Re: Jazbo8
"With SPICE, gm curve can be plotted easily, which ones you would like to see?"
If you have a model of E55L, it would be interesting to see how well it matches up with the datasheet at 125 V on g2 and Va, since that one does give the gm graph on page 5:
http://frank.pocnet.net/sheets/009/e/E55L.pdf
I'm already a little dubious as to how well the datasheet curves match up with the real tubes, and then the Spice models approximate those besides.....
Looking on the curve tracer, I often find the real curves only crudely matching the datasheet curves. The datasheet writers must have worn out a lot of French curve stencils. And then the screen grid current kinks in the plate curves almost routinely got edited out. Then the curve tracers used to get the curves often apparently had a 60/50 Hz ripple in the power supply. That made the curves displace slightly up or down for every other plate curve, giving a doubled up pair appearance.
Then there are the differences between different manufacturers for the same tube. Sometimes you would never think they were the same tube. I found recently that Sylvania 21LG6 tubes trace identical to Sylvania 24LQ6 tubes, just need about 10% more Vg2 on the LG6. Now those two tubes are on nearly opposite ends of the exponent list above! GE and RCA LG6s do however look like the datasheet. None of them look any too great as triodes unfortunately, I guess that's because of the very low g2 exponent. Difference between g1 and g2 exponents determines the Mu variation.
So I would take the above linearity order with a bit of reserve. That's why I would like to get a real gm tracer operating to see what's really out there.
Another issue is that the usual gm curves are done at a constant voltage (Vg2, Vp). But what one really wants is a gm plot along the chosen load line. Again, need a real gm tracer to do things right.
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"Where is 6CW5/EL86"
It's on my to do list.
g2 exponents are fairly easy to work out, since Vg2 = 0 gives 0 mA Ip. Choose a constant Vp voltage up around 250 V, to get out of the knee curvature, on the plate curve set for different screen voltages (instead of the usual Vg1 voltages). (And use Vg1 = 0 curves, that's normally the case on screen voltage curve sets.)
Just take (Vg2'/Vg2)^n = Ip'/Ip and solve for n (iterative, with a calculator, start at n = 1.4) If you are in a hurry, just take two not quite adjacent plate curves (skip over one, for better accuracy) that are in the operating current region of interest. (I typically do the whole curve set to make sure the results are convergent, but that is time consuming.)
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Finding the g1 exponent is rather more tedious. You need to know the -Vg1 that causes cutoff at the chosen constant Vg2 and Vp. Sometimes the datasheet does give this in the parameter list, so that is a big help. Otherwise, sometimes there are continuous curves of plate current versus -Vg1 given, and you can estimate. However, beware of curve sets that abruptly intersect the zero current border, these have been extended with a ruler from coarse data above, and are incorrect. The real curves bend around to become asymptotic to the baseline. Typically, for a given Vg2, the -Vg1 cutoff will be between 1.25 and 1.5 X Vg2/Mu. I have resorted to testing an actual tube at times.
Then re-scale the Vg1 making -Vcutoff into 0 V and 0V into +Vcutoff. Ie, re-scale so the Vg go from 0 to +Vcutoff while the Ip go from 0 to Ipmax together.
Now solve for n: (Vg1'/Vg1)^n = Ip'/Ip (iterative on the calculator again, start with n = 2.1). The g1 exponents vary with current noticeably in some cases, especially the two end points. I solve for the whole data set, to see if the exponents are stable in a central region. If not, then try moving the Vcutoff and start over............................... very Tedious.
You will often see alternating magnitude exponents for every other Vg1 data point too, this is the curve tracer error issue I mentioned. All you can do is average the set up evenly to cancel that error out.
"With SPICE, gm curve can be plotted easily, which ones you would like to see?"
If you have a model of E55L, it would be interesting to see how well it matches up with the datasheet at 125 V on g2 and Va, since that one does give the gm graph on page 5:
http://frank.pocnet.net/sheets/009/e/E55L.pdf
I'm already a little dubious as to how well the datasheet curves match up with the real tubes, and then the Spice models approximate those besides.....
Looking on the curve tracer, I often find the real curves only crudely matching the datasheet curves. The datasheet writers must have worn out a lot of French curve stencils. And then the screen grid current kinks in the plate curves almost routinely got edited out. Then the curve tracers used to get the curves often apparently had a 60/50 Hz ripple in the power supply. That made the curves displace slightly up or down for every other plate curve, giving a doubled up pair appearance.
Then there are the differences between different manufacturers for the same tube. Sometimes you would never think they were the same tube. I found recently that Sylvania 21LG6 tubes trace identical to Sylvania 24LQ6 tubes, just need about 10% more Vg2 on the LG6. Now those two tubes are on nearly opposite ends of the exponent list above! GE and RCA LG6s do however look like the datasheet. None of them look any too great as triodes unfortunately, I guess that's because of the very low g2 exponent. Difference between g1 and g2 exponents determines the Mu variation.
So I would take the above linearity order with a bit of reserve. That's why I would like to get a real gm tracer operating to see what's really out there.
Another issue is that the usual gm curves are done at a constant voltage (Vg2, Vp). But what one really wants is a gm plot along the chosen load line. Again, need a real gm tracer to do things right.
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"Where is 6CW5/EL86"
It's on my to do list.
g2 exponents are fairly easy to work out, since Vg2 = 0 gives 0 mA Ip. Choose a constant Vp voltage up around 250 V, to get out of the knee curvature, on the plate curve set for different screen voltages (instead of the usual Vg1 voltages). (And use Vg1 = 0 curves, that's normally the case on screen voltage curve sets.)
Just take (Vg2'/Vg2)^n = Ip'/Ip and solve for n (iterative, with a calculator, start at n = 1.4) If you are in a hurry, just take two not quite adjacent plate curves (skip over one, for better accuracy) that are in the operating current region of interest. (I typically do the whole curve set to make sure the results are convergent, but that is time consuming.)
