The AM Forum

Found a nice big transformer. I was able to measure alot of things and even do some calculation, but I really don't know how much power it will handle. At least 1KW CCS certainly.

Is there a formula for E-I type transformers where I can supply the core volume (LxWxH) and guage the power rating? The ARRL handybook only shows two dimensions (what about the third!) and the total weight is really not a good indicator, depending how much pig-iron is present.

core inside the windings:
H=3.875"
W=1.875"
L=2.5"

volume = 18.16 cubic inches.
-where the windings are would around the height dimension (although it should not matter).

Also, secondaries are 6VAC/~10A, 10V/~10A, and 2240V with a tap at 1280V.

_______________o 2240V - 180 Ohms
\
/     104 Ohms
\______________o 1280V - 76 Ohms
/
\
/      76 Ohms
\
/______________o 0V (common)

Sort of an interesting secondary..

anyway, any ideas on core volume vs VA rating?

There's a lot of literature on transformer weight vs. power; core size vs. power, etc. But as you probably already know, older ARRL handbooks had a section devoted to the capacity of old TV power transformers, economy power supplies and rewinding data. 

a.) Are the filament windings marked at 10 amps or does your tilde mean "about," or your guess? Curious that if fil. windings were marked why not the HV?

b.) If you have a 25k /100watt resistor, put it across the secondary and measure the voltage droop from the no load condition.
     Then take a 5k / 250 watt resistor and do the same quickly.  You do have those in your junk box? Convert the values to current.  Plot both on a piece of semi-log paper and that will show you current vs. voltage droop.  This is fun to do on about all your power transformers.  If you can measure more that two points, say four different loads it gets better.  You can see the 'knee', slope changes showing saturation, etc.

Sounds like you've had similar experiences; one of the suprises I learned as a novice was that a lot of big transfomers with hefty fil. windings coupled with a HV winding were for oscillosopes.  Lots of voltage, little current.  Oh, (thought) the DC resistance values you gave look like you might have a keeper.  An 800-0-800 ac volts plate transfomer I have rated at 500 ma. shows 140 ohms across the total secondary.

Yours looks like it would be good in a bridge circuit since it's not center tapped.  Curious to see what you finally measure.

Just some quick off-the-top-of-the-head answers for  now:

I've always used a rule of thumb based on some transformers here, lower than 1 KW, but 15 - 30 Watts per pound (range due to duty cycle and how conservative/reliability) for 60 Hz. iron-laminated transformers.

An electrical emperical test would be to have high value power resistors on hand and see what load across the HV secondary it takes to produce a 10 - 15% drop from no load versus loaded.

I will take a look at what transformer design notes I have, but as I recall we need the actual core cross-sectional area.  Is this what your H & W dimensions are?

Another BIG clue is to measure the primary inductance, no loads.   This is a fundamental design parameter.

From the Radiotron Designer's Handbook 4th Edition page 235:

core sized needed (for 60 Hz.)

 cross-sectional area in inches = (V*A)^1/2 / 5.58


So, solving for Volt*Amps, knowing core cross-sectional area:

 (5.58 x core area)² = Volt*Amps capacity

If your numbers are correct for the core cross-section, then your core area is 7.266 square inches yielding 1644 Volt*Amps.                                         

To analyze backwards from the measured primary winding inductance, we need to know the magnetic path length (download the drawing shown below) and the primary voltage.

If I look at the transformer so that the winding bobbin is a cylinder I'm theoretically looking through, the cross section area dimensions according to how I named them previously is
W=1.875" * L=2.5" for 4.6875 square inches.

Your writing made me think about this some more. The height of the coil/solenoid/bobbin does not matter because the magnetic lines have to fit inside the solenoid. The height (what I am calling the length of the solenoid bobbin) does not matter except to accommodate the wire. Thanks for that education.

