Selecting OT Primary Impedence from Load Lines

[kleptocracy]

So I know I'm new here, but I'm definitely enjoying it! There's a lot of a really interesting discussions in here. Anyway...

Ok, so I'm designing a PP amp based on the Sunn Model T....in other words, 4x6550, ~150W, UL with +505V B+ and -55V bias.

Basically, I want to know how to pick a load line and determine the OT primary impedance from that. I've searched the forums and done a pile of googling, and haven't come up with many clear examples of how to do this. Most people seem to just select operating conditions from the tube data sheet, but I want to understand a more in-depth approach.



My best guess...
based loosely on a method described in Richard Kuehnel's book on power amps, is to do the following:

Take the load line in the attached image, for example, (50mA & 505V --> 200mA & 0V), and the slope of that line should represent (for a 4 tube push-pull amp) twice the required primary impedence on each side. In other words:

(505V - 0V) / (.2A - .05A) = 3.4k Ohm Primary Z per side, for a pair of tubes. For two pairs, 3.4k/2 = 1.7k Ohm Primary Z per side, or 3.4k plate-to-plate.

Two questions:
1) Is any of this right? Even if it is, is there a better, more real-world-accurate method?
2) Is this a reasonable load line for a Model T, or any 6550-based power amp for that matter? I kind of just pulled it out of my a-hole, so to speak. I've got a big beefy supply, so I'm just trying to squeeze as much wattage as I can out of the 6550s without doing anything that would hurt them, or even cause them to be replaced more than average.

Thanks!

[tbryanh]

Check out "The Push-Pull Power Output Stage" by The Valve Wizard (http://www.freewebs.com/valvewizard1/pp.html)

It describes in detail how to design a PP stage, and the following quote indicates the info might be what you are looking for.

"Usually we have . . . an output transformer to hand already. . . . Of course, the whole method can be worked backwards to find a suitable transformer . . ."

[GibsonLover]

Here's some more of the same info as applied to you preamp stages

[Steve Conner]

One other thing to check is the screen current. Pentode and beam tube datasheets should have a chart that shows plate and screen current together. The left hand end of your loadline should hit the zero grid voltage line before it pokes into the region of sharply increasing screen current.

Also, where did you get the load point of 200mA and 0V from? No real tube can conduct clean down to 0V, they always drop 50V, 100V, 150V, whatever, when turned on as hard as possible.

And, I'd expect a 6550 to push a peak current of considerably more than 200mA. EL34s, 6L6GC and the like can all draw 300mA from B+ when reproducing a sine wave, and that implies peak cathode currents of 420mA per tube.

We can run the calculation backwards to get a rough sanity check: You have 505V and you want 150W. If you assume the tubes drop 105V when turned fully on, then there is 400V peak available to drive the OT. This must generate 300W, since instantaneous peak power of a sine wave is twice the average. By Ohm's law, P=V^2/R, therefore R=400^2/300 = 533 ohms, and the plate-plate impedance is four times that: 2.1k. And the peak cathode current will be 400/533= 750mA per pair of tubes, or 375mA per tube.

Fender got 300W out of four 6550s in their 300PS, so at 150W they're still more or less lying on the couch with a beer.

[kleptocracy]

Ah, that's exactly what I was looking for. Thanks so much, guys!

Steve, you're right. I'm not sure where that graph came from exactly (other than duncanamps.com), so I'm going to use the old GE 6550-A sheet for graphs. That gives much more reasonable values for voltage dropped across the tube at full conduction.

I ran through the Valve-Wizard's method, too, and it looks like the Hammond 1650T is pretty much perfect...assuming the quoted "1.9k center tapped" actually means 1.9k plate-to-plate with a center tap, and the 120W rated output power is really as conservative as it seems (-1db at 30Hz and 30kHz)

My only real concern then is the "design maximum values" on the data sheet. The old GE 6550-A data sheet lists the max cathode current at 190mA, which we seem to be far exceeding (at 375mA through the cathode).

It also lists the rated max plate dissipation (which I assume is the aforementioned dropped voltage times peak current) of 42W. With a peak draw of 375mA, the plate dissipation is 105V * 375mA = 39.3W, which is ok but way borderline. So borderline, in fact, that doing the same analysis the other way for the 1650T gives 400V/475Ohms = 842mA per pair of tubes or 421mA per tube. Then, with the same 105V across the tube: 105V*.421mA = 44.2W! Even worse is that the VW method has the load line crossing 0V at 400mA with 125V dropped across the tube, giving a plate dissipation of 50W! Am I missing something or do we crazy musicians just beat the piss out of our tubes?

[Old Tele man]

...the picture at the bottom of John Broskies' TUBECAD page 11 is worth its' weight in vacuum tube gold.

...and it's a proverbial "...picture worth a 1K words!"

[Steve Conner]

Yes, you are missing something. The datasheet maximum values for cathode current and dissipation aren't instantaneous, they're time averaged. And your method calculates the instantaneous values at the waveform peak.

Also, sometimes the "headline" figures on the datasheet front page are not even maxima, but a recommended operating point for single-ended Class-A use.

These two differences often trip up people used to designing with transistors.

[kleptocracy]

Oh, sweet. Well, there it is then. Thanks, Steve!