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Hoffman Amps Forum image Author Topic: Class AB 6L6 design help  (Read 3185 times)

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Offline Gregwor

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Class AB 6L6 design help
« on: October 26, 2023, 04:10:39 pm »
I've been banging my head over this for a while and decided it's time to reach out for some help from the more experienced tube lovers.

Power transformer = Hammond 372HX
     600 VCT
     230 mA
Output transformer = 1650NA
     4300 ohms
     60 watts

Full wave diode rectifier should put me around 411V unloaded. 382V loaded.

I'm using this calculator to check my load lines:
https://www.vtadiy.com/loadline-calculators/loadline-calculator/

Merlin Blencowe's wepage on designing a Class AB power amp states that when the tube enters class B operation and sees only 1/4 of the primary winding, 'This load line may cross the max dissipation curve over a portion of its length- this is allowable.'

Well, mine does, but I'm questioning how much is okay. So, I went and plotted a 50 watt Fender Hot Rod Deville to see how a tried and true design would fair. I've attached a screenshot of it. It is REALLY through the max dissipation zone.

The only way for me to get my load lines to live under/around the SOA would be to drastically lower my B+ or of course choose a quiet different output tranny. Most of my 50 watt amps run high 400s for B+ so why is my 382V not working?

I've studied so much on my own and everyone who builds amps around me seem to just make clones and don't actually understand theory so they can't answer my questions. Please help.

Thank you in advance!

Greg

Offline shooter

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Re: Class AB 6L6 design help
« Reply #1 on: October 26, 2023, 05:13:22 pm »
Quote
don't actually understand theory
theory is theoretical clones are factual, so play outside or do math? hmm  :icon_biggrin:
Went Class C for efficiency

Offline HotBluePlates

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Re: Class AB 6L6 design help
« Reply #2 on: October 26, 2023, 05:18:35 pm »
Merlin Blencowe's wepage on designing a Class AB power amp states that when the tube enters class B operation and sees only 1/4 of the primary winding, 'This load line may cross the max dissipation curve over a portion of its length- this is allowable.'

... I went and plotted a 50 watt Fender Hot Rod Deville ... It is REALLY through the max dissipation zone. ...

Look at the long distance moving from the Red Dot ("idle") to the left where the loadline crosses the "Vg=0v" curve (at about 96v plate and 393mA of plate current).  Think of this long distance as "a long time."

Now look at the very short distance moving from the Red Dot to the right where the loadline runs down to 0mA of plate current (at about 546v).  Think of this short distance as "a very short time" until the tube shuts off.

----------------------------------

If the tube stays on all the time ("Class A") then Best Practice is to keep the loadline always stay below the curve of max dissipation.

One the tube's grid-input causes its plate to move along the loadline to the right and cut off (at 0mA plate current), dissipation has fallen to zero.  The plate's temperature is falling.

If the tube gets to cutoff very fast (short distance, "short time") when swinging in that direction, then the plate cools off longer.  That allows us to push it to higher peaks and further over the max-dissipation-curve when the plate is moving along the loadline to the left.

------------------------------------

The time (here, distance along the loadline) the tube will take to move from Idle to the furthest excursion to the left will be similar to the time the tube will take to move from Idle to the furthest excursion to the right.

Plate voltage dropped from 485v to ~96v, a reduction of 389v.  Plate voltage will also rise from 485v to about (485v + 389v) = 874v.  But all the time the plate is swinging from 546v up to 874v (328v worth), the plate current and hence the plate dissipation is zero.

------------------------------------------------

"Class B" is when we cut off the tube exactly half the time.  Theoretically, the tube is biased to the grid-voltage that "just shuts off the plate current."

What this allows is using a higher supply voltage, a smaller transformer primary impedance (which then results in), a higher peak plate current, which then takes the loadline further above the max-dissipation-curve.  The additional plate-heat caused by straying above the max-dissipation curve is offset by the long time (exactly half the time) that plate current is cut off, plate dissipation is momentarily zero, and the plate's temperature is falling.

-----------------------------------------

The plate swing from its Idle voltage downward about 389v to peak plate current.  If it also swung upward 389v (to 485v + 389v = 874v) and was idled at/near cutoff, we would call it "Class B."

Since the tube idles closer to cutoff (in terms of both "current" and "time") than to halfway between the extremes of plate current & plate voltage swing, we might call this "deep Class AB."  As in, "deep into Class AB, and closer to Class B."

-----------------------------------------

What matters is the "Average plate dissiaption" over the entire signal cycle, and that can be painful to calculate by hand (and our handy website here doesn't give you most of the "in-between grid-voltages" you would need to calculate it properly).

The full method for calculating when using a loadline could be taken from RDH4 pages 563-565, 579-580 (especially the latter).  You would be looking at the product of instantaneous plate current & plate voltage at each of the % of grid-voltage noted, but you would be taking the current of a single tube (not tubes from each side of the OT) so you will need to figure values over 180º of the signal, not the 90º used when only trying to figure out power output.
« Last Edit: October 26, 2023, 05:54:57 pm by HotBluePlates »

Offline Gregwor

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Re: Class AB 6L6 design help
« Reply #3 on: October 26, 2023, 08:01:37 pm »
This is exactly what I needed to read! Thank you!

So correct me if I misunderstood, but as long as the class A part of the load line is nicely under the SOA line, I shouldn't worry about burning things up, yeah?

I'll take a peak at those RDH4 pages, but it does sound tedious at best!

Greg

Offline PRR

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Re: Class AB 6L6 design help
« Reply #4 on: October 26, 2023, 10:12:44 pm »
It is REALLY through the max dissipation zone.

