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Re: MOT Testing

Original poster: "Paul B. Brodie" <pbbrodie@xxxxxxxxxxxxx>

Mark,
Thanks a lot. You have been extremely helpful. While at the appliance repair shop, I grabbed a bunch of stove top heating elements and oven heating elements. All of them test good. I thought they would make great low ohm, high watt, low inductance resistors. Other than looks, is there any reason I shouldn't use these as resistors this way?

I saw that business about Z^2=Z1^2+Z2^2+Z3^2+... and I don't know where this came from. I checked my copy of The Art of Electronics by Paul Horowitz and Winfield Hill, second edition and on page 32 it says that series impedance is calculated as Z=Z1+Z2+Z3+.... Where did I miss the boat? I know you guys are way ahead of me, so I figure there is something I'm not taking in to account. The Electric Engineering professor at the local university told me that this book is just about the best electronics reference available, so I feel like I can trust it, especially on something as basic as this. I hope someone can explain this to me, please.
BTW, was "sum" discussion a pun?

Now, I'm reading full tilt about constructing inductors. I'm sure I'll find plenty in the archives.

Paul
Think Positive

----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Monday, April 25, 2005 11:21 AM
Subject: RE: MOT Testing

> Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>
>
>
> Paul:
>
> I agree with your approach. Not certain why you get a different result
> when you check the secondary as opposed to the primary. Maybe at these
> low test voltages the resistance of the secondary is affecting the
> results. Most of my MOT secondaries have about 60 ohms of resistance.
>
> For current limiting you indicate you want 15 amps (now this calc is for
> 1 MOT mind you, not a bank of 2 or 4) so your total impedence needs to
> be 120V/15amps = 8 Ohms.
>
> You have 4.75 Ohms already in the MOT so you need 3.25 Ohms of impedence
> from your current limiting inductor(Note - follow current limiting
> inductor thread discussion going on. There is sum discussion about
> whether Z=Z1+Z2 or Z^2=Z1^2+Z2^2. If the latter then you need 6.5 Ohms).
>
> Now for the moment assume R=0 for the inductor, then X^2 = Z^2 - R^2 so
> X = Z. Then L = X/(2*Pi*60).
>
> So you have L = 3.25/(2*Pi*60)=.00862 H or 8.62 mH for your cuurent
> limiting inductor.
>
> Note that in doing this the MOT primary voltage will be reduced so you
> won't get the full HV out. A big resistor will do the same thing and
> create a lot of heat as well.
>
> Mark
>
>
> Original poster: "Paul B. Brodie" <pbbrodie@xxxxxxxxxxxxx>
>
> Mark,
> OK, I have tested a couple of the MOT's as you suggested. The first one
> is
> one of the smaller ones. I connected my variac to the primary and at 10
> V
> in I got 200 V out, at 20 V in I got 400 V out, and at 30 V in I got 600
>
> out. So I feel like I can be sure it has a 1 to 20 turns ratio. At 120 V
>
> in, this MOT should have 2400 V out. BTW, I have a "true RMS" DMM. When
> I
> first checked this MOT, I connected it in reverse in order to deal with
> much lower and safer voltages. Checking it this way with power going in
> to
> the secondary, it checked as if it has a 25 to 1 turns ratio. I put 100
> V
> on the secondary and got 4 V on the primary. I checked again with 50 V
> and
> got 2 V out. I then checked with 25 V and got 1 V out. I don't
> understand
> why I get different results when checking it in reverse. Anybody have an
>
> explanation for this?
>
> I then checked the large MOT that is labeled 4000 Vac. At 10 V in I get
> 185
> V out, 30 V in 550 V out. This gives a turns ratio of approximately
> .055.
> So 120 V / .055 = 2180 Vac out. I measured the impedance of the primary
> by
> shorting the secondary and measuring the amps on the primary while
> feeding
> first 10 volts and then 20 volts. At 10 V I measured app. 2.1 A and at
> 20 V
> I measured app. 4.2 A. Z=V/I and Z=20/4.2 or Z=4.76 ohms. With no
> current
> limiting and a short on the secondary it should pull 120V/4.76ohms=25.2
> A.
> So evidently it is at least current limited a little. It shouldn't
> require
> too much current limiting to get it down to a more workable 15A.
>
> I'm somewhat fuzzy on current limiting. How do I determine the size of
> my
> current limiting inductor? I guess using resistance to limit the current
>
> would be highly inefficient and generate too much heat?
>
> If any of my math is wrong or if I'm using incorrect equations, will you
>
> please straighten me out? As if it is possible to straighten me out!!!
> {:-) Thanks. Paul Think Positive PS Is it possible to have too many
