Magnetic Core Crossover Inductor Distortion Testing
Part 3
How does the mH rating of an inductor or the load impedance change how much power/current an inductor can handle?
First I will start with the load impedance question.
Since inductors are current driven devices, that is to say the it's a change in current through a coil which generates the magnetic flux in the core. The core saturation caused by said flux should occur at roughly the same current level through the inductor regardless of the load impedance. So lets see how it plays out.
For this test I'm using the small 0.52mH ferrite core inductor since it's easily driven into saturation. I ran test sweeps of increasing level until the measured total harmonic distortion reached a peak of 1% with the inductor connected to the test load configured at 4, 8 and 16 ohms and recorded the sweep voltage for each which can be used to determine the current and power through the coil.
As you can see from the harmonic distortion graphs the recorded voltage at each load impedance when THD reaches 1% is as follows:
4 Ohms - 5.93v which taking the DCR of the coil into account results in a current of ~1.4 Amps, this gives a power of ~8.3w.
8 Ohms - 11.5v which taking the DCR of the coil into account results in a current of ~1.4 Amps, this gives a power of ~16.1w.
16 Ohms - 22.7v which taking the DCR of the coil into account results in a current of ~1.4 Amps, this gives a power of ~31.8w.
As expected the current required to reach saturation is not affected by the impedance of the load since the magnetic field strength/flux is purely current driven.
In terms of overall power handling since current is cut in half with the doubling of the impedance the voltage can also be doubled which will return current in the circuit to the point where saturation in the inductor occurs again. Basically doubling the the impedance means voltage can double for the same current level and the inductor can handle roughly twice the power before saturation.
The coil DCR adds a bit resistance to the circuit which is why the math didn't result in exact doubling of voltage/power as the overall resistance of the circuit was not exactly doubling.
Now for the second question: The effect of the inductance value on the current handling ability of a of inductor before saturation.
If a magnetic core inductor has a limit on the amount of flux in the core before saturation and said flux strength is determined by both the current running though the coil and the inductance value of the coil itself. It should be obvious that as the inductance value of a coil is increased the required current for the magnetic field strength to saturate the core will decrease as a result.
To test this I am using 4 inductors of the same type, Erse 18 gauge laminated steel I-Core in four different values 2.0, 3.0, 4.0 & 6.0mH.
Similar to the previous test I will increase the drive level of the test sweeps on each inductor using the 4 ohm load until measured total harmonic distortion reaches a peak of 0.5% then record the voltage of said sweep to calculate the current and power through each of the inductors at that point.
Erse 2.0mH 18awg I-Core - 0.5% THD reached at 66.5v - 4 Ohm load
Taking DCR into account (0.26 Ohms) this results in 15.6A of current or ~1037 Watts
Erse 3.0mH 18awg I-Core - 0.5% THD reached at 56v - 4 Ohm load
Taking DCR into account (0.33 Ohms) this results in 12.9A of current or ~724 Watts
Erse 4.0mH 18awg I-Core - 0.5% THD reached at 49.9v - 4 Ohm load
Taking DCR into account (0.39 Ohms) this results in 11.4A of current or ~569 Watts
Erse 6.0mH 18awg I-Core - 0.5% THD reached at 42v - 4 ohm load
Taking DCR into account (0.51 Ohms) this results in 9.3A of current or ~391 Watts
Now if the numbers are compared between the 2mH & 4mH and the 3mH & 6mH we can see what calculate roughly how much current or power an inductor needs to be derated when the value doubles.
For the 2.0 to 4.0mH comparison:
Comparing the voltage between the 2.0mH and 4.0mH (49.9V / 66.5V) = 0.75
Comparing the current between 2.0mH and 4.0mH (11.4A / 15.6A) = 0.73
In terms of overall power (569W / 1037W) = 0.55
For the 3.0 to 6.0mH comparison:
Comparing the voltage between the 3.0mH and 6.0mH (42V / 56V) = 0.75
Comparing the current between the 3.0mH and 6.0mH (9.3A / 12.9A) = 0.72
In terms of overall power (391W / 722W) = 0.54
Interestingly when inductance is doubled the voltage required to drive the inductor into saturation is almost exactly 75% the previous value. Due the differences in DCR when more wire is added to the coil the current ends up slightly less then that 75% when inductance is doubled. Since power is the two values combined it end up a just bit more then 9/16th the original power rating before saturation when inductance is doubled. You could likely round that down to 1/2 to be safe and for a quick rule of thumb, so double the inductance, half the power rating all else equal.
From that data we could extrapolate that the 1.0mH Erse 18 gauge I-Core would have saturated at ~88.7v which would be 21.3A or ~1887 Watts taking DCR into account.