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Re: MOT Testing



Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

Hi Paul,
          In a nutshell:

(1) Algebraically sum the inductive reactances assuming they are
connected in series and not coupled to each other
i.e. Xl(tot) = X1 + X2 +....+ Xn.
             = 2.PI.F(L1 + L2 +.....+ Ln)

(2) Algebraically sum the resistances (as in resistances in series)
Rtot = R1 + R2 +....+ Rn.

(3) Vectorially sum the results from step (1) & (2) to get the total
impedance a la Z = SQRT[Xl(tot)^2 + Rtot^2]  (i.e. Z^2 = X^2 + R^2).
(4) If you want the phase angle = tan^-1(X/R).
The above assumes you aren't interested in any capacitance which will
be present.

Malcolm

On 25 Apr 2005, at 12:47, Tesla list wrote:

> 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>t
> esl > a@xxxxxxxxxx> > > To: >
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>t
> esl > 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@wmwmeyer
> .co > 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:pbbrod
> ie@ > 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>t
> esl > a@xxxxxxxxxx> > > > To: >
> <<<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx>mailto:tesla@xxxxxxxxxx>t
> esl > 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:mi
> ke. > 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 > > > > > > >
>
>
>