Off-line Converter Design and Magnetics--Flyback Converter Magnetics

Polarity of Windings in a Transformer

In FGR. 1, the turns ratio is n = nP/nS, where nP is the number of turns of the primary winding, and nS is the number of turns of the secondary winding.

We have also placed a dot on one end of each of the windings. All dotted ends of a transformer are considered to be mutually "equivalent." All non-dotted ends are also obviously mutually equivalent. That means that when the voltage on a given dotted end goes "high" (to whatever value), so does the voltage on the dotted ends of all other windings.

That happens because all windings share the same magnetic core, despite the fact that they are not physically (galvanically) connected to each other. Similarly, all the dotted ends also go "low" at the same time. Clearly, the dots are only an indication of relative polarity.

Therefore, in any given schematic, we can always swap the dotted and non-dotted ends of the transformer, without changing the schematic in the slightest way.

In the flyback, the polarity of the windings is deliberately arranged such that when the primary winding conducts, the secondary winding is not allowed to do so. So when the switch conducts, the dotted end at the drain of the mosfet in FGR. 1 goes low. And therefore, the anode of the output diode also goes low, thereby reverse-biasing the diode.

We should recall that the basic purpose of a buck-boost (which this in fact also is) is to allow incoming energy from the source during the switch on-time to build up in the inductor (only), and then later, during the off-time, to "collect" all this energy (and no more) at the output. Note that this is the unique property that distinguishes the buck-boost (and the flyback) from the buck and the boost. For example, in a buck, energy from the input source gets delivered to the inductor and the output (during the on-time). Whereas, in a boost, stored energy from the inductor and the input source gets delivered to the output (during the off-time). Only in a buck-boost do we have complete separation between the energy storage and the collection process, during the on-time and the off-time. So, now we understand why the flyback is considered to be just a buck-boost derivative.

We know that every dc-dc topology has a so-called "switching node." This node represents the point of diversion of the inductor current - from its main path (i.e. in which the inductor receives energy from the input) to its freewheeling path (i.e. in which the inductor provides stored energy to the output). So clearly, the switching node is necessarily the node common to the switch, the inductor, and the diode. Further, we will find that the voltage at this node is always "swinging" - because that is what is required to get the diode to alternately forward and reverse-bias, as the switch toggles. But looking at FGR. 1, we see that with a transformer replacing the traditional dc-dc inductor, there are now, in effect, two "switching nodes" - one on each side of the transformer, as indicated by the "X" markings in FGR. 1 - one "X" is at the drain of the mosfet, and the other "X" is at the anode of the output diode. These two nodes are clearly "equivalent" because of the dots, as explained above. And since at both these nodes, the voltage is swinging, therefore both are considered to be "switching nodes" (of the transformer-based topology). Note that if we had, say, three windings (e.g. an additional output winding), we would have had three (equivalent) switching nodes.

Transformer Action in a Flyback, and Its Duty Cycle:

Classic "transformer action" implies that the voltages across the windings of the transformer, and the currents through each of them, scale according to the turns ratio, as described in FGR. 1. But it’s perhaps not immediately apparent why the flyback inductor exhibits transformer action.

When the switch turns ON, a voltage VIN (the rectified ac input) gets impressed across the primary winding of the transformer. And at the same time, a voltage equal to VINR = VIN/n ("R" for reflected) gets impressed across the secondary winding (in a direction that causes the output diode to get reverse-biased). Therefore, there is no current in the secondary winding when the primary winding is conducting.

Let us calculate what VINR is. This voltage translation across the isolation boundary follows from the induced voltage equation applied to each winding:

Note that both windings enclose the same magnetic core, so the flux f is the same for both, and so is the rate of change of flux df/dt for each winding. Therefore:

This above equation represents classic "transformer action" with respect to the voltages involved. But we also learn from the preceding equation that the Volts/turn for any winding (at any given instant) is the same for all the windings present on a given magnetic core - and this is what eventually leads to the observed voltage scaling.

Note also that voltage scaling in any transformer occurs irrespective of whether a given winding is passing current or not. That is because, whether a given winding is contributing to the net flux f present in the core or not, each winding encloses this entire flux, and so the basic equation V =-N × df/dt applies to all windings, and so does voltage scaling.

