help understanding split load PI's

[jdandry]

I'm working on a tweed Gibson amp and was wondering how this split load works and more pointed, how the values of the resistors around the tube effect the tone. Right now there is a 470K/1.5k/47k combo. I've noticed on a Tweed Deluxe the combo is 1M/1.5k/56k combo. What difference does this produce as far as driving the next stage (6V6's) and the gain of the stage (the tube distorting faster or slower). Thanks for any comments!- John

[Lanesplitter]

I would recommend keeping the couping cap at the PI input very small. The connection between the grid and cathode is a form of feedback, and gives the PI a very large input impedance of several Mohms. Too big a capacitor and you'll get a wonky bass sound. F3dB = (2piRC)^-1.

[jdandry]

I was wondering about resistor values. I did the Paul C mod last night, haven't played it yet. Bye the way, the amp is a Gibson GA-18t. On the original schematic, I guess I'm just wondering how the 470k/47k/1.5k and the 47k plate on the PI interacts with each other.

[Lanesplitter]

The Paul C Mod lowers the input impedance of the Phase Inverter by putting the 1Mohm resistor to ground. People claim that this increases the headroom, but it actually only decreases the bass by lowering the 'R' in the 'RC' time constant. This same effect could be achieved by lowering the value of the coupling capacitor.

[donnyjaguar]

Are you sure you are looking at a split load phase inverter?
These will have the same value resistor on the cathode and anode.
That doesn't sound like what you're describing.

[RedMan]

Are you sure you are looking at a split load phase inverter?
These will have the same value resistor on the cathode and anode.
That doesn't sound like what you're describing.


I think he's talking a 47K for the plate and a 47K for the cathode. A split load PI uses the small value resistor (1.5K or so) on the cathode to bias the tube correctly via the grid resistor. There's really no feedback there since the cathode is in phase with the grid and they're almost exactly the same voltage level. But the cathode resistor is of course unbypassed

[donnyjaguar]

Fair enough. I always considered this to be paraphase though. Cheers! DJ

[Wakarusa]

Fair enough. I always considered this to be paraphase though. Cheers! DJ

The other name for split-load is cathodyne. A paraphase uses two triodes with the non-inverting side driven off a voltage divider from the inverting side's output (so the non-inverting is really a double inversion). A 5D4 Super is a good example.