I wanted to try to design a buck-boost converter (instead of buying it) to drive an electromagnet. I have a 12 V source and I want to have an output of 24 V and at least 200 mA. As input I'm using a bench power supply of 300 W, which can deliver at most 10 A. The circuit has the usual and most simple topology for a buck-boost converter:

Buck–boost converter diagram

Image source: Wikipedia - Buck–boost converter (diagram by Cyril BUTTAY)

The thing is that I haven't got a large choice of components, so I want to settle with a 100 uH inductor. Now searching online, I've found different sources which give the relations to calculate inductance and capacitance with respect to the input and output requirements and with respect to the frequency. In my case, I need to do the inverse. The problem is that I don't care much to have a super stable output, thus I don't know which values to choose for ripple voltage and current either...

I can't figure it out: how could I determine the switching frequency and capacitance, given the inductance? (I have much more choice for the capacitors.)

P.S.: Thank you in advance for any eventual answer and excuse my English, I'm still practicing it!

  • \$\begingroup\$ If your input voltage is less than your output voltage then you need a boost converter not a buck-boost. \$\endgroup\$
    – Andy aka
    May 24 at 18:36
  • \$\begingroup\$ @Andyaka Well, apparently polarity doesn't matter, so a buck-boost will do. Making the switch work, is a bit tricky though; I might suggest a beginner do boost or SEPIC (once the similarity of the two is understood) for that reason. \$\endgroup\$ May 24 at 18:52
  • 1
    \$\begingroup\$ How much peak current can you inductor handle? If you start from zero and apply 12 V, how long does it take until that current? This is the absolute longest on-time you can ever have. Calculate the off-time needed to discharge the inductor. You now have minimum frequency. For the upper limit, calculate how much switching losses is your limit for the MOSFET. This sets you upper limit unless you have hit the self resonant frequency for the inductor. \$\endgroup\$
    – winny
    May 24 at 18:53
  • \$\begingroup\$ @TimWilliams more difficult to drive the MOSFET \$\endgroup\$
    – Andy aka
    May 24 at 19:04
  • \$\begingroup\$ @Andyaka exactly. \$\endgroup\$ May 24 at 19:05

1 Answer 1


Well, what's an inductor?

The fundamental equation of inductance is,

$$ V = L \frac{dI}{dt}$$

An applied voltage causes a change in current. The actual current (starting/ending) could be whatever; we only know how to change it, by applying voltage and waiting.

Picking an inductor, means fixing the ratio between voltage and current rate-of-change.

For a periodic signal of some frequency, the ratio of voltage, frequency, and current change is a constant.

In switching converters, we generally have frequency either fixed, or variable over a modest range (depending on control type and desired operating mode). In any case, it's at least quasi-periodic and steady-state, meaning we have the continuity condition that the current at the end of a cycle equals that at the start. Thus, we further have a relation between input/output voltages and the change in current (peak-to-peak or ripple). Further, we have square voltage waveforms, which give triangular current waveforms; integrating the above equation is trivial, by drawing the areas under rectangles and equating them to changes between points in time (that is: simply change \$d\$s to \$\Delta\$s).

Inductor ripple current is then integrated (approximately speaking) by the input or output filter capacitors, and thus either a rampy or sinusoidal voltage ripple results. This is usually made small enough not to worry about its exact shape, and so we have clean and easy-to-use power on both sides of the converter.

How much, then, to change the current by? To reiterate: the actual current could be whatever, meaning, we still don't care what DC input/output is, with respect to the inductor value. The change could be large or small, as long as it averages to the desired output. What's the catch?


Practical converters must be efficient, and compact, putting competing pressure on reducing inductance (thus energy storage, thus component size), and increasing it (the change in energy stored is what causes core losses).

So, for a basic demo, you need an inductor that can handle at least the DC requirement, with low enough voltage drop not to burn up. (Typically, inductors are rated for current at 40°C temp rise. Core loss is not included in the DC rating!) You should probably have saturation current at or above this as well (saturation is the effect where inductance decreases with rising current, i.e. \$L\$ is not constant, but a function of \$I\$, making things much more complicated). This is sufficient for zero current ripple, but you will have nonzero, so you should add the AC RMS current to the resistive loss calculation (and add core loss if coefficients are provided), and peak current to the saturation rating.

For a 12 to 24V, 200mA boost converter (ignoring the buck-boost for now), the input sees at least 400mA (conservation of power), and the inductor is in series with the input, so this equals the minimum rating. If current ripple is 10% of DC current, then Isat should be 440mA or more; if 100%, 800mA.

And at 12V and 100µH, a change of 400mA takes 3.3µs, so a switching frequency of 150kHz would be a quite reasonable starting place.

Some control schemes dictate the current ripple. In particular, peak current mode control requires "more than" 100% current ripple to function (this can be relaxed to 30-50% with slope compensation). Lower ripple fractions are typically used with other types: average current mode, fixed-on/off time, hysteretic, or voltage mode. Exactly 100% is used by BCM (or "quasi-resonant") control.

"100% current ripple" means inductor current fully discharges each cycle (to zero), then charges fully to twice IDC at the peak: this is boundary conduction mode (BCM). If the same peak current is held, but frequency is reduced, less power is delivered (the pulses contain the same energy, but come less often), and current remains at zero for some time between pulses -- discontinuous current mode (DCM). If the switch turns on before the inductor's fully discharged, some current remains, thus: continuous current mode (CCM).

For starting out, you might not worry about the control scheme, or conduction mode, at all, just hook up a PWM generator to a switch and play around with it. Beware, the output voltage will be unregulated, which particularly means for boost style converters, the unloaded output voltage can rise quite seriously! Always test with a load resistor connected.

Then you can use a controlled PWM generator, and close a feedback loop to regulate the output voltage. But beware: the peak current is completely uncontrolled this way -- you're just setting PWM from voltage, the current inbetween could be anything, including destructive for the switch! This is why current-mode schemes are most popular, of which I would recommend peak current mode as the simplest (example controller: UC3843).


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