Off-line Converter Design and Magnetics--Forward Converter Magnetics

The procedure presented in this section applies explicitly to the single-switch forward converter. However, the general procedure remains unchanged for the two-switch forward converter as well.

Duty Cycle

The duty cycle of a forward converter is ...

FGR. 6: The single-ended Forward Converter---Switch and Pri. winding; Magnetization current component.

Comparing this with the duty cycle of a buck, we see that the only difference is the term nS/nP. As mentioned, this is the coarse fixed-ratio step-down function available due to transformer action. We can therefore visualize that the input voltage VIN gets reflected to the secondary side. This reflected voltage VINR = VIN/n (where n = nP/nS) gets impressed at the secondary-side switching node. From there on, we have in effect a simple dc-dc buck stage, with an input voltage of VINR and an output of VO (see FGR. 6). Therefore the design of the forward converter's choke is not going to be covered here, as it’s designed using the same procedure as that of any buck inductor. However, the forward converter's transformer is another story altogether!

Note: Regarding choke design, we should keep in mind that for high-current inductors, as would be found in a typical forward converter, the calculated wire gauge may be too thick (and stiff) for winding easily over the core/bobbin. In that case, several thinner wire gauges may be twisted together to make the winding more flexible and easier to handle in production. Further, since choke and inductor design has usually little to do with high-frequency skin depth considerations, we can choose strands of almost any practical diameter, so long as we have enough net copper cross-sectional area to keep the temperature rise to within about 40 to 50fic.

Unlike a flyback transformer, the forward converter's secondary winding conducts at the same time as the primary winding. This leads to an almost complete flux cancellation inside the core. But there is one component of the primary current waveform which remains the same, irrespective of the load. This is the magnetization current component - shown in gray on the left side of FGR. 6. At zero load, this is the entire current through the primary winding and switch (assuming duty cycle remains fixed). As soon as we try to draw some load current, the secondary winding current increases, and so does the primary winding current. Each current increases proportionally to the load current, and so their increments too are mutually proportional - the proportionality constant being the turns ratio. But more significantly, they are of opposite sign - that is, looking at FGR. 6, we see that the current enters the dotted end of the transformer on the primary side, and on the secondary side, it leaves by the dotted end at the same time. Therefore, the net flux in the core of the transformer remains unchanged from the zero load condition (assuming D is fixed) - because the core just never "sees" any change in the net ampere-turns flowing through its windings. All conditions inside the core, i.e. the flux, the magnetic fields, the energy stored, and even the core loss, are only dependent on the magnetization current. Of course the windings themselves have a different story to tell - they bear the entire brunt, not only of the actual load current, but the sharp edges and consequent high-frequency content of the pulsed current waveforms.

The magnetization current component is not coupled by transformer action to the secondary.

In that sense, it’s like a "parallel leakage inductance." We need to subtract this component from the total switch current, and only then will we find that the primary and secondary currents scale according to the turns ratio. In other words, the magnetization current does not scale - it stays confined to the primary side.

But in fact, the magnetization current is the only current component that is storing any energy in the transformer. So in that sense, it’s like the flyback transformer! But, if we are to achieve a steady state, even the transformer needs to be "reset" every cycle (along with the output choke). But unfortunately, the magnetization energy is effectively "uncoupled," simply because of the output diode direction, and so we can't transfer it over to the secondary side. If we don't do anything about this energy, it will certainly destroy the switch by a spike similar to the leakage in a flyback. We don't want to burn it either, for efficiency reasons. Therefore, the usual solution is to use a 'tertiary winding' (or "energy recovery winding"), connected as shown in FGR. 6. Note that this winding is in flyback configuration with respect to the primary winding. It conducts only when the switch turns OFF, and thereby freewheels the magnetization energy back into the input capacitor. There is some loss associated with this "circulating" energy term, because of the diode drop and resistance of the tertiary winding. Note however, that any bona-?de leakage inductance energy also gets recycled back into the input by the tertiary winding. So we don't need an additional clamp for it.

For various subtle reasons, like being able to ensure the transformer resets predictably under all conditions, and also for various production-related reasons, the number of turns of the tertiary winding is usually kept exactly the same as the primary winding. Therefore by transformer action, the voltage at the primary-side switching node (drain of the mosfet) must rise to 2 × VIN when the switch turns OFF. Therefore, in a universal-input off-line single-ended (i.e. single-switch) forward converter, we need a switch rated for at least 800 V.