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Finding the g1 exponent is rather more tedious. You need to know the -Vg1 that causes cutoff at the chosen constant Vg2 and Vp. Sometimes the datasheet does give this in the parameter list, so that is a big help. Otherwise, sometimes there are continuous curves of plate current versus -Vg1 given, and you can estimate. However, beware of curve sets that abruptly intersect the zero current border, these have been extended with a ruler from coarse data above, and are incorrect. The real curves bend around to become asymptotic to the baseline. Typically, for a given Vg2, the -Vg1 cutoff will be between 1.25 and 1.5 X Vg2/Mu. I have resorted to testing an actual tube at times.
Then re-scale the Vg1 making -Vcutoff into 0 V and 0V into +Vcutoff. Ie, re-scale so the Vg go from 0 to +Vcutoff while the Ip go from 0 to Ipmax together.
Now solve for n: (Vg1'/Vg1)^n = Ip'/Ip (iterative on the calculator again, start with n = 2.1). The g1 exponents vary with current noticeably in some cases, especially the two end points. I solve for the whole data set, to see if the exponents are stable in a central region. If not, then try moving the Vcutoff and start over............................... very Tedious.
You will often see alternating magnitude exponents for every other Vg1 data point too, this is the curve tracer error issue I mentioned. All you can do is average the set up evenly to cancel that error out.
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If no model for the E55L is available, the WE 300B datasheet gives a gm curve to compare with Spice.
http://frank.pocnet.net/sheets/084/3/300B.pdf (page 5)
http://frank.pocnet.net/sheets/084/3/300B.pdf (page 5)
"but did not get the result like the datasheet, so it still needs some work..."
Gm, being the derivative of the datasheet curve data, is very jumpy with any small errors. Whether they be errors in the original data, or errors in reading the data, or in curve fitting the data. Takes a fair amount of work getting the g1 power exponents to numerically converge reasonably. I think it will take a tester dedicated to gm measurement to get good gm curve results.
I have the filter and sweep osc. here to set up a gm or gain tester, so that will be the next project after crunching the datasheet numbers for a few more tubes of interest.
The 6HB6 GE datasheet gives a gm curve set (unusual for a power tube), so another one (of maybe a half dozen around).
http://frank.pocnet.net/sheets/135/6/6HB6.pdf
I just crunched the grid exponent numbers for the 12HL7 frame grid tube, and it is interesting (g1 g2):
12HL7 2.03 1.5
Not too surprising to see a 2.0 range for the g1 exponent with a frame grid. What IS interesting here, is the return to a 1.5 exponent for the g2.
With a frame grid for g1, the g2 wires are not aligned with the g1 wires, so no wire to wire shielding effect here. My hypothesis of aligned grid wires causing anti-wire proximity effects for g2 seems confirmed (causing the g2 exponent to drop below 1.5 for aligned grid cases).
The EL34 also has g2 at 1.5 and the 300B too, with its solid plate. 6BQ5 is a bit puzzling at 1.2 for g2, but since some of these tubes were beam versions, maybe even the pentode versions had aligned grid wires. I know the (Russian) GU50 has aligned grid wires, but still calls itself a pentode on the datasheet. I guess I will check the GU50 datasheet next to see if it has a g2 curve set.
The 6LQ6 and I think it was the 6973 have g2 exponents approaching 1.0 (absolute linearity). And then there were some "shadow grid" tubes that had an aligned shield grid. The makings of how to make a truly linear tube are blowing in the wind.
Yep: take a look at page 6 of the 6973 for linear g2:
http://frank.pocnet.net/sheets/049/6/6973.pdf
Gm, being the derivative of the datasheet curve data, is very jumpy with any small errors. Whether they be errors in the original data, or errors in reading the data, or in curve fitting the data. Takes a fair amount of work getting the g1 power exponents to numerically converge reasonably. I think it will take a tester dedicated to gm measurement to get good gm curve results.
I have the filter and sweep osc. here to set up a gm or gain tester, so that will be the next project after crunching the datasheet numbers for a few more tubes of interest.
The 6HB6 GE datasheet gives a gm curve set (unusual for a power tube), so another one (of maybe a half dozen around).
http://frank.pocnet.net/sheets/135/6/6HB6.pdf
I just crunched the grid exponent numbers for the 12HL7 frame grid tube, and it is interesting (g1 g2):
12HL7 2.03 1.5
Not too surprising to see a 2.0 range for the g1 exponent with a frame grid. What IS interesting here, is the return to a 1.5 exponent for the g2.
With a frame grid for g1, the g2 wires are not aligned with the g1 wires, so no wire to wire shielding effect here. My hypothesis of aligned grid wires causing anti-wire proximity effects for g2 seems confirmed (causing the g2 exponent to drop below 1.5 for aligned grid cases).
The EL34 also has g2 at 1.5 and the 300B too, with its solid plate. 6BQ5 is a bit puzzling at 1.2 for g2, but since some of these tubes were beam versions, maybe even the pentode versions had aligned grid wires. I know the (Russian) GU50 has aligned grid wires, but still calls itself a pentode on the datasheet. I guess I will check the GU50 datasheet next to see if it has a g2 curve set.
The 6LQ6 and I think it was the 6973 have g2 exponents approaching 1.0 (absolute linearity). And then there were some "shadow grid" tubes that had an aligned shield grid. The makings of how to make a truly linear tube are blowing in the wind.
Yep: take a look at page 6 of the 6973 for linear g2:
http://frank.pocnet.net/sheets/049/6/6973.pdf
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6CZ5 too: (page 4)
http://frank.pocnet.net/sheets/137/6/6CZ5.pdf
Too bad gm2 is only 600 uMho for these.
1/2 of a 12AX7 section.
http://frank.pocnet.net/sheets/137/6/6CZ5.pdf
Too bad gm2 is only 600 uMho for these.
1/2 of a 12AX7 section.
A couple more tubes for grid exponents (g1 g2):
6973 2.43 1.16
6HB6 2.1 1.43
However, the 6HB6 tube had an actual gm graph to compare with, and using the -13V Vg1 cutoff from the datasheet, and the plate curves, does not give a power law compatible with their gm graph. After fiddling around with the cutoff value, I could get them to roughly agree using a -9.5V cutoff. However, they also give a graph of Ip versus Vg1 and clearly the cutoff shown there is at -13V. ??????
