Fairly straightforward math. But be careful when reading the datasheet graphs, since they can be plotted versus V or versus current I. Different curves result.
For "n" power law of current I versus V:
I = k x V^n
^ denotes "to the power"
So 3/2 power law would be I = k x V^1.5
Get gm by differentiating that formula:
I = k x V^n leads to dI/dV = gm = n x k x V^(n-1)
So 3/2 power law would give gm = 1.5 x k x V^0.5
and square law (n = 2.0) would give gm = 1 x k x V
which is a linear ramp of gm versus electrode voltage V.
You can sub the current law formula back into the gm law formula to get gm as a function of I or current:
For I = k x V^n and gm = n x k x V^(n-1) that would give
gm = n x (nth root of k) x I^[(n-1)/n]
notice (n-1)/n is always less than one, so never a linear ramp versus I.
So cubic law would give: I = k x V^3
and gm = 3 x k x V^2
or gm = n x (cube root of k) x I^(2/3)
Neither of which plots out as any easy to recognize graph on linear coordinates. But on log log coordinates you could check the slope as jhstewart9 mentioned. I don't think you will find that plotted on many datasheets though (and you would need to know the constant k as well).
The gm versus current I curves (using linear coordinates) always curve over/down, but get closer to a straight line as the power law n increases.
For exponential I versus V (bipolar transistors) the gm plot IS a linear ramp versus I. So they work well for the differential amplifier stage case above a CCS tail (the two device gm's summing to a constant versus complementary currents). That analysis is for voltage drive. Usually SS circuits are more readily obvious as current driven though. Complementary current drives with constant current gain Beta's still give linear results fortunately.
And the usual grounded cathode power tube P-P output stage works well for constant gm sum, since gm is approx. linear ramps versus grid voltage for typical square law devices. gm1 = k x Vg1
and gm2 = - k x Vg2 for complementary grid V drives. That assumes perfectly matched tubes. (works great in Spice simulations anyway)
Putting typical square law (approx.) power tubes above a CCS tail for P-P will give you an odd harmonic generator due to summed gm variation. However, putting a special resistance (typically less than 1/gm) in for the tail can null out 3rd harmonic and most of the other odd harmonics. (or can be tuned to null 5th harmonic and reduced others) The tail resistor allows the tail current to vary some symmetrically with drive voltage, so that the gm sum can be almost straightened out.
.
For "n" power law of current I versus V:
I = k x V^n
^ denotes "to the power"
So 3/2 power law would be I = k x V^1.5
Get gm by differentiating that formula:
I = k x V^n leads to dI/dV = gm = n x k x V^(n-1)
So 3/2 power law would give gm = 1.5 x k x V^0.5
and square law (n = 2.0) would give gm = 1 x k x V
which is a linear ramp of gm versus electrode voltage V.
You can sub the current law formula back into the gm law formula to get gm as a function of I or current:
For I = k x V^n and gm = n x k x V^(n-1) that would give
gm = n x (nth root of k) x I^[(n-1)/n]
notice (n-1)/n is always less than one, so never a linear ramp versus I.
So cubic law would give: I = k x V^3
and gm = 3 x k x V^2
or gm = n x (cube root of k) x I^(2/3)
Neither of which plots out as any easy to recognize graph on linear coordinates. But on log log coordinates you could check the slope as jhstewart9 mentioned. I don't think you will find that plotted on many datasheets though (and you would need to know the constant k as well).
The gm versus current I curves (using linear coordinates) always curve over/down, but get closer to a straight line as the power law n increases.
For exponential I versus V (bipolar transistors) the gm plot IS a linear ramp versus I. So they work well for the differential amplifier stage case above a CCS tail (the two device gm's summing to a constant versus complementary currents). That analysis is for voltage drive. Usually SS circuits are more readily obvious as current driven though. Complementary current drives with constant current gain Beta's still give linear results fortunately.
And the usual grounded cathode power tube P-P output stage works well for constant gm sum, since gm is approx. linear ramps versus grid voltage for typical square law devices. gm1 = k x Vg1
and gm2 = - k x Vg2 for complementary grid V drives. That assumes perfectly matched tubes. (works great in Spice simulations anyway)
Putting typical square law (approx.) power tubes above a CCS tail for P-P will give you an odd harmonic generator due to summed gm variation. However, putting a special resistance (typically less than 1/gm) in for the tail can null out 3rd harmonic and most of the other odd harmonics. (or can be tuned to null 5th harmonic and reduced others) The tail resistor allows the tail current to vary some symmetrically with drive voltage, so that the gm sum can be almost straightened out.