As soon as the transformer is reset (i.e. the current in the tertiary winding returns to zero), the drain voltage suddenly drops to VIN - that is, no voltage is then present across the primary winding - and therefore there is no voltage across the secondary winding either. The catch diode of the output stage (i.e. the diode connected to the secondary ground in FGR. 6) then freewheels the energy contained in the choke. Note that there is actually some ringing on the drain of the mosfet for a while, around an average level of VIN, just after transformer reset occurs. This is attributable to various undocumented parasitics (not displayed in the figure). The ringing however does contribute significantly to the radiated EMI.

Note that even prior to transformer reset, the secondary winding has not been conducting for a while - simply because the output diode (i.e. the one connected to swinging end of the secondary winding) has been reverse-biased during the time the tertiary winding was conducting.

Note also that the duty cycle of such a forward converter can under no circumstances ever be allowed to exceed 50%. The reason for that is we have to unconditionally ensure that transformer reset will always occur, every cycle. Since we have no direct control on the transformer current waveforms, we have to just leave enough time for the current in the tertiary winding to ramp down to zero on its own. In other words, we have to allow voltseconds balance to occur naturally in the transformer. However, because the number of turns in the tertiary winding is equal to the primary turns, the voltage across the tertiary winding is equal to VIN when the switch is ON, and is also equal to VIN (opposite direction) when the switch is OFF. Reset will therefore occur when tOFF becomes equal to tON. So, if the duty cycle exceeds 50%, tON would certainly always exceed tOFF, and therefore transformer reset would never be able to occur. That would eventually destroy the switch.

Therefore, just to allow tOFF to be large enough, the duty cycle must always be kept to less than 50%.

We realize that the forward converter transformer is always in DCM (its choke is usually in CCM, with an r of 0.4). Further, since the flux in the transformer remains unchanged for all loads, we can logically deduce that no part of the energy flowing through it into the output must be being stored in the transformer. So the question really is - what does the power-handling capability of a forward converter transformer depend on? We intuitively realize that we can't use any size transformer for any output wattage! So what governs the size? We will soon see that it’s determined simply by how much copper we can squeeze into the available 'window area' of the core (and more importantly, how well we can utilize this available area), without getting the transformer too hot.

Worst-case Input Voltage End

The most basic question in design invariably is - what input voltage represents the worst-case point at which we need to start the design of the magnetics (from the viewpoint of core saturation)? For the forward converter choke, this should be obvious - as for any buck converter, we need to set its current ripple ratio at around 0.4 at VINMAX. But coming to the transformer, we need some analysis before we can make a proper conclusion.

Note that the transformer of a forward converter is in discontinuous mode (DCM), but the duty cycle is determined by the choke, which is in CCM. Therefore, the duty cycle of the transformer also gets "slaved" at the CCM duty cycle of D = VO/VINR, despite the fact that it’s in DCM. This rather coincidental CCM + DCM interplay leads to an interesting observation - the voltseconds across the forward converter transformer is a constant, irrespective of the input voltage. The following calculation makes that clear, by the fact that VIN cancels out completely:

So in fact, the swing '?' of the current or the field is the same at high input or at low input, or in fact at any input (as long as the choke is in CCM). Since the transformer is in DCM, its peak is equal to its swing, and so the peak too does not depend on VIN. Of course, the peak switch current ISW_PK is the sum of the peak of the magnetization current IM_PK, and the peak of the secondary-side current waveform reflected onto the primary side that is, ...

So although the current limit of the switch must be set high enough to accommodate ISW_PK at VINMAX (since that is where the maximum peak of the reflected output current component occurs), as far as the transformer core is concerned, the peak current (and corresponding field) is just IM_PK, which does not depend on VIN! This is indeed an interesting situation.

Note also, that as far as the choke is concerned, the peak inductor current is no longer equal to the (reflected) peak switch current (as in a dc-dc buck topology), though the peak diode current still is. Yes, if we subtract the magnetization current from the switch current, and then scale (reflect) it to the secondary side according to the turns ratio, then the peak of that waveform will be equal to the peak inductor current.

