# High current in Coil/Inductor

I'm trying to build a 5V 8A buck converter and I chose the XL4016E1 IC for this operation (datasheet). There it says I need a 47uH 12A coil. Unfortunately i couldn't find a coild with such a high current rating.

So my question is, is it possible to for example place 3 4A coils in series/ parallel, without losing the 47uH inductance.

Second, what does the 105 value near the C2 and C1 mean?

• 105 = 10x10^5 pF = 1uF. – brhans Mar 26 '18 at 14:17

First, a quick search in digikey gave me multiple results for a 47uH inductor with a 12A minimum current rating. Try going to component sites and using the parameter search tools. There are some out there that meet your requirements.

Secondly, as for putting inductors in parallel and series, you can calculate the overall inductance the same as you would with resistors, so in series, you add them together (L1+L2+L3...... etc) and in parallel, once again, same as resistors (1/L1+1/L2+1/L3......) and you will also be able to calculate the current through each via the same method as resistors.

Thirdly, as brhans mentioned in his comment, the capacitor printed codes are in pF, with the 3rd number being the multiplier so 10 = 10 and 5 = x10^5, thus 105 = 10x10^5pF = 1uF

Further reading: Simple series and parallel inductors

• Is it possible to place 3 141uH 4A Coil in parallel to achieve the 12A 47uH? (three 141uH in parallel are 47uH overall) – Ribisl Mar 26 '18 at 14:52
• It's possible, sure. Just do the math. Is it preferable to just getting a 47uH 12A inductor? No. I'd just get the required part. It took me 2 minutes on digikey to find a number of them – MCG Mar 26 '18 at 15:05
• Sure it would be better, but my deliverer doesn't have the 47uH 12A inductor. – Ribisl Mar 26 '18 at 15:08
• Who is your 'deliverer'? Why can you not use another? There are plenty of specialised online retailers that can provide you with the component specs you need – MCG Mar 26 '18 at 15:14

You can wind your own 47uH inductor, using thick wire.

L(uH) = (a^2 * N^2) / (9*a +10*b)

where a = radius, inches, N = # turns, b = length, inches

For a 2" diameter coil (thus a = 1) with coil length also of 1 inch, our denominator is 18, and the numerator must near 1,000 which requires 30 turns of wire in this single-layer inductor.

Can you run 12 amps thru 1/32 inch wire? perhaps if air-cooled