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RE: More about MOT currents



Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>


Norman:

Interesting.  I've never seen an MOT with aluminum wire.  Must be a real
cheapy.

You testing has proven that you will never draw 10 amps with the
secondary open without hitting saturation.  Obviously, you will not want
to run saturated so maybe your question should be rephrased for 3 amps
(75 volts/25 ohms(Z) = 3 amps) on your aluminum primary MOT and how
about we use 1 amp for the big one corresponding to 100 volts and 100
ohms(Z)(not exactly what you got but in the ballpark).

Now, it is true that the amps are additive, but not quite in the way
that I think you are thinking.  The amps I have outline above are the
magnetizing current for the transformer and if we assume an ideal
inductor are near purely Reactive.  This means that this current is at a
phase angle of -90 degrees to the voltage.  As you add a load on the
secondary of the transformer the current will rise, but the current
being "added" is at a phase angle of 0 degrees with respect to the
voltage.

I am neglecting some losses here to simplify calcs:

Two examples:

1) Your alumimum MOT

Lets say you draw 5 amps under load.

We established the reactive portion above as -3j amps
Then the "real" portion is (5^2 - 3^2)^0.5 = 4 amps

So your Total Current is 4 - 3j amps or 5 amps @ -37 degrees.
I don't think you'll get that much thru it before hitting saturation.


2) Your Big MOT

How about 10 amps under load?
Reactive portion -1j amp.
So Real portion (10^2 - 1^2)^0.5 = 9.9

Total Current 9.9 -1j amps or 10 amps @ -5.8 degrees

BUT you'll never "make" 10 amps without saturating I don't think.  You
can do the calcs with the correct load #'s once you get them.

You can add capacitance to wipe out the reactive portion and then all
the calculations will be with real values because the reactive part will
be zero.

Problem is, you will still need to keep the voltage input down to
prevent saturation regardless because the cap will not prevent
saturation.

Mark


>Original poster: norman@xxxxxxxxxxxxxxxxxxxxxxxxx

>My thanks to all who took the time to answer my question about MOT
currents. Several
>suggestions were made which led me to make the following
>measurements:

>SMALL  MOT:
>I removed the secondary coil, clamped the transformer back together
again and measured the >primary current.  From 0 volts up to 75 volts
the primary impedance measured 25 ohms
>(primary voltage/primary current).  At 120 volts the impedance dropped
to 12 ohms and the >current was increasing very rapidly with
>voltage.   Thus the high primary current (no secondary) was due to the
core
>saturating.  They are clearly pushing these transformers to the limit!
(The MOT is
>2.75" x 3.25" x 3.75".  The primary  is 120 volts and the secondary is
about 1800 volts.  >The primary wire is aluminum and it came from a
1000W
>oven.)

>LARGE MOT:
 >From 0 volts to 60 volts the primary impedance was 120 ohms.  At 80
volts the impedance
>dropped to 108 ohms,  at 120 volts the impedance was 32 ohms, and at
125 volts it was 26
>ohms. (The MOT is 5.25" x 4.5" x 2.25" and the wire is
>copper.)   Thus all MOT are not created equal!

>NEW QUESTIONS.
>If  I draw current from the secondary, calculating the primary current
for an
>ideal transformer is straight forward.   I(pri) = I(sec)V(sec)/V(pri).
If  my
>MOT draws 10 Amps with no secondary current will the total primary
current for my MOT with >secondary current be I(pri) + 10 Amps.

>I assume that the problem of the core saturating cannot be solved by
using a power factor >correcting capacitor in parallel with the primary
because the primary current will remain >10 Amps even though the line
current will be reduced.  Also the nonlinear inductance due
>to core saturation should generate harmonics in the primary current.
Does anyone have a
>feel for how seriously core saturation and the resulting harmonics
reduce the
>effectiveness of the capacitor in reducing the line current?