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21st July 2007, 03:22 AM  #1 
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Join Date: Apr 2004
Location: Philly

My attempt: 5 element Pentode Spice model
This is one of my little projects. I am no spice guru, but I have gotten fairly savvy over the years. I have been frustrated by the lack of a true 5 element pentode model, which accurately reproduces the effect of the supressor grid on the plate transfer charactaristics. This would be helpfor for RF, modulation... and just plain tinkering.
I have successfully built up a 5 element macro, as well as a model of a 5 element pentode with a dummy supressor grid input (it is just a 7199 with two identical control grids). But I would like to get some accurate models together. If anyone can help me pound this into my brain... then please do. I also have had NO success finding data on the g3 grid for most tubes... but I have a few mutant tube g3 charts that I can use to cobble together a rough sketch. I am just going to walk through my logic from a diode model to a 5 element pentode model just in case it sparks any ideas. These are the simplest iterations that I could think of: just basic Child's law stuff. WHERE: Ip=plate current Ig2=screen grid current Ig3=supressor grid current µ=mu (amplification factor) K=perveance Kg1=perveance of control grid Kg2=perveance of screen grid Kg3=perveance of supressor grid Vpk=plate to cathode voltage potential Vgk=g1 (control grid) to catode voltage potential Vsk=g2 (screen grid) to cathode voltage potential Vg3k=g3 (supressor grid) to cathode voltage potential KVB=knee in the charactaristic curves @ VG=0 *DIODES: Ip=K*(Vpk)^1.5 *TRIODES: Ip=K*(Vpk+Vgk*u)^1.5 *TETRODES and PENTODES (with supressor grid tied to cathode): Ip=Kg1*((Vgk+Vsk/µ)^1.5)*(2/p)*arctan(Vpk/kVB) Ig2=Kg2*(Vpk+Vsk/µ)^1.5 This can be restated like this: Ip=(2/Kg1)(Vgk+Vsk/µ)^1.5*arctan(Vpk/KVB) Ig2=(Vgk+Vsk/µ)^1.5/Kg2 This should also work if we want to ditch the perveance: Ip+Ig2=K(Vgk+Vsk/µ1+Vpk/µ2)^1.5 As far as I know, the transfer functions just add up as long as out current is greater than zero, so maybe something like this would work: I=K(Vgk+Vsk+Vg3k/µ1+Vpk/µ2+Vg3k/µ3)^1.5 And then add the equations for G2 and G3. The mu of g3 should be swamped by the u of the others... but then again, I don't really know. What am I missing? Can anyone think of a similar way to get at the same thing, or maybe a different iteration using perveance?
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21st July 2007, 03:57 PM  #2 
diyAudio Member
Join Date: May 2005
Location: Stittsville, Ontario, Canada

multi grid tube models
Please excuse the length of this reply, you have pushed one of my buttons. For what its worth here are my comments and experience with this question :
"As far as I know, the transfer functions just add up as long as out current is greater than zero, so maybe something like this would work: I=K(Vgk+Vsk+Vg3k/µ1+Vpk/µ2+Vg3k/µ3)^1.5" Not quite correct, Vgk and Vsk should not have the same coefficient, after all the control grid is more sensitive than the screen grid. So Ik=K(Vgk/ug+Vsk/us+Vg3k/µ3+Vpk/µp)^1.5 should give the total cathode current. This is the equivalent diode model of a multigrid tube. It is based on the fact that electic fields add linearly. ( Note that this does not mean that currents add linearly, just the elctric field strength ) An electron coming off the cathode "knows" it is in a certain electic field but it has no way of "knowing " if it is due to one plate at a certain distance and voltage, or several plates at various distances and voltages. So assuming Ik =K(Vp)^1.5 was an accurate equation for diodes, the equation given above would be accurate for multigrid tubes. But there are 2 problems. 1) Ik= K(Vp)^1.5 is not a very accurate model even for simple diodes. Initial electron velocities, whether the cathode is cylindrical or rectangular, the ratio of the cathode diameter to the plate distance, and all kinds of stuff totally mess up the diffferential equations describing this stuff and make a simple mathematical solution impossible. 2) even if you do come up with an accurate equivalent diode model, you are still left with the problem of how to divide up the total current into plate current, screen current, and grid current. So what I have done for some of my own models is this : I observe that Korens triode models seem to work very well for triodes. I do not know exactly why, but they do. So I derive an equivalent triode model for the total cathode current of the pentode. I do this by entering real measured data into a spreadsheet and doing a regression of the data onto Korens equation using Solver. I dont know if data taken from spec sheet curves is accurate enough for this, maybe. But in any event I end up with a triode model for the total current. I find that these triode models are not too bad, not as accurate as Korens models generaly are for real triodes, but not bad. Where my method seems to fall down is in coming up with a current partioning formula, or how to divide the total current up into plate, screen, and grid currents. Nothing I have tried is very accurate, and despite everything I have tried, I keep coming back to the old arctan curve, which despite having no apparent basis in the physics of the situation, at least that I can see, seems to give the best result, of an admittedly poor lot. That being the case, what I generally end up doing is giving up on my "theory" based model, and going back to a model style based on Korens pentode model, with the "diode line" current division feature of Duncan Munros models added in, and tweaked for the specific tube. In other words, purely empirical. I dont know if that helps you, I hope it does not discourage.
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Robert McLean 
22nd July 2007, 12:24 AM  #3 
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Join Date: Apr 2004
Location: Philly