Well, I think the problem is the long slow cutoff tail this tube shows. If I use 1.26 * Vg2/Mu for the g1 cutoff voltage, which is what has often turned up for other tubes, I get the -9.5V figure. Using the stated -13V cutoff figure makes the g1 exponent jump to 2.4. I'm going to assume the gm graph is more accurate for gm, so the adjusted 2.1 exponent for g1 stands.
But now I'm wondering about some other previous tubes where I blindly accepted the stated Vg1 cutoff voltage listed. I didn't have a gm graph for them to back-check the results.
The 6JB5/6JC5 tube with the worst g1 exponent has a long cutoff tail and the stated cutoff does not at all agree with the heuristic 1.26 *Vg2/Mu figure. Same long tail for the runner up worst one: 6LG6, although the cutoff listed there does agree with my heuristic rule figure.
So I'm going to go back and check some of the already done tubes to see if the cutoff figure is reasonable. (close to 1.26 x Vg2/Mu) Probably have to revise some g1 exponent figures.
This may take a while.
Some tubes on my "to do" list yet are 6KG6/EL509, Gu50, 6JC6, D3a, 6JF6/6JG6, 6JR6/6JU6
6973 2.43 1.16
6HB6 2.1 1.43
However, the 6HB6 tube had an actual gm graph to compare with, and using the -13V Vg1 cutoff from the datasheet, and the plate curves, does not give a power law compatible with their gm graph. After fiddling around with the cutoff value, I could get them to roughly agree using a -9.5V cutoff. However, they also give a graph of Ip versus Vg1 and clearly the cutoff shown there is at -13V. ??????
Well, I think the problem is the long slow cutoff tail this tube shows. If I use 1.26 * Vg2/Mu for the g1 cutoff voltage, which is what has often turned up for other tubes, I get the -9.5V figure. Using the stated -13V cutoff figure makes the g1 exponent jump to 2.4. I'm going to assume the gm graph is more accurate for gm, so the adjusted 2.1 exponent for g1 stands.
But now I'm wondering about some other previous tubes where I blindly accepted the stated Vg1 cutoff voltage listed. I didn't have a gm graph for them to back-check the results.
The 6JB5/6JC5 tube with the worst g1 exponent has a long cutoff tail and the stated cutoff does not at all agree with the heuristic 1.26 *Vg2/Mu figure. Same long tail for the runner up worst one: 6LG6, although the cutoff listed there does agree with my heuristic rule figure.
So I'm going to go back and check some of the already done tubes to see if the cutoff figure is reasonable. (close to 1.26 x Vg2/Mu) Probably have to revise some g1 exponent figures.
This may take a while.
Some tubes on my "to do" list yet are 6KG6/EL509, Gu50, 6JC6, D3a, 6JF6/6JG6, 6JR6/6JU6
I think this is the problem, how do we know the datasheet's curves are correct? For example, with the 300B, while both the plate and transfer characteristics from the SPICE model match up with the datasheet, the gm characteristic does not. So it would be great as you said, to have an actual gm measurement that we can use to compare against the datasheet.
Your cut-off voltage example is another good one, it seems the long smooth decline we typically see on the transfer characteristic curves were just made up by a draftsman using a "french curve" and not really based actual measurement data. So if the SPICE model is based on false data to begin with, the errors just keep multiplying themselves.
Hi smoking-amp,
how did you manage to derive the exponents from the graphs? Which algorithm did you use? I respect your work, anyway!
Howsoever, all graphs in datasheets represent a bogey valve, i.e. the mean out of, let's say, one hundred individual valves or so. Being hand made in most of the important manufacturing (!) steps, this is an almost inevitable fact in vacuum tubes, I think.
Best regards!
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"how did you manage to derive the exponents from the graphs? Which algorithm did you use?"
Back in post 401 I gave some explanation. And in post 410 there is some explantion of its weakness. Basically I just take two adjacent plate curves (points taken at the same plate voltage, with stepped grid voltages, ie, a vertical line up through the plate curves) and fit the numbers into: Ip/Ip' = (Vg/Vg')^n and solve iteratively for n. 1st the negative Vg numbers get rescaled from Vcutoff, as zero on up through positive numbers.
So, for grid 1 voltages: with Ip cutoff at -22V, then -15V, -10V, -5v and 0V curves, these get re-scaled to 0V, 7V, 12V, 22V for the grid voltages. (Vg2 curves do not need the re-scale, since they are already positive ) Currents associated with them stay the same. Then start putting adjacent Vg and Ip pairs into the equation above. Solve iteratively for n each time. I do the whole set and then look for a region in the middle where the n's are reasonably stable, then average those.
This all assumes that the cutoff Vg1 stays the same for all data sets, which it would if the tube were a fixed n power law throughout. But it is not the case for real tubes with varying n, Vcutoff is corrupted for the higher currents.
So this works reasonably well for Vg2 ( no grid Vcutoff rescaling needed in this case with positive Vg2) where the power law is stable and confined to the 1.3 to 1.5 range.
For Vg2 data I just work through the data-set once and average the similar n's.
For Vg1 though, the tail portion of the plate current near cutoff is not the same power n as the rest of the curve. So it corrupts the Vcutoff for the upper currents. An error in Vcutoff however has much more effect at the lower currents (changing Vcutoff has a bigger effect on the smaller re-scaled Vg1's there). So if one adjusts Vcutoff (errh....., long and tedious doing the whole data set each time) the lower current computed n's will vary the most and I try to get the upper/middle n's to stablilise coherently the best I can. The lowest current n's will be wrong then, so I discard one or two of them which are usually low values by then. The 6HB6 tube mentioned above had a long cutoff current tail, and so it really messed things up with this method.
So the method is not foolproof by any means, in fact rather tricky, but it is the only one that seems to be getting somewhat reasonable n #s so far. What is really needed is some kind of curve fitting program, maybe just visually matching up power law curves on a graph, using something like Mathcad.
