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Author Topic: Questions on mod transformer ratio and Heising reactors  (Read 7008 times)
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K1JJ
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« on: November 11, 2005, 08:42:01 PM »

Hola Fine Men of AM!

I have two questions.

I need advice measuring a mod transformer's turns and impedance ratio.

I measured my RCA 1KW mod transformer using 122VAC 60 hz across the FULL primary [this includes the CT] and measured its output across the secondary.

I put 122 volts in and saw 101 volts out. Is this the proper way to measure the turns ratio or do I need to input from the CT to ONLY one side and measure its output at the secondary?

So anyway, I see 122/101 = 1: 1.20  turns ratio
1.20 X 1.20 =1.45 impedance ratio. Is this correct?

HOWEVER,
The transformer impedance on the case says 10.3K primary and 4.3K secondary. 10.3K/4.3K =  2.39 impedance ratio.

This doesn't add up. Please set me straight, someone.

I've decided to modulate a single 4-1000A with a pair of 833A's using it. I was told that a 2.39 :1 ratio would get me barely a little over 110% or so - using a common supply.

But, I have another smaller mod transformer that I used with a pair of 813's X 813's that measures 122V into the full primary and shows only 82V out at the secondary. This rig gave me 150%+ modulation using a common supply for finals and modulators. It has even a greater step down than the other one above.

I don't understand why this first transformer would not do better than 110% if the second one will do 150%.

Question #2.  I have a pair of Peter Dahl  150H at 300 ma chokes. I ordered them for audio heising work, but not sure what he sold me back in the early ninties. I want to put the pair in parallel to give 600 ma at 75 Henries.   I have a chance to pick up a REAL RCA Heising reactor that is 800 ma at 50H.  Is there any difference?  What I wonder about is if the Dahl chokes are really 60hz power supply chokes, will they do well at high audio frequencies like a well designed  broadcash heising choke?  Or, maybe a choke is a choke. I remember Tron saying something about this and laminations, but cannot remember the full story.

Thanks.
Tom, K1JJ
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« Reply #1 on: November 12, 2005, 12:13:50 PM »

Hey T,
I'm no expert on this but I'll add a penny or two to this.  I'd be curious to know what those numbers would be if you put in a wider range of frequencies.  I'm assuming you just put 60Hz on the primary.
B3
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« Reply #2 on: November 12, 2005, 08:39:57 PM »

Tom,
Look at the transformer Z as a series resistance. This will limit the power you can ram through it. Just like a pi network with excessive L. My numbers show it should have read about 1.5 :1 but don't forget you only had 120 volts across the primary and there is no load on the secondary. The ratings on the transformer reflect resistive  and core losses when the transformer is at its rated power. I suspect the modulator voltage will have to increase to handle the high input Z of the transformer. Then you have to look at the winding resistance to see if the wire will get too hot. If you are lucky the core won't saurate with extra voltage across the primary. Ask someone who has used the iron to see if it will plaay in your application. a smaller transformer with the correct ratio could saturate when you increase the current through it.
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Bacon, WA3WDR
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« Reply #3 on: November 12, 2005, 10:01:52 PM »

Something is seriously wrong, Tom.  Yes, that's how it's done; the voltage across the full primary is compared to the voltage across the secondary, at least the section of the secondary that you're going to use.  If you could not get much modulation with that transformer, then either your mod tubes can't provide the current required with a low load impedance, or there are some shorted turns in the transformer, or you had DC on the winding, and it couldn't take DC, or something like that.

120V on the primary and 101 on the secondary is 1.19:1 turns ratio and that produces a 1.42:1 impedance ratio.  I would expect this to produce about 140% mod with a common supply.

Each tube looking into the transformer will see 1/4 of the plate to plate impedance.  The peak voltage delivered to the load will be Eb minus about 150V minimum plate voltage, so the maximum peak plate current should be 4(Eb-150)/Zp-p.  So if you have 10K p-p primary impedance, and you have 2650V Eb, you'll have 2500V peak negative swing during conduction, and about 1A peak plate current.  The average plate current for two tubes will be about 0.71A plus resting current on a sine wave at full output.  Current on voice should be lower.  That minimum voltage may be higher than 150V depending on the tubes and the currents used.

What kind of DC current did you see on the modulators with each transformer?  Assuming the transformer with the higher primary impedance makes 100% mod, a transformer with a lower p-p impedance will pull more current for 100% mod, but it should be able to modulate beyond 100% if the tubes will provide the peak current. Of course a transformer with shorted turns will draw all kinds of current and not put out much audio.

Are you sure the primary and secondary windings had continuity?  Maybe one of the center taps was open, and you saw capacitive coupling or something.