.
Last edited:
An IMPORTANT UPDATE on the Twin/Crazy drive stuff.
After upgrading the TEK 576 curve tracer here to sweep to higher plate V (and more grid step curves), I have noticed a problem with the Mosfet follower Twin drive circuit recommended so far. See the 26LX6 Twin Drive curves below. (notice plate current pickup at the lowest 0V curve with high plate V).
When the plate swings to 2X B+ in the P-P type output circuit, the plate Rp (or gm_p = 1/Rp ) can turn the tube on somewhat if the drive voltages to grid2 and grid1 do not swing slightly negative to compensate. The Twin/Crazy drive linearization scheme still needs to have its divider R's connected to the cathode. So a resistor Rb (or Rbias) needs to be added from the Mosfet source output to a negative Vbias voltage. (see diagram below)
Looking at the relative gm's of the tube electrodes:
gm_p = 1/Rp and that gets pushed to 2X B+ during the turn-off phase
while effective gm of the grid2 and grid1 with the Twin drive R divider gives an effective gm of:
gm_drive = gm2 + gm1 x [Rg1k/(Rg2g1+Rg1k)] and that needs to be driven negative by -Vbias through Rbias. (gm_drive is likely going to be NEAR to 2 x gm2 typically, to simplify this mess)
note: gm2 = 1/Mu_internal x gm1
To keep the tube turned off when in the off phase:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] must be greater than gm_p x 2 x B+
So one can solve for -Vbias with a practical pull-down Rb.
Looking at the 26LX6 for example. Rp = 6000 so gm_p = 166 uMhos
and gm1 = 14000 and gm2 = 3500
so gm_drive is approx. = 7000 uMhos (can probably, typically, just simplify this to 2X gm2)
with 2 x B+ swinging up to say 900 V and using say 10K Ohms for Rbias, one would need about -60V for -Vbias. One would still want to test the final circuit to make sure the tubes are turning off properly.
After upgrading the TEK 576 curve tracer here to sweep to higher plate V (and more grid step curves), I have noticed a problem with the Mosfet follower Twin drive circuit recommended so far. See the 26LX6 Twin Drive curves below. (notice plate current pickup at the lowest 0V curve with high plate V).
When the plate swings to 2X B+ in the P-P type output circuit, the plate Rp (or gm_p = 1/Rp ) can turn the tube on somewhat if the drive voltages to grid2 and grid1 do not swing slightly negative to compensate. The Twin/Crazy drive linearization scheme still needs to have its divider R's connected to the cathode. So a resistor Rb (or Rbias) needs to be added from the Mosfet source output to a negative Vbias voltage. (see diagram below)
Looking at the relative gm's of the tube electrodes:
gm_p = 1/Rp and that gets pushed to 2X B+ during the turn-off phase
while effective gm of the grid2 and grid1 with the Twin drive R divider gives an effective gm of:
gm_drive = gm2 + gm1 x [Rg1k/(Rg2g1+Rg1k)] and that needs to be driven negative by -Vbias through Rbias. (gm_drive is likely going to be NEAR to 2 x gm2 typically, to simplify this mess)
note: gm2 = 1/Mu_internal x gm1
To keep the tube turned off when in the off phase:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] must be greater than gm_p x 2 x B+
So one can solve for -Vbias with a practical pull-down Rb.
Looking at the 26LX6 for example. Rp = 6000 so gm_p = 166 uMhos
and gm1 = 14000 and gm2 = 3500
so gm_drive is approx. = 7000 uMhos (can probably, typically, just simplify this to 2X gm2)
with 2 x B+ swinging up to say 900 V and using say 10K Ohms for Rbias, one would need about -60V for -Vbias. One would still want to test the final circuit to make sure the tubes are turning off properly.
Attachments
Last edited:
For most incandescent light bulbs the slope on a log-log graph is about 0.8. That includes the kind we find in the HP200 Series & some Heathkit Audio Oscillators.
See the attachment, 6L6GC Heater, 1157 Automotive Bulb, 33 vacuum tube filament & 3W bulb as used in an HK Audio Oscillator.
All worth knowing if one is a serious designer.
Last edited by a moderator:
Whoops,
I left out the (gm_drive) term above (in post #875 ) in the equation:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] must be greater than gm_p x 2 x B+
it should read:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] x (gm_drive)
must be greater than (gm_p) x 2 x B+
I think I got it right in the schematic diagram equations.
