DC-DC Converter Design and Magnetics--AC, DC and Peak Currents; Defining the "Worst-case" Input Voltage

Understanding the AC, DC and Peak Currents

We have seen that the ac component (IAC = delta I/2) is derivable from the voltseconds law.

From the basic inductor equation V = L dI/dt, we get:

2 × IAC = delta I = voltseconds / inductance

So the current swing IPP = delta I, can be intuitively visualized as "voltseconds per unit inductance". If the applied voltseconds doubles, so does the current swing (and ac component). And if the inductance doubles, the swing (and ac component) is halved.

Let us now consider the dc level again. Note that any capacitor has zero average (dc) current through it in steady-state, so all capacitors can be considered to be missing altogether when calculating dc current distributions. Therefore, for a buck, since energy flows into the output during both the on-time and off-time, and via the inductor, therefore the average inductor current must always be equal to the load current. So: IL = IO (buck)

On the other hand, in both the boost and the buck-boost, energy flows into the output only during the off-time, and via the diode. Therefore, in this case, the average diode current must be equal to the load current. Note that the diode current has an average value equal to IL when it’s conducting (see the dashed line passing through the center of the down-ramp in the upper half of Fgr. 3). If we calculate the average of this diode current over the entire switching cycle, we need to weight it by its duty cycle, that is, 1 - D. Therefore, calling 'ID' the average diode current, we get ...

---- Fgr. 3: Visualizing the AC and DC Components of the Inductor Current as Input Voltage Varies:

At VINMAX, the ac component decreases, dc component decreases even more, and so peak current decreases VINMAX corresponds to DMIN VINMIN corresponds to DMAX VINMAX corresponds to DMIN VINMIN corresponds to DMAX At VINMAX, the ac component increases, dc component remains the same, and so peak current increases Buck-Boost Buck

----

.... solving:

IL = IO / 1 - D (boost and buck-boost)

Note also, that for any topology, a high duty cycle corresponds to a low input voltage, and a low duty cycle is equivalent to a high input. So increasing D amounts to decreasing the input voltage (its magnitude) in all cases. Therefore, in a boost or buck-boost, if the difference between the input and output voltages is large, we get the highest dc inductor current.

Finally, with the dc and ac components known, we can calculate the peak current using

I_PK = I_AC + I_DC = delta I ^ 2 + IL

Fgr. 4: Plotting How the AC, DC, and Peak Currents Change with Duty Cycle

Defining the "Worst-case" Input Voltage

So far, we have been implicitly assuming a fixed input voltage. In reality, in most practical applications, the input voltage is a certain range, say from 'VINMIN'to'VINMAX'. We therefore also need to know how the ac, dc, and peak current components change as we vary the input voltage. Most importantly, we need to know at what specific voltage within this range we get the maximum peak current. As mentioned, the peak is critical from the standpoint of ensuring there is no inductor saturation. Therefore, defining the "worst-case" voltage (for inductor design) as the point of the input voltage range where the peak current is at its maximum, we need to design/select our inductor at this particular point always. This is in fact the underlying basis of the "general inductor design procedure" that we will be presenting soon.

We will now try to understand where and why we get the highest peak currents for each topology. In Fgr. 3, we have drawn various inductor current waveforms to help us better visualize what really happens as the input is varied. We have chosen two topologies here, the buck and the buck-boost, for which we display two waveforms each, corresponding to two different input voltages. Finally, in Fgr. 4 we have plotted out the ac, dc, and peak values. Note that these plots are based on the actual design equations, which are also presented within the same figure. While interpreting the plots, we should again keep in mind that for all topologies, a high D corresponds to a low input. The following analysis will also explain certain cells of the previously provided Tbl. 2, where the variations of delta I and IDC, with respect to D, were summarized.

a) For the buck, the situation can be analyzed as follows:

++ As the input increases, the duty cycle decreases in an effort to maintain regulation. But the slope of the down-ramp delta I/tOFF cannot change, because it’s equal to VOFF/L, that is, VO/L, and we are assuming VO is fixed. But now, since tOFF has increased, but the slope delta I/tOFF has not changed, the only possibility is that delta I must have increased (proportionally). So we conclude - that the ac component of the buck inductor current actually increases as the input increases (even though the duty cycle decreased in the process).

++ On the other hand, the center of the ramp IL is fixed at IO, so we know the dc level does not change.

++ So finally, since the peak current is the sum of the ac and dc components, we realize it also increases at high input voltages (see relevant plot in Fgr. 4).

Therefore, for a buck, it’s always preferable to start the inductor design at VINMAX (i.e. at DMIN).

b) For the buck-boost, the situation can be analyzed as follows:

++ As the input increases, the duty cycle decreases. But the slope of the down-ramp delta I/tOFF cannot change, because it’s equal to VOFF/L, that is, VO/L, and VO is fixed (same situation as for the buck). But since tOFF has increased, delta I must also increase to keep the slope delta I/tOFF unchanged. So we see that the ac component (delta I/2) increases as the input increases (duty cycle decreasing). Note that up till this point, the analysis is the same as for the buck -- traced back to the fact that in both these topologies VOFF = VO.

++ But now coming to the dc level IL of the buck-boost, we will find it must change for this topology (though it remained fixed for the buck). Note that the shaded portion of the waveform in the upper half of Fgr. 3 represents the diode current. The average value of this during the off-time is the square dashed line passing through its center, that is, IL. So the average diode current, calculated over the entire switching cycle, is IL × (1 - D). And we know this must equal the load current IO. So, as the input increases and duty cycle decreases, the term (1 - D) increases. So the only way IL × (1 - D) can remain constant at the value IO is if IL decreases correspondingly. We therefore realize that the dc level decreases as the input increases (duty cycle decreasing).

++ Further, since the peak current is the sum of the ac and dc components, it also decreases at high input voltages (see relevant plot in Fgr. 4).

Therefore, for a buck-boost, we should always start the inductor design at VINMIN (i.e. at DMAX).

c) For the boost, the situation is a little trickier to understand. On the face of it, it’s quite similar to the buck-boost, but there is a notable difference - and that is the reason why we did not even try to include it in Fgr. 3.

++ Once again, as the input increases, the duty cycle decreases. But the difference here is that the slope of the down-ramp delta I/tOFF must decrease - because it’s equal to VOFF/L, that is, (VO - VIN)/L (magnitudes only) - and we know that VO -VIN is decreasing. Further, the required decrease in the slope delta I/tOFF can come about in two ways - either from an increase in tOFF (which is already occurring as the duty cycle decreases), or from a decrease in delta I. But in fact, delta I may actually increase rather than decrease (as we increase the input). For example, if tOFF is increasing more the increase in delta I – then delta I/tOFF will still decrease as required. And in practice, that is what actually does happen in the case of the boost. With some detailed math, we can show that delta I increases as D approaches 0.5, but decreases on either side (see Tbl. 2 and Fgr. 4).

++ It’s therefore also clear that in either case above, the increase/decrease in the ac level does not dominate, and therefore, the peak current ends up being dictated only by the dc component. But we already know that the dc level of a boost changes in exactly the same way as for the buck-boost (discussed above) - it decreases as the input increases (duty cycle decreasing).

++ So we conclude that the peak current for the boost also decreases at high input voltages (see relevant plot in Fgr. 4).

Therefore, for a boost, we should always start the inductor design at VINMIN (i.e. at DMAX).