I do have a numerical formula that can calculate the local n power law from 3 adjacent points (equally spaced grid voltages) without reference to the Vcutoff:
1/n = I2/(I3-I2) - I1/(I2-I1)
This avoids the issue of the drifting Vcutoff with variable n power law. (and does not require iteration to solve) And it works perfectly on a data-set generated using Ip = k(Vg)^n as long as the numbers are accurate and reasonably closely spaced. But when I try it on actual tube data sets, it gives junk. It is too sensitive to # accuracy and requires reasonably close datapoints if the n law is varying (it assumes constant power law across the 3 data points). Tube curve sets are too spaced out (big grid V steps) and are hard to read or get accurate numbers.
I tried it on the WE300B data-curves (page 3, bottom) around -55V, -60V, and -65V Vg1 for a 250 V plate V. That spot shows as a flat curve section on page 5, top, in the gm curve set. So it should be close to a power law of 2.0 in the middle at that -60V point for 250 Vp. But the formula gives 3.33. No good obviously. The formula works robustly well on constant n datasets though.
http://frank.pocnet.net/sheets/084/3/300B.pdf
The one remaining approach I see is to combine the two methods heuristically. Choosing 3 adjacent (equally Vg spaced) datapoints, use the 1st formula:
Ip/Ip' = (Vg/Vg')^n to set: Ip2/Ip1 = Ip3/Ip2 by adjusting Ip2 slightly. In other words, make the 3 points n power consistent. (one problem remaining is that only non constant, proportioned, Vg1 steps give this nice fit for a constant power law.) (so work in progress) Then apply the 2nd formula above (1/n = .....) (using the Ip1, Ip2', Ip3) to get n. This is where I'm at now, further testing required. I think this should work well, but it is still going to be sensitive to errors in the data, so the whole data-set (all the plate curves) will still need to be done to check for smoothly varying n.
Once a cleanly working method is arrived at, it may also be quite useful to use it along a chosen load line, sloping across the plate curves rather than a vertical stripe. This would give an indication of linearity in real operation. I consider the present vertical stripe analysis mainly as just a test of the tube's grid linearity, for comparison with other tubes.
Back in post 401 I gave some explanation. And in post 410 there is some explantion of its weakness. Basically I just take two adjacent plate curves (points taken at the same plate voltage, with stepped grid voltages, ie, a vertical line up through the plate curves) and fit the numbers into: Ip/Ip' = (Vg/Vg')^n and solve iteratively for n. 1st the negative Vg numbers get rescaled from Vcutoff, as zero on up through positive numbers.
So, for grid 1 voltages: with Ip cutoff at -22V, then -15V, -10V, -5v and 0V curves, these get re-scaled to 0V, 7V, 12V, 22V for the grid voltages. (Vg2 curves do not need the re-scale, since they are already positive ) Currents associated with them stay the same. Then start putting adjacent Vg and Ip pairs into the equation above. Solve iteratively for n each time. I do the whole set and then look for a region in the middle where the n's are reasonably stable, then average those.
This all assumes that the cutoff Vg1 stays the same for all data sets, which it would if the tube were a fixed n power law throughout. But it is not the case for real tubes with varying n, Vcutoff is corrupted for the higher currents.
So this works reasonably well for Vg2 ( no grid Vcutoff rescaling needed in this case with positive Vg2) where the power law is stable and confined to the 1.3 to 1.5 range.
For Vg2 data I just work through the data-set once and average the similar n's.
For Vg1 though, the tail portion of the plate current near cutoff is not the same power n as the rest of the curve. So it corrupts the Vcutoff for the upper currents. An error in Vcutoff however has much more effect at the lower currents (changing Vcutoff has a bigger effect on the smaller re-scaled Vg1's there). So if one adjusts Vcutoff (errh....., long and tedious doing the whole data set each time) the lower current computed n's will vary the most and I try to get the upper/middle n's to stablilise coherently the best I can. The lowest current n's will be wrong then, so I discard one or two of them which are usually low values by then. The 6HB6 tube mentioned above had a long cutoff current tail, and so it really messed things up with this method.
So the method is not foolproof by any means, in fact rather tricky, but it is the only one that seems to be getting somewhat reasonable n #s so far. What is really needed is some kind of curve fitting program, maybe just visually matching up power law curves on a graph, using something like Mathcad.
I do have a numerical formula that can calculate the local n power law from 3 adjacent points (equally spaced grid voltages) without reference to the Vcutoff:
1/n = I2/(I3-I2) - I1/(I2-I1)
This avoids the issue of the drifting Vcutoff with variable n power law. (and does not require iteration to solve) And it works perfectly on a data-set generated using Ip = k(Vg)^n as long as the numbers are accurate and reasonably closely spaced. But when I try it on actual tube data sets, it gives junk. It is too sensitive to # accuracy and requires reasonably close datapoints if the n law is varying (it assumes constant power law across the 3 data points). Tube curve sets are too spaced out (big grid V steps) and are hard to read or get accurate numbers.
I tried it on the WE300B data-curves (page 3, bottom) around -55V, -60V, and -65V Vg1 for a 250 V plate V. That spot shows as a flat curve section on page 5, top, in the gm curve set. So it should be close to a power law of 2.0 in the middle at that -60V point for 250 Vp. But the formula gives 3.33. No good obviously. The formula works robustly well on constant n datasets though.
http://frank.pocnet.net/sheets/084/3/300B.pdf
The one remaining approach I see is to combine the two methods heuristically. Choosing 3 adjacent (equally Vg spaced) datapoints, use the 1st formula:
Ip/Ip' = (Vg/Vg')^n to set: Ip2/Ip1 = Ip3/Ip2 by adjusting Ip2 slightly. In other words, make the 3 points n power consistent. (one problem remaining is that only non constant, proportioned, Vg1 steps give this nice fit for a constant power law.) (so work in progress) Then apply the 2nd formula above (1/n = .....) (using the Ip1, Ip2', Ip3) to get n. This is where I'm at now, further testing required. I think this should work well, but it is still going to be sensitive to errors in the data, so the whole data-set (all the plate curves) will still need to be done to check for smoothly varying n.
Once a cleanly working method is arrived at, it may also be quite useful to use it along a chosen load line, sloping across the plate curves rather than a vertical stripe. This would give an indication of linearity in real operation. I consider the present vertical stripe analysis mainly as just a test of the tube's grid linearity, for comparison with other tubes.
