how to calculate load impedance for parallel se

[ALBATROS1234]

i understand how to look up in my rca book the load resistance . for instance a 6v6 is 8500 ohms @315v so my output transformer primary impedance should be roughly 8.5k @315v for single ended. for push/pull the ot should also be 8.5k from brown to blue wire .so this means technically each tube sees 4.25 k if i understand right? but ot are classified with the overall on the primary. what about parallel single ended? i have seen schematics for such amps but for the most part ot pri impedance is not included on the ones i have seen. the only one i saw was the angela super single ended which states pri imp should be 2.5 to 3k and it states b+ of 335vdc. so how do you come up with 2.5 to 3k? i want to know how this works and how to calculate it for other tubes. for instance i have a pair of nos tung sol 6bk5s which states a 250v max plate voltage and load resistance of 6.5k. i want to know how to calculate this myself for any tube i choose to use. i even eventually want to try and build a 2 watt 6sn7 pp which does not even give you effective load resistance only plate resistance. there must be a wat to derive a load resistance from plate resistance. how? i have looked around and have seen many threads talking about parallel single ended but not much good info on how to figure out yourself from tube data on obscure tubes. and i hate to just buy random shit hoping it will work out perfect. sometimes it works out ok like when i got a cheap 6G6/G tube and stuck it in a 6V6 amp of course i am exceeding the plate dissipation by 100% but it sounds really great although i know the little guy isnt going to last in those conditions. in the case of the 6G6 it has a 180vmax plate voltage and 12k load resistance. but yet sounds phenominal in a silvertone se 6v6 r2r amp i reworked. i would however like to try to build a 6G6 amp within spec eventually. even after many times googling stuff when i continually crap out i come here to ask the pros. thank you in advance for any responses.

[shooter]

short answer, divide by 2.
longer version, since SE is more "forgiving" (to a point), each time you tweak plate volts or cathode R, henrys get a little skewed, but then when you hammer it with huge drive signal, or even smaller distorted drive, simple math gets really complicated.

[jjasilli]

Albatross, you seem to be asking 2 different questions:


1.  How to calculate load resistance when you double the number of tubes?  Shooter gave the answer.  Divide by the number of tubes, in this case, two.


2.  How to calculate load impedance from plate impedance? 


a)  you don't.  The tube manufacturer does that for you.
b)  If you must, the simple thing is that for a voltage source (as opposed to a current source), the input impedance should be 10X the output impedance. Even a power tube may be considered a voltage source.  It takes an OT to transform the power tube's output into a decent amount of current.  So, to the OT primary, the power tube looks like a voltage source.  Hence, in theory the OT primary impedance should be 10X the output impedance of the power tube(s).  (Of course, the OT's primary impedance is a reflection of the speaker load impedance back through the OT's windings.)


But it's not that simple in practice, because sound is involved.  Distortion is another factor in play.  In the tube charts, the manufacturer is also giving you a distortion figure along with plate voltage and load impedance figures.  Measuring distortion requires specialized equipment and expertise.


Meanwhile, measuring plate impedance is essentially impossible for the ordinary mortal.  We need the manufacturer to specify this for us.

[PRR]

> Albatross, you seem to be asking 2 different questions:

Yes. I am confused.

If one 6BK5 *at datasheet conditions* likes 6,500, then two 6BK5s like 3,250 and ten 6BK5s like 650 Ohms. (But there are better ways to get 35 Watts.)

12AU7 is not rated as a "power tube". *For triodes*, the usual starting point is to find the plate resistance at a fairly high plate voltage and dissipation near max, then load it with twice that. 12AU7 has a suggestion for 250V 10.5mA, which is pretty near its max. rp is 7,700. Load it with 15,400 and it will make "power". How much? This really needs plotting. But expect that if it could make "WOW!" power for the price, they would boast about it; likely it won't even make 25% of Pdiss so much less than a Watt.

[ALBATROS1234]

thanks gents ,yea there were a few questions in that post, it started with parallel single ended which ,like i said, the only data i could find was from the angela super se which says use a 2.5k but 3k will work. at 315vdc on plates the rca manual said ,as stated above, 8.5k so if its "divide by 2" that would be 4.25k not 2.5k.

second i was wanting to know how to find load resistnce from a data sheet that only includes plate resistance. it seems on "popular" normal power tubes rca manual tells you the load resistance. sometimes it tells you the load resistance at a few different voltages. on other power tubes and triodes it only gives plate resistance so i wondered if there was a way to get at least a ball park figure. i have seen that if the ot primary is to far off your get a weak sound.

 i do however want to make a 1 to 2 watter that i can crank and get nice saturation that i get from my other builds without the neighbors down the street calling the cops. i know some guys like preamp overdrive but i am a fool for power amp overdrive. there is just something sweet about cranking your amp to that sweet spot and strumming an open G. when i was a younger man i did that with a late 70s jmp 50 watt, but i have had many amps through the years most of which were lost in hurricane  katrina along with several guitars a early 80s korg polyphonic synth ,pa equipment, about 30 of my paintings,etc,etc. anyhoo thanks gents thats basically what i needed to verify. still dont get why angela uses a 2.4k when clearly the books say double that is what is optimum

[PRR]

If you can see your neighbors, a small part of a Watt full-bore may be ample.

[ALBATROS1234]

precisely prr. thank you as always for the helpful response above.