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Need help understanding Aikido Preamp

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D

dromichet

Member
Joined 2021
#1
Hello:

A while back I've built an Aikido preamp using 6N2P and 6N1P tubes. This preamp has a lot of gain which is desired since it's used with a custom DAC output of about 75mV. I've played around with different Rk values as well as several different octal tubes and come back to this tube combination. For my system it sounds the best and I have no complaints.

My problem is that I don't fully understand the schematic as it's not the traditional tube schematic.

When trying to establish operating points of the tube in the first stage, do I use V3.1 or V3.2? Does tube V3.1 act as the plate resistor to V3.2?

Thanks
AikidoCircuit.JPG
 
#3
I've read his article as well as the documentation but I believe my question is more basic as in how does the first stage compare or translate to a standard triode schematic.

Do I not have to worry about calculating the working point of the triodes? If yes, which triode do I go by, top or bottom. One will have 290V at the plate and the second will see 145v at its plate. They both will have same cathode resistor and same current will flow through them. Looking at he 6N2P curve graph that puts me on 2 different spots.

Am I seeing this totally wrong? Basically I need to understand how to select a working point for these tubes in this schematic.
 
#4
The circuit is symmetric, so the upper and lower sections will split the B+ equally. For each, Vpk = 145V.
Remember the upper section's grid is referred to the center node, not to ground.

So just find the plate current, using the 390R cathode resistor, at half of the B+ (145V) as the Vpk,
use the equations Vgk = - Ip x 390R, and Vpk = 145V, and find the intersection of the two lines.

Plot Ip for each of the grid lines on the characteristic curves ( Vgk = 0, -1, -2, -3, etc.), and draw a line through them,
finding where that line intersects the vertical line, Vpk = 145V. That is the Q point for the current through the tubes.
 
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#9
The stage is push-pull , plain and simple . If you use power tubes you will have the simplest SEPP stage used in cheap OTL ( 800ohm ) TVs and radios . Of course for hi-fi usage they abandoned the self inverting concept and used a dedicated phase inverter , like in any OTL modern schematic
 
#10
Maybe this is just a language usage issue, but no polarity inversion happens in the load; both the load's grid and cathode are in the same signal polarity as the active stage's anode. As a two terminal device it could not invert polarity without having negative resistance, could it? And, it doesn't invert anything, so ?

All good fortune,
Chris
 
#11
If the lower device is drawing more current ( on the positive cycle of the sinewave , grid more positive ) , the upper will have more negative bias across the grid-cathode resistor , grid more negative , so it will conduct less . And of course the other way around . This is self inverting mechanism , in principle exactly the same as if the the two grids would be driven by a dedicated phase inverter .
 
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#12
Let's agree that the active load can be a larger resistor than a simple ordinary resistor with same static terminal Volts and Amps. We can just disagree about polarity inversion; just a matter of convention and where one places their scope probes. Your point that a differential scope across the "top" grid to cathode would show a polarity inversion is a good perspective. I would only argue that it is washed away as a two terminal device, but that's only a different perspective.

All good fortune,
Chris
 
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