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The most foolproof method for solving a varying power law (ie, like for grid 1) I think, may be to just plot the gm curve from the Vg & Ip datset.
gm = delta Ip / delta Vg
Calculate and plot gm versus Vg for each adjacent data-points. This will likely be somewhat noisy since it is using differences, but one knows the general shape of a tube gm1 curve, roughly S shaped, so some smoothing may be done after the plotting.
The varying n power law can then be obtained by integration of the area between the gm curve and the bottom zero gm axis, versus Vg1. (so one can readily see why a constant ramp function for gm gives a square law n = 2.0 )
However......this method does NOT give a true local power law derivation, it gives the average power law from Vcutoff up to the given point integrated up to. So this is still not very impressive. One would need to edit the gm curve back to Vcutoff to give it the same general curved shape everywhere for a correct local "n" power figure.
Seems like one might be able to condense this process down to a formula (if non-local "n" is acceptable enough). Needs some automatic rule for smoothing the gm curve. More work.....
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My above posted suggestion of a hybrid combination of the iterative and Vcutoff sensitive formula:
Ip/Ip' = (Vg/Vg')^n
with the Vcutoff independent and non-iterative formula:
1/n = I2/(I3-I2) - I1/(I2-I1)
has run into a problem. To re-scale one of the Ip accurately, to get 3 "n" consistent data-set points for the 2nd equation, still requires either knowledge of the specific data-point relevant Vcutoff, or of the actual "n" at that data-point. So that approach is stuck in the mud.
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There should be a math Fields Metal for solving this stupid problem!!
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Ahhh, maybe this will do:
Plot the gm curve and smooth it manually, or by some automated process. Then one should be able to get a local power law by using 2 or 3 adjacent gm points on the graph to get Ip numbers again for use in one of the formulas. I'll work out a new formula using just the gm's.
Heh, simple, just use 1/x = gm2/(gm3-gm2) - gm1/(gm2-gm1) and then the power law is n = x+1
That will have to be a darn smooth and accurately readable gm graph though.
Maybe something using the iterative: Ip/Ip' = (Vg/Vg')^n would be better, I'll have to convert it to a gm's and Vg's form. Stay tuned.....
gm = delta Ip / delta Vg
Calculate and plot gm versus Vg for each adjacent data-points. This will likely be somewhat noisy since it is using differences, but one knows the general shape of a tube gm1 curve, roughly S shaped, so some smoothing may be done after the plotting.
The varying n power law can then be obtained by integration of the area between the gm curve and the bottom zero gm axis, versus Vg1. (so one can readily see why a constant ramp function for gm gives a square law n = 2.0 )
However......this method does NOT give a true local power law derivation, it gives the average power law from Vcutoff up to the given point integrated up to. So this is still not very impressive. One would need to edit the gm curve back to Vcutoff to give it the same general curved shape everywhere for a correct local "n" power figure.
Seems like one might be able to condense this process down to a formula (if non-local "n" is acceptable enough). Needs some automatic rule for smoothing the gm curve. More work.....
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My above posted suggestion of a hybrid combination of the iterative and Vcutoff sensitive formula:
Ip/Ip' = (Vg/Vg')^n
with the Vcutoff independent and non-iterative formula:
1/n = I2/(I3-I2) - I1/(I2-I1)
has run into a problem. To re-scale one of the Ip accurately, to get 3 "n" consistent data-set points for the 2nd equation, still requires either knowledge of the specific data-point relevant Vcutoff, or of the actual "n" at that data-point. So that approach is stuck in the mud.
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There should be a math Fields Metal for solving this stupid problem!!
--------------------------------
Ahhh, maybe this will do:
Plot the gm curve and smooth it manually, or by some automated process. Then one should be able to get a local power law by using 2 or 3 adjacent gm points on the graph to get Ip numbers again for use in one of the formulas. I'll work out a new formula using just the gm's.
Heh, simple, just use 1/x = gm2/(gm3-gm2) - gm1/(gm2-gm1) and then the power law is n = x+1
That will have to be a darn smooth and accurately readable gm graph though.
Maybe something using the iterative: Ip/Ip' = (Vg/Vg')^n would be better, I'll have to convert it to a gm's and Vg's form. Stay tuned.....
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One last devious thought, using the gm graph, a 2.0 power law (for Ip from Vg) maps as a linear straight line on the gm graph. A typical tube gm curve will curve upward at low current, straighten, then curve the other way at high current. So one can easily see where 2.0 exists at least. What could be done, is make a transform so that n (selectable, say 1.0 up to 3.0) will be straightened out when graphed. Then one could sweep the n value, while looking at a Mathcad plot, and find the selected n power at any selected point by observing where the curvature inflects (straightens) from one way to the other. Just an idea.
I'm probably driving everyone nuts by now. Sorry!
I'm probably driving everyone nuts by now. Sorry!
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Update:
gm/gm' = (Vg/Vg')^(n-1)
So one can solve for the power law "n" iteratively using just two points on the gm graph.
(Unfortunately, the positive Vg values used here have an impicit Vcutoff still encoded in them, where zero is, but maybe the gm graph will give some obvious clue where the appropriate Vcutoff should be for some selected analysis point. Maybe a tangent from the gm curve at the analysis point down to the zero gm axis?? Tangent works for n = 2.0 anyway. ehhh, looks like tangent is only working for 2.0 )
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On the gm graph approach, with integrating the area under it. All one has to do for a local power law integral is integrate the trapezoid area right around the analysis point, and re-scale the area by the Vg1-Vg1' interval. Ie, just the usual integral problem with limits. This looks promising now. Maybe can derive a formula for the total process even.
gm/gm' = (Vg/Vg')^(n-1)
So one can solve for the power law "n" iteratively using just two points on the gm graph.
(Unfortunately, the positive Vg values used here have an impicit Vcutoff still encoded in them, where zero is, but maybe the gm graph will give some obvious clue where the appropriate Vcutoff should be for some selected analysis point. Maybe a tangent from the gm curve at the analysis point down to the zero gm axis?? Tangent works for n = 2.0 anyway. ehhh, looks like tangent is only working for 2.0 )
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On the gm graph approach, with integrating the area under it. All one has to do for a local power law integral is integrate the trapezoid area right around the analysis point, and re-scale the area by the Vg1-Vg1' interval. Ie, just the usual integral problem with limits. This looks promising now. Maybe can derive a formula for the total process even.
