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Old 19th February 2010, 03:32 AM   #1
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Default Please don't stone me. . . .

Scenario ; OPT with 6600 ohm primary winding and 40% UL taps.

If I were to use the UL taps as my Plate feeds, would this mean my primary load would be 3960 ohms ?

Is this a feasible thing to do ? I realise that my B+ will go up .

Now to compound this scenario : If the above works and is feasible , then could I use the 4 ohm tap on the secondary side , wired to an 8 ohm speaker to achieve a primary load of 7920 ohms ?

Any possible downfalls to this ?

The reason I ask is that I have an OPT with the above primary winding of 6600 ohms, and I am currently using it with a pair of triode tied 6V6's in PP . I have read quite a few posts that recommend 8-10k primary load for triode tied 6V6's , and wondered if I might try this with any success.


Thanks for your help........................Blake
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Old 19th February 2010, 03:51 AM   #2
tomchr is offline tomchr  United States
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Blake,

3960 ohms sounds off. Remember impedances on a transformer scale according to N^2, where N is the turns ratio. So to figure the impedances on each part of the primary winding, you run the following through your calculator:

6600 * sqrt(0.4) = 4174 ohm (for the 40 % part of the primary; B+ to UL taps)
6600 * sqrt(0.6) = 5112 ohm (for the 60 % part; UL to Plate taps)

The second part of your question is correct. If you wire the 8-ohm speaker on the 4-ohm secondary of the transformer, the plate load will double.

~Tom
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Old 19th February 2010, 03:55 AM   #3
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Thanks Tom.

So I would be at about 8348 ohms ?


..........................Blake
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Old 19th February 2010, 04:00 AM   #4
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"6600 * sqrt(0.4) = 4174 ohm (for the 40 % part of the primary; B+ to UL taps)
6600 * sqrt(0.6) = 5112 ohm (for the 60 % part; UL to Plate taps)"

I just reread this. It doesn't make sense to me. If the impedance from the B+ to the 40% tap is 4174 ohms, and the impedance from the UL tap to the Plate tap is 5112 ohms, then wouldn't my OPT be 9286 ohms ?

Wouldn't you just multiply .4x6600 ohms to find what the 40% impedence would be ? 40% is 40%, no ?
.............................Blake
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Last edited by Nihilist; 19th February 2010 at 04:03 AM.
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Old 19th February 2010, 06:22 AM   #5
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Quote:
Originally Posted by Nihilist View Post
If the impedance from the B+ to the 40% tap is 4174 ohms, and the impedance from the UL tap to the Plate tap is 5112 ohms, then wouldn't my OPT be 9286 ohms ?

Wouldn't you just multiply .4x6600 ohms to find what the 40% impedence would be ? 40% is 40%, no ?
.............................Blake
No, that doesn't work. Take the simpler example of a center tapped winding: 50% tap. Say the turns ratio between the entire primary and the secondary equals N. At the secondary, the generated power is Vsec*Isec, and the impedance is Zsec = Vsec/Isec. Now, ignoring losses, if you use the entire primary winding you get Vpri = Vsec*N and Ipri = Isec/N. Therefore, the impedance of the entire primary is Zpri = Vpri/Ipri = Zsec * N^2.

OK, so far so good.

Now if you use only half of the primary, you have Vpri = Vsec*N/2; Ipri = Isec / (N/2); therefore, Zpri = Zsec*N^2 / 4. The impedance of half the primary is only one fourth of the entire. So you see the sum of the impedances of both halves of the primary does not equal the impedance of the entire primary!

Hope this helps...
Kenneth
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Last edited by kavermei; 19th February 2010 at 06:27 AM.
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Old 19th February 2010, 06:25 AM   #6
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Quote:
Originally Posted by tomchr View Post
6600 * sqrt(0.4) = 4174 ohm (for the 40 % part of the primary; B+ to UL taps)
6600 * sqrt(0.6) = 5112 ohm (for the 60 % part; UL to Plate taps)
Tom, I think it should be:

6600 * 0.4 * 0.4 = 1056 ohm
6600 * 0.6 * 0.6 = 2376 ohm

Can you double-check that?
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Old 19th February 2010, 06:40 AM   #7
m6tt is offline m6tt  United States
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Math aside, which is apparently not my strong suit ,

I think I read correctly you want an a-a load above 8k, as opposed to the 6.6k you have. Wouldn't it be easier to run 12 ohms on the 8 ohm tap and use the transformer as designed, or 6 ohms on the 4 ohm tap? Both are 8-9k. Even 6.6k may be alright depending on the operating mode, I imagine.

Please note also that the actual wires that connect to the taps in an ultralinear output transformer may not be designed to pass full plate current. Any of the transformer wrangling you do will probably change the Q of the transformer, as well.
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Old 19th February 2010, 07:09 AM   #8
Yvesm is offline Yvesm  France
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Quote:
Originally Posted by kavermei View Post
Tom, I think it should be:

6600 * 0.4 * 0.4 = 1056 ohm
6600 * 0.6 * 0.6 = 2376 ohm

Can you double-check that?
Checked and approved !

Remember: screen tap value is ALWAYS given as a VOLTAGE ratio with the plate output, not as an impedance ratio . . . and the IMPEDANCE ratio is the VOLTAGE ratio SQUARED.

Yves.
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Old 19th February 2010, 07:46 AM   #9
godfrey is offline godfrey  South Africa
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Quote:
Originally Posted by kavermei View Post
Tom, I think it should be:

6600 * 0.4 * 0.4 = 1056 ohm
6600 * 0.6 * 0.6 = 2376 ohm

Can you double-check that?
Agreed, so with the original plan to connect an 8 ohm load to the 4 ohm tap, you'd get 2376 * 2 = 4752 ohm primary.
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Old 19th February 2010, 04:36 PM   #10
DougL is offline DougL  United States
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I read this yesterday and it was a math problem.

I have a comment on your engineering solution.
You would be better served to use the complete windings.
This will give you more primary inductance.
A load of 13,200 on a tube that textbook says 8k to 10k may have 20% less power but 30% less distortion. And most "8 ohm" speakers are 6 ohms at some frequency, that puts the toughest load in the 8k-10k range.

HTH

Doug
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