So effectively IM has the property of input voltage rejection. We can intuitively understand this in the following way - as the input increases, it tries to increase the slope of the transformer current ramp and thereby get ?I to increase. However, the output choke, sensing a higher VINR, decreases its duty cycle, and therefore also that of the transformer, and in effect tries to thereby reduce the current swing in the transformer. Coincidentally, these two opposing forces virtually counterbalance each other perfectly, and so there is no net change in the resultant current swing in the transformer.

As a corollary, the core loss in the transformer is independent of the input voltage. The copper loss, on the other hand, is always worse at low inputs (except for the dc-dc buck) - simply because the average input current has to increase so as to continue to satisfy the basic power requirement PIN = VIN × IIN = PO.

Though we can pick any specific input voltage point for assuring ourselves that the core does not saturate anywhere within its input range, since the copper loss is at its worst at VINMIN, we conclude that the worst-case for a forward converter transformer is at VINMIN. For the choke, it’s still VINMAX.

FGR. 7: An ETD-34 Bobbin Analyzed

FGR. 8: The Area Physically Occupied by a Round Wire, and a "Square Wire" of the Same Conducting Cross-sectional Area as a Round Wire Window Utilization

Looking at a typical winding arrangement on an 'ETD-34' core and bobbin in FGR. 7,we see that the plastic bobbin occupies a certain part of the space provided by the core - thus reducing the available window 'Wa' from 171 mm2 to 127.5 mm2 - that is, by 74.5%.

Further, if we include the 4 mm "margin tape" that needs to be typically provided on either side (to satisfy international safety norms regarding clearance and 'creepage' requirements between primary and secondary sides), we are left with an available window of only 78.7 mm2 - that's a total reduction of 78.7/171 = 46%. In addition to this, looking at the left side of FGR. 8, we see that for any given wire, only 78.5% of the square area it "physically occupies" (or will occupy in the transformer) is actually conducting (copper). So in all, this leads to a total reduction of the available window space by 0.46 × 0.785 = 36%.

We realize some more space will also be lost to interlayer insulation (and any EMI screens if present), and so on. Therefore, finally, we estimate that perhaps only 30 to 35% of the available core window area will actually be occupied by copper. That is the reason why we need to introduce a 'window utilization factor' K (later we will set it to an estimated value of 0.3). So ...

Here ACU is the cross-sectional area of one copper wire, and Wa is the entire window area of the core (note that for EE, EI types of cores this is only the area of one of its two windows!).

Relating Core Size to Its Power Throughput

We remember that the original form of the voltage dependent equation is ...

Substituting for N, the number of primary turns, we get ...

Performing some manipulations, ...

Let us do some substitutions here. Assuming a typical current density of 600 cmil/A, utilization factor K of 0.3, ?B equal to 1500 gauss, we get the following fundamental core-selection criterion:

Note: In a typical forward converter, it’s customary to set the swing in the B-field of the transformer at ?B ˜ 0.15 teslas. This helps reduce core loss, and usually also leaves enough safety margin for avoiding hitting BSAT under say power-up condition at high line. Note that in a flyback, the core loss tends to be much less, because ?I is a fraction of the total current (40% typically). But since the transformer of a forward converter is always in DCM, therefore the swing in B is now more significant - equal to its peak value, i.e. BPK = ?B. So, if we set the peak field at 3000 gauss, ?B would be 3000 gauss too, roughly twice that of a flyback set to the same peak. That is why we must reduce the peak field in a forward converter to about 1500 gauss.

Worked Example (8)-Designing the Forward Transformer

We’re building a 200 kHz forward converter for an ac input range of 90 to 270 Volts. The output is 5 V at 50 A, and the estimated efficiency is 83%. Design its transformer.

Input Power

We have ...

Selection of Core

We use the criterion calculated previously:...

This is in theory, probably a little larger than required. Bu it’s the closest standard size in this range. Later we will see it’s in fact just about adequate.

Note that this could have turned out to be significantly different from an integer. In that case, we would round it off to the nearest (higher) integer, and then recalculate the primary turns, the new flux density swing, and the core loss - similar to what we did for the flyback. But at the moment, we can simply use ....

