Hey guys! I searched first, but couldn't find any applicable topics, so I'll start a new one. Briefly, I've begun a very custom Quickie 1.1, point-to-point build (as the kit is no longer available). That said, I've designed and built working power supplies off the mains: one 45v B+ choke fed, and two 2.8v lm317 regulated + 10,000uf filtered filament supplies (discrete). OK, here's where I am...I was simply going to attach the 2.8v heater supply to pins 1 and 7 and then pin five goes the auto-biasing cathode resistor. Makes sense, but, the datasheet says if you run series filament heaters, then you must install a 'shunt' resister between pins 1 and 5 to prevent current 'back flowing' or something? Then, on https://www.radiomuseum.org/tubes/tube_3s4.html it clearly shows simply hooking up the 2.8v supply as I have. Does anyone have an opinion, or, has anyone tried this and succeeded/failed? Thanks in advance!
Custom Quickie + PJCCS Build with 2.8v Series Heater Questions
- Thread starter IndyNate
- Start date
Interesting. I've never heard that, but now I see the data sheets do mention it. Since that resistor would only be across half of the filament, I can't really see how it helps (might help the filament section between pins 1 and 5, but what about between pins 5 and 7?). Sadly I don't have an answer for you so I'll be interested in any replies. I'll also ask some tube expert friends to see what they say and reply back here if I get an answer I can make sense of.
I should have the build done tomorrow for listening tests so I'll report if it even works. After some continuing studies, I too can't understand what the 'shunt' resistor would accomplish or prevent. They also fail to give a value for it, and I wouldn't know how to calculate it. I'm also looking forward to more responses/thoughts/forbodings 
Oddly, the performance of the 3s4 suffers with a higher heater voltage (THD, gain). But, sometimes none of that matters either!
Oddly, the performance of the 3s4 suffers with a higher heater voltage (THD, gain). But, sometimes none of that matters either!Paul Joppa
Moderator
The grid is referred to pin 1, which is where the cathode bias resistor attaches. Any plate current that flows to the pins 5-7 filament section must also flow through the 1-5 section before it can get to the cathode bias resistor, adding to the 50mA filament power current. According to the Tung-Sol data sheet, you only have to worry about that if the total plate current is greater than 5.5mA - which it is not with a 45 volt B+ supply.
Actually, the positive half of the series filament (pins 5-7) is nearly cut off with series filaments, so most of the current goes to the pins 1-5 section.
Actually, the positive half of the series filament (pins 5-7) is nearly cut off with series filaments, so most of the current goes to the pins 1-5 section.
And there we have it! The CCS I would assume would also prevent any overdraw of current by the plate? The maths of direct heated pentodes are a little daunting to me. The guys at Emission Labs have center tapped cathode DHTs, but again, triodes. Thanks Paul! Can't wait to get this little FrankenQuickie fired up!
UPDATE: Just finished the build with 2.8v heater circuits strapped directly to pins 1 and 7. No problems. In fact, the pre sounds amazing. Such a simple but wonderful circuit and SLEEPER bottle. LOVE BOTTLEHEAD!!!! If you haven't experimented with an AC power supplied Quickie, you should!
UPDATE: Just finished the build with 2.8v heater circuits strapped directly to pins 1 and 7. No problems. In fact, the pre sounds amazing. Such a simple but wonderful circuit and SLEEPER bottle. LOVE BOTTLEHEAD!!!! If you haven't experimented with an AC power supplied Quickie, you should!
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Paul Joppa said:
Paul, that helps, but does that mean the plate current only flows between the mid-point (ground pin for parallel) and the "exit" (or entrance depending on which view you take), and not the other half of the filament? If so, what's the point of the other half, other than to burn off 1.4v? How does that provide electron flow to the plate if it doesn't contribute to plate current? Maybe that's what you mean by that half being nearly cut off... But then why doesn't plate current flow through the entire filament since in series wiring there is no inherent purpose of pin 5 (it's not grounded or anything else)? This is all a bit puzzling but also very interesting.
Ignoring that, how do we figure out the value of that resistor? Do we have to assess how much over the 5.5mA plate current limit we will be and choose a resistor to bring it back to below 5.5mA?
Thanks,
Carl
Paul Joppa
Moderator
This is hard to explain in a post. The plate current comes from the entire length of the filament/cathode. That current flows to the cathode bias resistor through pin 1. The plate current that comes from near pin 1 just exits. The portion that comes from near pin 5 (the center tap) must go through the portion of filament between 5 and 1 must go through the 1-5 section to get out. The portion from near pin 7 has to go through the entire filament. So the portions closer to pin 1 have more of the total current than those closer to pin 7. It's like a river with many tributaries - the volume of water gets greater as you go downstream.
If each piece of the filament contributed the same current per unit length, you can get a good guess of the distribution without diving into calculus. In a more detailed model, each point along the filament has a different voltage relative to the grid, and therefor a different contribution to the total plate current. For that you're probably going to need to do integrals or use a computer. This idea was compressed into my comment about cutoff.
Once you know these things, you can calculate the effective heating power delivered to each half-filament, and shunt aside enough of the current in the pin 1-5 section so that it gets the same heating power as the 5-7 segment.
If each piece of the filament contributed the same current per unit length, you can get a good guess of the distribution without diving into calculus. In a more detailed model, each point along the filament has a different voltage relative to the grid, and therefor a different contribution to the total plate current. For that you're probably going to need to do integrals or use a computer. This idea was compressed into my comment about cutoff.
Once you know these things, you can calculate the effective heating power delivered to each half-filament, and shunt aside enough of the current in the pin 1-5 section so that it gets the same heating power as the 5-7 segment.
Somehow I'm only just seeing this. Your comments make sense, Paul. And while I'd be very interested in doing some calculus, it's been a very long time (since the early 80s probably) and I can't even think through the equations much less calculating them. Though were someone to explain it, I might well smack my forehead and say "Duh!"