I left out the (gm_drive) term above (in post #875 ) in the equation:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] must be greater than gm_p x 2 x B+
it should read:
Vbias x [(Rg2g1+Rg1k)/(Rb+Rg2g1+Rg1k)] x (gm_drive)
must be greater than (gm_p) x 2 x B+
I think I got it right in the schematic diagram equations.
Just schemed up a way to reduce the power dissipation in the new Rbias (source to -Vbias) resistor. If one puts a diode in series with Rg1k, then there is no voltage division between Rbias and (Rg2g1+Rg1k) during tube turn off. (see diagram, D1)
So Rbias can be a much higher value now, and -Vbias does not need to be as large.
Only need Vbias x (gm_drive) to be greater than (gm_p) x 2 x B+ now.
Or approx. Vbias x 2 gm2 > gm_p x 2 x B+
or just:
****** Vbias x gm2 > B+/Rp ******* so: Vbias > B+/(gm2 x Rp) ******** gm2 = gm1/Mu_internal
a nice simple design rule now!
D1 should be a Schottky diode to minimize forward voltage drop. (like 0.1 Vf, and max V_reverse greater than Vbias)
So for the 26LX6 example now:
gm2 = 3500 uMho, Rp = 6000 Ohm, and say B+ = 450V
so Vbias x 3500/1000000 > 450/6000
or Vbias should be greater than -22 Volts.
and Rbias could be a 33K Ohms 1/2 Watt resistor (allowing for +100V drive V swing current). Very practical!
****** Edit:
Since now no current goes through Rg2g1 and Rg1k during turn-off (so no V division for grid1), the effective gm is more like gm2+gm1. So even less -Vbias is needed. So rule is revised to approx.
******* Vbias > B+/(gm1 x Rp) ******
so -Vbias for the 26LX6 case would be more like -5.5 Volts.
So Rbias can be a much higher value now, and -Vbias does not need to be as large.
Only need Vbias x (gm_drive) to be greater than (gm_p) x 2 x B+ now.
Or approx. Vbias x 2 gm2 > gm_p x 2 x B+
or just:
****** Vbias x gm2 > B+/Rp ******* so: Vbias > B+/(gm2 x Rp) ******** gm2 = gm1/Mu_internal
a nice simple design rule now!
D1 should be a Schottky diode to minimize forward voltage drop. (like 0.1 Vf, and max V_reverse greater than Vbias)
So for the 26LX6 example now:
gm2 = 3500 uMho, Rp = 6000 Ohm, and say B+ = 450V
so Vbias x 3500/1000000 > 450/6000
or Vbias should be greater than -22 Volts.
and Rbias could be a 33K Ohms 1/2 Watt resistor (allowing for +100V drive V swing current). Very practical!
****** Edit:
Since now no current goes through Rg2g1 and Rg1k during turn-off (so no V division for grid1), the effective gm is more like gm2+gm1. So even less -Vbias is needed. So rule is revised to approx.
******* Vbias > B+/(gm1 x Rp) ******
so -Vbias for the 26LX6 case would be more like -5.5 Volts.
Attachments
Last edited:
Oops,
I forgot the (2 x B+) factor in the last formula using gm1 only.
The Vbias rule should be approx.:
*************** Vbias > 2 x B+ / (gm1 x Rp) *****************
so the 26LX6 case would need a minimum of -11V for Vbias for 450V B+. Rbias still around 33K at 1/2 Watt to allow for the positive grid2 drive phase current through it (ie, allowing for +100V drive).
Guess I better get my 6HJ5 Twin Drive amp finished up now. Need to prove this out.
For 6HJ5: for B+ at 450V, gm1 = 10,000 uMho, Rp = 5000
so -Vbias = 18 Volts
note:
gm1 and Rp need only be specified at the same current, not necessary to use the actual design current. We are just figuring a relative Mu between grid1 and the plate for the shut-off function.
(so datasheet values are fine for this)
..
I forgot the (2 x B+) factor in the last formula using gm1 only.
The Vbias rule should be approx.:
*************** Vbias > 2 x B+ / (gm1 x Rp) *****************
so the 26LX6 case would need a minimum of -11V for Vbias for 450V B+. Rbias still around 33K at 1/2 Watt to allow for the positive grid2 drive phase current through it (ie, allowing for +100V drive).
Guess I better get my 6HJ5 Twin Drive amp finished up now. Need to prove this out.