We know that energy is built up in the transformer during the on-time. When the switch turns OFF, this stored energy (and its associated current) needs to flyback/freewheel. We also know that the voltages will automatically try to adjust themselves in any possible way, so as to make that happen. So we can safely assume the diode will somehow conduct during the switch off-time. Now, assuming we have reached a "steady state," the voltage on the output capacitor has stabilized at some fixed value VO. Therefore, the voltage at the secondary-side switching node gets clamped at VO (ignoring the diode drop). Further, since one end of the secondary winding is tied to ground, the voltage across this winding is now equal to VO. By transformer action, this reflects a voltage across the primary winding, equal to VOR =VO×n.

But the switch is OFF during this time. Therefore, under normal circumstances, the voltage at the primary-side switching node would have settled at VIN. However, now this reflected output voltage VOR, coming through the transformer, adds to that. Therefore, the voltage at the primary-side switching node eventually goes up to VIN +VOR (for now, we are ignoring the turn-off spike encircled in FGR. 1).

===

Table1: Derivation of dc transfer function of flyback Primary Winding Secondary Winding

===

Note: During the on-time, the primary side is the one determining the voltages across all the windings. And during the off-time, it’s the secondary winding that gets to "call the shots"! We can calculate duty cycle from the most basic equation (from voltseconds law)

We have the option of performing this calculation, either on the primary winding, or the secondary winding. Either way, we get the same result, as shown in Table 3-1.

We should be always very clear that transformer action applies only to the voltages across windings. And "voltage across" is not necessarily "voltage at"! To measure the voltage at a given point, we have to consider what the reference level (i.e. "ground" by definition) is, with respect to which its voltage needs to be measured, or stated. In fact, the reference level (i.e. by definition, "ground") is called the "primary ground" on the primary side and the "secondary ground" on the secondary side. Note that these are indicated by different ground symbols in FGR. 1.

To find out the (absolute) voltage at the swinging end of any winding, we can use the following level-shifting rule:

To get the absolute value of the voltage at the swinging end of any winding, we must add to the voltage across the winding, the dc voltage present at its "non-swinging" end.

So, For example, to get the voltage at the drain of the mosfet (swinging end of primary winding), we need to add VIN (voltage at other end of winding), to the voltage waveform that represents the voltage across the primary winding. That is how we got the voltage waveforms shown in FGR. 1.

Coming to the question of how currents actually reflect from one side of the transformer to the other, it must be pointed out that even though the final current scaling equations of a flyback transformer are exactly the same as in the case of an actual transformer, this is not strictly "classic transformer action." The difference from a conventional transformer is, that in the flyback, the primary and secondary windings don’t conduct at the same time. So in fact, it’s a mystery why their currents are related to each other at all! The current scaling that occurs in a flyback actually follows from energy considerations. The energy in a core is in general written as ...

We know the windings of our flyback conduct at different times, but the energy associated with each of the current flows must be equal to the energy in the core, and must therefore be equal to each other (we are ignoring the ramp portion of the current here for simplicity).

Therefore, .. where LP is the inductance measured across the primary winding - with the secondary winding floating (no current), and LS the inductance measured across the secondary winding - with the primary winding floating. But we also know that where AL is the inductance index, defined previously. Therefore in our case we get ...

Substituting in the energy equation, we get the well-known current scaling equations ...

We see that analogous to the Volts/turns rule, the ampere-turns also need to be preserved at all times. In fact, the core itself doesn't really "care" which particular winding is passing current at any given moment, so long as there is no sudden change in the net ampere-turns of the transformer. This becomes the "transformer-version" of the basic rule we learned in Section 1 - that the current through an inductor cannot change discontinuously. Now we see that the ampere-turns of a transformer cannot change discontinuously.

Summarizing, transformer action works as follows - when reflecting a voltage from primary to secondary side, we need to divide by the turns ratio. When going from the secondary to the primary side, we need to multiply by the turns ratio. The rule reverses for currents - so we multiply by the turns ratio when going from primary to secondary, and divide in the opposite direction.

