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Re: [TCML] Mot power supply question
First of all it seems the ballast is causing quite the loss in power.
I measure 124.2 out of the wall but only 98 at the input of the
This seems reasonable, the purpose of the ballast is to create a reactance in series to limit the current draw, as a consequence there must be a voltage drop across that reactance. A ballast makes the load highly inductive as well so that the current from the wall socket and the voltage are no longer in phase and the real power delivered is less than the E * I measured on the meters
But the second part of this is running the pack from 240,
if I parallel the primaries I should get 120 per leg but my
understanding of this is that it will increase the current output of
each of the mots, am I correct in this? Then keeping in line with
this, how do I ballast it correctly? Do I need a matching ballast on
each leg, or can I just ballast one side of my 240? My thinking on
this and please by all means correct me if I am wrong, is that I can
just use the 2 opposite phases of the 240 with no neutral and the
centre grounding to the rod I buried just for the transformers.
I live in 240 v 50 Hz land and access to two phases in a STAR mode (480V) is not common in a home situation. Two MOT primaries will need to be in series for this mode and a ballast anywhere in the circuit will limit the current through the entire circuit. No neutral will need to be connected for this to work but if you did then two ballasts would be required, one in each phase input. This two phase drive arrangement is desirable for you as it reduces the wall current for a given power draw from the mains and at these power levels this must be a concern.
Also, in determining power levels, while I understand that in
transformers as you step up the voltage you step down the current, so
if you have a 10:1 transformer and put 1 volt in at 1 amp you get 10
volts out at 10ma. How does this hold true in a stack? Does this same
rule still apply? According to my testing I have basically a 1:123
transformer, I confirmed this by running 2 volts in and testing my
output, this also matches what each of them measure by them selves and
added up. Now running this at my measured 98 in I am getting 12054
out. This also means that if this rule holds true I am drawing 39 amps
which gives me 317mA out. Or is the current only determined by the
centre two? Something I read on the list had me questioning this.
Hang on a minute, 1:123??? a MOT is 120v primary and about 2kV secondary, each MOT is about 20 to 1 turns ratio and yes the current ratio is the inverse as power is conserved. A MOT stack is the same essentially given the power input is efficiently transferred. A point to watch however is power factor which is about the phase of the current compared to the voltage. This reduces the real available power at the load and if power factor correction is added in the primary it will reduce the current drawn from the wall socket as the power factor correction ensures that the primary load is resistive. You probably don’t need to worry about this BUT it is very likely that the true power actually consumed by your machine is somewhat less than the input current multiplied by the input voltage
Take care this machine is deadly and I suspect from your questions you may not be fully familiar with AC theory and issues associated with this machine and you are working with meters on live circuits, no second chances at these voltage and power levels
Ted in NZ
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