Polarity of Windings in a Transformer

In Figure 3.1, the turns ratio is n=N_P/N_S, where N_P is the number of turns of the Primary winding, and N_S is the number of turns of the Secondary winding.

Figure 3.1: Voltage and currents in a flyback.

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. This 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 all the dotted and non-dotted ends of the transformer simultaneously, without changing the schematic in any way.

In a flyback, the relative 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 Figure 3.1 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 start to understand why the flyback is considered to be just a Buck-Boost derivative. More on this in Chapter 5.

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., by which the inductor receives energy from the input) to its freewheeling path (i.e., by 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. We thus 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 Figure 3.1, 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 Figure 3.1 — 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, 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 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 Figure 3.1. But it is perhaps not immediately apparent why the flyback inductor exhibits transformer action since the windings do not conduct at the same time.

When the switch turns ON, a voltage V_IN (the rectified AC input) gets impressed across the Primary winding of the transformer. And at the same time, a voltage equal to V_INR=V_IN/n (“R” stands 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 V_INR is. The 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 ϕ is the same for both, and so is the rate of change of flux dϕ/dt for each winding. Therefore,

or

Also,

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 ϕ present in the core or not, each winding encloses this entire flux, and so the basic equation V=−N×dϕ/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 V_O. Therefore, the voltage at the Secondary-side switching node gets clamped at V_O (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 V_O. By transformer action, this reflects a voltage across the Primary winding, equal to V_OR=V_O×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 V_IN. However, now this reflected output voltage V_OR, coming through the transformer, adds to that. Therefore, the voltage at the Primary-side switching node eventually goes up to V_IN+V_OR (for now, we are ignoring the leakage spike encircled in Figure 3.1).

Note: During the on-time, the Primary side is the one determining the voltages across all the windings. And during the off-time, it is 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 on the Secondary winding. Either way, we get the same result, as shown in Table 3.1.

Table 3.1. Derivation of DC Transfer Function of Flyback.

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 describe 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 Figure 3.1.

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

To get the value of the voltage at the swinging end of any winding, we must add the voltage across the winding to the DC voltage present at its “nonswinging” end.

So, for example, to get the voltage at the Drain of the MOSFET (swinging end of Primary winding), we need to add V_IN (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 Figure 3.1.

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 do not conduct at the same time. So, in fact, it seems 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, so the energy associated with each of them 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 L_p is the inductance measured across the Primary winding with the Secondary winding floating (no current), and L_S the inductance measured across the Secondary winding with the Primary winding floating. But we know that

where A_L 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:

or

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 Chapter 1 — that the current through an inductor cannot change discontinuously. Now we see that the net Ampere-turns of a transformer cannot change discontinuously.

Summarizing, transformer action works as follows — when reflecting a voltage from Primary side to Secondary side, we need to divide by the turns ratio. When going from the Secondary side 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 procedures 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 — such as the capacitor, diode, and switch currents — 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 reflect everything either to the Primary side or to the Secondary side. We thus get the two equivalent Buck-Boost models as shown in Figure 3.2. 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.

Figure 3.2: The equivalent Buck-Boost models of the flyback.

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” V_OR is one of the most important parameters of a flyback. As the name indicates, V_OR is effectively the output voltage as seen by the Primary side. In fact, if we compare the switch waveform of the flyback in Figure 3.1 with that of a Buck-Boost, we will realize that to the switch, it seems as if the output voltage is really V_OR. See Figure 3.2.

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 V_OR 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 V_OR 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 is a simple DC–DC Buck-Boost — delivering an output voltage of V_OR with a load current of I_OR.

As mentioned, the only difference between a transformer-based flyback that “thinks” it is providing an output of V_OR with I_OR and an inductor-based version that really is providing an output of V_OR with I_OR 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 Figure 3.1 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 C_O in the original flyback was charged up to a value of V_O. So, its stored energy was . In the Primary-side Buck-Boost model, the output of the converter is V_OR, that is, V_O×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 C_O/n². Notice also from Figure 3.2 how the inductance reflects. This is consistent with the fact that L∝Ν².

The Current Ripple Ratio for the Flyback

Looking at the equivalent Buck-Boost models in Figure 3.2, the center of the ramp on the Secondary side (average inductor current, “I_L” or I_DC) must be equal to I_O/(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 “I_LR,” where I_LR=I_L/n. Equivalently, it is equal to I_OR/(1–D), where I_OR is the reflected load current, that is, I_OR=I_O/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 are 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 try to set it to around 0.4 in most cases.

The value of r for a flyback is the same for both the Primary and the 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 “I_PKP,” 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 Figure 3.1). 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 leakage energy, the 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 burn it (dissipation). The latter approach is usually preferred for the sake of simplicity. It is commonly accomplished by means of a straightforward “zener diode clamp,” as shown in Figure 3.1. Of course, the zener voltage must be chosen according to the maximum voltage the switch can tolerate. Note that for several reasons, in particular efficiency, it is preferable to connect this zener across the Primary winding, as shown (via a blocking diode in series with it). An alternative method is to connect it from the switching node to Primary ground.