The Equivalent Buck-boost Models:

Because of the many similarities, and also because of the way voltages scale in the transformer, it becomes very convenient (most of the time) to study the flyback as an equivalent dc-dc (inductor-based) buck-boost. In other words, we separate out the coarse fixed-ratio step-down ratio and incorporate it into equivalent (reflected) voltages and currents. We thereby manage to reduce the flyback transformer into a simple energy-storage medium, just like any conventional dc-dc buck-boost inductor. In other words, for most practical purposes, the transformer goes "out of the picture." The advantage is that almost all the equations and design procedure we can write for a conventional buck-boost now apply to this equivalent buck-boost model. One exception to this is the leakage inductance issue (and everything related to it - the clamp, the loss in efficiency due to it, the turn-off voltage spike on the switch, and so on). We will discuss this exception later. But other than that, all other parameters - like the capacitor, diode, and switch currents for example, can be more readily visualized and calculated if we use this dc-dc model approach.

The equivalent dc-dc model is created essentially by reflecting the voltages and currents across the isolation boundary of the transformer to one side. But again, as in the case of the duty cycle calculation (see Table 3-1), we have two options here - we can either reflect everything to the primary side, or everything to the secondary side. We thus get the two equivalent buck-boost models as shown in FGR. 2. We can use the primary-side equivalent model to calculate all the voltages and currents on the primary side of the original flyback and the secondary-side equivalent model for calculating all the currents and voltages on the secondary side of the original flyback.

FGR. 2: The Equivalent Buck-Boost Models of the Flyback---Primary-side equivalent model; Secondary-side equivalent model:

We know that voltages and currents reflect across the boundary by getting either multiplied or divided by the turns ratio. In fact, the 'reflected output voltage,' VOR, is one of the most important parameters of a flyback, as we will see. As the name indicates, VOR is effectively the output voltage as seen by the primary side. In fact, if we compare the switch waveform of the flyback in FGR. 1 with that of a buck-boost, we will realize that to the switch, it seems as if the output voltage is really VOR.

As an example, suppose we have a 50 W converter with an output of 5 V at 10 A, and a turns ratio of 20. The VOR is therefore 5× 20 = 100 V. Now, if we change the set output to say 10 V and reduce the turns ratio to 10, the VOR is still 100 V. We will find that none of the primary-side voltage waveforms change in the process (assuming efficiency doesn't change). Further, if we have also kept the output power constant in the process, that is, by changing the load to 5 A for an output at 10 V, all the currents on the primary side will also be unaffected. Therefore, the switch will never know the difference. In other words, the switch virtually "thinks" that it’s a simple dc-dc buck-boost - delivering an output voltage of VOR at a load current of IOR.

As mentioned, the only difference between a transformer-based flyback that "thinks" it’s providing an output of VOR at the rate of IOR, and an inductor-based version that really is providing an output of VOR at the rate of IOR, is the 'leakage inductance' of the flyback transformer. This is that part of the primary side inductance that is not coupled to the secondary side and therefore cannot partake in the transfer of useful energy from the input to the output. We can confirm from FGR. 1, that the only portion of the primary-side (switch) voltage waveform that "doesn't make it" to the secondary side is the spike occurring just after the turn-off transition. This spike comes from the uncoupled leakage inductance, as we will soon see.

Note that in the equivalent buck-boost models, the reactive component values also get reflected - though as the square of the turns ratio. We can understand this fact easily from energy considerations. For example, the output capacitor CO in the original flyback was charged up to a value of VO. So its stored energy was 1/2 COVO2. In the primary-side buck-boost model, the output of the converter is VOR, that is, VO × n. Therefore, to keep the energy stored in this capacitor invariant (in the dc-dc model, as in the flyback), the output capacitance must get reflected to the primary side according to CO/n2. Note also from FGR. 2 how the inductance reflects. This is consistent with the fact that L ? N2.

The Current Ripple Ratio for the Flyback:

Looking at the equivalent buck-boost models in FGR. 2, the center of the ramp on the secondary side (average inductor current, "IL") must be equal to IO/(1 - D), as for a buck-boost (because the average diode current must equal the load current). This secondary-side "inductor" current gets reflected to the primary side, and so the center of the primary-side inductor current ramp is "ILR," where ILR = IL/n. Equivalently, it’s equal to IOR/(1 - D), where IOR is the reflected load current, that is, IOR = IO/n. Similarly, the current swings on the primary and secondary sides are also related, via scaling (turns ratio n). Therefore, we see that the ratio of the swing to the center of the ramp is identical on both sides (primary- and secondary-side dc-dc models). We’re thus in a position to define a current ripple ratio r for the flyback topology too - just as we did for a dc-dc converter. We just need to visualize r in a slightly different manner this time -in terms of the center of the ramp (switch or diode), rather than the dc inductor level (because there is no inductor present really). And as for dc-dc converters, we should normally try to set it to around 0.4.