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CORE mmf (amp-turns) portions

Note: If nS = 6 turns (1 layer), we would get p = 1/2 for this layer, by interleaving in the manner shown

Current going IN + INTERLEAVED NON-INTERLEAVED

Current coming OUT Example:

nP=12 turns, nS=12 turns

FGR. 9: How Proximity Losses Are Reduced by Interleaving

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FGR. 10: Finding the Lowest AC Resistance, as the Thickness of a Foil Is Varied

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Secondary Foil Thickness and Losses

The concept of skin depth presented earlier actually represents a single wire standing freely in space. For simplicity, we just ignored the fact that the field from the nearby windings may be affecting the current distribution significantly. In reality even the annular area we were hoping was fully available for the high frequency current, is not. Every winding has an associated field, and when this impinges on nearby windings, the charge distribution changes, and eddy currents are created (with their own fields). This is called the 'proximity effect.' It can greatly increase the ac resistance and thus the copper losses in the transformer.

The first thing we need to do to improve the situation is have opposing flux lines cancel each other. In a forward converter, that is in fact something that tends to happen automatically, because the secondary windings pass current at the same time as the primary, and in the opposite direction. However, even that can prove totally inadequate, especially at the higher power levels that a forward converter is more commonly associated with. So a further reduction in these proximity losses is achieved by interleaving as shown in FGR. 9.

Basically, by splitting the sections, and trying to get primary and secondary layers adjacent to each other as much as possible, we can increase cancellation of local adjoining fields.

In effect, we are trying to prevent the ampere-turns from cumulating as we go from one layer to the next. Note that the ampere-turns are proportional to the local fields that are causing the proximity losses. However, it’s impractical to interleave too much - because we will need several more layers of primary-to-secondary insulation, more terminations, and also more EMI screens at every interface (if required) - all of which will add up to higher cost and eventually lead to possibly higher, rather than lower, leakage. Therefore, most medium-power off-line supplies just split the primary into two sections, one on either side of a single-section secondary.

The other way to reduce losses is to decrease the thickness of the conductor. But there are several ways we can do this. If, For example, we take a winding made up of single-strand wire, and split the wire into several paralleled finer strands in such a way that the overall dc resistance does not change in the process, we will find that the ac resistance goes up first before it reduces. On the other hand, if we take a foil winding, and decrease its thickness, the ac resistance falls before it rises again.

In FGR. 9, we have also defined 'p,' the layers per portion. Note how p gets reassigned when we interleave.

But how do we go about actually estimating the losses? Dowell reduced a very complex multidimensional problem into a simpler, one-dimensional one. Based on his analysis, we can show that there is an optimum thickness for each layer. Expectedly, this turns out to be much less than 2 × d, where d is the skin depth defined earlier.

Note: In the flyback, we had ignored the proximity effect for the sake of simplicity. But in any case, since the primary and secondary windings don’t conduct at the same time, interleaving won't help. But interleaving is still carried out in the flyback, in a manner similar to the forward converter. However, the purpose then is to increase coupling between primary and secondary, and thereby reduce the leakage inductance. However, this also increases the capacitive coupling - unless grounded screens are placed at the primary-secondary interface. Screens are in general helpful in reducing high-frequency noise from coupling into the output, and suppressing common-mode conducted EMI. But they also increase the leakage inductance, which is of great concern particularly in the flyback. Note also that screens must be very thin, or they will develop very high eddy current losses of their own. Further, the ends of a primary-secondary screen should not be connected together, or they will constitute a shorted turn for the transformer.

In FGR. 10, we have plotted out Dowell's equations in a form applicable to a square current waveform (unidirectional) in a transformer with foil windings. Note that the original Dowell curves actually plot FR versus X. But we have plotted FR/X versus X, where ...and ...

X = h / d

… h being the thickness of the foil. The reason why we have not plotted FR versus X is that FR is only the ratio of the ac to dc resistance. It’s not FR, but RAC that we are really interested in minimizing. So the "optimum RAC" point need not necessarily be the point of the lowest FR.

Let us try to understand this for a stand-alone foil (similar to what we did in FGR. 3).

If we slowly increase the thickness of the foil, once the foil thickness exceeds 2d, the ac resistance won't change any further, since the cross-sectional area available for the high-frequency current remains confined to d on each side of the foil. But the dc resistance continues to decrease as per 1/h - and as a result FR will increase. So the relationship between RAC and FR is not necessarily obvious. Therefore, since FR = RAC/RDC, with RDC ? 1/h, we get RAC ? FR/h. And this is what we really need to minimize (for a foil).

