First fire up of a 6BK4 pulser - Page 1
Forum section: electronics / projects

author: Qmavam
author_rank: Regular Contributor
timestamp: June 04, 2023, 03:46:56 pm
content: 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

author: TimFox
author_rank: Super Contributor
timestamp: June 04, 2023, 04:47:46 pm
content: 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.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 04, 2023, 05:27:29 pm
content: Quote from: TimFox on June 04, 2023, 04:47:46 pmI think that means, raise the B+ and see what happens?QuoteI'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

author: TimFox
author_rank: Super Contributor
timestamp: June 04, 2023, 05:42:20 pm
content: Yes, of course the cut-off grid-cathode voltage is negative.For an ideal triode, the plate current is given byIP=Gx (VPK/mu+VGK)3/2wheremuis the amplification factor (electrostatic screening of grid) andGis 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 forIP= 0 isVGK= - (VPK/mu), independent ofG.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 withVbb.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 04, 2023, 06:51:06 pm
content: Quote from: TimFox on June 04, 2023, 05:42:20 pmPeople on the DIY audio forum, have it, they ran all the sims for me.The subckt is in the 5th post, here:https://www.diyaudio.com/community/threads/hv-pulse-amplifier-help-6bk4.399314/QuoteSorry, 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, MikekP.S. Or, is there a flaw with the design?

author: TimFox
author_rank: Super Contributor
timestamp: June 04, 2023, 07:44:01 pm
content: Go to the actual datasheet for the 6BK4 or 6BK4C:https://frank.pocnet.net/sheets/107/6/6BK4C.pdf1. Note that the absolute maximum plate current rating is 1.6 mA.2. With an 180 k\$\Omega\$ resistor fromVbbto the plate, and another 180 k\$\Omega\$ resistor toVbbfrom plate to ground gives a Thévenin-equivalent circuit of 90 k\$\Omega\$ fromVbb/2 to the plate.3a. Use your Spice model to do a .DC (Spice syntax) analysis of that model with this simple load, varying the grid voltage from past cut-off to zero.3b. Or, estimate the perveanceGfrom the graphs in the datasheet and use the ideal equation in my earlier reply. Do the algebra with that equation to see what grid voltage is required to obtain the plate current, when the plate voltage isVP=Vbb/2 -IPx 90 k\$\Omega\$ (or the equivalent circuit for other values of resistors). See what you can obtain withVgk= 0. Simple algebra gives youImax=Gx (VPK/mu)3/2withVgk= 0

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 04, 2023, 09:22:40 pm
content: What do you mean by "pulse" here? What duration, rise/fall time, load resistance (or capacitance), etc.?Datasheet doesn't specify low Va region so it's a guess how low it can actually go, or how high Ia can peak at. Likely it's not much more than a couple mA, even with Vg > 0. Likely "hFE" (Ia/Ig) is less than 1, given the very high mu ( = tightly spaced grid wires), so don't expect much if any improvement for higher Vg (which so far is the observation).There's also types like 6HV5, which I tested to a couple amperes peak; they're as underwhelming in pulsed operation as they are steady-state. They only go up to 6 or 8kV, somewhere around there, of course.Tim