usually expressed in µJ (micro-Joules). This must be greater than or equal to the energy it needs to store in the application, 1 2 × LIPK2.

Note that the "L" used in this equation carries with it information about the number of turns too, since L ? N2, where N is the number of turns.

Question---23: What really determines the current rating of an inductor?

Answer---There are two limiting factors here. One is the heat developed (I^2R losses), which we should ensure isn't excessive (usually 50 deg. C or less). The second is the magnetic field it can withstand without saturating. So most ferrites allow a maximum B-field of about 3000 gauss before saturation starts.

Question---24: Does the maximum allowable B-field depend on the air gap used?

Answer---When designing (gapped) transformers, we need to remember that first, the B-field present within the core material (e.g. ferrite) is the same as the B-field in the air gap. It does not change. Second, though by changing the air gap we can end up decreasing the existing B-field, the maximum allowable B-field depends only on the core material used - it remains fixed, for example, at about 3000 gauss for ferrites. Note that the H-field is defined as H = B/µ, where µ is the permeability of the material. So since the permeability of ferrite is much higher than that of air, and since the B-field is the same in both, therefore the H-field is much lower in the ferrite than in the air gap.

Question---25: Why is it commonly stated that in a flyback transformer, the "air gap carries most of the stored magnetic energy"?

Answer---We can intuitively accept the fact that the energy stored is proportional to the volume of the magnetic material. And because of that, we also tend to think the ferrite must be carrying most of the energy, since it occupies the maximum volume - the amount of air enclosed between the ends of the ferrite being very small. However, the stored energy is also proportional to B × H, and since the H-field in the gap is so much larger, it ends up storing typically two-thirds of the total energy, despite its much smaller volume.

Question---26: If air carries most of the stored energy, why do we even need the ferrite?

Answer---An air-cored coil would seem perfect as an inductor, especially since it would never saturate. However, the number of turns required to produce a given inductance would be impractically large, and so we would get unacceptable copper losses. Further, since there isn'thing to "channel" (constrain) the flux lines, the air-cored inductor would spew electromagnetic interference (EMI) everywhere.

The ferrite is useful, because it's the very means by which we can create such high magnetic fields in the first place - without an excessive number of turns. It also provides us the "channel" for flux lines that we had been looking for.

cont. to part 2 >>








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