DC-DC Converter Design and Magnetics--DC Transfer Functions, etc.

DC Transfer Functions

When the switch turns ON, the current ramps up in the inductor according to the inductor equation VON = L × delta ION/tON. The current increment during the on-time is:

delta ION = (VON × tON)/L

When the switch turns OFF, the inductor equation VOFF = L × delta ION/tOFF leads to a current decrement delta IOFF = (VOFF × tOFF)/L.

The current increment delta ION must be equal to the decrement delta IOFF, so that the current at the end of the switching cycle returns to the exact value it had at the start of the cycle - otherwise we wouldn't be in a repeatable (steady) state. Using this argument, we can derive the input-output (dc) transfer functions of the three topologies, as shown in Tbl. 1.Itis interesting to note that the reason the transfer functions turn out different in each of the three cases can be traced back to the fact that the expressions for VON and VOFF are different.

Other than that, the derivation and its underlying principles remain the same for all topologies.

Tbl. 2: How varying the inductance, frequency, load current, and duty cycle influence _I and IDC --- delta indicates it increases and decreases over the range

* maximum at D = 0.5

'×' indicates no change

? (=) indicates - IDC is increasing and is equal to IO

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time Changing Inductance does not change IDC (D constant)

IDC IDC Changing Frequency does not change IDC (D constant) current ; current; current voltage voltage time Fgr. 1: If D and IO Are Fixed, IDC Cannot Change IDC = IO 1 - D (boost and buck-boost)

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The DC Level and the "Swing" of the Inductor Current Waveform

From V = L dI/dt, we get delta I = V delta t/L. So the "swinging" component of the inductor current "delta I" is completely determined by the applied voltseconds and the inductance.

Voltseconds is the applied voltage multiplied by the time that it’s applied for. To calculate it, we can either use VON times tON (where tON = D/f), or VOFF times tOFF (where tOFF = (1 - D)/f ) - and we will get the same result (for that is how D gets defined in the first place!). But note also, that if we apply 10 V across a given inductor for 2 µs, we will get the same current swing delta I, if we apply say, 20 V for 1 µs, or 5 V for 4 µs, and so on.

So, for a given inductor, either talking about the voltseconds or about delta I is effectively one and the same thing.

What does the voltseconds depend on? It depends on the input/output voltages (duty cycle) and the switching frequency. Therefore, only by changing L, f, or D can we affect delta I.

Nothing else! See Tbl. 2. In particular, changing the load current IO does nothing to delta I.

IO is therefore, in effect, an altogether independent influence on the inductor current waveform. But what part of the inductor current does it specifically influence/determine? We will see that IO is proportional to the average inductor current.

The inductor current waveform is considered to have another (independent) component besides its swing delta I - this is the dc (average) level "IDC," defined as the level around which the swing delta I takes place symmetrically - that is, delta I/2 above it, and delta I/2 below it.

See Fgr. 1. Geometrically speaking, this is the "center of the ramp." It’s sometimes also called the "platform" or "pedestal" of the inductor current. The important point to note is that IDC is based only on energy flow requirements - that is, the need to maintain an average rate of energy flow consistent with the input/output voltages and desired output power. So if the "application conditions," that is, the output power and the input/output voltages, don’t change, there is in fact nothing we can do to alter this dc level - in that sense, IDC is rather "stubborn". In particular

++ Changing the inductance L doesn't affect IDC.

++ Changing the frequency f doesn't affect IDC.

++ Changing the duty cycle D does affect IDC - for the boost and buck-boost.

To understand the last bullet above, we should note the following equations that we will derive a little later

IDC = IO (buck)

The intuitive reason why the above relations are different is that in a buck, the output is in series with the inductor (from the standpoint of the dc currents - the output capacitor contributing nothing to the dc current distribution), and therefore the average inductor current must at all times be equal to the load current. Whereas, in a boost and buck-boost, the output is likewise in series with the diode, and so the average diode current is equal to the load current.

Therefore, if we keep the load current constant, and change only the input/output voltages (duty cycle), we can affect IDC - in all cases except for the buck. In fact, the only way to change the dc inductor current level for a buck is to change the load current. Nothing else will work! In the buck, IDC and IO are equal. But in the boost and buck-boost, IDC depends also on the duty cycle. That makes the design/selection of magnetics for these two topologies rather different from a buck. For example, if the duty cycle is 0.5, their average inductor current is twice the load current. Therefore, using a 5A inductor for a 5A load current may be a recipe for disaster.

One thing we can be sure of is that in the boost and buck-boost, IDC is always greater than the load current. We may be able to cause this dc level to fall and even approach the load current value if we reduce the duty cycle close to 0 (i.e. a very small difference between the input and output voltages). But then, on increasing the duty cycle toward 1, the dc level of the inductor current will climb steeply. It’s important we recognize this clearly and early on.

