GU-50 Push-Pull AB OT Impedance/Load Line Questions

[defaced]

Hola, MEF! Long time no see. I’ve been pondering a project that I’d like to resurrect and need a sanity check. I’ve been doing the design work for a push-pull GU-50 amp. I’ve never really done the whole “design an output stage from scratch” thing before, so I figured this would be a good time to learn. I’m trying to work out what reflected impedance I need for a pair of tubes operating in a push-pull AB1 output stage. After I get impedance down, I’ll then ask questions about musical output power and power consumption. Unfortunately since this is my first time doing this, I’m a little fuzzy on the rules and customs for selecting an OT impedance, and the research I’ve done is somewhat confusing, so I’m looking for some comment. For the sake of discussions, assume I’m designing around:

800v anode
250v screen
Load line selected for maximum power
Fixed bias



Here’s how I approached drawing this load line:
Start at 800v, 0ma and draw a line directly through the middle of the knee of the 0v grid line. This is the Class B load line (1/4 Za-a) and reflected impedance during the Class B portion of the operating cycle - I name it Blue.
Calculate the Za-a, then calculate 1/2 Za-a
Plot a load line for Class A portion of the operating cycle - I name it purple.
And finially, shift the Class A line up to the desired idle bias point to construct the full AB load line - I name it green.

Now here’s where I start to have questions. I understand that in Class AB we’re able to gain efficiency because the tube is not conduction during the full 360 deg cycle. Now as I look at these load lines, I see that for the most part, they’re violating the 40w max dissipation curve – significantly. Further, it looks like I’d need to drop the B+ to ~450v to cross the same point on the 0v grid line while not violating the 40w max dissipation curve. These two observations seem to contradict one another, so could someone help me out here? From my reading violating the max dissipation curve is allowed, but something tells me that this is too much. If it is indeed too much, are there any negatives to increasing the reflected impedance load to reduce the slope of the load line? If I eyeball what I think is acceptable (just a SWAG), this is what I come up with:

[Malcolm Irving]

Hi,
I think you might have named the ‘green’ and ‘purple’ lines the wrong way round in your text – but no matter.
If you reduce B+, the blue line will come down ‘in parallel’, because the intercept with the Ia axis also comes down (assuming you keep the load resistance the same). That would mean you would only need to come down to around B+ = 650V or so. Power output would be reduced though.
I think it’s OK to violate the power dissipation limit somewhat in push-pull, as each tube is ‘cooling down’ while it is in cut-off.
Your second design looks safer, but watch out for screen dissipation, as you now go further below the knee. (Screen stopper resistors are advisable.)
Malcolm

[es345]

Hi,

Assuming that the operation is AB with only a "small portion" of A you basically can use the dissipation curve of 2x40W due to the push pull operation mode.
Here is the link to the datasheet of the FL152 http://www.mif.pg.gda.pl/homepages/f...18/e/EL152.pdf.
The FL152 and the GU50 are both copies of the LS50, different glass and pins but functionally identical.
On page 9 you will find some audio PP configurations, e.g. 120W output power with Raa 8KOhm at Ub=800V.

Hans- Georg

[trobbins]

Your class B line indicates the peak current that the driven tube could get to. Typically that loadline is placed at 'idle' supply voltage conditions.

The power supply rails will sag when the tube currents get higher. With increasing over-drive, each tube spends more time conducting current at/near the peak current. As such, the class B loadline may start at idle 'alignment', but may quickly and significantly move during overdrive, and in a dynamic manner depending on supply impedance/capacitance, and duration of overdrive signal.

[es345]

...to illustrate that a little bit : These are the load lines of my bassamp (6xGU50), Idle with 30 mA and full load. The tubes have vented cooling.
They are now working in this config without any drift since more than one year.

[defaced]

Thanks for all the feedback guys. After I made this post I continued digging around on the web and found a reference to the FL152 datasheet and Page 9 that es345 mentioned above (I also found your 6 output tube bass project, but hadn't worked out its load line for a pair yet). I plotted the load lines for the Page 9 120w conditions and noted the same massive violation of the 40w dissipation curve. What I wasn't realizing was that because the tubes are mainly operating in Class B, the efficiency gained also changes the max dissipation. This explains how the FL152 page 9 datasheet conditions are acceptable. There are also reports on the web that this tube has been able to dissipate 100w and still operate properly afterward. Knowing its original use in military applications (robust design) and that the efficiency gained by operating is mainly in Class B gives increased efficiency, I'm a bit more comfortable with what I'm seeing on the load lines. I think I'm going to be boring and just build a datasheet output stage.

The ultimate goal of this project is to be used in a high gain amp for metal, so power tube distortion will not be a part of the sound of the amp. With a pair putting out 120w, I'll have plenty of power on tap.

I plotted the FL152 Page 9 120w conditions on its data sheet and on the GU-50 datasheet. I also noticed that the GU-50 has two dissipation ratings, one for normal service (55w) and one for extended service (40w). So three dissipation curves are plotted on each (40w, 55w, and 80w).

FL152 plot:



GU-50 Datasheet:


This brings me to my next question: power transformer current rating. Looking at the FL152 Page 9 datasheet conditions, it gives the following:

Idle: 2x50 ma anodes, 2x0.9 ma screens
Full modulation: 2x120 ma anodes, 2x13 ma screens.
62.5% efficiency

It seems to me those are RMS values. If I run the math for 800v, 120w and 62.5% efficiency [120/.625)/800], I get a max current of 240 ma per tube. It looks reasonable based on the load line and the conditions stated in the table of values on Page 9 (-51v bias, 36v signal swing, so that means at full output g1 is around -15v, which is right around 200 ma). Does that sound about right? If so, I'll probably spec the PT at 300 ma DC.