For less than half the time. We rarely plot the full "other side" of the wave.

If you are driving motors, as for variable-speed film transfer, you may want to check more. But audio rarely swings exactly to full voltage. If it falls short OR clips the dissipation goes -down-. Most music is transient intermittent.

Build it. Beat it. In a dark room. Does it glow?

Offline pdf64

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Re: Class AB 6L6 design help
« Reply #5 on: October 27, 2023, 05:51:09 am »
Bear in mind that at peak anode current, screen grid current will shoot up, increasing from 1 to 2mA at idle to many 10s of mA. Given resistance in the feed to the screen grids, screens grid voltage will momentarily drop 10s of volts. Thereby reducing anode current and hence anode dissipation.
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Offline PRR

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Re: Class AB 6L6 design help
« Reply #6 on: October 27, 2023, 12:13:16 pm »
....Thereby reducing anode current and hence anode dissipation.

If anode current is less, plate voltage will rise. By E*I, anode dissipation will rise?

But remember all the Greats lived in days before online calculators. Figuring dynamic dissipation of an audio AB amp on slide-rule is HARD. If you are in-sight of datasheet suggestions (derived from junior engineer tests), you are probably fine.

Offline HotBluePlates

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Re: Class AB 6L6 design help
« Reply #7 on: October 27, 2023, 04:13:34 pm »
The full method for calculating when using a loadline ... would be looking at the product of instantaneous plate current & plate voltage at each of the % of grid-voltage noted ... over [the signal cycle].
I'll take a peak at those RDH4 pages, but it does sound tedious at best!
... remember all the Greats lived in days before online calculators. Figuring dynamic dissipation of an audio AB amp on slide-rule is HARD. If you are in-sight of datasheet suggestions (derived from junior engineer tests), you are probably fine.

So here's a demonstration of the tedium...  :laugh:

From the loadline you provided for the Hot Rod Deville, bias voltage was -56.7v, so let's assume we fully-drive this amp with a signal to the 6L6GC grids that peaks at 56v.

   -  We figure the grid voltage at each 10º increment of a sine wave from 0º through 360º:
   -  Sine of angle (like sin(10)) yields a % that is multiplied by the 56v peak signal; add this value to the bias voltage to get Grid-Volts.

   - I attempted to estimate the plate voltage & plate current at each value of grid voltage found from the previous step.  Kinda hard because the online loadline only provides definite figures in 5v increments of grid voltage.

   - I had to estimate where plate voltage would swing when the tube is cut off.  But plate current here being zero reduces the error of the next step.

   - Multiply the momentary plate volts by momentary plate current to get momentary plate dissipation.

   - Add all the individual dissipations and divide by the number of data points.

----------------------------------------------------

I came up with an Average Dissipation of 22.48 watts.  So as bad as the Hot Rod Deville might have looked, it had room for even more with 6L6GC.  OR you could also look at it as running as close as possible to the "23 watts" the various modern Russian substitutes, to extract maximum power without risking failures.

Turns out the maximum instantaneous plate dissipation happens part-way between idle and peak plate current.  And the tube spends most of its time above "30 watts" when it is conducting, but also has so much time below that level to appreciably cool off the tube.

"Plate Volts" is lumpy around the transition to cut-off for a few reasons (and may not be completely accurate).  List of estimated data points are in the table at the bottom.




----------------------------------------------------

Code: [Select]
Degrees Grid-Volts Plate Volts Plate Current Instant. Dissipation
00 -56.7 485 0.03 14.55
10 -46.98 425 0.06 25.5
20 -37.5 380 0.105 39.9
30 -28.7 315 0.165 51.975
40 -20.7 252 0.23445 59.0814
50 -13.8 186 0.285 53.01
60 -8.20 138 0.34 46.92
70 -4.078 105 0.375 39.375
80 -1.55 100 0.385 38.5
90 -0.7 96 0.393 37.728
100 -1.55 100 0.385 38.5
110 -4.078 105 0.375 39.375
120 -8.20 138 0.34 46.92
130 -13.8 186 0.285 53.01
140 -20.7 252 0.23445 59.0814
150 -28.7 315 0.165 51.975
160 -37.5 380 0.105 39.9
170 -46.98 425 0.06 25.5
180 -56.7 485 0.03 14.55
190 -66.4 498 0.01847 9.19806
200 -75.85 516 0.0126 6.5016
210 -84.7 525 0.01 5.25
220 -92.7 641 0 0
230 -99.6 784 0 0
240 -105.2 832 0 0
250 -109.3 865 0 0
260 -111.89 870 0 0
270 -112.7 874 0 0
280 -111.85 870 0 0
290 -109.3 865 0 0
300 -105.2 832 0 0
310 -99.6 784 0 0
320 -92.7 641 0 0
330 -84.7 525 0.01 5.25
340 -75.9 516 0.0126 6.5016
350 -66.4 498 0.01847 9.19806
360 -56.7 485 0.03 14.55

Offline pdf64

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Re: Class AB 6L6 design help
« Reply #8 on: October 27, 2023, 04:37:35 pm »
....Thereby reducing anode current and hence anode dissipation.

If anode current is less, plate voltage will rise. By E*I, anode dissipation will rise?

Well, it would depend on the relative magnitudes of the fall and the rise  :icon_biggrin:

But whatever, I don’t see it like that.
For a typical load line, peak anode current will occur on the g1=0V plot.

From that, as Vg2 reduces, left of the knee, anode voltage also reduces (a bit).
So for those peaks, lower Vg2 should reduce anode current and possibly anode voltage a bit, so dissipation should fall.
« Last Edit: October 27, 2023, 04:41:00 pm by pdf64 »
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