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Old 20th January 2016, 10:50 PM   #1
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Default Heater cathode elevation

Is it possible to run the heaters of a indirectly heated rectifier tube and the other tubes in the circuit off the same winding without tying the cathode to the heater on the rectifier as long as you elevate the "center tap" reference to ground? Say, I have a B+ of around 450V, but no additional filament winding.

Will this work?
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Old 20th January 2016, 11:41 PM   #2
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Don't do it! The filament winding associated with a directly heated rectifier is at B+ potential.

SS rectify that B+ rail. Directly heated rectifiers turn on almost as fast as SS diodes do. Install a CL-130 inrush current limiter between the SS diodes and the PSU filter and you'll do fine.
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Old 20th January 2016, 11:48 PM   #3
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The 6BY5 is a indirectly heated rectifier isn't it?

I just figured applying the same heater elevation method as used for small signal tubes might work.
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Old 20th January 2016, 11:58 PM   #4
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http://www.nj7p.org/Tubes/PDFs/Frank/hb3/6BY5GA.PDF
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Old 21st January 2016, 12:03 AM   #5
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The 6BY5 is fully indirectly heated, with cathodes electrically separate from the heater. It can share a filament supply with signal tubes, when the B+ rail voltage is modest. Look at the published heater to cathode potential limits. Remember, the cathodes are at B+ potential.
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Old 21st January 2016, 12:08 AM   #6
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Thanks Eli!

So, theoretically, to close the heater to cathode potential I could reference the heater center tap at 1/4 B+?
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Old 21st January 2016, 11:16 AM   #7
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Put a bypass capacitor from the elevation point to ground as well.
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Old 21st January 2016, 12:05 PM   #8
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Hi,

So, you think it will work? What value cap, and what wattage resistors? On line stages, I usually use 1W resistors.

Thanks!
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Old 21st January 2016, 12:57 PM   #9
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Arbitrary 'large' value, say 22F or 47F. A smaller would suffice, but no matter.

Wattage of resistors can be calculated using ohm's law. Let's say the resistors are 220k and 33k, and B+ is 300V.

Total current of thru resistor string is I = 300 / (220k + 33k)

Then calculate voltage drop over 220k resistor U = I x 220k

Power dissipation is then U x I, so around 0.3 W. 1W resistors are fine.
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Old 21st January 2016, 01:13 PM   #10
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As to the concept in general; on paper, it should work. Whether or not it's a good idea I cannot say. When I used tube rectifiers, I always had a separate winding for their heater.

EZ80 seems to have 500V cathode insulation. There might be some other recs that have more safety headroom.
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