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Registered Member #3599
Joined: Mon Jan 10 2011, 05:50PM
Location:
Posts: 15
if not can you pls point out my mistakes and help me! thanks:)
to be precise: i have 235-237v in my wall outlet i have 4 MOTs rated 1kw ea with 120v input, 2100v out i am using a voltage doubler circuit (pls see the schematic for the values)
Now, i am calculating my output current to my voltage doubler as i solved earlier, here's what i did.
My MOT output volage, to be precise, is 1833V each I used series parallel as you can see in the schematic therefore the voltage now is 3666
Using capacitive reactance formula Xc=1/(2*pi*f*C) = 1000000/(2*pi*60hz*1.05uF) Xc=2526
Using ohms law
I=V/R =3666/2526 Iout=1.45A
output power is: P=IV =1.45*3666 Pout=5320w
but my supply is only capable of handling 4000w of 4 mots(my theory)
to my conclusions, when i use these calculations:
1.Am i overpowering my MOTs? or should i say, did the MOTs suck more than it should be like it is unballasted? or should say it is capacitively ballasted but not enough to limit the current?
2.If yes, it should get a little hotter in temperature than its normal when operating, right?
3.If i were to remove the 2 MOTs leaving only 2 seriesed MOTs, does the power output change? if not, it should be hotter than the 4 MOTs connected, is this right? if the power output changes, can you pls explained why is that?
4. Does the voltage doubler also limits the current?
5. Can I get more than 4KW from these MOTs? like 5 or 6KW? (Cuz ive seen people running only 2 MOTs but in 20KVA by arcing and putting more capacitor in parallel like 4 to 6 MOCs)
in a nutshell, my real plan is making a DC resonant charging coil with 4KW supply (total of 4 MOTs in series-parallel)
Sorry for dumb questions, cuz i will not plug this device until i solve this problem of mine.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
You want to use a 2.1 kV capacitor for 3666 V rms = 5185 V peak .. overstressed ! (though I've seen reports of MOC at 5 to 6 kV it seems risky to me)
Your calculation using Xc is sort of valid during initial power-up, once running with no load there will be negligible current (leakage currents) in the MOT secondaries, diodes and capacitors as the capacitors will be charged to the peak voltage. Of course once you draw current with a load a different set of calculations is required.
There is quite a lot of inductive ballasting in a MOT (if you haven't removed the shunts) and the doubler adds capacitive ballasting but I think that the total is not enough to ensure protection for the components.
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Let's consider just one pair of transformers, with their primaries and secondaries in series as you have drawn.
If the AC source voltage is the value for which the MOT's were designed, each transformer will get half the primary voltage and generate half the secondary voltage. So the series-series pair will generate 1833 volts RMS, same as a single transformer, or about zero volts if the phases are backwards.
The no-load primary current will be much less than half of what you would have with a single transformer, because at half voltage we are no longer close to saturation. But the total leakage inductance (which limits the output current) will be roughly doubled, so the short-circuit current from the pair will be only about half of what you would get from a single transformer.
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is my calculations correct?
