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May I make a suggestion? Use higher power cathode resistors. The power dissipation of over 6 watts is going to make those 5 watt resistors extremely hot even with two 5 watt resistors in series. I used two 20 watt resistors in parallel (40 watts total capacity) and they still get very hot.
I understand your concern. I chose the 25 series Ohmite resistors because of their excellent derating factor. These resistors derate from 100% to 0% from +25˚C to +350˚C. Because all the thermal design for the amp is based on a max internal temperature of +75˚C, these 5W rated resistors actually derate to 4.23W at the max temperature. Their actual dissipation is ~3.16W (0.095A^2 * 350Ω = 3.16W) so this represents a design margin of (4.23-3.16)/4.23*100% = 25.3%. I understand that they may get warm in this configuration but they are still well within their ratings. And, since they are contained on the inside of the chassis, I didn't think this would be an issue. If we really wanted to keep these resistors cool, I would recommend going to the Ohmite 825J series which have the screw tabs for mounting to a piece of metal as a heat sink.
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What about not using regulators on the 300B heaters. What about just reticiation with good filtering. I can see the problem the regulators have with the differential between the output voltage and the input voltage being very small. This is not why I don't want to use regs, I'm wondering can we obtain good quiet DC heater current without regulation. It does reduce complexity.
I agree with reducing complexity and the totally passive approach was my first idea. The problem I encountered was that because of the very small source resistance of the supply, I couldn't find a way to get anywhere near the target ripple level of 2.2mV. Using the 6.3V transformer the total drop voltage we can see in a filter is 6*sqrt(2)-1-5 = 2.48v. The source resistance is something very small (~.25Ω) for the winding plus 1V/1.2A=0.83Ω for the diode bridge for a total around 1.1Ω. Even with the current 6800µf primary filter cap, the input ripple to the regulator is still on the order of 1v. Theoretically, to get it down to the target ripple we could use a 2Ω series dropping resistor but we would need a ~0.4F capacitor (Or we could use four 0.25Ω resistors and four 100,000µF capacitors).
You begin to see my problem? In the interest of trying to reduce complexity the filters are getting very complex. After some consideration, I thought that the low dropout linear regulator was the best option.
There is a possibility of using a combination of the hum-pot configuration with a filtered DC filament supply. In this instance we could assume 70dB of filtering by the hum-pot and get the additional filtering via passive filtering on the DC filament supply. In this case the ripple on the DC line would only have to be reduced by -20dBv to approximately 1/2v. The cathode resistors would need to be reduced to offset the additional 2.5V bias voltage, but that should be manageable. What do you think? Of course the other option is to go with the AC heater/Hum-pot configuration and live with the -70dB hum level as opposed to the -90dB target value.
Anyone else have any thoughts on this subject?
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FAQ
Didn't you mention that you were thinking about going that way?
Remember that a multi-meter puts out about 2 to 3 V when measuring resistance. This could have something to do with getting zapped.
On the filament schematic posted above, I said that the mains supply was 240 (234) but I wired it like it was a 120v main. So here is the corrected schematic with the proper primary side wiring.
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