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Not the slope of the gm characteristic, that would be a 2nd derivative and so would be extremely noisy from actual tube data.
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I'm just trying to find the power exponent "n" for the Ip = k(Vg)^n equation. The nominal n from textbooks is 1.5, but for real tubes the g1 (grid 1) exponent varies some from maybe 2 up to 3.0+ due to grid wire proximity effects (to the cathode). For the grid 2, the range is more limited. For actual plates (gm = 1/Rp) the exponent is right around the 1.5 per the textbook. For grid 2 the n is typically 1.5 for a non aligned g1-g2. But aligned g1-g2 grid wires tend to drop the exponent down to a 1.1 to 1.4 range for grid 2.
(my hypothesis is this is due to grid1 wire shielding of grid2 wires, causing a kind of anti-proximity effect for g2) The opposite is the case for grid 1, where proximity of the g1 wires to the cathode causes the power law exponent to increase from 1.5 up to the 2.0 to 3.0+ range.)
If you have the Ip = k(Vg)^n relation with the "n" known, then you can calculate the gm versus Vg
(or versus Ip with a little re-arranging) from:
gm = (delta Ip)/(delta Vg) or gm = DIp/DVg in calculus speak.
So for n in the Ip = k(Vg)^n equation, simple calculus gives gm = DIp/DVg = nk(Vg)^(n-1)
so for example: for =2 gives gm = 2k(Vg)^1 or just gm = 2kVg
So a square law Ip versus Vg power relation gives a linear gm versus Vg relation.
Since an n=2 Ip power law will give an n=1 or linear gm law, that was why I mentioned earlier about if you see a linear ramp for gm, then the Ip power law mst be 2.0 in that location.
The 300B datasheet shows gm ramping (top, page 5) linearly in the middle of the gm curves graph, so one should be able to find n=2.0 from the Ip data in that range (bottom, page 3). Note the parabolic (square law, or n=2) type shape of the plate current Ip versus Vg1 (page 3, bottom), so this is approximately confirmed.
http://frank.pocnet.net/sheets/084/3/300B.pdf
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new Vcutoff formula!
(Vcutoff is the negative Vg1 voltage that drops Ip or plate current to zero. )
I was looking back at the derivation of the
1/n = I2/(I3-I2) - I1/(I2-I1)
formula I found earlier, and realized that half way through the calculations, the option of solving for "n" -- OR -- for Vcutoff is available. I had solved for "n" before. So this time I solved for Vcutoff.
Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ]
Now these formulas are valid for constant "n" data-sets (Ip = k(Vg)^n ), with small increments between the Vg = V1,V2,V3 and associated Ip = I1,I2,I3 currents.
When I used the 1/n= .... formula on real tube Ip data, I got results like 67% too big for the "n" values (using 300B data sets). Which I would ascribe to the large Vg1 steps in the tube data, and the additional effect of "n" varying some as well through the data list.
Since Vcutoff varies relatively slowly, it occurred to me that maybe I could use this new Vcutoff formula to get the Vcutoff needed at some local point for the earlier method (Ip'/Ip = (Vg'/Vg)^n ) to get "n" from just two data points (a more localized method).
The zero current point (Vg1 = Vcutoff) in the actual tube Ip versus Vg1 graphs is valid for the low current data derivations of "n", but is not quite accurate for the higher current data point derivations (generally with lower "n").
Explanation: A lower "n" exponent for the higher current data causes the extrapolated grid cutoff to occur at a less negative voltage, if one were to use that "n" consistently to extrapolate back from that point to a zero current using that (consistent) "n" power law curve.
This is simply working backwards from the chosen datapoint (at higher Ip current), and using the determined "n" power law exponent, to find where a power law curve (using that "n" value consistently) would intersect the zero current axis (ie, at what cutoff -Vg1).
The reason for determining the specific Vcutoff, applying to a specific data point in the Vg Ip dataset, is important, is that the simple Ip'/Ip = (Vg'/Vg)^n formula uses an implied third data point, the zero current point or Vg1 = Vcutoff point.
So my latest approach is to calculate the Vcutoff applying to a given data-set point of interest, then apply the localized Ip'/Ip = (Vg'/Vg)^n formula to determine the "n" value for that data point.
So to make this a little more real and obvious, lets do a 300B calculation:
Using the page 3 data, we make a Vg1 and Ip data list for the curve with Vp always = 250V
The -71V cutoff is just a visually estimated Vg1 from the graph for zero Ip. The rescaled Vg1 then use the -71V as if that were now 0. So the re-scaled Vg1 and Ip track in a simple relation that can be described by an (near) power law.
Vg1 -- Vg1(rescaled for a -71V cutoff) -- Ip
-71 00 0
-60 11 13
-55 16 28
-50 21 52
-45 26 80.5
-40 31 113
-35 36 150
-30 41 190
1st, compute the Ip'/Ip ratios between adjacent data points.
Then using the simple Ip'/Ip = (Vg'/Vg)^n formula, one iteratively solves for a "crude" n value between each of the data points:
00 0
11 13
Ip'/Ip = 2.1538 n= 2.048 (from (16/11)^2.048 = 2.1538)
16 28
Ip'/Ip = 1.8571 n = 2.277 (from (21/16)^2.277 = 1.8571)
21 52
Ip'/Ip = 1.5480 n = 2.046 (from (26/21)^2.046 = 1.5480)
26 80.5
Ip'/Ip = 1.4037 n = 1.928 (etc...)
31 113
Ip'/Ip = 1.3274 n = 1.894
36 150
Ip'/Ip = 1.2666 n = 1.818
41 190
average of the n values: 2.0018
Thus far we have just used the original Ip'/Ip = (Vg'/Vg)^n scheme to solve for crude n values. This is not taking into account any variation of Vcutoff for the different data points, so there is some inherent error in the analysis of "n" for the higher current data-points. But note the n = 2.048 found between Vg1 of 11 and 16 (or -60V and -55V original Vg1) Looking on page 5, top, of the 300B data we see the gm curve (at Vp 250V) straightens around Vg1 = -58V. So the n power law must be 2.0 there since the gm power law is 1.0 or linear. Not bad at all, we agree. But the selected Vcutoff should be good for the 1st data point anyway, so no surprise it worked well there.
