> ## Content Index > Fetch the complete content index at: https://sreekars-journal.labvion.com/llms.txt > Use this file to discover other available public pages before exploring further. # One change I can actually follow: the switching frequency - URL: https://sreekars-journal.labvion.com/dc07-changing-only-the-switching-frequency/ - Published: 2025-11-13T12:30:00.000Z - Updated: 2026-10-29T08:26:53.000Z - Description: I’m keeping the ideal question fixed so I can follow one change: switching frequency. The calculation is 12 V input, 5 V output and 0.9 A peak-to-peak inductor-current ripple. With those conditions held still, the duty ratio is 5/12, and the report's equation gives about 18.0 microhenries at 18 - Author: Sreekar - Tags: Converters, #Import 2026-10-29 10:08 I’m keeping the ideal question fixed so I can follow one change: switching frequency. The calculation is 12 V input, 5 V output and 0.9 A peak-to-peak inductor-current ripple. With those conditions held still, the duty ratio is 5/12, and the report's equation gives about 18.0 microhenries at 180 kHz and 9.53 microhenries at 340 kHz. Those are calculated inductances for an ideal comparison, not parts selected or tested on a module. That distinction makes the exercise more useful, because I can explain the direction of the change before thinking about a catalogue. A shorter cycle gives the current less time to rise or fall. The next question is how that model relates to the other limits of a real inductor. The ideal duty ratio is 5/12, about 0.417\. For this simplified buck-converter case, the inductor expression is L = Vout(1 − D)/(ripple × switching frequency). I can type that quickly. Before substituting, I want to understand why the frequency is in the denominator. During each switching interval, the voltage across the inductor makes its current change. If the cycle is shorter, the current has less time to rise or fall. So, with the other conditions fixed, a higher switching frequency can give the same ripple using a smaller inductance. The formula follows an idea I can picture, rather than being a separate rule to remember. At 180 kHz, the calculation gives about 18.0 microhenries. At 340 kHz, it gives about 9.53 microhenries. The second frequency is nearly twice the first, and the inductance is correspondingly smaller. Holding the rest of the example still makes that relationship easy to see. The units are worth slowing down for. The numerator has volts. Ripple times frequency has amperes per second. Dividing gives volt-seconds per ampere, which is a henry. In practice these answers are small fractions of a henry, so the microhenry label matters. Dropping a prefix would turn an ordinary-looking number into a very different component. Another useful check is to reverse the question. If I held the inductor fixed instead and increased the switching frequency, the simplified expression predicts less current ripple. If I loosened the ripple requirement at a fixed frequency, I could calculate a smaller inductance. Those are two different ways of changing the answer. Keeping them apart helps me understand what a comparison is actually showing. It would be tempting to stop here and say that the higher-frequency option is better because the inductor is smaller. But I haven't selected a physical inductor yet. I have calculated one property under one set of conditions. A real part also has a current limit, winding resistance, core behaviour, tolerances and a package that has to fit the board. The peak current is one reason the ripple definition matters. If the current is a symmetrical triangular ripple around an average value in this model, half the peak-to-peak ripple is above the average. With 0.9 A peak-to-peak ripple, that half is 0.45 A. So, for a 3 A average current, the peak would be about 3.45 A. It shows why average current alone isn't enough when checking the part. I also don't want to mix current ratings that refer to different things. A limit related to core saturation isn't simply interchangeable with a rating based on heating. The inductor has to behave properly in the circuit as well as avoid becoming too hot. This is the point where a catalogue value needs more reading rather than a quick match to the calculated microhenries. Higher frequency brings its own questions. The switch has to change state more often, and losses associated with switching can matter. The board layout and the fast-changing signals deserve attention too. That doesn't cancel the smaller-inductor advantage. It means I need to keep more than one consequence in view before making a practical choice. What I find satisfying about this exercise is that I can separate the stages. First, I decide the case. Second, I derive what the simplified model predicts. Third, I check how a real component differs from the ideal symbol. If I jump straight to the third stage, I get lost in specifications without knowing what I'm looking for. If I stop at the second, I can mistake a calculated value for a finished design. I want my mentor to check the assumptions around this example, particularly the current waveform and operating mode. I can explain why changing frequency changes the inductance now. The next thing is knowing when this equation is an appropriate starting point and which real limits I have to add afterwards. I’m keeping both sets of units in the notes so the comparison survives the next time I reopen the page. The units help keep this calculation attached to its question. In Vout(1−D)/(ripple×frequency), volts divided by amperes per second gives volt-seconds per ampere: henries. The numerical answer is in microhenries because the switching rate is high. The 180 kHz and 340 kHz values are the frequencies used in the comparison, while 0.9 A is a shared ripple allowance chosen for this paper exercise. Each number has a different job in explaining why the two calculated inductances differ. The result on paper is two small values: 18.0 and 9.53 microhenries. The useful result for me is being able to explain the direction of the change before reaching for a calculator. That's the part I want to keep when the numbers in the next example are different.