The value of r for a flyback is the same for either the primary and secondary dc-dc equivalent models.

The Leakage Inductance:

The leakage inductance can be thought of as a parasitic inductance in series with the primary-side inductance of the transformer. So just at the moment the switch turns OFF, the current flowing through both these inductances is "IPKP," that is, the peak current on the primary side. However, when the switch turns OFF, the energy in the primary inductance has an available freewheeling path (through the output diode), but the leakage inductance energy has nowhere to go. So it expectedly "complains" in the form of a huge voltage spike (see FGR. 1). This spike (or a scaled version of it) is not seen on the secondary side, simply because this is not a coupled inductance, like the primary inductance.

If we don't make any effort to collect this energy, the induced spike can be very large, causing switch destruction. Since we certainly can't get this energy to transfer to the secondary side, we have just two options - either we can try to recover it and cycle it back into the input capacitor, or we can simply burn it (dissipation). The latter approach is usually preferred for the sake of simplicity. It’s commonly accomplished by means of a straightforward 'zener diode clamp,' as shown in FGR. 1. Of course the zener voltage must be chosen according to the maximum voltage the switch can tolerate. Note that for several reasons, in particular that of efficiency, it’s usually considered preferable to connect this zener across the primary winding (via a blocking diode in series with it). The alternative is to connect it from the switching node to primary ground.

We can ask - where does the leakage inductance really reside? Most of it’s inside the primary winding of the transformer, though some of it lies in the PCB trace sections and transformer terminations, especially with those associated with the secondary winding, as we will see below.

Zener Clamp Dissipation:

If we burn the energy in the leakage, it’s important to know how this affects the efficiency. It’s sometimes intuitively felt the energy dissipated every cycle is just 1/2× LLKPIPK2, where IPK is the peak switch current, and LLKP is the primary-side leakage. That certainly is the energy residing in the leakage inductance (at the moment the switch turns OFF), but it’s not the entire energy that eventually gets dissipated in the zener clamp on account of the leakage.

The primary winding is in series with the leakage, so during the small interval that the leakage inductance is trying, in effect, to reset, by freewheeling into the zener, the primary winding is forced to follow suit and continue to provide this series current. Though the primary winding is certainly trying (and managing partly) to freewheel into the secondary side, a part of its energy also gets diverted into the zener clamp - until the leakage inductance achieves full reset (zero clamp current). In other words, some energy from the primary inductance gets literally "snatched away" by the series leakage inductance, and this also finds its way into the zener, along with the energy residing in the leakage itself.

A detailed calculation reveals that the zener dissipation actually is ...

So the energy in the leakage 1/2× LLK × IPK2 gets multiplied by the term VZ/(VZ - VOR) (this additional term from the primary inductance).

Note that if the zener voltage is too close to the chosen VOR, the dissipation in the clamp goes up steeply. VOR therefore always needs to be picked with great care. That simply means that the turns ratio has to be chosen carefully!

Secondary-side Leakages also Affect the Primary Side:

Why did we use the symbol "LLK" in the dissipation equation above? Why didn't we identify it as the primary-side leakage ("LLKP")? The reason is that LLK represents the overall leakage inductance as seen by the switch. So, it’s partly LLKP - but it also is influenced by the secondary-side leakage inductance. This is a little hard to visualize, since by definition, the secondary-side leakage inductance is not supposed to be coupled to the primary side (and vice versa). So how could it be affecting anything on the primary side? The reason is that just as the primary-side leakage prevents the primary-side current from freewheeling into the output immediately (and thereby causes an increase in the zener dissipation), any secondary-side inductance also prevents the freewheeling path from becoming available immediately (following switch turn-off). Basically, the secondary-side inductance insists that we ("politely" and) slowly build up the current through it - respecting the fact that it’s an inductance after all! However, until the current in the bona

?de freewheeling path can build up to the required level, the primary-side current still needs to freewheel somewhere (because the switch is turning OFF)! The path the inductor current therefore seeks out is the one containing the zener clamp (being the only path available).

The zener can therefore see significant dissipation, even assuming zero primary-side leakage.

In brief, the secondary-side leakage has created much the same effect as a primary-side leakage.

