Boost converter: using smaller inductance than "typical" application?

Question

I need to boost 3.3V all the way up to 12.0V.

Current requirements are rather small, I only need to power an SSD1309 OLED driver (from pg7):

Common maximum sink current: 40mA

Almost all of the available high-quality boost converters are providing way more current than this, so I decided to use the smallest converter I can get: MP3438:

The MP3438 is a highly integrated boost converter with a 1.2MHz fixed frequency and a wide input voltage (VIN) range. The MP3438 starts up from a VIN as low as 2.7V, and can support up to a 2A switching current limit with integrated, low RDS(ON) power MOSFETs.

This is their recommended typical application converting 3.3V to 12V:

There is a section on pg16 "Selecting the Inductor":

An inductor is required to transfer the energy between the input source and the output capacitors. A larger-value inductor results in less ripple current and a lower peak inductor current, which reduces the stress on the power MOSFET. However, a larger-value inductor is physically larger, has a higher series resistance, and has a lower saturation current. For most designs, the inductance can be estimated with Equation (6):

L = Vin*(Vout-Vin)/(Fsw*Vout*dIl)

Where dIl is the inductor ripple current.

Choose the inductor ripple current to be approximately 20% to 50% of the maximum inductor average current.

Question: what is "the maximum inductor average current"? Is it the 40mA? Or RMS Current (Irms) which is 4.4A for the recommended inductor?

Calculating the above formula with the recommended 3.3uH inductor:

dIL = Vin*(Vout-Vin)/(Fsw*Vout*L)
dIL = 3.3*(12-3.3)/(1200000*12*3.3/1000000)
dIL = 0.60417

This means ripple is 604mA?

Answer 1

Question: what is "the maximum inductor average current"?

It is the average inductor current when the current loading on the 12 volt output is greatest. Maybe the boost calculator from my basic website might help a tad: -

I've set the load resistance to 300 Ω to match the 40 mA load current stated in the question.

So, it's running in DCM (discontinuous conduction mode) and, the average inductor current is 145 mA as expected for an ideal circuit. The calculator tells you that the ripple current is 419 mA p-p. The reason why the ripple current value disagrees with the data sheet guidance is because the data sheet assumed the chip would be used in CCM mode.

There's always a more ripply current in DCM because of the inductor current depleting to zero.

_{The calculator is for an ideal circuit; there will be switching losses, diode volt drops and inductor resistance losses.}

Answer 2

"the maximum inductor average current" should be the output current (+\$I_{R1}\$) divided by the fraction of time inductor current flows in the direction of the output. Display current may be 128*320 μA, about 40 mA.
I'd assume current limiting to be dissipative and aim for a voltage low in the range of 7-16 V.
For 12(8) V, 40 mA the average current should not be much higher than \$\frac{12(8)}{3.3}\times 40 mA\$: 145(97) mA.

The computation of the ripple current looks correct, note how it is about 30 % of the MP3438's rated current.

With the current as low as this, it may be hard to find an inductor that allows operation in continuous conduction mode (CCM).
This would need a greater inductance than typical application, which may make it bigger in spite of not needing to be rated for 2 A.
Dis-continuous operation (DCM) shouldn't hurt much, but makes eye-balling \$I_{RMS}\$ and \$I_{peak}\$ harder: simulate!