For 6HJ5: for B+ at 450V, gm1 = 10,000 uMho, Rp = 5000
so -Vbias = 18 Volts
note:
gm1 and Rp need only be specified at the same current, not necessary to use the actual design current. We are just figuring a relative Mu between grid1 and the plate for the shut-off function.
(so datasheet values are fine for this)
..
Last edited:
Interesting observation for Twin/Crazy Drive.
If you set the -Vbias to a level that -just- turns the tubes off when 2 x B+ is on the plates, during the off phase. Then the tubes will start to conduct if the plate voltage exceeds that level for any reason. Such as no speaker connected! An automatic safety function for the OT. I would still put some over-voltage protection devices (like spark gap devices or MOVs) across the OT, like for any tube amplifier. But nice to know the tubes will also perform that safety clamping here if all else fails.
If you set the -Vbias to a level that -just- turns the tubes off when 2 x B+ is on the plates, during the off phase. Then the tubes will start to conduct if the plate voltage exceeds that level for any reason. Such as no speaker connected! An automatic safety function for the OT. I would still put some over-voltage protection devices (like spark gap devices or MOVs) across the OT, like for any tube amplifier. But nice to know the tubes will also perform that safety clamping here if all else fails.
Very interesting reading!
A interesting resolution then results for a hypotetical power of one device ("linear power law"):
gm = n x (nth root of k) x I^[(n-1)/n]
gm = 1 x (k ^ 1/1) x I^[(1-1)/1]
gm = 1 x k x 1
gm = k (?!
)About PP with common special value cathode resistor, some publication in 195x talked about it, and in 2013 I made some experiments. Yes, for a fixed loadline (aka. resistor), I'm able to reduce a lot of 3H from a 6EM5 pair when I carefully adjust the common resistor. But if I change the load, the effect sometimes vanishes, or reduces a lot. So this nulling is current-dependent, perhaps...
You always want a current flowing through the mosfet. It needs to be a few mA or more.
I learned this back when working on screen drive amps. The mosfet has a bunch of transconductance. It effectively drops to zero as the current through it ceases. It snaps back rather abruptly as the current begins to flow again, and is often the time that RF oscillation bursts will occur.
In a normal P-P driver circuit the driver voltage will attempt to cut one output tube off while the other is driven towards saturation. If the mosfet gets cut off, it may emit a burst of RF or high frequency oscillation as it returns to conduction. Sometimes this had no effect on sound quality, but makes nearby TV's and radios go nits, other times it imparts a harshness to the treble. Even when driving the screen alone, I returned the screen to the negative rail.
Note the burst on the scope as the amp is driven to clipping with some Depeche Mode cranked up to 11.
I learned this back when working on screen drive amps. The mosfet has a bunch of transconductance. It effectively drops to zero as the current through it ceases. It snaps back rather abruptly as the current begins to flow again, and is often the time that RF oscillation bursts will occur.
In a normal P-P driver circuit the driver voltage will attempt to cut one output tube off while the other is driven towards saturation. If the mosfet gets cut off, it may emit a burst of RF or high frequency oscillation as it returns to conduction. Sometimes this had no effect on sound quality, but makes nearby TV's and radios go nits, other times it imparts a harshness to the treble. Even when driving the screen alone, I returned the screen to the negative rail.
Note the burst on the scope as the amp is driven to clipping with some Depeche Mode cranked up to 11.
Attachments
About cut-off in some tubes...
...for types like PL509, some negative bias is needed for g1 or g2 for turn-off them (I noted in my measurements, too), since they conduct some current*** with Vg1=0 and Vg2=0 and anode with several 100's volts (just when need to cut-off in output...), so for Twin Drive I imagine some "heavy" negative bias to turn-off completely for these tubes, more than the smoking-amp examples (time to calculate).
I can return to playing with this next week
LATE EDIT:*** or I can say: a LOT of current for PL509...
...for types like PL509, some negative bias is needed for g1 or g2 for turn-off them (I noted in my measurements, too), since they conduct some current*** with Vg1=0 and Vg2=0 and anode with several 100's volts (just when need to cut-off in output...), so for Twin Drive I imagine some "heavy" negative bias to turn-off completely for these tubes, more than the smoking-amp examples (time to calculate).
I can return to playing with this next week
LATE EDIT:*** or I can say: a LOT of current for PL509...
Last edited:
"A interesting resolution then results for a hypothetical power of one device ("linear power law"):"
"gm = 1 x k x 1
gm = k (?!)"