When both primary- and secondary-side leakages are present, we can find the effective primary-side leakage (as seen by the switch and zener clamp) as [...]

So, like any other reactive element, the secondary-side leakage also reflects onto the primary side according to the square of the turns ratio, where it adds up in series with any primary-side leakage present.

For a given VOR, if the output voltage is "low" ( For example 5 V or 3.3 V), the turns ratio is much greater. Therefore, if the chosen VOR is very high, the reflected secondary-side leakage can become even greater than any primary-side leakage. This can become quite devastating from the efficiency standpoint.

Measuring the Effective Primary-side Leakage Inductance:

The best way to know what LLK really is, is by measuring it! Commonly, a leakage inductance measurement is done by shorting the secondary winding pins and then measuring the inductance across the ends of the (open) primary winding. By shorting, we virtually cancel out all coupled inductance. And so what we measure is just the primary-side leakage inductance in this case.

However, the best method to measure leakage is actually an in-circuit measurement - so that we include the secondary-side PCB traces in the measurement. The recommended procedure is as follows.

On the given application board, a thick piece of copper foil (or a thick section of braided copper strands), with as short a length as possible, is placed directly across the diode solder pads on the PCB. A similar piece of conductor is placed across the output capacitor solder pads. Then, if we measure the inductance across the (open) primary winding pins, we will measure the effective leakage inductance LLK (not just LLKP).

We will find that the contribution from the secondary-sides traces can in fact make LLK several times larger than LLKP.LLKP can of course be measured, if desired, by placing a thick conductor across the secondary pins of the transformer.

The PCB used in the above procedure can be just a bare board with no components mounted on it, other than the transformer. Or it can even be a fully assembled board (though sometimes, we may need to cut the trace connecting the drain of the mosfet to the transformer).

If we want to mathematically estimate the inductance of the secondary-side traces, the rule-of-thumb we can use is 20 nH per inch. But here, we need to include the full electrical path of the high-frequency output current - starting from one end of the secondary winding, returning to its other end, through the diode and output capacitor(s). We will be surprised to calculate or measure, that even an inch or two of trace length can dramatically decrease the efficiency by 5 to 10% in low output voltage applications.

Worked Example (7)-Designing the Flyback Transformer A 74 W universal input (90 VAC to 270 VAC) flyback is to be designed for an output of 5 V @ 10 A and 12 V @ 2 A. Design a suitable transformer for it, assuming a switching frequency of 150 kHz. Also, try to use a cost-effective 600 V-rated mosfet.

Fixing the VOR and VZ

At maximum input voltage, the rectified dc to the converter is

...

With a 600 V mosfet, we must leave at least 30 V safety margin when at V_IN_MAX. So in our case, we don’t want to exceed 570 V on the drain. But from FGR. 1, the voltage on the drain is VIN + VZ. Therefore ...

Note that if we plot the zener dissipation equation presented earlier, as a function of VZ/VOR, we will discover that in all cases, we get a "knee" in the dissipation curve at around VZ /VOR = 1.4. So here too, we pick this value as an optimum ratio that we would like to target. Therefore ...

Turns Ratio

Assuming the 5 V output diode has a forward drop of 0.6 V, the turns ratio is:

Note that the 12 V output may sometimes be regulated by a post-linear-regulator. In that case, we may have to make the transformer provide an output 3 to 5 V higher (than the final expected 12 V) - to provide the necessary "headroom" for the linear regulator to operate properly. This additional headroom not only caters to the dropout limits of the linear regulator, but in general also helps achieve a regulated 12 V under all load conditions.

However, there are also some clever cross-regulation techniques available that allow us to omit the 12 V linear regulator, particularly if the regulation requirements of the 12 V rail are not too "tight," and also if there is some minimum load assured on the outputs. In our example, we are assuming there is no 12 V post regulator present. Therefore, the required turns ratio for the 12 V output is 128/(12 + 1) = 9.85, where we have assumed the diode drop is 1 V in this case.

Maximum Duty Cycle (Theoretical)

Having verified the selection of VZ and VOR at highest input, now we need to get back to the lowest input voltage, because we know from the previous discussions about the buck-boost (see the "general inductor design procedure" in the previous section) that VINMIN is the worst-case point we need to consider for a buck-boost inductor/transformer design.

The minimum rectified dc voltage to the converter is ...

