I recently built a 5e4a tweed super. As part of the build I reversed the triodes on the pi tube installed a 12dw7, lowered the plate load, and cathode resistor to 22k, and lastly change the pi to fixed bias using a 2.2m resistor off the pi node of the ps. While the amp sounds really good I think the pi still has a little more to give. Here are my voltages on the pi.
Plate 201.2
Cathode 95
Grid 87.1
Optimizing the cathodyne pi
Moderators: pompeiisneaks, Colossal
Re: Optimizing the cathodyne pi
I think your question isn't really that clear with respect to what you want to get done. Maybe there is something useful from Merlin here: http://www.freewebs.com/valvewizard/cathodyne.html
Re: Optimizing the cathodyne pi
I am concerned that my pi is not properly biased. I have read the vw site over and over and I am a little confused.
Re: Optimizing the cathodyne pi
It might help to post more complete information. For example, I'm assuming you are using the AU7 side of the 12DW7, but I think your reference to it is ambiguous. You didn't post the value of the plate load resistor or the B+ voltage. All these things can play into diagnosis. I suggest posting a full schematic with voltages if you can. Even if you take a stock schematic and do a few cross outs, that could really help. If not, then a hand drawn schematic of the PI with voltages would be OK, too!
Are you hearing something you don't think is right, or what? Try to give us more to go on.
The AU7 half, IMO (I don't know much; take my opinion with a grain or two of salt), is a good choice for a cathodyne inverter, which is more about current than voltage. With that said, to see if it's balanced (not actually a bias issue, per se) the current on the plate and the cathode should be about the same. Just use Ohm's law. Check the voltage drop across the plate load resistor and the cathode resistor. Do the math: I=V/R. Post the calculation.
Are you hearing something you don't think is right, or what? Try to give us more to go on.
The AU7 half, IMO (I don't know much; take my opinion with a grain or two of salt), is a good choice for a cathodyne inverter, which is more about current than voltage. With that said, to see if it's balanced (not actually a bias issue, per se) the current on the plate and the cathode should be about the same. Just use Ohm's law. Check the voltage drop across the plate load resistor and the cathode resistor. Do the math: I=V/R. Post the calculation.
Re: Optimizing the cathodyne pi
Ok
Plate resistor 22k with a 95vdc drop across it 4.31ma
Cathode resistor 22k with 95vdc drop across it 4.31ma
Grid voltage 87vdc
Yes I am using the au7 side of the dw7.
The reason I think it has more to give is I believe it is clipping prior to power tube saturation.
Plate resistor 22k with a 95vdc drop across it 4.31ma
Cathode resistor 22k with 95vdc drop across it 4.31ma
Grid voltage 87vdc
Yes I am using the au7 side of the dw7.
The reason I think it has more to give is I believe it is clipping prior to power tube saturation.
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gingertube
- Posts: 531
- Joined: Mon Nov 14, 2011 2:29 am
- Location: Adelaide, South Oz
Re: Optimizing the cathodyne pi
From the voltages you supplied the B+ for the PI must be 296V - call it 300V.
For that voltage the PI is currently running a little hot (too much current).
Remember that the anode and cathode voltages move toward each other as the grid signal goes positive - squeezing the voltage across the tube down, and away from each other as the grid signal swings negative.
For best operation of a cathodyne, the cathode voltage needs to be at about B+ divided by 5. You are currently running at about B+ divided by 3. That operating point severely reduces voltage across the tube on those positive signal peaks and low anode cathode voltage promotes high grid current which can disturb your operating point still further.
You want to get that cathode voltage down to say 60V. To do that you need to adjust the grid voltage to about 67 - 68 volts.
I'm also a bit worried about the "fixed biased by use of a 2M2 from The PI supply node" comment. You need a voltage divider (2 resistors) to set that grid voltage.
Cheers,
Ian
For that voltage the PI is currently running a little hot (too much current).
Remember that the anode and cathode voltages move toward each other as the grid signal goes positive - squeezing the voltage across the tube down, and away from each other as the grid signal swings negative.
For best operation of a cathodyne, the cathode voltage needs to be at about B+ divided by 5. You are currently running at about B+ divided by 3. That operating point severely reduces voltage across the tube on those positive signal peaks and low anode cathode voltage promotes high grid current which can disturb your operating point still further.
You want to get that cathode voltage down to say 60V. To do that you need to adjust the grid voltage to about 67 - 68 volts.
I'm also a bit worried about the "fixed biased by use of a 2M2 from The PI supply node" comment. You need a voltage divider (2 resistors) to set that grid voltage.
Cheers,
Ian
Re: Optimizing the cathodyne pi
Ian thanks the grid is sitting at a voltage divider 2.2m And 1m.
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gingertube
- Posts: 531
- Joined: Mon Nov 14, 2011 2:29 am
- Location: Adelaide, South Oz
Re: Optimizing the cathodyne pi
Jon,
should be easy to change that bottom resistor (lower value) in your 2M2 + 1M divider chain to get the cathode down to that B+/5 point.
B+/4 would probably be OK but B+/3 is not.
My "back of the envelope" calcs suggest that reducing the 1M at teh bottom of the bias divider to 680K will get you close eneough to the ideal bias point.
Cheers,
Ian
should be easy to change that bottom resistor (lower value) in your 2M2 + 1M divider chain to get the cathode down to that B+/5 point.
B+/4 would probably be OK but B+/3 is not.
My "back of the envelope" calcs suggest that reducing the 1M at teh bottom of the bias divider to 680K will get you close eneough to the ideal bias point.
Cheers,
Ian