PI questions

[Ghetto_Soundwave]

Just finished up a push-pull 6LY8 build http://www.digikey.com/schemeit/#cpz and have a few questions. I based my build on this amp http://paulrubyamps.com/Schematic6gw8.gif Hammond organ conversion by Paul Ruby http://paulrubyamps.com/6gw8.html Unlike Mr. Ruby's amp I chose to leave out the 1K grid resistor that is between the Vol pot 3rd lug (typically the ground lug) and the 2.7K + 100K split at the PI cathode.

1) Is the 1K there just for protection in case the pot fails?

2) Mine uses a 500KA pot.  If I inserted a 500K resistor where the 1K resistor is in Mr. Ruby's amp would it effectively make it a 1 Meg in total load resistance?

3) Using 100K resistors for the PI's plate and cathode versus using 56K's increases or decreases gain?

I am happy how this amp turned out so far. I wanted a clean amp 5 watt amp that maximizes output tube distortion, is linear throughout output and distorts from 3/4 volume onwards. This amp hit those marks and doesn't (need) use grid stoppers, doesn't swirl or swoosh. One con is it doesn't sag/compress as much as I wanted so I may insert a higher ohm sag resistor and/or adjust PSU setup with smaller uF caps / higher resistor values.

[stratele52]

Good to know digikey schemeit .

[PRR]

> increases or decreases gain?

No. (The cathodyne is unity-gain.)

> If I inserted a 500K resistor where the 1K resistor is

You couldn't turn-down to zero.

The value of that pot is NOT critical.

[HotBluePlates]

2) Mine uses a 500KA pot.  If I inserted a 500K resistor where the 1K resistor is in Mr. Ruby's amp would it effectively make it a 1 Meg in total load resistance?

Because the grid resistance represented by your 500kΩ pot is bootstrapped in the split-load inverter, the apparent impedance presented to the previous stage is already several-Megohms. We could prove that with some math, but take my word it's already high enough that it's not a factor in decreasing the gain of the first stage.

What I mean is that while you can calculate a loadline for the 220kΩ plate load of the first stage, the total effective load with an a.c. signal is the 220kΩ plate resistor in parallel with whatever the input impedance is of the following stage. Since the bootstrapped split-load inverter input is already several-Meg, it has almost no effect of reducing the apparent load resistance of the first stage. Therefore, nothing to be gained by trying to increase the input resistance of the inverter.

If I inserted a 500K resistor where the 1K resistor is in Mr. Ruby's amp ...

What's the volume like when you set your Volume control halfway up? If you add the 500kΩ resistor, that will be the volume you get when you turn your Volume control all the way off.

One con is it doesn't sag/compress as much as I wanted so I may insert a higher ohm sag resistor ...

That's one brute-force way of doing it.

Or, you could add screen resistors to the output tubes. Experiment with values, and make them larger until you get the sag you want at a predetermined volume level.

How it works:
If screen voltage drops, the tube passes less plate current. Output stages designed for maximum clean power want the screen voltage rock-steady to get every last watt. Because screen current typically increases with increased output power (either a little or a lot more screen current depending on tube type), series screen resistance is generally held to a minimum, because Ohm's Law says the screen voltage would drop with increased screen current.

You can purposely make screen resistance over-size to cause screen voltage sag with increasing output, which you'll hear/interpret the same as sag from a rectifier tube, small power supply, series B+ resistor, etc. The advantage is you don't waste B+ power in a small-value high-wattage resistor (which just makes the chassis hot).

The bigger you make the screen resistor, the lower power output (and apparent volume level) at which the sag effect will kick in. Remember, even very saggy amps have some lower-volume clean range where there's no obvious sag.

Try it!

[Ed_Chambley]

Or, you could add screen resistors to the output tubes. Experiment with values, and make them larger until you get the sag you want at a predetermined volume level.

How it works:
If screen voltage drops, the tube passes less plate current. Output stages designed for maximum clean power want the screen voltage rock-steady to get every last watt. Because screen current typically increases with increased output power (either a little or a lot more screen current depending on tube type), series screen resistance is generally held to a minimum, because Ohm's Law says the screen voltage would drop with increased screen current.

You can purposely make screen resistance over-size to cause screen voltage sag with increasing output, which you'll hear/interpret the same as sag from a rectifier tube, small power supply, series B+ resistor, etc. The advantage is you don't waste B+ power in a small-value high-wattage resistor (which just makes the chassis hot).

The bigger you make the screen resistor, the lower power output (and apparent volume level) at which the sag effect will kick in. Remember, even very saggy amps have some lower-volume clean range where there's no obvious sag.

Try it!
[/quote]
Is there a point at which increasing screen resistance will cause the power tube to cease current flow.  In other words, can the screen become a wall?  I like the idea and have never tried it, but will like to experiment.

[HotBluePlates]

Is there a point at which increasing screen resistance will cause the power tube to cease current flow.  In other words, can the screen become a wall?  I like the idea and have never tried it, but will like to experiment.

Yes. If you could make the voltage at the screen nearly equal to the cathode voltage.

Let's say you've figured that at your proposed idle conditions, the screen passes 5mA and is at 400vdc (relative to the cathode).

To get the screen voltage to equal the cathode voltage, the screen resistor would have to drop 400v. So 400vdc/5mA = 80kΩ. Therefore if you use an 80kΩ screen resistor, the tube current should be shut off.

Realistically though, you won't be able to turn tube current off completely with a too-big screen resistor. Why? If there is no tube current, there is also no screen current, and no way for the screen resistor to drop voltage. With no voltage drop across the screen resistor, screen voltage must be higher than 0v, and the tube would have to pass some amount of current.

Compare to what you know about cathode biasing and a cathode follower. Thinking simply, if you increase the cathode resistor, tube current decreases. But the same situation exists in that a too-big cathode resistor has a hard time turning off tube current. if there's no current, there's no voltage across the cathode resistor to turn off tube current. So while 1kΩ might be normal biasing for a preamp tube, even a 100kΩ cathode resistance doesn't shut off tube current.

Next you'll point out the cathode follower doesn't have a resistance in the plate circuit, and that an open-circuit is infinite resistance, so therefore some extremely-large resistance must turn off tube current. And you'd be right on both counts; however, you're now way, way beyond any resistance you're likely to use in series with your screen.

So give it a try. A reasonable upper bound might be 5kΩ or 10kΩ. The lower bound would be 0Ω (a piece of wire). I think Tubenit has found that around 3kΩ often works in his amps to induce some sag without killing all power output.