We can ask — where does the leakage inductance reside? Most of it is inside the Primary winding of the transformer, though some of it lies in the printed circuit board (PCB) trace sections and transformer terminations, especially with those associated with the Secondary winding, as we will see further.

Zener Clamp Dissipation

If we intend to burn the energy in the leakage, it is important to know how this affects the efficiency. It is sometimes intuitively felt the energy dissipated every cycle is , where I_PK is the peak switch current and L_LKP is the Primary-side leakage. That certainly is the energy residing in the leakage inductance (at the moment the switch turns OFF), but it is not the entire energy that 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 attain reset by freewheeling into the zener, the Primary winding is forced to follow suit and continues to provide this in-series current. Though the Primary winding is certainly trying (and managing partially) to freewheel into the Secondary side, a part of its energy also gets diverted into the zener clamp — continuing till the leakage inductance achieves full reset (zero clamp current). In other words, some energy from the Primary inductance gets virtually plucked out by the series leakage inductance, and this energy also finds its way into the zener, along with the energy residing in the leakage itself. A detailed calculation (see Figure 7.10) reveals that the zener dissipation actually is

So, the energy in the leakage gets multiplied by the term V_z/(V_z−V_OR) (this additional term is from the Primary inductance).

Note that if the zener voltage is too close to the chosen V_OR, the dissipation in the clamp goes up steeply. V_OR 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 “L_LK” in the dissipation equation above? Why didn’t we identify it as the Primary-side leakage (“L_LKP”)? The reason is that L_LK represents the overall leakage inductance as seen by the switch. So, it is partly L_LKP — 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 following the turn-off transition (thereby causes an increase in the zener dissipation), any Secondary-side inductance also prevents the freewheeling path from becoming available immediately. Basically, the Secondary-side inductance insists that we (“politely” and) slowly build up the current through it — respecting the fact that it is an inductance after all! However, until the current in the bona fide freewheeling path can build up to the required level, the Primary-side current still needs to freewheel somewhere! The path the inductor current therefore seeks out is the one containing the zener clamp (that 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 calculate 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 (see Figure 3.2).

For a given V_OR, if the output voltage is “low” (e.g., 5 V or 3.3 V), the turns ratio is much greater. Therefore, if the chosen V_OR 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 L_LK 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 L_LK (not just L_LKP).

We will find that the contribution from the Secondary-side traces can in fact make L_LK several times larger than L_LKP. L_LKP 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-described 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 5–10% in low-output voltage applications.

Worked Example (7) — Designing the Flyback Transformer

A 74-W Universal Input (90–270 VAC) flyback is to be designed for an output of 5 V at 10 A and 12 V at 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.

Selecting the Wire Gauge and Foil Thickness

In an inductor, the current undulates relatively smoothly. However, in a flyback 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 skin depth considerations.

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 Figure 3.3). This crowding worsens with frequency as per . There is thus the possibility that though we may be using thick wire in an effort to reduce 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 Figure 3.3). This is a function of frequency because so is the skin depth. Therefore, instead of wasting precious space inside the transformer and losing efficiency, 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.

Figure 3.3: Skin depth and AC resistance explained.

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 at 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 is 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 “δ” 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 ∞ is equal to the area of a rectangle passing through its 1/e point (see Figure 3.3 and also Figures 5.21 and 5.22).

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 Figure 3.4, 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 in 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-carrying capability of 1.5 A (which is slightly better than what we need).

Figure 3.4: Nomogram for selecting wires and foil thicknesses, based on skin depth considerations.

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 is 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.

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

One “cmil” is equal to 0.7854 square mils (sq.mils). Therefore, 400 cmils is 400×0.7854=314 sq.mils (note π/4=0.7854). So, for 23 A, we need 23×314=7,222 sq.mils. But the thickness of the foil is 14 mils. Therefore, we need the copper foil to be 7,222/14=515 mils wide, that is, about half an inch. Looking at a bobbin for the EI-30 in Figure 3.5, 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 (these are EE cores with the “R” indicating a round centerpost). 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.

Figure 3.5: Checking to see if a 23-A foil can be accommodated on an EI-30 bobbin.

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

Comparing this with the duty cycle of a Buck, we see that the only difference is the term N_S/N_P. As mentioned, this is the coarse fixed-ratio step-down function available due to transformer action. We can therefore visualize that the input voltage V_IN gets reflected to the Secondary side. This reflected voltage (V_INR=V_IN/n where n=N_P/N_S) 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 V_INR and an output voltage of V_O (see Figure 3.6). Therefore, the design of the Forward converter’s choke is not going to be covered here, as it is designed using the same procedure as that of any Buck inductor. However, the Forward converter’s transformer is another story altogether!

Figure 3.6: The single-ended Forward converter.

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 40–50°C.