Further, since we always like to write any frequency-dependent dimension in reference to the skin depth, we have plotted not FR/h, but FR/X, versus X, in FGR. 10.

Note that in FGR. 10, the p = 1 and p = 0.5 curves don’t really have an "optimum." For these, the FR/X (ac resistance) can be made even smaller as we increase X (thickness). FR will in fact become much greater than 1. However, we see that for p = 1 For example, no significant reduction in ac resistance occurs if X exceeds about 2, that is, thickness of foil equal to twice the skin depth. We can make it thicker if we want, but only for marginal improvement in the secondary winding losses. Further, in the process, we may also take away available area for the primary windings (and any other secondary windings), and that can lead to higher overall losses. Though we are also cautioned not to fill up all "available space" with copper, especially when we come to (round) wire windings. That can be shown, not only to increase FR, but RAC too.

Now let us apply what we have learned to our ongoing numerical example. We start by taking a copper foil wound twice on the ETD-34 bobbin - to form the 5 V secondary winding. Since this is interleaved with respect to the primary, only one turn "belongs" to each split section. So the layers per portion for the secondary is p = 1. We will calculate the losses, and if acceptable, we will stay with the resulting arrangement.

We can start with a reasonable current density (about 400 cmils/A should suffice here).

We use

h = IO × Jcmils/A × 10^2

width × 197,353 mm

...where h is the foil thickness in mm, IO is the load current (50 A in our example), and 'width' is the width available for the copper strip (20.9 mm for the ETD-34).

If the losses are not acceptable, we may need to look for a bobbin that will allow a wider width of foil. Or we can consider paralleling several thinner foils to increase p. For example, if we take four paralleled (thinner) foils in parallel (each insulated from the others), we will get four effective layers for the secondary, and the layers per portion will then become 2.

Primary Winding and Losses For the secondary, we have finally chosen copper foil of thickness 0.4625 mm (i.e. 0.4625× 39.37 = 18 mil). Let us assume each foil is covered on both sides by a 2 mil thick mylar tape. Since 1 mil is 0.0254 mm, we have effectively added 4 × 0.0254 mm to the foil thickness. In addition there will be three layers of tape between each of the two primary-secondary boundaries. So in all, the thickness occupied by the secondary and the insulation, h ,is ...

The ETD-34 has an available height inside the bobbin of 6.1 mm. That now leaves 6.1 - 1.434 = 4.67 mm. Therefore each section of the split primary has an available winding height of 2.3 mm only. We should ultimately check that we can accommodate the primary winding we decide on, within this space.

Note that for the primary, the available width is only 12.9 mm (since there is 4 mm margin tape on each side - for the secondary, since we have a foil with tape wrapped over it, we don’t need the margin tape). We need to find how best to accommodate eight turns into this available area, with minimum losses.

Note: It’s not mandatory to use a particular thickness of insulating tape, provided it’s safety-approved to withstand a specified voltage. We can, For example, use 1 mil approved tape or even 1/2 mil, if it suits our production, helps lower the cost, and/or improves performance in some way.

Let us first understand the basic concept for winding wires here. For a stand-alone wire, as in FGR. 3, as we increase the diameter of the wire, the cross-sectional area available for the high-frequency current is (p × d) × d. And since resistance is inversely proportional to cross-sectional area, we get RAC ? 1/d. Similarly, RDC ? 1/d2.SoFR ? d. Therefore, RAC ? 1/FR. This actually means that a higher FR (bigger diameter) will decrease the ac resistance! That is not surprising, because the annulus available for the high-frequency current does increase if the diameter increases. However, this is not the way to go when dealing with "non-stand-alone" wire. Because, by increasing the diameter, we will inevitably move on to higher number of layers, and Dowell's equations then tell us that the losses will increase, not decrease.

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FGR. 11: Understanding the process of "Subdivision" - Keeping the DC Resistance Unchanged, and How the Equivalent Foil Transformation Process Takes Place

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On the top left side of FGR. 11, we have Dowell's original curves, which show how FR varies with respect to X (i.e. h/d). The parameter for each curve is layers per portion (i.e. p).

