Lifting 12ax7 cathode

[Chuck H]

Sure it would. Not even using charts or formulas I know that a cathodyne type PI with a pair of 100k resistors will have something like 25% of the B+ on it's cathode. So for a B+ node with 250V you'll have about 50V on the cathode with a 100k plate and a 100k cathode. With the grid remaining at 0V I think. So if you figure for a 220k cathode, plate impedance not withstanding (I think) since we're dealing with DC, and all else being equal, I guess we'll have about 110V on the cathode? Not too high at all. But what if we have a 400+HV node? Couldn't that exceed the heater to cathode differential? With the plate also above max. And the grid at 0V but of a high impedance.

[trobbins]

Best to measure across the resistors, as the meter resistance may only be 1Meg, and even if it was 10Meg then it could easily 'colour' the voltage one could measure across a high impedance anode-cathode. But essentially the values will be on the loadline.

[J M Fahey]

Sure it would. Not even using charts or formulas I know that a cathodyne type PI with a pair of 100k resistors will have something like 25% of the B+ on it's cathode.

I beg to differ.
A cathodine may have the cathode at 50/100/200V from ground, yes, but with the grid also around 50/100/200 above it.
A cathode follower too.
It's not what we are discussing here: a "regular" tube gain stage, with its grid referred to ground through a high value resistor (100k to 1 M) which operates normally and then gets its cathode switched open.

So for a B+ node with 250V you'll have about 50V on the cathode with a 100k plate and a 100k cathode. With the grid remaining at 0V I think.

Well, think again.
50V into 100K mean 0.5mA .
Do you think a 12AX7 will pass 0.5mA with -50V on its grid?

So if you figure for a 220k cathode, plate impedance not withstanding (I think) since we're dealing with DC, and all else being equal, I guess we'll have about 110V on the cathode? Not too high at all. But what if we have a 400+HV node? Couldn't that exceed the heater to cathode differential? With the plate also above max. And the grid at 0V but of a high impedance.

Same considerations.
Here you assume 110V into a 220K cathode resistor, still 0.5mA, grid at 0V from ground but -110V from cathode.
No way.
That tube will be absolutely switched off under those biasing conditions.

Yes, I know it can get confusing at first sight, because we jump to compare this with other different circuits ... but similarity is only superficial; what really matters is grid to cathode voltage .
Since we are not trying to amplify (or oscillate or whatever) but to cut this tube off, we need to know it's cutoff bias voltage.
Which can be deduced from the datasheets, but even without doing so, I think it will be in the order of 10V .
Of course, for a power tube such as 6L6, we know the cutoff voltage is over 50V.
Even in that case (such as the Mesa Mark I 6L6 cathodes switched off), the grids *continue* having the around -50 V bias voltage, so to get into cutoff the cathodes will not need to rise *that* much above ground, since we already have those around -50V working for us.
That's why rather than jumping to conclusions through analogies , I suggest somebody simply measure actual cathode voltage and post it here.
Yes, I think a 1M input impedance multimeter will do.