We’re ignoring the voltage ripple on the input terminals of the converter, and therefore we will take this as the dc input to the converter stage. So the duty cycle at minimum input voltage is ...

This is clearly a "theoretical" estimate - implying 100% efficiency. We will in fact ignore this value ultimately, as we will be estimating D more accurately by another trick.

Note however, that this is the operating D_MAX. When we "power down" our converter For example, the duty cycle will actually increase further in an effort to maintain regulation (unless current limit and/or duty cycle limit is encountered along the way). Then depending upon the number of missing ac cycles for which we may need to ensure regulation (the 'holdup time' specification), we will need to select a suitable input capacitance and also the maximum duty cycle limit, DLIM, of our controller. Typically, DLIM is set around 70%, and the capacitance is selected on the basis of the 3 µF/W rule-of-thumb. For example, for our 74 W supply with an estimated 70% efficiency at low line, we will draw an input power of 74/0.7= 106 W. Therefore we should use a 106 × 3 = 318 µF (standard value 330 µF) input capacitor. However, note that the ripple current rating of this capacitor (and its life expectancy) must be verified.

Effective Load Current on Primary and Secondary Sides:

Let us lump all the 74 W output power into an equivalent single output of 5 V. So the load current for a 5 V output is ...

On the primary side, the switch "thinks" its output is VOR and the load current is IOR, where...

Duty Cycle

The actual duty cycle is important because a slight increase in it (from the theoretical 100% efficiency value) may lead to a significant increase in the operating peak current and the corresponding magnetic fields.

The input power is ...

The average input current is therefore ...

The average input current tells us what the actual duty cycle 'D' is, because IIN/D is also the center of the primary-side current ramp, and must equal ILR, that is, ...

We thus have a more accurate estimate of duty cycle.

Actual Center of Primary and Secondary Current Ramps

The center of the secondary-side current ramp (lumped) ...

The center of the primary-side current ramp is ...

Peak Switch Current

Knowing ILR, we know the peak current for our selected current ripple ratio

We may need to set the current limit of the controller, For example, based on this estimate.

Voltseconds

We have at VINMIN

Primary-side Inductance

Note that when we come to designing off-line transformers, for various reasons like reducing high-frequency copper loss, reducing size of transformer, and so on, it’s more common to set r at around 0.5. So the primary-side inductance must then be (from the "L× I" rule).

Selecting the Core:

Unlike made-to-order or off-the-shelf inductors, when designing our own magnetic components, we should not forget that adding an air gap dramatically improves the energy storage capability of a core. Without the air gap, the core could saturate even with very little stored energy.

Of course, we still need to maintain the desired L, corresponding to the desired r! So if we add too much of a gap, we will also need to add many more turns - thus increasing the copper loss in the windings. At one point, we will also run out of window space to accommodate these windings. So a practical compromise must be made here, one that the following equation actually takes into account (applicable to ferrites in general, for any topology):

...

where f is in kHz.

In our case we get

...

We start looking for a core of this volume (or higher). We find a candidate in the EI-30.

Its effective length and area are given in its datasheet as

...

So its volume is

Ve = Ae × le = 5.8 × 1.11 = 6.438 cm^3

which is a little larger than we need, but close enough.

Number of Turns The voltage dependent equation

B = LI

NA tesla

...connects B to L. However we also know that a statement about r is equivalent to a statement about L - for a given frequency (the "L× I equation"). So combining these equations, and ...also connecting the swing in the B-field to its peak (through r), we get a very useful form of the voltage dependent equation, in terms of r (expressed in MKS units):

(voltage dependent equation, any topology) So even with no information about the permeability of the material, air gap, and so on, we already know the number of turns required on a core with area Ae that will produce a certain B-field. We also know that with or without an air gap, the B-field should not exceed 0.3 T for most ferrites. So solving the equation for N (N is nP here, number of primary turns), ...

We will have to verify that this can be accommodated in the window of the core - along with the bobbin, tape insulation, margin tape, secondary windings, sleeving, and so on.

Usually, that is no problem for a flyback.

Note that if we want to reduce N, the only possible ways are -to allow for a larger r, or decrease the duty cycle (i.e. pick a lower VOR), or allow for a higher B (new material!?), or increase the area of the core - the latter, hopefully, without increasing the volume, because that would amount to over-design. But certainly, just playing with the permeability and air gap is not going to help!

The number of secondary turns (5 V output) is....

