Boost Power supply - Adafruit calculator

Other Hardware General Electronics
1

I'm looking at the options for getting a boost power supply for some VFD type displays I have purchased.
Of course, one of the options is to buy a step up converter to provide the required 27 to 30 volt at max. 25mA. The other is to make something myself using an existing design.
Adafruit publishes a basic design together with a calculator for adjusting the component values to suit the required input and output values. It's on The Calculator | DIY DC/DC Boost Calculator | Adafruit Learning System
I've tried using this calculator to verify existing designs I have found and it seems, on that basis, OK.
However, I have a problem understanding one aspect. That is, the higher the required output current in mA, the lower the value of the inductor (microHenries).
This is, for me, counter-intuitive. I'd have thought that a higher output current would require a beefier inductor if the other input parameters stayed the same.
Has anyone an explanation ?

The calculation behind the Adafruit calculator is here ( in Javascript):

<script language="JavaScript">

function calcBoost(form) {
vinmax=eval(form.vinmax.value);
vinmin=eval(form.vinmin.value);
voutmax=eval(form.voutmax.value);
voutmin=eval(form.voutmin.value);
freq=eval(form.freq.value);
iout=eval(form.iout.value);
vripple = eval(form.vripple.value);

// calculate duty cycle
minduty = 100*(1-vinmax / voutmin);
maxduty = 100*(1-vinmin / voutmax);
form.minduty.value = minduty;
form.maxduty.value = maxduty;

// calculate inductor values: D*Vin * (1-D) / (freq * 2 * Iout ) &lt; L
// for vinmin and voutmin
D1 = (1-vinmin / voutmin);
L1 = D1 * vinmin * (1-D1) / (freq * 2 *iout);
// calculate for vinmin and voutmax
D2 = (1-vinmin / voutmax);
L2 = D2 * vinmin * (1-D2) / (freq * 2 *iout);
// calculate for vinmax and voutmin
D3 = (1.0-vinmax / voutmin);
L3 = D3 * vinmax * (1-D3) / (freq * 2 *iout);
// for vinmax and voutmax
D4 = (1.0-vinmax / voutmax);
L4 = D4 * vinmax * (1.0-D4) / (freq * 2.0 *iout);

L = Math.max(L1, L2, L3, L4);
form.minL.value = L * 1000000;

// Calculate peak current Ipk = (Vin * D)/(f * L)
// for vinmin and voutmin
Ipk1 = vinmin * D1 / (freq * L);
// for vinmin and voutmax
Ipk2 = vinmin * D2 / (freq * L);
// for vinmax and voutmin
Ipk3 = vinmax * D3 / (freq * L);
// for vinmax and coutmax
Ipk4 = vinmax * D4 / (freq * L);

Ipk = Math.max(Ipk1, Ipk2, Ipk3, Ipk4);
form.ipk.value = Ipk;
2

It does, but a beefier inductor doesnt mean a higher inductance, but a higher maximum current rating before the inductor saturates,so that means a physically bigger inductor.
Boost converters are critical in their design as if you get the inductance value wrong, the inductor saturates, and this instantly destroys your switching Fet.
Id simply buy one.

3

Thanks for your reply. In the mean time, I have found another calculator and an application note from Texas Instruments which I am still studying.

Boost Design Calculator: Boost Converter
It is similar to the Adafruit one but not directly comparable because it appears to assume a fixed 1:3 duty cycle from the values I have tried it with. It also does a nice graphic representation of the waveforms for a specific design.

If I try a home made version, in view of your advice, I'll order a few spare mosfets.