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Hoffman Amps Forum image Author Topic: Firefly powertube eval using the RC30 method for class A amplifiers  (Read 2369 times)

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Offline drgonzonm

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The firefly output tubes were evaluated.  A simplified approach was taken.

The cornell note format was used to present the method.  In the right hand column, the RC30 write up was typed in.  The right hand column evaluates the RCA 2A3 tube.  The left hand column evaluates the 12AU7, using the same method.  

The following software used to put together the pages.

OST, (One Screen Takeoff), used to convert downloaded pdf current plate voltage info. from pdf to TIF.  a free program, that I use evaluate architectural drawings.
Draftsight 5.0, a freeware 2D CAD program, similar in application to AutoCAD LT.  
dopdf V7, a pdf printer.  
MS Word 2003.  
   The TiFF is imported as a raster drawing and stored in raster space.  This allows the drawing to be scaled.  Plot lines are overlaid in paperspace,  Line width, colors can then be modified.  If the tiff is converted in black and white, transparency mode can also be used.  
The drawing is plotted (printed) using dopdf.

The pdf is imported as a picture, formatted, cropped to remove the extra space.  

From Word, the final result is again printed using dopdf.  

Results, The output wattage of the firefly using the 12AU7, using posted voltages, is about 1 watt.  
« Last Edit: February 14, 2014, 06:20:38 pm by drgonzonm »

Offline sluckey

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Re: Firefly powertube eval using the RC30 method for class A amplifiers
« Reply #1 on: February 14, 2014, 04:54:57 pm »
Might be better if the chart was intelligible.
A schematic, layout, and hi-rez pics are very useful for troubleshooting your amp. Don't wait to be asked. JUST DO IT!

Offline HotBluePlates

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Re: Firefly powertube eval using the RC30 method for class A amplifiers
« Reply #2 on: February 14, 2014, 06:46:35 pm »
There were some minor errors, mostly related to the fact the original example used a 2A3 and made some deviations for that tube type.

Regardless, I think I see the idea is the reference material in one column, and your desired calculations in a second column. The drawback is it will take a lot of formatting work to make matching items from both columns appear on the same lines so the average observer can follow what's happening.

Offline dbishopbliss

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Re: Firefly powertube eval using the RC30 method for class A amplifiers
« Reply #3 on: February 15, 2014, 07:06:36 am »
I like the idea of what you are doing with having the reference material and the calculations on the same page, but it is a little hard to read.  I think having your calculations wrapping in the left-hand column is one of the issues. 

Here's a thought... what if you change the Cornell Format slightly (I read about it) as follows: Put only the formulas and very brief descriptions of what they are for in the left column and the reference material in the right column.  Also, highlight the key points from the reference material in the right column.  Then, use the section at the bottom for your actual calculations.  This way the graphs could be larger and you could write out the formulas with having them wrap lines.

Of course... what I just suggested may be hard to read as well feel free to ignore.   :icon_biggrin:
Check out my blog for more details. Bliss Amplifiers

Offline HotBluePlates

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Re: Firefly powertube eval using the RC30 method for class A amplifiers
« Reply #4 on: February 15, 2014, 04:10:19 pm »
3.  The cornell note format calls for the left hand column to be 2-1/2 wide, This column width could be violated, but then we have to call it "modified Cornell Note Format"  :icon_biggrin:

If the format is being used to aid in understanding, then slavish adherence to the format when it impedes understanding seems counterproductive.

A second problem is that tube manual examples are carefully crafted to match optimum conditions previously worked out. When you re-apply the process to real-life situations, you can correctly get an answer which turns out to not be the "best" or "right" answer, even though the stesp were all executed correctly. So you might have to make multiple iterations, and these slop-up the note format further.

Which brings up the 2A3 issue:

I lost a reply when I added this.

Known sources of error in the posting.

1.   RC-30 would make a n bias volt adjustment, for the 2A3, if filament voltage is ac.  If you argue, that the 12AU7, was an indirect heated, and did not need the adjustment.  I can see your argument.

Yep, that's the big one. For others following, the 2A3 is a filamentary, or "directly-heated," power triode which means its filament is also its cathode. They are not separate structures. The 12A_7 series is instead a heater/cathode type tube, where these are separate structures.

Technically, if you see "filament" it implies a tube that is directly-heated like the 2A3. These are very uncommon today, so heater and filament are often interchangeable. The problem is if a directly-heated tube is fed a.c. to light the filament, it may interfere with bias since the filament is also the cathode. So these circuits would use a filament winding with a center-tap, and the total bias had to be increased by half of the filament voltage as it was inseparable from the cathode voltage.

Now the impact of this discrepancy is bigger than you'd think. The 2A3 has a filament voltage of 2.5v, which means a half-voltage increase of bias is a change of 1.25v. You can see from data sheets (and the example here) that bias is in the range of -40 to -45v, or about a 3% change in bias.

The example adjusted the 12AU7 bias for a.c. by adding 1.7v of bias (unrelated to the tube's a.c. heater voltage of 6.3v?) to go from -13.3v to -15v. That's a 12.7% change of bias, as we're talking about a tube with a much smaller bias voltage (because its mu is well above that of the 2A3).

I'd also argue the calculated mu should be different. We have to cheat from the answers to know, but from the Firefly schematic we see that 12AU7 plate current should be 0.5 * 8.8v / 440Ω = 10mA. The Firefly also shows B+ as 265v, with 8.8v across the cathode resistor, or ~250v as the supply voltage. Your example cited ET-T880A, and we should focus on the graph on page 3 of the G.E. 12AU7A data sheet. It shows that mu changes with both supply voltage and especially plate current, so using the Eb=250v curve and 10mA of plate current, mu is given as ~16.8.

 The result of those 2 changes is that bias should be ~-10.1v, which raises the loadline closer to the plate dissipation curve. That the line was so far from the plate dissiaption curve should have been a clue that the calculation couldn't be correct for optimum power output. However, even my calculated numbers don't jive with the reality of the Firefly, which then suggests that the formula just gives a starting point, which the user must refine. It's too bad it doesn't sync well with the 12AU7.

 


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