Electronic – 20AWG wire inductor and 8A, possible

amperageinductorwire

I have a inductor made up of ~5cm of 20AWG wire. Switching frequency of the inductor is 500kHz. And the duty cycle is somewhere between 10-30%. Maximum current through the inductor is 8A at around 13.5V.

I am wondering can this "relatively small" wire, take this amount of current? How could I calculate it out?

If it cant, will simply tying together more wires work? How many would I have to tie together?

EDIT:

Since it isn't possible, I am thinking of buying one. Would this be suitable? Its supposebly high-frequency rated for 9A.

Best Answer

No, not realistically.

AWG 20 would be marginal for 8A RMS (it's the RMS current that matters, not the peak) even at DC, however the maximum frequency for 100% skin depth on AWG 20 is in the low tens of kHz. 500kHz is way above that and the waveform your question implies will have substantial energy at higher harmonics like 1.5MHz. Google Litz(endraht). You'll probably end up with some dozens of AWG34 wires (in parallel). See here, for example.

Edit: Skin effect is caused by eddy currents circulating within the wire. The current only goes so deep into the wire at high frequencies (it only flows through the skin), so the inner part of the wire is effectively doing nothing (it actually falls off exponentially rather than suddenly ceasing at some depth- we refer to the depth where it's fallen to 1/e \$\approx\$ 0.368 as the skin depth). So, as the wire diameter increases, the resistance only decreases proportionally to the diameter (the circumference) rather than the square of the diameter (cross sectional area) as you'd expect at DC.

The exact number of wire gauge(s) you need will depend on how hot your inductor will run (including self heating from copper and core losses as well as ambient) and what temperature your wire insulation is capable of. Allowable maximum resistance or power loss may also be concerns.