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 Figure 3.6. 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 Figure 3.6, 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, that is, the flux, the magnetic fields, the energy stored, and even the core loss, are dependent only on the magnetization current. Of course, the windings themselves have a different story to relate — they bear the entire brunt, not only of the actual load current, but also of 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 current. In that sense, it is 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 is like the flyback transformer! But, if we are to achieve a steady state, even a transformer needs to be “reset” every cycle (along with the output choke). But unfortunately, the magnetization energy is effectively “uncoupled,” 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 Figure 3.6. Note that this winding is in flyback configuration with respect to the Primary winding. It conducts only when the switch turns OFF, and thereby it 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 fide leakage inductance energy also gets recycled back into the input by the tertiary winding. So, we don’t need an additional clamp for it in a traditional Forward converter.

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×V_IN 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 V_IN — 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 Figure 3.6) then freewheels the energy contained in the choke. Note that there is actually some ringing at the Drain of the MOSFET for a while, around an average level of V_IN, just after transformer reset occurs. This is attributable to various undocumented parasitics. The ringing contributes significantly to the radiated electromagnetic interference (EMI).

Note that even prior to transformer reset occurring, the Secondary winding has not been conducting for a while — simply because the output diode (i.e., the one connected to the 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 V_IN when the switch is ON and is also equal to V_IN (opposite direction) when the switch is OFF. Reset will therefore occur when t_OFF becomes equal to t_ON. So, if the duty cycle exceeds 50%, t_ON would certainly always exceed t_OFF, and therefore transformer reset would never be able to occur. That would eventually destroy the switch. Therefore, just to allow t_OFF 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 discontinuous mode (DCM) (its choke, i.e. inductor L, is usually in continuous conduction mode (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 is 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 V_INMAX. 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 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=V_O/V_INR, despite the fact that it is 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 V_IN cancels out completely:

So, in fact, the swing of the transformer current (or its field) is the same at high input or at low input, or in fact at any input (as long as the choke remains in CCM). Since the transformer is in DCM, its peak is equal to its swing, and so the peak too does not depend on V_IN. Of course, the peak switch current I_{SW_PK} is the sum of the peak of the magnetization current I_{M_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 I_{SW_PK} at V_INMAX (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 I_M__PK, which does not depend on V_IN! 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 (free wheeling) 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 I_M has the property of input voltage rejection. We can understand this in the following way — as the input increases, the slope of the transformer current increases, and ΔI therefore tends to increase. However, the output choke, sensing a higher V_INR, decreases its duty cycle and therefore also the on-time of the transformer, and that tends to reduce its current swing. Coincidentally, these two opposing forces counterbalance each other perfectly, and so there is no net change in the current swing of the transformer.

As a corollary, the core loss in the transformer is independent of the input voltage too. 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 P_IN=V_IN×I_IN=P_O.

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 V_INMIN, we conclude that the worst case for a Forward converter transformer is at V_INMIN. For the choke, it is still V_INMAX.

Window Utilization

Looking at a typical winding arrangement on an “ETD-34” core and bobbin in Figure 3.7, 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 (separation through air) and “creepage” (separation over surface of insulator) requirements between Primary and Secondary sides), we are left with an available window of only 78.7 mm² — that is a total reduction of 78.7/171=46%. In addition to this, looking at the left side of Figure 3.8, 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%. See also Figure 5.21.

Figure 3.7: An ETD-34 bobbin analyzed.

Figure 3.8: The area physically occupied by a round wire and a “square wire” of the same conducting cross-sectional area as a round wire.

We realize some more space will be lost to interlayer insulation (and any EMI screens if present), and so on. Therefore, finally, we estimate that perhaps only 30–35% 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,

and

Here A_CU 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,

where is the current density in A/m² and “AP” is called the “area product” (AP=A_e×Wa). Let us now convert into CGS units for greater convenience. We get

where AP is also in cm² now. Finally, converting the current density into cmils/A by using

we get

Solving for the area us do some substitutions here. Assuming a typical current density of 600 cmil/A, utilization factor K of 0.3, and ΔΒ equal to 1,500 G, we get the following fundamental core-selection criterion:

Note: In a typical Forward converter, it is customary to set the swing in the B-field of the transformer at ΔB ≈ 0.15 T. This helps reduce core loss and usually also leaves enough safety margin for avoiding hitting B_sat under say power-up condition at high line. Note that in a flyback, the core loss tends to be much less because ΔΙ is a fraction of the total current (40% typically). But since the transformer of a Forward converter is always in DCM, the swing in B is now more significant — equal to its peak value, that is, B_PK=ΔΒ. So, if we set the peak field at 3,000 G, ΔΒ would be 3,000 G 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 1,500 G.

Worked Example (8) — Designing the Forward Transformer

We are building a 200-kHz Forward converter for an AC input range of 90–270 V. The output is 5 Vat 50 A, and the estimated efficiency is 83%. Design its transformer.

Thermal Resistance

An empirical formula for EE-EI-ETD-EC types of cores is

where V_e is in cm³. Therefore, since V_e=7.64 cm³, for the ETD-34