The later data points will have a slightly varied Vcutoff due to varying n power effects. But variation of Vcutoff does not have much affect on analyzing the higher current points, since changing a volt or two of Vcutoff does not much affect the larger re-scaled Vg1.
To get accurate results for n at any given data point pair, one would like to know the accurate applicable Vcutoff to use there. So lets now try the Vcutoff formula at Vg1 -60V, -55V:
Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ]
putting in V1 = -60, V2 = -55, I1 = 13, I2 = 28, I3 = 52
gives Vcutoff = -74.44 Volts instead of the -71 Volts we used. Not looking real good. Just to check this out, I re-did the crude "n" calc using Vcutoff = -74.44 and I now get n = 2.581 instead of 2.048
If I put the I1 = 13, I2 = 28, I3 = 52 data into the 1/n = I2/(I3-I2) - I1/(I2-I1) formula, I get n = 3.333 even worse!!
The computed Vcutoff is only off by 5% from the observed one. This I'm sure is due to the large increments in the Vg1 used. If the data were more closely spaced, then the Vcutoff should converge to the correct value for data-point each case. Or maybe one can just use 95% of the computed Vcutoff everywhere for typical tube data curves???? More analysis with actual tube data needed to see if some heuristic rule can be applied..................
Dinner time. Gee, I started at breakfast.
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I'm just trying to find the power exponent "n" for the Ip = k(Vg)^n equation. The nominal n from textbooks is 1.5, but for real tubes the g1 (grid 1) exponent varies some from maybe 2 up to 3.0+ due to grid wire proximity effects (to the cathode). For the grid 2, the range is more limited. For actual plates (gm = 1/Rp) the exponent is right around the 1.5 per the textbook. For grid 2 the n is typically 1.5 for a non aligned g1-g2. But aligned g1-g2 grid wires tend to drop the exponent down to a 1.1 to 1.4 range for grid 2.
(my hypothesis is this is due to grid1 wire shielding of grid2 wires, causing a kind of anti-proximity effect for g2) The opposite is the case for grid 1, where proximity of the g1 wires to the cathode causes the power law exponent to increase from 1.5 up to the 2.0 to 3.0+ range.)
If you have the Ip = k(Vg)^n relation with the "n" known, then you can calculate the gm versus Vg
(or versus Ip with a little re-arranging) from:
gm = (delta Ip)/(delta Vg) or gm = DIp/DVg in calculus speak.
So for n in the Ip = k(Vg)^n equation, simple calculus gives gm = DIp/DVg = nk(Vg)^(n-1)
so for example: for =2 gives gm = 2k(Vg)^1 or just gm = 2kVg
So a square law Ip versus Vg power relation gives a linear gm versus Vg relation.
Since an n=2 Ip power law will give an n=1 or linear gm law, that was why I mentioned earlier about if you see a linear ramp for gm, then the Ip power law mst be 2.0 in that location.
The 300B datasheet shows gm ramping (top, page 5) linearly in the middle of the gm curves graph, so one should be able to find n=2.0 from the Ip data in that range (bottom, page 3). Note the parabolic (square law, or n=2) type shape of the plate current Ip versus Vg1 (page 3, bottom), so this is approximately confirmed.
http://frank.pocnet.net/sheets/084/3/300B.pdf
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new Vcutoff formula!
(Vcutoff is the negative Vg1 voltage that drops Ip or plate current to zero. )
I was looking back at the derivation of the
1/n = I2/(I3-I2) - I1/(I2-I1)
formula I found earlier, and realized that half way through the calculations, the option of solving for "n" -- OR -- for Vcutoff is available. I had solved for "n" before. So this time I solved for Vcutoff.
Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ]
Now these formulas are valid for constant "n" data-sets (Ip = k(Vg)^n ), with small increments between the Vg = V1,V2,V3 and associated Ip = I1,I2,I3 currents.
When I used the 1/n= .... formula on real tube Ip data, I got results like 67% too big for the "n" values (using 300B data sets). Which I would ascribe to the large Vg1 steps in the tube data, and the additional effect of "n" varying some as well through the data list.
Since Vcutoff varies relatively slowly, it occurred to me that maybe I could use this new Vcutoff formula to get the Vcutoff needed at some local point for the earlier method (Ip'/Ip = (Vg'/Vg)^n ) to get "n" from just two data points (a more localized method).
The zero current point (Vg1 = Vcutoff) in the actual tube Ip versus Vg1 graphs is valid for the low current data derivations of "n", but is not quite accurate for the higher current data point derivations (generally with lower "n").
Explanation: A lower "n" exponent for the higher current data causes the extrapolated grid cutoff to occur at a less negative voltage, if one were to use that "n" consistently to extrapolate back from that point to a zero current using that (consistent) "n" power law curve.
This is simply working backwards from the chosen datapoint (at higher Ip current), and using the determined "n" power law exponent, to find where a power law curve (using that "n" value consistently) would intersect the zero current axis (ie, at what cutoff -Vg1).
The reason for determining the specific Vcutoff, applying to a specific data point in the Vg Ip dataset, is important, is that the simple Ip'/Ip = (Vg'/Vg)^n formula uses an implied third data point, the zero current point or Vg1 = Vcutoff point.
So my latest approach is to calculate the Vcutoff applying to a given data-set point of interest, then apply the localized Ip'/Ip = (Vg'/Vg)^n formula to determine the "n" value for that data point.
So to make this a little more real and obvious, lets do a 300B calculation:
Using the page 3 data, we make a Vg1 and Ip data list for the curve with Vp always = 250V
The -71V cutoff is just a visually estimated Vg1 from the graph for zero Ip. The rescaled Vg1 then use the -71V as if that were now 0. So the re-scaled Vg1 and Ip track in a simple relation that can be described by an (near) power law.
