Wikipedia article on Leakage inductance provides a formula for the leakage:
But I'd like a formula that does not neglect the "associated winding resistances", as the transformer of which I have measured the primary inductance using a multimeter for the secondary open and shorted is 1:800 and has a very high secondary resistance. With the formula as above I'm getting k≈0.5 which seems far from reasonable (even though I know this transformer is intended to be run with current limiting in series with the primary; it's a dental X-ray transformer).These equations can be developed to show that, neglecting associated winding resistances, the ratio of a winding circuit's inductances and currents with the other winding short circuited and at no-load is as follows,[7]
One way to measure the leakage inductance is to short the primary and resonate the secondary with a known capacitance. If the known capacitance is MUCH bigger than the transformer's secondary self-capacitance then the error introduced from ignoring the transformer's self-capacitance is small.
If you inject an impulse and measure the frequency of resonant oscillation you can calculate (2*pi*f)=1/sqrt(L*C). You know C, you know pi, you measured f, thus solve for L.
It might be a good investment of 10 minutes to read about "Quasimodo ExtraLight" here on DIYA. (this) is one post which discusses it. Poke around for more.
If you inject an impulse and measure the frequency of resonant oscillation you can calculate (2*pi*f)=1/sqrt(L*C). You know C, you know pi, you measured f, thus solve for L.
It might be a good investment of 10 minutes to read about "Quasimodo ExtraLight" here on DIYA. (this) is one post which discusses it. Poke around for more.
even high winding resistance does not much modify inductance;
k=0.5 may not be so unreasonable; after all the load is a x-ray tube which is a vacuum diode driven forward - in other words not far from a dead short; high leakage inductance may therefore be a feature, not a defect, further limiting current; and might add an additional level of safety in case the current limiter front up fails ...
similar case is microwave oven; a xfrm working into a magnetron tube, also a vacuum diode; u-wave xfrm is deliberatley fitted with an internal magnetic shunt to increase leakage inductance; here leakage inductance is the only current limiting factor and it is fail-safe ...
or think of the good old times when welding was done with mains transformers, rather than with inverters; the arc is a dead short and the xfrms were intentionally built with high leakage.
k=0.5 may not be so unreasonable; after all the load is a x-ray tube which is a vacuum diode driven forward - in other words not far from a dead short; high leakage inductance may therefore be a feature, not a defect, further limiting current; and might add an additional level of safety in case the current limiter front up fails ...
similar case is microwave oven; a xfrm working into a magnetron tube, also a vacuum diode; u-wave xfrm is deliberatley fitted with an internal magnetic shunt to increase leakage inductance; here leakage inductance is the only current limiting factor and it is fail-safe ...
or think of the good old times when welding was done with mains transformers, rather than with inverters; the arc is a dead short and the xfrms were intentionally built with high leakage.
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Looks like my initial measurement was invalid, as my meter uses a 1 kHz test signal on the mH range. That's not appropriate for a laminated steel core. I remeasured with the H range which makes the meter use 200 Hz, and got a higher coupling, though still not as high as typical power transformers.
One correction: an X-ray tube isn't a dead short — your radiologist controls the tube current by changing heater power. You could technically do it with a standard vacuum diode, except it's only designed to not take damage for a narrow range of heater power.
One correction: an X-ray tube isn't a dead short — your radiologist controls the tube current by changing heater power. You could technically do it with a standard vacuum diode, except it's only designed to not take damage for a narrow range of heater power.
I would not trust measuring leakage inductance with an L-meter as long as winding resistance is not cancelled out.
Why not? Mine (Agilent U1733 handheld) reports the imaginary part (inductance) AND the real part (winding resistance) simultaneously. Maybe you're thinking of super cheap meters that don't measure both the in-phase and the quadrature-phase impedances?
Yes, I didn't know you use an adequate L-meter. Anyway I wonder about the frequency dependency you observed as I do not expect significant contribution of the core to leakage inductance.
Wouldn't it be sufficient for a meter to account for resistance by simply superimposing some DC current on the AC impedance measuring signal and then subtracting the resistance calculated from the DC offset?
In any case, my primary winding resistance is a fraction of an ohm, so I doubt it affects the measurement for this transformer.
In any case, my primary winding resistance is a fraction of an ohm, so I doubt it affects the measurement for this transformer.
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