using Q factor of inductor for power loss calculation

That may help you.

Assuming you actually know the Q at the operating frequency and power level, just remember that Rseries=3DXl/Q, and power=3Di^2*R. Of course you need to account for the DC component separately; that obviously will be the DC current squared times the DC resistance. You really need to make the Q measurement at the operating frequency and power level (including the DC); Q measured at a frequency less than a tenth the operating frequency tells you practically nothing about the Q at the operating frequency. Have a look at the Q versus frequency curves in the info available from Micrometals or one of the ferrite manufacturers to get an idea how much it can vary with frequency.

Cheers, Tom

So the formula: Q = 2pi * (energy stored / energy dissipated per cycle)

for a Q of 90, that is about a 14:1 ratio of the energy stored to the energy dissipated in the inductor per cycle.. so now just need to know the energy stored in the inductor per cycle, E = 1/2(L*I^2) so I think that is 0.00036 joules (assuming an average of 12amps of DC current in the inductor) so then we would lose about 1/14 of that per cycle, and then multiply that by 200,000 cycles per second, is 5.14 joules per second, or 5.14watts. That assumes we have a Q of 90 at 200kHz, but it will be less for my inductor as I used a 15.6kHz measurement frequency.

Is this correct? I think it is more losses than actual, maybe I should be using the AC current in the inductor instead of the DC current in the calculation.

cheers, Jamie

Ok, I see now that the AC current at 200kHz should be used with the Q factor, the DC current is not important except for inductor saturation and DC wiring losses, so the inductor losses will be far less than

cheers, Jamie

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Huh?? Are you talking about your 260 turn coil?? If 10 turns is 5uH, then 260 turns should be something like 3.3 millihenries. At 200kHz, that's about 4200 ohms reactance. At a Q of 100, that's 42 ohms effective series resistance. At 1 amp RMS, that's 42 watts. Even if I'm off in total by an order of magnitude in my assumptions, it will be a far cry higher than 36 milliwatts. Heck, even the 10A DC current will cause a higher dissipation than that in 14AWG wire corresponding to 260 turns: 2.5 milliohms per foot is 250 milliwatts per foot at 10 amps, and I'm sure that 260 turns will be a lot more than a foot.

Hi,

Woops.. I meant to say 260 strands of 38AWG litz wire, not turns.. the two inductors I made both have 10 turns, one is litz wire, the other is normal stranded copper wire, both are 14AWG equivalent.

cheers, Jamie

Hi Jamie, Why don't you measure the Q of that 5uH inductor at the 100KHz and 200KHz operating frequency. Just need a signal generator driving the 5uH with