Average inductor current

The average inductor current in a Buck delivering a load current of 70 is 70. 

But in a Boost or Buck-Boost, the average inductor current is equal to 70/(l - D). Further, the peak current in all cases is typically 20 to 30% higher... 

This is the relationship between the average inductor current and the load current. Note that in Figure 1-13, in the embedded table, we have asked for an inductor rated for 1.2 x I0/(l — D). The factor 1.2 comes from the fact, that by typical design criteria, the peak of the inductor current waveform is about 20% higher than its average. So we need to look for an inductor rated at least for a current of 1.2 x II. 

Average Input Current equals Average Inductor Current A verage Output Current is equal to average Diode Current... 

Since the average current from the output capacitor is zero, therefore, for the buck, the average inductor current must be equal to the load current (where else can the current come from ). Therefore... 

We will see that Io is proportional to the average inductor current. 

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. 

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 5 A inductor for a 5 A load current may be a recipe for disaster. 

We should also note from Figure 2-2 that II = 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 Id, but they are actually synonymous. In particular we... 

For the buck, as the input voltage is raised, the duty cycle falls, and because the average inductor current II remains fixed at lo, the average diode current increases. That means we get the worst-case diode current (and dissipation) at Vinmax for a buck. So we can just use the numbers we already have derived from carrying out the general inductor design procedure (at Vinmax)-... 

For the boost and the buck-boost, as the input is raised, D decreases, but the average inductor current also falls, thereby keeping Id always fixed at lo- (We should remember that the boost and the buck-boost are unique in the sense that all the output current must pass... 

Looking at the equivalent buck-boost models in Figure 3-2, the center of the ramp on the secondary side (average inductor current, II ) must be equal to Io/(l — D), as for a buck-boost (because the average diode current must equal the load current). This secondary-side inductor current gets reflected to the primary side, and so the center of the primary-side inductor current ramp is Ilr, where Ilr = Ii/n. Equivalently, it is... 

Answer In a buck, the average inductor current ( II ) is equal to the load current ( Io )> that is, II = Io- But in a boost and a buck-boost, this average current is equal to Io/(l - D). Therefore, in the latter two topologies, the inductor current is a function ofD (duty cycle) — and therefore indirectly a function of the input voltage too (for a given output). 

This gives us the inductance in Henries, when f is in Hz. Note that Von is the voltage across the inductor when the switch is ON. It is therefore equal to Vin — Vo for a buck, and Vin for a boost and a buck-boost. Also, II is the average inductor current, equal to Io for a buck, and I0/(l - D) for a boost and a buck-boost. 

Question 31 Why are the equations for the average inductor current of a boost and a buck-boost exactly the same, and why is that equation so different from that of a buck ... 

Answer This is simply the converse of the previous question. For the buck, the average output current equals the average inductor current. For the boost and buck-boost, it is equal to the average diode current. 

So the average input current is equal to the average inductor current — which we know is Io/(l — D) for the boost. Let us again do a check in terms of power... 

Question 34 How is the average inductor current related to the input and/or output currents for the three topologies ... 

Answer For the buck, we know that average inductor current is equal to the output current, that is, II = Io- For the boost we know it is equal to the input current, that is, II = Iin- But for the buck-boost it is equal to the sum of the (average) input current and the output current. Let us check this assertion out... 

For a buck, we know that at turn-on, the instantaneous switch (and inductor) current is Io x (1 — r/2), where r is the current ripple ratio, and Io is the load current of the dc-dc converter. At turn-off, the current is Io x (1 + r/2). Usually, we can ignore the current ripple ratio and take the current as Io for both the turn-on and the turn-off analysis. So the load current of the dc-dc converter, Io, becomes the same as the Io used so far in the switching loss analysis. Similarly, in a boost and buck-boost, the current Io in our switching loss analysis, is actually the average inductor current Io/(l — D). 

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