Note that Dowell's curves talk in terms of foils only. They don't care about the actual number of turns in the primary or secondary (i.e. from the electrical point of view), but only the effective layers per portion (from the field point of view). So, when we consider a layer of round wires of diameter 'd,' we need to convert this into an equivalent foil. Looking back at the right side of FGR. 8, we see that that this amounts to replacing a wire of diameter d with a foil slightly thinner (i.e. with the same amount of copper, but in a square shape).

Alternatively, if we want to get a foil of X = 4 For example, we need to start with a wire of diameter 1/0.886 = 1.13timesX. Finally, as indicated, all these copper squares then merge (from the field point of view), to give an equivalent layer of foil.

In FGR. 11, we are also conducting a certain "experiment" - as an alternative way of laying out wires optimally. Suppose we have several round wires laid out side by side with a diameter 1.13 × 4d. Suppose also, that this constitutes one layer per portion in a given winding arrangement. This is therefore equivalent to a single-layer foil of thickness 4d, that is, X = 4. Now using Dowell's curves, the FR is about 4 (points marked "A" in FGR. 11).

Suppose we then divide each strand into four strands, where each strand has a diameter half the original. Therefore, the cross-sectional area occupied by copper remains the same because...

.

However, the equivalent foil thickness is now half of what it was -2d (i.e. X = 2). And we also now have two layers per portion from Dowell's standpoint. Consulting Dowell's curves, we get an FR of about 5 now (marked "B"). Since we are keeping RDC fixed in the process, RAC ? FR. Therefore now, decreasing FR is a sure way to go, to decrease RAC. So an FR of 5 is decidedly worse than an FR of 4. We now go ahead and subdivide once more, in a similar manner. So we then get four layers per portion, each with X = 1, and FR has gone down to about 2.6 (points marked "C"). We subdivide once more, and we get eight layers per portion, with X = 0.5. This gives us an FR of about 1.5 (marked "D"). This is an acceptable value for FR.

Note that all these steps have been collected and plotted out in FGR. 11 on the right side, with the horizontal axis being the number of successive subdivision steps (in each step we subdivided each wire into four of the same dc resistance). These steps are being called "sub" (for subdivision step), where sub goes from 0 (no subdivision) to 1 (1 subdivision), 2 (2 subdivisions), and so on. We then also realize that with each step, X and p change as per....

For example, after four subdivision steps, the foil thickness will drop by a factor of 16, and the number of layers will increase by the same factor. We can then look at Dowell's curves to find out the new FR.

However, there are a few problems with directly applying Dowell's curves to switching power regulators. For one, the original curves only talked about the ratio of the thickness to the skin depth - and we know skin depth depends on frequency. So implicitly, Dowell's curves provide the FR for a sine wave. Further, Dowell's curves don’t assume the current has any dc value. So engineers, who adapted Dowell's curves to power conversion, would usually first break up the current waveform into its ac and dc components, apply the FR obtained from the curves to the ac component only, compute the dc loss separately (with FR = 1), and then sum as follows:

...

However, in our case, we have preferred to follow the more recent approach of using the actual (unidirectional) current waveform, splitting it into Fourier components, and summing to get the effective FR. The losses are expressed in terms of the thickness of the foil as compared to the d at the fundamental frequency (first harmonic). We also include the dc component in computing this effective FR. That is the reason, when calculating the secondary winding losses, we were able to use the simple equation ...

In that case, the FR was actually the effective FR (computed for a square wave with dc level included), though not explicitly stated. However, note that the graphs in FGR. 11 are still based on the original sine-wave approach, and the purpose here was only to demonstrate the subdivision technique through the original curves.

But in FGR. 12, we have finally modified Dowell's original sine-wave curves. Fourier analysis has been carried out while constructing these curves, and so the designer can apply them directly to the typical (unidirectional) current waveforms of power conversion. We will now use these curves to do the calculations for the primary winding of our ongoing numerical example.

But one question may still be puzzling the reader - why are we not using the previous FR/X curves (see FGR. 10) that we used for the secondary? The reason is the situation is different now. The curves in FGR. 10 are Dowell's curves for a square wave, except that on the vertical axis we have used FR/X, not FR. That is useful only when we are varying h and seeing when we got the lowest RAC. But for the primary windings, we are going to fix the height of the windings in each step of the iterations that follow. We will be using the subdivision technique in each iteration, and therefore keep the dc resistance constant. So now the minimum RAC (for a given iteration step) will be achieved at the minimum FR, not at the minimum FR/X.

