Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: Blooze on April 06, 2025, 06:09:23 pm
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In my GA-18T build there is a 1K5 resistor followed by a 47K resistor to ground on V2A. The voltage I measured at the cathode of the 12AX7 was 36.5V. When calculating the actual current for that half of the triode would I use the resistors in series in this case? I typically only see a single cathode resistor with a parallel cap.
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In my GA-18T build there is a 1K5 resistor followed by a 47K resistor to ground on V2A. The voltage I measured at the cathode of the 12AX7 was 36.5V. When calculating the actual current for that half of the triode would I use the resistors in series in this case?
yes
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Thanks. I assumed so, but not having seen a configuration like that I wanted to be sure.
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Thanks. I assumed so, but not having seen a configuration like that I wanted to be sure.
It's very common. You see it in the cathodyne PI like your GA18T. It's also in the LTP PI.
Why are you calculating the cathode current in that circuit?
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I’m not super familiar with different designs. I was just curious where the 12AX7 was running at, and thought it was curious that there was no bypass cap there.
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I’m not super familiar with different designs. I was just curious where the 12AX7 was running at, and thought it was curious that there was no bypass cap there.
The cathodyne PI has one signal output taken from the plate and another output taken from the cathode. There cannot be a cathode bypass cap because that cap would kill the signal on the cathode. Read about cathodynes here...
https://www.valvewizard.co.uk/cathodyne.html
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... not having seen a configuration like that ...
Concertina = Cathodyne = Split Load ---> phase inverter type
A 12AX7 often has a 100kΩ load resistor in the plate circuit.
A "split-load hase inverter" breaks that load resistor into 2 parts, then puts one in the cathode circuit and one in the plate circuit. 100kΩ / 2 = 50kΩ =~ 47kΩ
This circuit has no gain, but the single tube-current flows through both load resistors. That means 2 equal outputs are developed by this single-current flowing through 2 equal loads. The voltage developed across each load resistor is opposite because with a given direction of current, we use opposite sides of each load as the "reference" for the voltage signal.
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Thanks to the both of you for the info and direction. I’ve only built a few basic amps over the years, so plenty to learn.
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The voltage developed across each load resistor is …
… the same for each resistor (because the resistance of each resistor is the same and) because voltage = current x resistance. :-)
The signal at the cathode follows the signal at the grid, whereas the signal at the plate is inverted. And seeing as how both voltages are equal, and the signal phase is inverted, gain is cancelled out.