But we want an integral number of turns. Further, approximating this to one full turn is not a good idea since there will be more leakage. We therefore prefer to set....

So, with the same turns ratio (i.e. VOR unchanged) ...

The number of turns for the 12 V output is obtained by the scaling rule .....where we have assumed the 5 V diode has a drop of 0.6 V and the 12 V diode has a drop of 1 V.

Actual B-field

So now we can use the voltage dependent equation again, to solve for B ...

But in fact we don't have to use this equation anymore! We realize that BPK is inversely proportional to the number of turns. So if, with a calculated 35.5 turns, we had a peak field of 0.3 T, then with 46 turns we will have (keeping L and r unchanged!)

Note that in CGS units, the peak is now 2315 gauss, and the ac component is half the swing, that is, 463 gauss (since r = 0.5).

Note: If we start with a B-field target of 0.3 T, we are likely to reach a lesser B-field after rounding up the secondary turns to the nearest higher integer, as we did above. That of course is not only expected, but acceptable. However note that on power-up or power-down, For example, the B-field will increase further, as the converter tries to continue regulating. That is why we need to set the maximum duty cycle limit and/or current limit accurately, or the switch can be destroyed due to inductor/transformer saturation. Cost-effective flyback designs with fast-acting current limit and fast switches (especially those with an integrated mosfet) generally allow for a peak B-field of up to 0.42 T, so long as the operating field is 0.3 T or less.

Air Gap

Finally, we need to consider the permeability of the material! L is related to permeability by the equation ...

Here z is the 'gap factor' ...

Note that z can range from 1 (no gap) to virtually any value. A z of 10, For example, increases the energy-handling capability of an ungapped core set by a factor of 10 (its AL value falls by the same factor, and so does its effective permeability - µe = µµ0/z). So large gaps certainly help, but since we are still interested in maintaining L to a certain value ...based on our choice of r, we will have to increase the number of turns substantially. As mentioned, at some point, we just may not be able to accommodate these windings in the available window, and further, the copper loss will also increase greatly. So z in the range of 10 to 20 is a good compromise for gapped transformers made out of ferrite material. Let us see what it comes out to be, based on our requirements

Note: In general, if we use a center-gapped transformer, the total gap in the center must be equal to the above calculated value, whether each center limb has been ground or not. But if spacers are being inserted on both side limbs (say on an EE or EI type of core), the thickness of the spacer on each outer limb must be half of the above-calculated value, because the total air gap is then as desired.

FGR. 3: Skin Depth and AC Resistance Explained

Selecting the Wire Gauge and Foil Thickness

In an inductor, the current undulates relatively smoothly. However, in a transformer, the current in one winding stops completely, to let the other winding take over. Yes, the core doesn't care (and doesn't even know) which of its windings is passing current at a given moment, as long as the ampere-turns is maintained - because only the net ampere-turns determine the field (and energy) inside the core. But as far as the windings themselves are concerned, the current is now pulsed - with sharp edges, and therefore with significant high-frequency content. Because of this, 'skin depth' considerations are necessary for choosing the appropriate wire thickness of the windings of a flyback transformer.

Note: We had ignored this for dc-dc inductors, but in high-frequency dc-dc designs too (or with high r), we may need to apply these concepts there too.

At high frequencies, the electric fields between the electrons become strong enough to cause them to repel each other rather decisively, and thereby cause the current to crowd on the exterior (surface) of the conductor (see exponential curve in FGR. 3). This crowding worsens with frequency as per v f . There is thus the possibility that though we may be using thick wire in an effort to reduce the copper loss, a good part of the cross-section of the wire (its "innards") just may not be available to the current. The resistance presented to the current flow is inversely proportional to the area through which the current is flowing, or is able to flow. So this current crowding causes an increase in the effective resistance of the copper (as compared to its dc value). The resistance now presented to the current is called the 'ac resistance' (see lower half of FGR. 3). This is a function of frequency, because so is the skin depth. Instead of thus wasting precious space inside the transformer and losing efficiency too, we must try to use more optimum diameters of wire - in which the cross-sectional area is better-utilized. Thereafter, if we need to pass more current than the chosen cross-sectional area can handle, we need to parallel several such strands.

So how much current can a given wire strand handle? That depends purely on the heat buildup and the need to keep the overall transformer with an acceptable temperature rise.

