# Buck-Boost Converter Inductor Value

I am designing a typical buck-boost converter using the LTC3530EMS#PBF IC from Analog Devices.

My design specs are as follows:

Vin: 3.0V - 4.2V Vout: 3.3V Iout(max): 300mA Fsw: 1MHz

In the datasheet, it gives the below equations for selecting the inductor value.

Given that my maximum output current is so small, is it okay to have my inductor ripple current (delta IL) be larger than 40% of my maximum output current?

For example, if I chose my inductance to be 3.3uH, my ripple current would be ~210mA. This would be ~70% of my maximum output current, and that value is way outside the suggested range of 20% - 40%. However, if I arbitrarily changed my maximum output current to be 1A without changing anything else, that ripple current of 210mA would suddenly become "acceptable", even though nothing about the circuit has changed.

For example, if I chose my inductance to be 3.3uH, my ripple current would be ~210mA. This would be ~70% of my maximum output current, and that value is way outside the suggested range of 20% - 40%.

This is what the data sheet says: -

It seems as though you are confusing maximum inductor current with maximum desired output current. Of course your ripple current can be high compared to your load current; it wouldn't be a good regulator if this wasn't true. The data sheet mentions maximum inductor current and, that is both a rating of the inductor and the worst-case maximum load current used. So, if the design could deliver (say) 500 mA then your inductor needs to be appropriately rated (even if your expected full load current is smaller).

If you need any further convincing, look at page 1: -

As you can see, that particular design works from a load current of 400 mA all the way down to 1 mA and, the only reason they stopped showing lower values is because the efficiency is rubbish (naturally) at this low output level.

Sure, if your load current maximum is only 300 mA it won't be a problem.

• If I am understanding you correctly, you are saying that in this case the maximum inductor current would be Ioutmax (300mA) + 0.5*Ioutripple (210 * 0.5 = 105), so maximum inductor current would be 405mA, making the Iout ripple of 210mA be around 50% of the maximum inductor current, is that correct? Nov 4, 2022 at 15:20
• The maximum inductor current is what is stated in its data sheet or what the design can achieve if you pushed it. Probably didn't make myself very clear so I'll make an amendment. @RGBEngineer Nov 4, 2022 at 16:16
• @RGBEngineer are we done here now? Feb 21 at 10:48