simple question about cathodyne PI

hpupo · 25th May 2011, 04:45 AM

i still dont fully understand the cathode follower (an thus the cathodyne PI), but i am not looking for an explanation on this thread. My question is: since the gain is almost unity, does it mean i am limited to more or less 4Vpp on the input (for a 12ax7)? Or the whole essense of this circuity is that it allows the input voltage to go higher (of course to the limitations of voltage swing available for both resistors). lets say i have, for example a V supply of 300V, which half would be for the tube (150) and half divided on the 2 resistor (75V), could actually (assuming correct bias) input 75Vpp on the tube grid?

Thanks

artosalo · 25th May 2011, 06:34 AM

Quote:

My question is: since the gain is almost unity, does it mean i am limited to more or less 4Vpp on the input (for a 12ax7)?

No. The input voltage swing can be essentially higher since the cathode potential "follows" the the grid voltage variations.

Quote:

Or the whole essense of this circuity is that it allows the input voltage to go higher

Exactly so.

Quote:

lets say i have, for example a V supply of 300V, which half would be for the tube (150) and half divided on the 2 resistor (75V), could actually (assuming correct bias) input 75Vpp on the tube grid?

Yes. The typical available output swing (peak-to-peak) of a cathodyne phase splitter is some 0,25 x supply voltage.

aardvarkash10 · 25th May 2011, 06:39 AM

read the valvewizard on this

hpupo · 25th May 2011, 02:14 PM

Thanks, actually the valve wizard article was what got me thinking....

Palustris · 25th May 2011, 03:35 PM

If you look at the example on the valvewizard web site you will see the plate curves for a 12AX7. As it says, input voltage swing will be from 0V to 4V peak to peak. That's it. Any further input will cause the tube to go into cutoff and saturation.

The maximum input voltage to the cathodyne will be the difference in voltage between the cathode and the grid. In the example above it is 2V peak or 1.4VRMS.

Don't confound grid voltage and grid to cathode voltage.

hpupo · 25th May 2011, 03:41 PM

Ok, now im confused again. As palustris said, let's say the max Vpp is in fact 4V. how can we drive a power tube that would require at least 10Vpp? One could argue that we could use a driver in between, but all the circuits ive seen have a voltage amplifier PROCEEDING the cathodyne. Actually that's what the valvewizard says on the article....

DF96 · 25th May 2011, 04:13 PM

No, voltage swing on the grid with respect to the cathode is 4V. Voltage swing on the grid with respect to ground is much higher. If post #5 confuses you, ignore it. It confuses me too! He has confused grid voltage (wrt ground) and grid to cathode voltage, the very thing he says he is not confusing.

25th May 2011, 04:45 AM	#1
hpupo diyAudio Member Join Date: Feb 2011	simple question about cathodyne PI i still dont fully understand the cathode follower (an thus the cathodyne PI), but i am not looking for an explanation on this thread. My question is: since the gain is almost unity, does it mean i am limited to more or less 4Vpp on the input (for a 12ax7)? Or the whole essense of this circuity is that it allows the input voltage to go higher (of course to the limitations of voltage swing available for both resistors). lets say i have, for example a V supply of 300V, which half would be for the tube (150) and half divided on the 2 resistor (75V), could actually (assuming correct bias) input 75Vpp on the tube grid? Thanks

25th May 2011, 06:39 AM	#3
aardvarkash10 diyAudio Member Join Date: Mar 2007 Location: Auckland, NZ	read the valvewizard on this __________________ Yes, conservatism thrives on low intelligence and poor information. But the liberals in politics... continue to back off, yielding to the supremacy of the stupid. It's turkeys all the way down. - George Monbiot, guardian.co.uk, 6 Feb 2012

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25th May 2011, 02:14 PM	#4
hpupo diyAudio Member Join Date: Feb 2011	Thanks, actually the valve wizard article was what got me thinking....

25th May 2011, 03:35 PM	#5
Palustris diyAudio Member Join Date: Aug 2006 Location: Cape Cod	If you look at the example on the valvewizard web site you will see the plate curves for a 12AX7. As it says, input voltage swing will be from 0V to 4V peak to peak. That's it. Any further input will cause the tube to go into cutoff and saturation. The maximum input voltage to the cathodyne will be the difference in voltage between the cathode and the grid. In the example above it is 2V peak or 1.4VRMS. Don't confound grid voltage and grid to cathode voltage.

25th May 2011, 03:41 PM	#6
hpupo diyAudio Member Join Date: Feb 2011	Ok, now im confused again. As palustris said, let's say the max Vpp is in fact 4V. how can we drive a power tube that would require at least 10Vpp? One could argue that we could use a driver in between, but all the circuits ive seen have a voltage amplifier PROCEEDING the cathodyne. Actually that's what the valvewizard says on the article....

25th May 2011, 04:13 PM	#7
DF96 diyAudio Member Join Date: May 2007	No, voltage swing on the grid with respect to the cathode is 4V. Voltage swing on the grid with respect to ground is much higher. If post #5 confuses you, ignore it. It confuses me too! He has confused grid voltage (wrt ground) and grid to cathode voltage, the very thing he says he is not confusing.

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