[tedmich]

Df you've seen the Wavebourn Pyramid right?
www.wavebourn.com ? View topic - GU-50 push-pull amp
I guess UL does not work for these Frankenstein looking bottles..
www.wavebourn.com ? View topic - PP UL GU50

AGRO! Really looks like it should be filled with blue Barbicide and have a couple little combs in it...

[jazbo8]

It looks reasonable based on the load line and the conditions stated in the table of values on Page 9 (-51v bias, 36v signal swing, so that means at full output g1 is around -15v, which is right around 200 ma). Does that sound about right? If so, I'll probably spec the PT at 300 ma DC.

You pretty much got it right, but page 9's recommended operating conditions, are for Eg2=300V not 250V, so you need to re-draw the curves and the loadlines. WRT the loadlines, the "Class A 4k" line should not pass through Ea, like the "4K 70%" loadline, it has to be shifted up, but it is really not a big deal since we are more interested in the Class B loadline in anycase. The grid swing is an RMS value, so 36Vrms means Vg-g = 102V or Vg = 51V, therefore the input voltage swings fully to the Eg1 = 0V curve.

[es345]

Here are the curves for Ug2=250V and Ug2=300V from the original LS 50 document.

One additional hint: there is an error in the FL152 spec, the corrected page is attached. Interesting to know that this error is already in the original LS 50 doc...

[defaced]

Ted, yes, I have seen the work done by wavebourn. Running the numbers for his work, and comparing it to the "how to" on Merlin's site is what prompted this thread. I couldn't reconcile the large violation of the max power line that I kept seeing against what appeared to be a very polite example given on Merlin's page.

You pretty much got it right, but page 9's recommended operating conditions, are for Eg2=300V not 250V, so you need to re-draw the curves and the loadlines. WRT the loadlines, the "Class A 4k" line should not pass through Ea, like the "4K 70%" loadline, it has to be shifted up, but it is really not a big deal since we are more interested in the Class B loadline in anycase. The grid swing is an RMS value, so 36Vrms means Vg-g = 102V or Vg = 51V, therefore the input voltage swings fully to the Eg1 = 0V curve.

Duh, I should have known that about the Grid voltage being reported as an RMS value. Regarding the Class A load lines, I'm following the instructions on the Valve Wizard page: http://www.valvewizard.co.uk/pp.html. I'm showing both steps of drawing the Class A load line for my own completeness, so in reality, the line titled Class A 4K is just sketch work, the 70% line is what I'm really looking at. That said, if I messed up drawing them, I'd like to understand what I did incorrectly. This thread is an exercise in increasing my understanding of how to design power stages.

Thanks for the additional curves es345. I'll take a look at those and I'm sure things will start to be clearer for me.

[jazbo8]

That makes sense, so the "Class A 4k" is just an interim step, then just put them on the Eg2=300V curve that es345 provided and you are all set.

[es345]

I guess UL does not work for these Frankenstein looking bottles..

Well there is a little bit "exotic" possibilty to go to UL, if you use an OT with a separate G2 winding feed with 300V.
I have made an evaluation circuit and published the results here About GU50 PP in UL with separate G2 winding and supply - diyAudio

The drawback is the price for the nonstandard OT.

[loudthud]

Plate dissipation is a rating for long term average, not instantaneous as in a semiconductor. Below shows a 6L6 hitting about 160W with an inductive load. Pic was taken of a 5F6a re-issue just looking at cathode current measured with a 1 ohm resistor and a very high votlage probe on the plate.

[defaced]

Plate dissipation is a rating for long term average, not instantaneous as in a semiconductor. Below shows a 6L6 hitting about 160W with an inductive load. Pic was taken of a 5F6a re-issue just looking at cathode current measured with a 1 ohm resistor and a very high votlage probe on the plate.

That's a really cool series of scope traces. Let me see if I'm understanding them correctly:

Left plot is the conduction cycle of the tube with an inductive load. The bright horizontal line at the bottom of the trace is when the tube is not conducting. The faint diagonal line up to the grid curve is the transition between off and conduction. The line tracing the grid curve is full conduction. And the bright diagonal line coming back down to the bright horizontal line is the transition between full conduction to off. Max power is the point furthest "up" the 0v grid curve line. I eyeball it at about 350v and 430 ma which agrees well with your 160w power output. The diagonal lines are the load lines for the on and off transition part of the cycle, yes?

So average power is something I'm intimately familiar with. I'm a welding engineer and we pulse currents all the time to control heat buildup in the part while still realizing the penetrating power of high currents. So in my world of square waves understanding average power is easy. Here, not so much. So it seems to me that if we represent the inductive conduction curve on a straight line graph it'll be easier to see the on/off times as compared to the entire cycle, thus making visualizing power output easier. So the area under the faint triangle added to the area under the 0v grid curve triangle shape type thing is the total power dissipated during conduction, yes? So then the average of this composite curve is the dissipation we're trying not to have exceed the max dissipation rating of the tube, correct?



The middle plot is self explanatory.

The right plot shows a resistive load which can be compared to the inductive load on the left. So here the power output is going to be reduced as compared to an inductive load because the tube can't ride up the 0v grid line. I think of the interaction of an output stage and a speaker as a spring, so using a resistive load basically replaces the spring with a fixed rod. We gain consistency but lose output power.
_______

Just to keep the load lines up to date, here is the LS-50, 300v screens with 8k load lines and applicable dissipation curves.

[es345]

The idle current is 50 mA, you need to shift the line, added in red in the attachment.
With increasing output power the supply voltage will decrease and the load line will move to the blue one.
How much it will move depends on the PT and the rectifier architecture and filtering.