Generally a mod transformer with about a 3.3:1 impedance ratio will give 100% mod with a common supply.  (4:1 would do it, if the modulator tubes could pull all the way to zero plate volts on peaks.)  Yes, I think that a transformer with an impedance ratio of 2.39:1 should give you about 110% mod with a common supply.  Are you sure you didn't have that baby turned around backwards?  Or tapped down on the primary?

If the DC resistance of those two 150 Hy chokes is nearly equal, the current will split about evenly, and you should get 75Hy with 600mA capability with the two of them in parallel.  But yes, the laminations may introduce eddy current loss at higher frequencies if they are too thick.  A power choke would have thick laminations, because it is for low audio frequencies, but you would have to try it to see if it worked.  It would load the modulator heavier at high audio frequencies, but if it's not too bad, and the tubes can provide the current, you can compensate.  Also you could put a lower inductance audio choke in series for higher frequency audio, if it can take the current.
 
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« Reply #4 on: November 12, 2005, 10:44:00 PM »

Thanks for the well thought out info, Bacon.

Well, I haven't put the RCA 1 KW transformer in service yet with the 833A modulators. I'm just trying to figure out what I have and what to expect.

I  measured the primary and secondary for resistance and get 55 ohms for the pri and 47 ohms for the sec.  The transformer is marked "primary" so the orientation is correct.

The metal ID tag on it says: 10,300 ohms pri and 4,300 ohm secondary. Yet, I still measure the 122 > 101 volts AC giving a 1.19 turns ratio = 1: 1.41 step down imp ratio, as you said.

Yes, shorted turns would certainly give a bogus reading like this.. good point.

BTW, I just sent an email out to Peter Dahl providing him with the ID tags from the chokes. I axed him what service they were wound for. That shud help based upon what you told me.

I guess the only thing left to do is try the mod transformer and chokes in service and run the tones thru from 30 hz to 7kc and see what happens. I should have this new rig working in a month or two.

Thanks again for the info, OM.

73,
Tom, K1JJ
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« Reply #5 on: November 12, 2005, 11:38:04 PM »

Interesting Tom. I have the exact same transformer that is spec-ed out for my 250 watt RCA rig, (1KW transformer). Get a megger and check dem windings! Any transformer I get I usually use a megger or insulation tester upto 8KV. You will get a good indication if any windings are shorted before you dump the B+ to it. I have usually used 50Meg ohm as good isolation. You have to admit, its a good way to avoid bangs, flames and dangs.
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« Reply #6 on: November 12, 2005, 11:49:40 PM »

Hi Mike,

Interesting that you have the same transformer.

Some questions if you don't mind:

So, how well does it sweep 30hz - 7kc when in the rig?   

What kind of heising reactor are you using, and what value heising coupling cap?

The number on mine is: RCA   450217-1  94404    1 KVA

Does yours also say 10.3K to 4.3K on the ID tag?
If so, the magic test would be to see what voltage YOU get out of yours with 120AC on the full primary. That may tell a lot, assuming they are identical.

I was told that this transformer was supposedly taken out of a working unit that was being scrapped. So, I doubt it has shorted turns, but ya never know.

If it's toast, no big deal - I'll PDM the rig.... screw it -   Grin

T

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« Reply #7 on: November 13, 2005, 11:17:47 AM »

I just realized that the secondary has THREE taps... marked: "Low, Tap and High."

Across the entire Low-High, I get the 101V.  But across the Tap-High, I get 87V. This equates to a 1.95:1 imp ratio. Not 2.3:1 as the ID tag says, but closer than before.

Maybe they labeled the ID tag for the tap connection, dunno.

Bottom line is when using the full winding, I get the 1.5:1 step down ratio. That's PERFECT for the job, assuming there's no shorted turns in there... Shocked

BTW, Interesting solution to put a smaller choke in there for the highs, Bacon. I actually have a 10H? 1A choke that might suit the job. But cornsidering the human voice, maybe THAT wud be enuff for the whole job. I know most guys like to run 30H , usually.

T

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« Reply #8 on: November 13, 2005, 01:36:00 PM »

Tom, from what you state the iron you have is signifcantly different than mine. Mine is an RCA 900763 and is toasted. I would like to find another one since it was in my Lend-Lease rig. Currently I have a VM-4 in there but would like to stay original. Since simplicity was the name of the game, they didn't include a Heising reactor with it. Hell, it is a single source B+ so that you are running 1500VDC on the 813's and the 805's. Still I'm glad you were able to figure out the hook-up configuration. I can't stress the importance of checking the transformer with a HIPOT or Megger before putting it in service. I wish I had one to take to the fests so that I could determing I'm buying a good transformer or truly a boatanchor.
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« Reply #9 on: November 13, 2005, 05:07:28 PM »

one way to ck the xfmr wud be a plate xfmr into one side and another out the other es put a few lightbulbs on the end fer a load
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