> MOT's??!!
>
>
>
>
> ----- Original Message -----
> From: "Tesla list" <<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>
> To: <<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>
> Sent: Friday, April 22, 2005 10:02 AM
> Subject: RE: Expensive hobby
>
> > Original poster: "Mark Dunn"
> <<mailto:mdunn@xxxxxxxxxxxx>mdunn@xxxxxxxxxxxx>
> >
> >
> > Paul:
> >
> > Short the Secondary and put 5 - 10 VAC @ 60 HZ on the primary. Make
> > sure your variac or power supply is good for min 5 amps for this
> test(Z > will likely be 3 to 4 Ohms so at 10 VAC you will draw 2.5 to
> 3.5 amps). > If not go with even lower voltage or put a resistor in
> series to drop > the voltage down. You can work in the mV range and
> still get good > results. Then measure current and voltage across the
> primary and you > can compute impedence as Z=V/I. You can then choose
> your current > limiting based on this. > > If you want to do full
> tranformer analysis, you can measure secondary > current as well. Then
> you need to repeat the testing with the secondary > open(Obviously, sec
> I = 0) for this. >From all this data you can figure > leakage
> inductance, k, etc. > > I have approx 20 MOT's and none are current
> limited by the shunts even > though the shunts are in place. I have
> tested MOT's momemtarily across > 120 volt mains. They will pull 30 to
> 40 amps easy with the secondary > shorted. If you try be prepared to
> weld the breaker contacts or weld > the plug into the wall. > > Mark
> > > > > Original poster: "Paul B. Brodie"
> <<mailto:pbbrodie@xxxxxxxxxxxxx>pbbrodie@xxxxxxxxxxxxx>
> >
> > Mark,
> > Thanks for the info. How are you going about measuring the impedance
> of > the > primary? Do your MOT's have shunts? All of mine do. If so,
> are there > just > not enough to limit current like the NST's do?
> Thanks. > Paul > Think Positive > > > > ----- Original Message
> ----- > From: "Tesla list"
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>tesl
> a@xxxxxxxxxx>
> > To:
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>tesl
> a@xxxxxxxxxx>
> > Sent: Thursday, April 21, 2005 1:56 PM
> > Subject: RE: Expensive hobby
> >
> > > Original poster: "Mark Dunn"
> >
> <<<mailto:mdunn@xxxxxxxxxxxx>mdunn@xxxxxxxxxxxx>mailto:mdunn@xxxxxxxxxxx
> m>mdunn@xxxxxxxxxxxx>
> > >
> > >
> > > Paul:
> > >
> > > I have many MOT's. A number are marked 4000 Volt. They are not.
> > All > of my MOT's have a ratio between 16:1 and 20:1. Most of my
> MOT's > have > an impedence with the secondary shorted of about 3 to 4
> Ohms. > Thus they > will pull 30 to 40 amps from 120 Volt mains with
> the > secondary shorted > -Don't try that. You test at 10 VAC with
> the > secondary open I have done > many times to verify ratio. Note
> you will > be reading 160 to 200 VAC on > the open secondary. Hook up
> meter > before applying power so you avoid > shock risk. > > Mark >
> > > > Original poster: "Paul B. Brodie" >
> <<<mailto:pbbrodie@xxxxxxxxxxxxx>pbbrodie@xxxxxxxxxxxxx>mailto:pbbrodie@
> bellsouth.net>pbbrodie@xxxxxxxxxxxxx>
> > >
> > > Mike,
> > > I doubt it because this MOT is substantially larger than the other
> > coils > > and it has a lot more turns on the secondary. Also, the
> 4000 > V is > labeled > right on the transformer with the
> manufacturer's > data. Since the > manufacturer doesn't know how the
> end user is going > to wire the > transformer, they wouldn't put the
> 4000 V assuming it is > going to be > driving a voltage doubler or
> anything else, for that > matter. > > I'm curious, where did you get
> the 1650 vac figure? Almost > everything > I've > read on this list
> and on countless web sites say > that the majority of > MOT's > are
> 2000 vac and the heavy duty ones > 4000 vac. I am going to drive them
> > with my variac set to 10 vac and > measure the output from the
> secondary. > Then, I can extrapolate the > output at 100 vac on the
> primary. Anyone > have a > better idea of > determining the secondary
> voltage on MOT's?? > > Paul > Think > Positive > > ----- Original
> Message ----- > From: "Tesla list" >
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>tesl
> a@xxxxxxxxxx>
> > > To:
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>tesl
> a@xxxxxxxxxx>
> > > Sent: Wednesday, April 20, 2005 6:31 PM
> > > Subject: Re: Expensive hobby
> > >
> > > > Original poster: "Mike"
> >
> <<<mailto:mike.marcum@xxxxxxxxxxxx>mike.marcum@xxxxxxxxxxxx>mailto:mike.
> marcum@xxxxxxxxxxxx>mike.marcum@xxxxxxxxxxxx>
> > > >
> > > > Odds are the 4000v is dc after the 1650vac or so from the mot
> is > > rectified > and doubled under the load of the magnetron. > >
> Mike > > > > >
>
>