The heading says 12 V down to 5 V at a 3 A load. Then the MP2307 calculation uses 3.3 V, while the XL4015 calculation uses 5 V. The current-ripple allowances change too: 1.5 A in one example and 0.9 A in the other. Before I compare the final inductor values, I need to notice that the questions have changed.
That’s an interesting kind of mistake to work through because the arithmetic can look tidy. Output voltage changes the ideal duty ratio, and both duty and ripple enter the inductance equation. A smaller answer therefore cannot be credited to frequency alone when several inputs changed with it. I’m putting the conditions ahead of the results so I can explain which difference each calculation actually shows.
At first, the smaller inductor value looks like an interesting difference between the converters. Then I look at what produced it. The output voltage affects the duty ratio, and the duty ratio appears in the inductor calculation. The allowed ripple appears there too. If those things change, the answer changes even before I decide how much of the difference belongs to the switching frequency or the particular converter.
I'm putting the conditions ahead of the answers this time. Input voltage. Output voltage. Average load current. Allowed peak-to-peak ripple. Switching frequency. Operating assumptions. It’s not an impressive-looking list, but it stops the numbers underneath from pretending they describe the same job when they don't.
The distinction between a shared question and shared hardware is useful here. I can compare two circuits under the same input, output and load conditions while allowing their own switching frequencies to differ. That might be the relevant comparison for choosing between those devices. If I want to isolate the effect of frequency itself, I need an idealised calculation where the other terms are kept fixed. Those are related exercises, but they don't answer exactly the same thing.
A calculation makes the problem easier to see. At 12 V in, an ideal 5 V output gives a duty ratio of about 0.417. A 3.3 V output gives about 0.275. In the simplified inductor expression, the numerator is Vout × (1 − D). That’s about 2.92 for the 5 V case and about 2.39 for the 3.3 V case, before dividing by ripple and frequency. Even with the same frequency, the conditions have already changed the calculated inductance.
Their job is to show why a seemingly small change at the top of a column reaches all the way to the answer. If I also allow different ripple, I have introduced another reason for the answers to differ. Looking only at the final values would hide both changes.
Ripple itself needs a clear label. Peak-to-peak ripple means the difference between the highest and lowest current in the switching cycle. It isn't the average load current. It also isn't automatically the same percentage of that current unless I have explicitly chosen it that way. Writing only a number beside the word ripple makes it too easy to forget which definition I used.
For instance, suppose I choose a ripple target as a fraction of a load current. If I later change the load but keep the fraction, the ripple in amperes changes. If I instead hold the ripple in amperes fixed, that fraction changes. Both choices can be deliberate. Trouble starts when I carry one number into another calculation without saying which choice I made.
The same habit affects loss comparisons. A diode loss at one load current and a transistor loss at another current can't show the effect of replacing the path. A higher output voltage can also change how a fixed loss looks as a fraction of useful output power. This isn't a reason to avoid comparing things. It’s a reason to keep the question attached to the numbers.
I'm separating two parts of the correction. One is fixing the mismatched example. The other is being more careful about the conclusion. It establishes what follows from a particular set of assumptions. The next operating point may make different losses important.
This is something I want to discuss with my mentor using the actual sheet. I can point to where the conditions change and explain which terms are affected. That should make it possible to decide whether the examples should be matched, or whether both should stay with clearer labels because they illustrate different requirements.
What bothers me most is that neither column looked obviously wrong. The equations and units could all be tidy. That made the mismatch easier to overlook than an arithmetic error. A calculator would have happily repeated the same answers without asking whether I meant to compare them.
The different conditions reach the loss examples too. Their off fractions are 0.584 and 0.725, corresponding to different duty ratios. The printed diode and low-side transistor losses therefore contain a change in conducting time as well as a change in the current path. I can still learn from each example. What I should not do is treat the difference between their watt values as an isolated measurement of the benefit of replacing the diode.
For now, I'm leaving the conditions at the top where I can't miss them. It’s a small change in the notes, but a fairly large change in what I look at first. Before asking why two answers differ, I want to know whether I actually asked the same question twice.