Thank you for the reply and especially the correction to the equasion.
I don't usually use simple child's law equasions for models... I was just trying to simplify the math as much as possible. I can never get Korean's equasions to converge properly in my simulator, so I usually use Rydel's equasions, but I can't figure out how to adapt them to include the supressor grid. If anyone else can help as well, please do!
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22nd July 2007, 03:57 AM  #4 
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Join Date: Dec 2001
Location: Hickory, NC

Once you have a workable triode model to calculate cathode emission, it should be extendable to multigrids by including Mu factors for each grid and summing since the E field is linear.
Then you need a current partition formula. Check Spangenberg's book on pages 225 to 227. For a simple model, equation 9.12 should work for the screen grid. Calculate current partition and adjust plate and screen current accordingly. The suppressor grid normally doesn't draw much current itself if kept negative or zero volts, but its affect on emission is partly thru returning plate bound current back to the screen grid. Only thing I can think of to calculate this effect would be to model the K,g1,& g2 as a new equiv. triode's Kathode, and the suppressor as the new equiv. triode g1. Difference between the original model plate current and the equiv. model plate current would then be what is additionally returned back to the screen grid. For positive suppressor grid voltage, equation 9.12 could be used on the original (kathodescreen) current to calculate suppressor current, which would then subtract further from the original calculated plate current. I haven't tried this. But it looks like a place to start anyway. Adjust the cdf parameters ("&" in formula 9.12 for screen and suppressor cases ) to see if the model can then match real tubes. Don
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If it isn't broken, take it apart and mod it. And then, what is the "speed of dark"? Hint: plug it in to find out. 
22nd July 2007, 05:14 AM  #5 
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Join Date: Dec 2001
Location: Hickory, NC

Notes:
When modeling the equivalent triode (K,g1 and g2 as the new Kathode, g3 as the new g1), keep in mind that voltages are now relative to the screen voltage. So a zero volt suppressor and +150 volt screen looks like a 150V g1 in the new equivalent triode. Plate voltage is relative to screen V also. I would also use a new transconductance factor for the suppressor in the new equiv. model, say gm5, rather than the gm3 used in the initial triode kathode current calculation. (one more parameter to adjust in the final model = a better chance of successful modeling) Similarly, the plate gets a new transconductance factor, gm6, for the equiv. model since tube geometry is different when centered on the screen grid as kathode. Gm5 and gm6 will have to be adjusted to give new model plate current somewhat less than the original model plate current. (the difference being the current returned to the screen grid portion) Spangenberg also gives some pointers on how to measure the partition factors for real tubes on page 226, useful for starting values to tweak around. So the final composite model should end up with gm1, gm2, gm3, gm4 = 1/rp, gm5, gm6, &scrn, and &supp parameters. Lots of adjustments to control the model. Don
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If it isn't broken, take it apart and mod it. And then, what is the "speed of dark"? Hint: plug it in to find out. 
22nd July 2007, 05:47 PM  #6 
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Join Date: Dec 2001
Location: Hickory, NC

Looks like you need Spangenberg's equation 9.13 also (for Vp < 0.8 Vscrn), besides eqn. 9.12 (for Vp > 0.8 Vscrn) to get the left side (low Vp) of the characteristic curves correctly modeled.
Screen current apparently really takes off below 0.8 to 0.7 Vscrn in figure 9.18. This Vp versus Vscrn equation switching parameter (0.9 to 0.7 or so range) could be useful to have adjustable for modeling too. And as the figure 9.18 suggests, it may be a modest factor of screen voltage. Could try a linear model like Vswitch = 0.9  0.02*Vscrn. This will give the staggered or tilted left boundary seen on many tube curve sets. The 0.02 factor becoming yet another adjustable parameter. These equations, 9.12 and 9.13, could also be used to extend the pentode model for positive grid1 voltage and current too, which is the general case Spangenberg treats. A more complicated model again. Don (the Spangenberg book is available, scanned, at several sites on the Web. Pete Millett's site being one.)
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If it isn't broken, take it apart and mod it. And then, what is the "speed of dark"? Hint: plug it in to find out. 
23rd July 2007, 07:33 AM  #7 
diyAudio Member
Join Date: Apr 2004
Location: Philly

Thanks for all the input Don. I am working my way through the Spangenberg equations, and also the RCA "Vacuum tube design" text. It will take a while... as well as a lot of trial and error, so hopefully it works out.
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4th July 2013, 10:52 PM  #8 
diyAudio Member

Spang book is available online here:
Technical books online 
30th September 2013, 02:19 AM  #9 
diyAudio Member

Will ask the moderators to merge this thread with the one I started a day ago (mea culpa):
Pentode SPICE 
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