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Old 25th May 2011, 04:45 AM   #1
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Default 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?

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Old 25th May 2011, 06:34 AM   #2
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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.

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

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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.
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Old 25th May 2011, 06:39 AM   #3
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read the valvewizard on this
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Old 25th May 2011, 02:14 PM   #4
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Thanks, actually the valve wizard article was what got me thinking....
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Old 25th May 2011, 03:35 PM   #5
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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.
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Old 25th May 2011, 03:41 PM   #6
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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....
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Old 25th May 2011, 04:13 PM   #7
DF96 is offline DF96  England
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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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