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| | #1 |
| Junior Member Join Date: Aug 2009
Posts: 15
| Determining OT specs?
What is the process for determining the OT impedance? I have a couple of OT's that I am unsure of. One is a Pioneer SM500 that used 8R and 16R tap and has an additional unused tap. Not sure of which is which.
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| | #2 |
| Member Join Date: Jun 2006 Location: Seattle
Posts: 55
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Take a small transformer, say 12V and put the 12V secondaries on the OT primary leads. You should be able to determine which wires are connected to each other with your multimeter based on continuity and resistance to determine middle tap. Careful though, cause if you do connect the 12V to the OT secondary you could get a few hundred volts on the primary side so use caution. So with 12V on the OT primary, measure the AC on the secondaries. Now divide input voltage by secondary voltage and square that value. This is your ratio. So if you had 1V on the secondary and 12v on the primary, you have an impedance ratio of 144:1. |
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| | #3 |
| Junior Member Join Date: Aug 2009
Posts: 15
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Hi Mike, What is the best way to determine which tap is 4/8/16 ? The white goes to ground, so that is good. Then there is a Blue (I think is 4R, was not hooked up in original Pioneer SM500) A Yellow one and Purple one. I think the Yellow is 8R. Just want to be sure. The output tubes being used are 7189's, so more than likely 8K tranny. Thinking of alternate output tubes that will work. 6V6 are in this range, but dealing with 450V is high for 6V6's. Yes I can turn down the B+.........just interested in all big bottles that can work with OT. The PT is huge and anything is fair game powerwise. Also have another PT that will do over 700V for my next build. It powered a 42 tube organ. OT ran a pair of 6L6's........hmmmm smell a EL34 Plexi there! I could do a Plexi, TW and a AB763 together with their own tubes and controls into a single PP amp! Wouldn't that be crazy............... interesting switch network on that deal huh? Thanx for your input Brother. Any additional info is welcome and appreciated |
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| | #4 |
| Supporting Member Join Date: Jun 2008 Location: Italy
Posts: 993
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This subject has been dealt with deeply enough in the past, so I won't go too much into details here, do a little search and you'll find all the info you need. The key point is : transformers DON'T have an inherent impedance of their own. They have RATIOS, so they REFLECT whatever impedance they "see" at the primary on the secondary, according to the formula : Ratio=Sqrroot(Zin/Zout), or, if you like it more, Ratio^2=Zin/Zout where : Ratio= the tranformer's winding ratio Zin=the impedance seen at the transformer's input ( the output stage's characteristic impedance ) Zout=the impedance at the transformer's output ( the speaker's impedance ). a numeric example, if your output stage consists of four EL84s ( 6BQ5, 7189 etc. ) in push-pull, the impedance is around 4 K; if you have an 8 Ohm speaker handy, the needed ratio will be : Ratio=Sqrroot(4000/8)=22.36 To match the same output stage to a 4 Ohm load, the ratio will have to be be : Sqrroot(4000/4)=31.62 About testing the thing, the rule of thumb is that the primary will exhibit a higher DC resistance, and this will help you in determining, as you said, "which is which". Once you have determined which is the primary winding, a good method is to inject a 1 V, 1 Khz signal ( the standard mains frequencies might be too low for some transformers to let them pass freely ) at the secondary ( for multi-tap xformers, start with one of the taps then repeat for the other ones ), then measure the signal amplitude at the primary. Being the signal you fed 1 V, the measured amplitude at the primary will be equal to the transformer's ratio. By "manipulating" the above formula, you' ll be able to determine the "best match". Zin=Zout*((Ratio)^2) Zout=Zin/((Ratio)^2) Oh, and the wire to ground is ( most likely ) connected to an internal screen. Hope this helps Best regards Bob
__________________ Foolproof equipment is hard to design, because fools are VERY ingenious... Last edited by Robert M. Martinelli; 10-23-2009 at 06:12 AM. |
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| | #5 | |
| Junior Member Join Date: Oct 2006 Location: Nuernberg, Germany
Posts: 15
| Quote:
0-8 ohms usually is 4 times the inductivity between 0-4 0-16 ohms usually is 4 times the inductivity between 0-8 That's an ideal case, in reality it's about 3.2...3.6 times of the previous. Larry
__________________ The fault almost always is sitting in front of the amp | |
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| | #6 |
| Senior Member Join Date: Oct 2007 Location: Tucson, AZ
Posts: 431
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...I've found the "impedance ratio" the easiest method to use and explain to others, but it DOES require some 'a prioria' knowledge about the OT, typically its Zpp:Zout numbers, ie: 5.6Kpp:4 ohms, etc. ...once you know this information for just ONE tap, everything else is easy.
__________________ ...and the Devil said: "...yes, but it's a DRY heat!" |
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| | #7 |
| Junior Member Join Date: Aug 2009
Posts: 15
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I know the primary leads and the secondary common, just not the 3 taps. The original SM500 (stereo integrated amp > 2 OT's each driving a PP 7189 pair 8K) used 8R and 16R taps, and the remaining was unused. Test gear other than my Beckman DVM here, is out of state at this time.....O-scopes and Freq Gen etc. I usually deal with known value Transformers. And can determine if there is a short fault with what I have here. This tranny is fine.....just 3 secondary wires to be certain of. The schemos for the Pioneer SM83/SM500 unit shows part numbers and values/voltage but no wire specific info. Not much luck with Pioneer on more info. That is why I wanted to know how to decifer secondary identities. Thanx guys |
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