For this a good guideline/rule-of-thumb for the current density of flyback transformers is 400 circular mils ('cmils') per ampere, and that is our goal too in the analysis that follows.

Note: Expressing "current density" in the North American way of cmils/A needs a little getting used to.

It’s actually area per unit ampere, not ampere per unit area (as we would normally expect a "current density" to be)! So a higher cmils/A value actually is a lower current density (and vice versa) - and will produce a lower temperature rise.

We define the skin depth 'd' as the distance from the surface of a conductor at which the current density falls to 1/e times the value at the surface. Note that the current density at the surface is the same as the value it would have had all through the copper, were there no high-frequency effects. As a good approximation to the exponential curve, we can also imagine the current density remaining unchanged from the value at the surface, until the skin depth is reached, falling abruptly to zero thereafter. This follows from an interesting property of the exponential curve that the area under it from 0 to 8 is equal to the area of a rectangle passing through its 1/e point (see FGR. 3).

Therefore, when using round wires, if we choose the diameter as twice the skin depth, no point inside the conductor will be more than one skin depth away from the surface. So no part of the conductor is unutilized. In that case, we can consider this wire as having an ac resistance equal to its dc resistance - there is no need to continue to account for high-frequency effects so long as the wire thickness is chosen in this manner.

If we use copper foil, its thickness too needs to be about twice the skin depth.

In FGR. 4 we have a simple nomogram for selecting the wire gauge and thickness. The upper half of this is based on the current carrying capability as per the usual requirement of 400 cmil/A. But the readings can obviously be linearly scaled for any other desired current density. The vertical grid on the nomogram represents wire gauges. An example based on a switching frequency of 70 kHz is presented within the figure. In a similar manner, for our previous worked example, we see that for 150 kHz operation, we should use AWG 27. But its current carrying capacity is only 0.5 A at 400 cmils/A (and only 0.25 A at a lower current density of 800 cmils/A!). Therefore, since the center of the primary current ramp was iterated and estimated to be 1.488 A, we need three strands of AWG 27 (twisted together) to give a combined current capability of 1.5 A (which is slightly better than what we need).

Coming to the secondary side of the worked example, we remember we had lumped all the current as a 5 V equivalent load of 15 A. But in reality it’s only 10 A, two-thirds of that.

So the center of its current ramp, which we had calculated was about 34 A, is actually (2/3) × 34 = 22.7 A. The balance of this, that is, 34 - 22.7 = 11.3 A, reflects as (5.6/13)× 11.3 = 4.87 A into the 12 V winding. So the center of the 12 V output's current ramp is 4.87 A. We can choose the 12 V winding arrangement using the same arguments we present below for the 5 V winding.

FGR. 4: Nomogram for Selecting Wires and Foils Thicknesses, Based on Skin Depth Considerations

For the 5 V winding, we can consider using copper foil, since we have only two turns, and we need a high current capability. The center of the 5 V secondary-side current ramp is about 23 A. The appropriate thickness (2d) at this frequency is found by projecting downward along the AWG 27 vertical line. We get about 14 mils thickness. But we still don't know if the current through it will follow our guideline of 400 cmil/A, since it’s a foil.

We need to check this out further.

One circular mil ('cmil') is equal to 0.7854 square mils. Therefore 400 cmils is 400 ×

0.7854 = 314 sq. mils (note p/4 = 0.7854). So for 23A we need 23 × 314 = 7222 sq. mils.

But the thickness of the foil is 14 mils. Therefore we need the copper foil to be 7222/14 = 515 mils wide, that is, about half an inch. Looking at a bobbin for the EI-30 in FGR. 5, we see it can accommodate a foil 530 mils wide. So this is just about acceptable. Note that if the available width is insufficient, we would need to look for another core altogether - one with a "longer" (stretched out) profile. Cores like that are available as American "EER" cores. Or we can again consider using several paralleled strands of round wire. The problem is that a bunch of 46 twisted strands (of AWG 27) is going to be bulky, difficult to wind, and will also increase the leakage inductance. So we may like to use say 11 or 12 strands of AWG 27 twisted together into one bunch, and then take four of these bunches (all electrically in parallel), laid out side by side to form one layer of the transformer. For a two-turns secondary, therefore, we would wind two layers of this.

FGR. 5: Checking to See if a 23 A Foil Can Be Accommodated on an EI-30 Bobbin: Foil windings.