Vg1 -- Vg1(rescaled for a -71V cutoff) -- Ip
-71 00 0
-60 11 13
-55 16 28
-50 21 52
-45 26 80.5
-40 31 113
-35 36 150
-30 41 190
1st, compute the Ip'/Ip ratios between adjacent data points.
Then using the simple Ip'/Ip = (Vg'/Vg)^n formula, one iteratively solves for a "crude" n value between each of the data points:
00 0
11 13
Ip'/Ip = 2.1538 n= 2.048 (from (16/11)^2.048 = 2.1538)
16 28
Ip'/Ip = 1.8571 n = 2.277 (from (21/16)^2.277 = 1.8571)
21 52
Ip'/Ip = 1.5480 n = 2.046 (from (26/21)^2.046 = 1.5480)
26 80.5
Ip'/Ip = 1.4037 n = 1.928 (etc...)
31 113
Ip'/Ip = 1.3274 n = 1.894
36 150
Ip'/Ip = 1.2666 n = 1.818
41 190
average of the n values: 2.0018
Thus far we have just used the original Ip'/Ip = (Vg'/Vg)^n scheme to solve for crude n values. This is not taking into account any variation of Vcutoff for the different data points, so there is some inherent error in the analysis of "n" for the higher current data-points. But note the n = 2.048 found between Vg1 of 11 and 16 (or -60V and -55V original Vg1) Looking on page 5, top, of the 300B data we see the gm curve (at Vp 250V) straightens around Vg1 = -58V. So the n power law must be 2.0 there since the gm power law is 1.0 or linear. Not bad at all, we agree. But the selected Vcutoff should be good for the 1st data point anyway, so no surprise it worked well there.
The later data points will have a slightly varied Vcutoff due to varying n power effects. But variation of Vcutoff does not have much affect on analyzing the higher current points, since changing a volt or two of Vcutoff does not much affect the larger re-scaled Vg1.
To get accurate results for n at any given data point pair, one would like to know the accurate applicable Vcutoff to use there. So lets now try the Vcutoff formula at Vg1 -60V, -55V:
Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ]
putting in V1 = -60, V2 = -55, I1 = 13, I2 = 28, I3 = 52
gives Vcutoff = -74.44 Volts instead of the -71 Volts we used. Not looking real good. Just to check this out, I re-did the crude "n" calc using Vcutoff = -74.44 and I now get n = 2.581 instead of 2.048
If I put the I1 = 13, I2 = 28, I3 = 52 data into the 1/n = I2/(I3-I2) - I1/(I2-I1) formula, I get n = 3.333 even worse!!
The computed Vcutoff is only off by 5% from the observed one. This I'm sure is due to the large increments in the Vg1 used. If the data were more closely spaced, then the Vcutoff should converge to the correct value for data-point each case. Or maybe one can just use 95% of the computed Vcutoff everywhere for typical tube data curves???? More analysis with actual tube data needed to see if some heuristic rule can be applied..................
Dinner time. Gee, I started at breakfast.
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An interesting observation:
From the 300B analysis above at Vg1 between -55V and -60V (Vp = 250V) we got 3 values of n [for use in the power law Ip = k(Vg1'/Vg1)^n) equation ] from the 3 different approaches:
1) Using the (Ip'/Ip = (Vg'/Vg)^n ) equation, with the visual Vcutoff of -71V, gave n = 2.048, which is very close to the real figure 2.00.
2) Using the Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ] equation to get the Vcutoff for (impicit, hidden in the re-scaled Vg's) use in the (Ip'/Ip = (Vg'/Vg)^n ) equation gave an n = 2.581
3) And the 1/n = I2/(I3-I2) - I1/(I2-I1) equation using just the 3 nearest Ip currents gave n = 3.333
Notice this: 2.581 * (2.581/3.333) = 1.9987 very close to the real n
Now this might just be a coincidence.
But if we inspect the equations giving these results, the 1st has no difference terms. The 2nd has difference terms in both the numerator and the denominator. And the 3rd has difference terms in just the denominator. Difference terms are very susceptible to errors in the data, or in this case, large steps between the Vg voltages given. So possibly the 2nd equation is canceling some of the error by having difference terms both top and bottom, while the 3rd equation sees the full impact with them only on the bottom. And so the errors between these equations may be predictable and usable, as just seen, to get the real answer (using only the Vg and Ip values near the point of interest).
(ie, one does not need to know the visual Vcutoff with this approach, if it really works, and the visual Vcutoff is often quite nebulous on datasheet curves.)
Some testing on real tube data needed now to see if this heuristic fix-up continues to work.......
From the 300B analysis above at Vg1 between -55V and -60V (Vp = 250V) we got 3 values of n [for use in the power law Ip = k(Vg1'/Vg1)^n) equation ] from the 3 different approaches:
1) Using the (Ip'/Ip = (Vg'/Vg)^n ) equation, with the visual Vcutoff of -71V, gave n = 2.048, which is very close to the real figure 2.00.
2) Using the Vcutoff = [ V2*I1*(I3-I2) - V1*I2*(I2-I1) ] / [ I1*(I3-I2) - I2*(I2-I1) ] equation to get the Vcutoff for (impicit, hidden in the re-scaled Vg's) use in the (Ip'/Ip = (Vg'/Vg)^n ) equation gave an n = 2.581
3) And the 1/n = I2/(I3-I2) - I1/(I2-I1) equation using just the 3 nearest Ip currents gave n = 3.333
Notice this: 2.581 * (2.581/3.333) = 1.9987 very close to the real n
Now this might just be a coincidence.
But if we inspect the equations giving these results, the 1st has no difference terms. The 2nd has difference terms in both the numerator and the denominator. And the 3rd has difference terms in just the denominator. Difference terms are very susceptible to errors in the data, or in this case, large steps between the Vg voltages given. So possibly the 2nd equation is canceling some of the error by having difference terms both top and bottom, while the 3rd equation sees the full impact with them only on the bottom. And so the errors between these equations may be predictable and usable, as just seen, to get the real answer (using only the Vg and Ip values near the point of interest).
(ie, one does not need to know the visual Vcutoff with this approach, if it really works, and the visual Vcutoff is often quite nebulous on datasheet curves.)
Some testing on real tube data needed now to see if this heuristic fix-up continues to work.......
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