# using Q factor of inductor for power loss calculation

Discussion in 'Electronic Design' started by Jamie Morken, Sep 28, 2008.

1. ### Jamie MorkenGuest

Hi,

I made some ferrite rod inductors, about 10 turns of 14 AWG wire on the
rod gives about 5uH and a Q factor of ~30 or so, while 260 turns of
38AWG litz wire (14AWG equivalent) gives a Q approaching 90 or so, both
at a measurement frequency of 15.6kHz. This inductor will have about 4
amps of current ripple through it at about 200kHz, (rough estimation)
and about 12Amps DC current. How can the power losses of the inductor
be calculated from the Q factor and this AC current information?

cheers,
Jamie

2. ### JamieGuest

http://en.wikipedia.org/wiki/Q_factor

http://webpages.charter.net/jamie_5"

3. ### Tom BruhnsGuest

Assuming you actually know the Q at the operating frequency and power
level, just remember that Rseries=Xl/Q, and power=i^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

4. ### Jamie MorkenGuest

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

5. ### Jamie MorkenGuest

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
5.14watts, as the AC ripple is about 1Amp maximum at 200kHz, so about
36mW of AC losses in the inductor (if Q was 90 at 200kHz).

cheers,
Jamie

6. ### Tom BruhnsGuest

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.

7. ### Jamie MorkenGuest

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

8. ### HarryDGuest

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
0.50uf cap in parallel for 100KHz and 0.125uF for 200KHz parallel resonance.
Measure the bandwidth at +/-45d phase shift or 0.707 amplitude. Q= Fo/BW.
I'm betting the Q @ 200KHz < 10, @100KHz<20. This is what you need for power
dissipation calculations.

Cheers,
Harry