# Boost converter inductor calculation

I am designing a boost converter with the following parameters:

• Vin: 3.0 - 3.7 V
• Vout: 30 V (nominal)
• Iout(max): 10 mA
• Fsw: 600 kHz

TI's application note has the following equations for boost converter inductor selection:

When I plug my numbers into the equation, I get a very large value of ~2250 uH. This does not make sense to me. Why do I need such a large inductor to power such a small load?

Here are my numbers:

• Vin: 3.0 V (worst case scenario)
• Vout: 30 V
• delta Il 2 mA (20% of 10 mA)
• Fs: 600 kHz
• I get a different answer. What numbers did you plug in? Jun 3, 2022 at 18:41
• Vin: 3.0V, Vout: 30V, delta I: 2mA (20% of 10mA load), Fs: 600KHz Jun 3, 2022 at 18:43
• @jonk, you got the same answer I did Jun 3, 2022 at 18:53
• @RGBEngineer Yes. Using your figures, I did. Note that the key reason likely relates to the low ripple. You don't need much load current and 20% of small current is even smaller current.
– jonk
Jun 3, 2022 at 18:55
• It's just V=L*di/dt. You want a very small ripple during the on time of a 600 kHz switching period, so you need a very large inductor to get there. The formula is targeted at higher current outputs, you can get away with much more ripple current to get a smaller inductance, as long as you can meet your output voltage ripple requirements. Jun 3, 2022 at 18:58

The delta I is missing a factor of Vout/Vin which is ten so you get ten times larger inductance.

• yep, I just noticed that. Is it usual to get that large of an inductor value for a boost converter? Jun 3, 2022 at 19:19
• @RGBEngineer That factor actually reduces the size of the inductor you need. Jun 4, 2022 at 14:07

Continuous conduction mode requires a minimum inductance of 33.5 μH: -

Image from my basic website. Note that the boundary power uplift is just below the power required by the load hence, the circuit will operate in CCM and produce an inductor ripple current of 162 mA p-p (for example).

If you factored-in a real diode on the output, the minimum inductance would be a tad smaller so, maybe you could get away with 33 μH if you had a constant load.

• If minimum load current was (say) 1 mA, you can avoid DCM by with a 330 μH inductor.
• This would also naturally produce a ripple current of 16 mA p-p.
• 3300 μH would naturally produce a ripple current of 1.6 mA p-p.
• That result uses a lot higher current ripple than the original calculation (if I understand correctly, 80mA vs. 2mA).
– jpa
Jun 4, 2022 at 6:22
• @jpa the picture in my answer implies a ripple p-p of 162 mA. If inductance rose by ten to 330 uH ripple would be 16 mA and, if it rose by ten again (3300 uH) ripple would be 1.6 mAp-p. I've amended my answer to make that clearer. Jun 4, 2022 at 8:41
• Thanks for the confirmation, this makes sense. I guess the 20% rule-of-thumb for ripple current is not necessarily a good choice for very low loads.
– jpa
Jun 4, 2022 at 8:47

The bigger issue is finding a regulator rated for such a low current, in terms of its control range. Most are made for upwards of 200mA, and expect an inductor of corresponding value for stable operation. This is a fixed property of the IC, defined by internal current limit threshold and slope compensation (and to a lesser extent, by size of the switch, which then wastes relatively a lot of power charging its capacitance).

So the high value is because you're asking for a very small one, and indeed a 10mA 2mH inductor will be pretty modestly sized.

I don't know why appnotes suggest inductor values in terms of output range; I mean, it's correct for general application certainly, but they're almost never a general application, they're following on from some very specific, internally tuned chip. Chock it up as yet another shortcoming appnotes frequently have.