Another thing we can conclude with certainty is that in all the topologies, the dc level of the inductor current is proportional to the load current. So doubling the load current For example (keeping everything else the same), doubles the dc level of the inductor current (whatever it was to start with). So in a boost with a duty cycle of 0.5 For example, if we have a 5 A load, then the IDC is 10 A. And if IO is increased to 10 A, IDC will become 20 A.

Summarizing, changing the input/output voltages (duty cycle) does affect the dc level of the inductor current for the boost and the buck-boost. But changing D affects the swing delta I in all three topologies, because it changes the duration of the applied voltage and thereby changes the voltseconds.

++ Changing the duty cycle affects IDC for the boost and the buck-boost.

++ Changing the duty cycle affects delta I for all topologies.

Note: The off-line forward converter transformer is probably the only known exception to the above logic.

We will learn that if we For example double the duty cycle (i.e., double tON), then almost coincidentally, VON halves, and therefore the voltseconds does not change (and nor does delta I). In effect, delta I is then independent of duty cycle.

Based on the discussions above, and also the detailed design equations, we have summarized these "variations" in Tbl. 2. This table should hopefully help the reader eventually develop a more intuitive and analytical "feel" for converter and magnetics design, one which can come in handy at a later stage. We will continue to discuss certain aspects of this table, in more detail, a little later.

Defining the AC, DC, and Peak Currents

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Boost, Buck-Boost; Example: Buck If load current is 1A IL is 1A.

So if r=0.4 peak-to-peak current ('_I') is 0.4A and the peak current is 1.2A Boost/Buck-Boost If load current is 1A D=0.5 IL is 2A.

So if r=0.4, peak-to-peak current ('_I') is 0.8A and the peak current is...

IL is the same as IDC

= Average Inductor Current

Fgr. 2: The AC, DC, Peak, and Peak-to-Peak Currents, and the Current Ripple Ratio 'r' Defined

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In Fgr. 2, we see how the ac, dc, peak-to-peak, and peak values of the inductor current waveform are defined. In particular we note that the ac value of the current waveform is defined as

I_AC = delta I/2

We should also note from Fgr. 2 that IL = IDC. Therefore, sometimes in our discussions that follow, we may refer to the dc level of the inductor current as "IDC," and sometimes as the average inductor current "IL," but they are actually synonymous. In particular we should not get confused by the subscript "L" in "IL." The "L" stands for inductor, not load.

The load current is always designated as "IO." Of course, we do realize that IL = IO for a buck, but that is just happenstance.

In Fgr. 2 we have also defined another key parameter called 'r,' or the 'current ripple ratio.' This connects the two independent current components IDC and delta I. We will explore this particular parameter in much greater detail a little later. Here, it suffices to mention that r needs to be set to an "optimum" value in any converter - usually around 0.3 to 0.5, irrespective of the specific application conditions, the switching frequency, and even the topology itself. That therefore becomes a universal design rule-of-thumb. We will also learn that the choice of r affects the current stresses and dissipation in all the power components, and thereby impacts their selection. Therefore, setting r should be the first step when commencing any power converter design.

The dc level of the inductor current (largely) determines the I^2R losses in the copper windings ('copper loss'). However, the final temperature of the inductor is also affected by another term - the 'core loss'- that occurs inside the magnetic material (core) of the inductor. Core loss is, to a first approximation, determined only by the ac (swinging) component of the inductor current (delta I), and is therefore virtually independent of the dc level (IDC or "dc bias").

We must pay the closest attention to the peak current. Note that in any converter, the terms 'peak inductor current', 'peak switch current' and 'peak diode' current are all synonymous.

Therefore, in general, we just refer to all of them as simply the 'peak current' IPK where...

IPK = IDC + IAC

The peak current is in fact the most critical current component of all - because it’s not just a source of long-term heat buildup and consequent temperature rise, but a potential cause of immediate destruction of the switch. We will show later that the inductor current is instantaneously proportional to the magnetic field inside the core. So at the exact moment when the current reaches its peak value, so does this field. We also know that real-world inductors can 'saturate' (start losing their inductance) if the field inside them exceeds a certain "safe" level - that value being dependent on the actual material used for the core (not on the geometry, or number of turns or even the air-gap, For example). Once saturation occurs, we may get an almost uncontrolled surge of current passing through the switch - because, the ability to limit current (which is one of the reasons the inductor is used in switching power supplies in the first place), depends on the inductor behaving like one.

Therefore, losing inductance is certainly not going to help! In fact, we usually cannot afford to allow the inductor to 'saturate' even momentarily. And for this reason, we need to monitor the peak current closely (usually on a cycle-by-cycle basis). As indicated, the peak is the likeliest point of the inductor current waveform where saturation can start to occur.

Note: A slight amount of core saturation may turn out to be acceptable on occasion, especially if it occurs only under temporary conditions, like power-up For example. This will be discussed in more detail later.