Electronics > Projects, Designs, and Technical Stuff

First fire up of a 6BK4 pulser

(1/2) > >>

Qmavam:
Hi all I have the following circuit.

I have applied a 150V B+ and have a squarewave input to drive the grid of +2V, -10V.
The output is a 48V pulse riding on top of 60Vdc.
I can go more positive on the grid, but it makes no difference on the output waveform DC level.
The plan is go to 8kV. Is 150V B+ enough to make this circuit act the way it will at 8kV?
Any ideas why I have the DC level on the output?
I would experiment with higher voltages to see is I stay at a 60 V DC level, but I'm waiting for parts.

Thanks, For your input, Mikek

TimFox:
The upper level of the output is when the tube is "OFF", meaning cut off and passing negligible plate current.
The lower level on the output pulse is when the tube is "ON", meaning passing its maximum current.
If you look at the published curves for triode plate current vs. plate-cathode voltage (set for different grid voltages), you see that with negative or zero grid voltages there is a limit to how much current you can get at low plate voltage.
What you need for calculations is to place the appropriate load lines (including the one due to the resistance in the cathode that affects the grid-cathode voltage) on the plate curves.
Unfortunately, for these tubes (which were designed to operate at constant plate voltage in shunt regulator operation), the only published curves are for constant plate voltage (rather than grid voltage).

By the way, you should note that the first-approximation grid-cathode cut-off voltage is the plate-cathode voltage divided by the amplification factor mu, which is very high for this class of tube, nominally 2000.
Therefore, if you have only 150 V on the plate, the tube will be cut off at very low grid voltage.

Qmavam:

--- Quote from: TimFox on June 04, 2023, 04:47:46 pm ---
What you need for calculations is to place the appropriate load lines (including the one due to the resistance in the cathode that affects the grid-cathode voltage) on the plate curves.
Unfortunately, for these tubes (which were designed to operate at constant plate voltage in shunt regulator operation), the only published curves are for constant plate voltage (rather than grid voltage).
--- End quote ---

I think that means, raise the B+ and see what happens?

--- Quote ---By the way, you should note that the first-approximation grid-cathode cut-off voltage is the plate-cathode voltage divided by the amplification factor mu, which is very high for this class of tube, nominally 2000.
Therefore, if you have only 150 V on the plate, the tube will be cut off at very low grid voltage.

--- End quote ---
I'm trying to understand, plate to cathode = 8kV / 2000 = 4v grid-cathode cut-off voltage, Is that correct? Wouldn't it be -4V?
I still don't understand, will the DC level to pulse level ratio get better at higher B+ voltages?
The LTspice sim shows about 200Vdc with a 7kv pulse.
This my first tube circuit, I'm a better builder than my electronics knowledge! I've had a lot of help to get this far.
Thanks, Mikek

TimFox:
Yes, of course the cut-off grid-cathode voltage is negative.
For an ideal triode, the plate current is given by

IP = G x (VPK/mu + VGK)3/2  where mu is the amplification factor (electrostatic screening of grid) and G is the "perveance" of the tube due to its geometry (electrode spacing and area), in the normal operating condition of "space-charge limit".

so the solution for   IP = 0  is  VGK = - (VPK/mu), independent of G .

Where did you get a Spice model for the 6BK4?
If you keep all the resistors the same, the low-voltage high-current case won't change much but the high-voltage zero-current case will go up with Vbb .

Qmavam:

--- Quote from: TimFox on June 04, 2023, 05:42:20 pm ---
Where did you get a Spice model for the 6BK4?
--- End quote ---

People on the DIY audio forum, have it, they ran all the sims for me.
The subckt is in the 5th post, here:

--- Quote ---If you keep all the resistors the same, the low-voltage high-current case won't change much but the high-voltage zero-current case will go up with Vbb .

--- End quote ---
Sorry, I changed the load for the reduced B+. 150Vdc. (Load was 510kΩ for previous comments.)

Restart;
I now have 180kΩ dropping resistor and a 180kΩ load resistor.
So if I remove the Plate cap, the voltage divider gives me 75vdc where the resistors connect. Then I reconnect the cap and with grid drive I have a 25V pulse riding on 49Vdc.
The current is just under 0.5ma.
When the circuit is at full voltage, I'm expecting at least 1.3ma.
I expect a few changes to the diagram I posted. One, I found I need to raise my power supply to 12kV, the dropping resistor will be changed to 6MΩ and the load resistor is 12MΩ. The gives me 8kV on the load when the tube is off.
Should I go ahead and adjust the resistors for 1.3ma? (at 150Vdc)
Or just wait until I receive my P.S. parts so I can start raising the voltage?
Thanks, Mikek

P.S. Or, is there a flaw with the design?