The subdivision method was originally presented in FGR. 11, except that now we will use the modified curves in FGR. 12.

FGR. 12: Dowell's Curves Modified for Square Current Waveforms, and the Corresponding FR Curves for the Subdivision Method

First Iteration:

Let us plan to try and ?t eight turns on one layer. Lesser number of layers will usually be better. We remember that we have 12.9 mm available width on the bobbin. So if we stack eight turns side by side (no gap between them) we will require each of these eight round wires to have a diameter of ...

We can check that the available height of 2.3 mm is big enough to accommodate this diameter of wire. The penetration ratio X is (using the equivalent foil transformation) ...

The p is equal to 1. From either of the graphs in FGR. 12, we can see that the FR will be about 10 in this case (marked "A"). Further, from the graph on the left side, we can see that we need to subdivide the "X = 7.7" curve (imagine it close to the X = 8 curve) seven times to get the FR below 2. That would give strands of diameter ...

The corresponding AWG can be calculated by rounding off ....

So we get ...

But this is an extremely thin wire, and may not even be available! Generally, from a production standpoint, we should not use anything thinner than 45 AWG (0.046 mm).

Second Iteration:

The problem with the first iteration is that we started with a very thick wire, with a very high FR. So this demanded several subdivisions to get the FR to fall below 2. But what if we start off with a wire of lesser diameter than 1.6125 mm? We would then need to introduce some wire-to-wire spacing so we can spread the eight turns evenly across the bobbin.

However, that would be wasteful! We should remember that if a layer is already assigned and present, we might as well use it to our full advantage to lower the dc resistance - the problem only starts when we indiscriminately increase the number of layers. Therefore in our case, let us try paralleling two thinner wires to make up the primary. We still want to keep to one layer (without spacing). That means we will now have 16 wires placed side by side in one layer. We then define a 'bundle' as the number of wires paralleled to make the primary winding (we will be subdividing each of these further). So in our case ...

The p is still equal to 1. From both the graphs in FGR. 12, we can see that the FR will be about 5.3 in this case (marked "B"). Further, from the graph on the left side, we can see that we need to subdivide five times to get the FR below 2. That would give strands of diameter ....

This is still thinner than the practical AWG limit of 0.046 mm.

So finally, the primary winding consists of four bundles in parallel, each bundle consisting of 64 strands, side by side in one layer, with an FR of about 1.8.

We can continue the process if we want to get a slightly lower FR. But at some point we will find the FR will start to go up again. For our purpose, we will take an FR of less than 2 as acceptable to proceed with the loss estimates.

Note that further tweaking will always be required since when we bunch wires together to form a bundle, they will "stack" in a certain manner that will affect the dimensions from what we have assumed. Further, the diameter of the wire we used was for bare wire, and is slightly less than the coated diameter. Note that in general, if after winding several layers evenly, we are left with a few turns that seem to need another layer to complete, we are better off reducing the primary number of turns and sticking to the existing completed layers, because even a few turns extra will count as a new layer from the field point of view, and increase proximity losses.

We can now calculate the losses for the two primary sections combined, since they can be considered to be identical and with the same FR. The ac resistance in ohms of the entire ....

Had we gone further and divided the primary into five bundles and then subdivided three times, we would get eight layers with 64 strands of 0.04 mm diameter wire per bundle, and an FR of 1.65 - which seems better than the 1.8 we got in the last step. But since the wires are so thin to start with, the dc resistance now goes up, and the dissipation will rise to 1.26 W.

Total Transformer Losses

The estimated temperature rise ...

What we are seeing is a typical practical situation! The temperature rise is 15fic higher than we were expecting! However, 55fic is perhaps still acceptable (even from the standpoint of getting safety approvals without special transformer materials). Admittedly, there is room for more optimization. However, the next time we do the process, we must note that the core loss is only a third the total loss, not half, as we had initially assumed.

Note also that methods in related literature may predict a smaller temperature rise. But the fact is that these are usually based on the sine-wave versions of Dowell's equations, and we know that will typically underestimate the losses significantly.