About the drawing, the widing is on the center element. The top and bottom elements are about 1/2 the thickness of the center one, so I suppose half the lines go through each? I will get the transformer soon and try to check the inductance. It's reserved for me. I left it till later because I already had a full truck.

I attached a drawing to show the dimensions and how I measured them. In the drawing, the top of the core is invisible.

So, am I right to say:
 
(5.58 * (4.6875))^2 = (V*A)

26.15625^2 = (V*A)

26.15625^2 = 684 watts. ?? the transformer seems much bigger. Am I mistaken?

The hi-volt winding wire I am calling '0' is closest to the inner side, near the low voltage secondaries. The current ratings of the LV secondaries are guesses based on the wire size which looks like 14 guage. I will re-check them.

PJ

You guys already know this but don't forget to subtract the filament winding VA's from the total core capable wattage calculated to get the HV's VA's.  Sounds like you already have to subtract, what?, 160 watts?

I didn't think Opcom wanted precise calcuations... already a lot of "partial knowns."
And don't forget wire size, circular mil cross section....  yeah, lots of just plain unknowns.

Try it out ; jumper it into an existing bridge circuit if nothing else.  The "true" ham method will yield some very good data points.  ;D
This method automatically converges all the unknowns into at least one good solution.

Personally I have more fun running stuff into know loads and plotting up the results on semi-log paper.    Might be fun to rig up an ADC into a trusty PC and plot it.  But then we could study the thing to death along with voltage dividers down to the ADC's level, programming and all the rest of 'study it to death' fun.

"Hook it up.
 Hook it up. " 
(we need a musical note thingy / icon to make this a jingle.)

I remember now; the jingle is from a wall wort air freshener that later was found to burn down houses.
"Plug it in.
 Plug it in !"

Y'all are forgetting the basic T.I.T. (Timtron Institute of Technology) formula for calculating transformer power handling capacity. Its kinda simple: Weigh the thing and figger 10w per pound of weight. This is kinda conservative, but it will never get you into trouble. I have used this for years without any problems. If it weighs 10 lbs it is easily good for 100w of power handling capacity. (this is total including fils and B+) Why make it harder on yourself than you have to??

                                                       The Slab Bacon

So that means my 225 pound 3.5kv ct iron should be sufficient to run legal limit in the scream modulated 4-1000!  The final is completed and I just started building the power supply.

225LBS! that is respectable. This one is not that heavy at all.

Ok.. I'll get it, weigh it, measure the primary inductance if possible, and measure where the 10% drop point is. Geting it will be the main point.

The reason I labeled the bottom end of the secondary as "0" is because the tap at 1280V is not a center tap, and the DC resistances of the two sections of secondary are not equal. This tells me it was for some kind of special purpose. It would not seem to lend itself to a bridge and still use both secondaries.

The leads for the 1280V and 2240V have extra sleeve of insulation on them, whereas the "0" lead wire is not especially well insulated. Since the 'lower' half of the secondary has so much lower ohms per volt (0.06) than the high voltage section (0.11), evidently the lower half carries all the current for both voltages, it makes sense.  The only thing I can think of this being from is some kind of self-rectifying oscillator like a RF medical instrument or ultrasonic cleaner?

Timtron's formula I think is pretty much for traditional E-I laminated iron.  Old tv xfors, Stancor Plate Xfor's, older BC iron, etc.

Silicon steel, tape wound cores like P Dahl has sold is much better.   Not sure of the equivalance but might be twice as good. Have that info somewhere...   might have been an article in "Q Short T" some time ago.

I have a 3000v 1A CT CCS/ 240 v. primary P Dahl that weighs 40 lbs.
And it never gets hot.  I'll bet it's good for at least 1.5A ICAS.  I'll go find my voltage vs. current plot of the beast and scan it into this site.  It really sings into a cap. input filter of 100uf for a lineeaaarr, SSB service.

And then there's three phase transformers, also more efficient, but I digress. - getting off topic.

***********
Found it but has kids' scribbles all over it; will clean it up and report back.