scope CRT blanking, capacitive coupling

[Martinn]

Hi all,

I am working on a scope clock. I started with the deflection amplifier by TurboTom https://www.eevblog.com/forum/projects/maybe-risk-a-guess-what-thats-gonna-be/msg1810052/#msg1810052 which works nicely, HV supply also should work (boost stage+cascade) and I think I understood astimatism and focusing correction.

Remains the control grid and blanking. Controlling blanking is a bit nasty as cathode and control grid are typically 1-1.5 kV away from the deflection amplifiers. There are many solutions for this like optocouplers (+extra supply) or 3 kV capacitors combined with a brightness potentiometer.

Now here is what I am struggling with: TurboTom (see link above) couples his blanking amplifier to G1 (control grid) via a (1 nF 1 kV) capacitor - without any DC reference (like usually 1M to a brightness potentiometer). So my question to the vacuum tube gurus:

What potential would result on a purely capacitively coupled grid? The datasheet specifies -50 V for cutoff and 10 V for normal 10 uA beam current, so with AC coupled drive the potential should somehow land in there.
What is the current into the grid? How much of the beam from the electron gun lands on the control grid? I attached the datasheet of the tube I'd like to use (D10-160).

EDIT I just did an experiment. With only the heater connected (no HV, no other connection - not risking to fry my instruments by a kV spark) I connected the cathode and G1 to a power supply. With G1 positive, current drops from a few uA to 0 (hurray! We just discovered a vacuum diode). With G1 at +2 V w.r.t cathode, current into G1 is 2 mA. 
Of course this is without the main acceleration voltage so I wonder how many electrons make it out there. But as beam current is rated at 10 uA, it can't be much. I wonder what current flows into G1 in regular operation?

Thanks - Martin

[TimFox]

I have never considered floating a CRT grid, but I it might float negative (with respect to the cathode) until the cathode current was cut-off.
Unless there is secondary emission from the grid, in which case it floats positive until something blows up.
That's why a resistive circuit, including a brightness pot, is a better idea.

[Martinn]

I have never considered floating a CRT grid, but I it might float negative (with respect to the cathode) until the cathode current was cut-off.

Could the potential be somehow self adjusting? On a floating grid, I assume electrons would accumulate on the grid, driving the potential that far into the negative region until the current is cut off? Could this mechanism work as self-biasing?
After all, I believe Tom's solution works.

EDIT: Another experiment. With the heater active (and nothing else), I connected my DMM to cathode and grid 1. DMM set to high impedance (datasheet says > 10 GOhms) grid voltage stabilizes at around -2 V.

[TimFox]

I have never considered floating a CRT grid, but I it might float negative (with respect to the cathode) until the cathode current was cut-off.

Could the potential be somehow self adjusting? On a floating grid, I assume electrons would accumulate on the grid, driving the potential that far into the negative region until the current is cut off? Could this mechanism work as self-biasing?
After all, I believe Tom's solution works.

EDIT: Another experiment. With the heater active (and nothing else), I connected my DMM to cathode and grid 1. DMM set to high impedance (datasheet says > 10 GOhms) grid voltage stabilizes at around -2 V.

That -2V is similar to what one would get with a high (> 1 megohm) resistor from grid to cathode on, say, a 12AX7 triode (often called "contact-potential" biasing in small-signal audio).  From your data sheet, -2V is not sufficient to cut off the CRT, but will cut off a high-mu triode at reasonable plate voltage.
It might well work, but I haven't done that with a high-voltage tube.
If the signals through the capacitor are positive pulses with a very low duty factor, then the DC (mean) level on the grid will stay at high negative potential (black), but going positive on the peaks to get local white.

[Martinn]

If the signals through the capacitor are positive pulses with a very low duty factor, then the DC (mean) level on the grid will stay at high negative potential (black), but going positive on the peaks to get local white.

As it's a scope clock I'd assume the opposite - long drawing periods with short moves to another starting point (so high duty factor). Tom uses a different tube, a B7S2 where g1 cutoff voltage  is specified at -25...-60 V (quite similar to the -50 V of my tube). Interestingly he designed a blanking amplifier with adjustable output from 0-250 V (but maybe just because no other voltage was around).
So it would be interesting to find out what potential g1 stabilizes at.
I was thinking maybe I just follow your recommendation and add a brightness potentiometer setting the bias. There's jsut one nasty side effect of that: As the cutoff voltage is even more negative than the cathode (being already at -1.5 kV), I'd need an extra supply on top of that (maybe one more multiplier stage) just for driving that brightness pot.

Hameg added an extra 30 V winding on the mains transformer on their HM203 scope (IIRC) just for the blanking and brightness circuitry (of course they have more functions in there, Z modulation input and an optocoupler). Maybe just letting the grid float would be a workaround to avoid that extra supply.

[TimFox]

Could you put a Zener diode in series with the cathode so that you have a more negative voltage at the other end of the Zener, avoiding another transformer winding?

[Martinn]

Could you put a Zener diode in series with the cathode so that you have a more negative voltage at the other end of the Zener, avoiding another transformer winding?

Thanks, sounds like a good idea! I'd have to add some buffering for when the trace is off. The question is what voltage is needed (is it really 50 V) to blank the beam completely? I'd rather not try this out with my lab supply (I doubt it would appreciate having 1.5 kV between its channels). Seems I need another test PCB to try this out.
Do you think the CRT might be damaged if g1 potential is too high? I wonder why the grid can be pushed to +10 V at all - without acceleration voltage, at 2 V I already get a current of 2 mA into the grid.
Can vacuum tubes act like a diode if the grid is biased too high (sorry, total tube newbie)? Probably the grid is always at some safe working point (like the -2 V you mentioned for the 12AX7).

[TimFox]

Yes, if you bias the grid positive with respect to the cathode, the grid-cathode diode will conduct in the same way as a small vacuum diode (6AL5, 6H6, etc.).
Some tubes were intended for such operation, and their data sheets show the I-V curve for the grid.
Grids are fragile, and should not be subjected to high currents:  the data sheets often have maximum ratings.