-
Yes, for a linear device, gm is a constant. k
-------------------
"You always want a current flowing through the mosfet. It needs to be a few mA or more. "
-
Agreed, the Mosfet needs some operating current at all times as you said. Which the Rb from the Mosfet source terminal to a negative Vbias will provide. And the tube also needs a minimum negative V on g2 and g1 to fully turn it off, when plate V is 2 x B+. So the question is how much?
the formula (from post # 879):
Vbias > 2 x B+ / (gm1 x Rp)
is based on the Mu factor between plate and grid1 staying constant down to 0 current. (which it doesn't quite stay at, see below)
-------------------
"for types like PL509, some negative bias is needed for g1 or g2 for turn-off them (I noted in my measurements, too), since they conduct some current*** with Vg1=0 and Vg2=0 and anode with several 100's volts (just when need to cut-off in output...), so for Twin Drive I imagine some "heavy" negative bias to turn-off completely for these tubes, more than the smoking-amp examples (time to calculate)."
-
The formula I derived above (post # 879 ) assumes a constant Mu factor between plate and grid1 (& grid2) for the pentode, down to 0 current. It only considers grid1, since grid2 is 1/Mu less effective.
Vbias > 2 x B+ / (gm1 x Rp)
Plugging in the 6KG6/PL509 #'s
at 250 mA: Rp = 3600 and gm1 = 16666 uMhos
So for 500V B+
the formula gives Vbias > 1000/(0.01666 x 3600) = 16.66 Volts (minus)
However, looking at the 6KG6/PL509 datasheet curves, the Mu factor is changing with current as the 0.25 power. So its average value down to zero current is going to be something less, so grid1 will be slightly less effective toward cut-off. So grid1 will have to go further negative than the simple formula estimated. If I integrate the 0.25 power rule for Mu (down to zero current) it comes out with an average of 0.8 of the initial Mu. So the simple rule needs to be modified to: [note: 1/0.8 = 1.25]
Vbias > 1.25 x 2 x B+ / (gm1 x Rp)
giving Vbias > 20.8 Volts (minus) for the PL509 with 500V B+
but then if we take grid2 into consideration also (Mu = 4), then the combined grid1 and grid2 at -Vbias will be 1.25 x as effective. So we can reduce Vbias back to:
******* Vbias > 2 x B+ /(gm1 x Rp) ********** the original simple formula!
so Vbias is back to -16.66 V for the PL509 with +500V for B+
Another check on this is to look at page 2 of the PL509 datasheet for CUT-OFF VOLTAGE. Working those numbers down to Vg2 = 0 gives -115V on grid1 for 7000 V on the plate, as cut-off. So for 1000 V on the plate, that would give approx. 1/7 that, or -16.43 V on grid1 for cutoff. Agreeing well with the simple formula above.
However this does not take into consideration that grid2 is also going down to -Vbias. With a Mu of 4 for the above #'s, that would give an effectiveness of 1+1/Mu = 1 + 0.25. So we could drop Vbias (for grid1 and grid2 combined) to 16.42/1.25 = -13 Volts.
we now have Vbias (grid1 and grid2) at -13V, -16.66V, -16.43V
The truth may be out there somewhere!
Let us know what the PL509 actually needs on grid1 and grid2 combined to shut it off, with +1000 V on the plate. (usually spec'd at 1 mA)
https://frank.pocnet.net/sheets/010/p/PL509.pdf
https://frank.pocnet.net/sheets/084/6/6KG6.pdf
..
"gm = 1 x k x 1
gm = k (?!)"
-
Yes, for a linear device, gm is a constant. k
-------------------
"You always want a current flowing through the mosfet. It needs to be a few mA or more. "
-
Agreed, the Mosfet needs some operating current at all times as you said. Which the Rb from the Mosfet source terminal to a negative Vbias will provide. And the tube also needs a minimum negative V on g2 and g1 to fully turn it off, when plate V is 2 x B+. So the question is how much?
the formula (from post # 879):
Vbias > 2 x B+ / (gm1 x Rp)
is based on the Mu factor between plate and grid1 staying constant down to 0 current. (which it doesn't quite stay at, see below)
-------------------
"for types like PL509, some negative bias is needed for g1 or g2 for turn-off them (I noted in my measurements, too), since they conduct some current*** with Vg1=0 and Vg2=0 and anode with several 100's volts (just when need to cut-off in output...), so for Twin Drive I imagine some "heavy" negative bias to turn-off completely for these tubes, more than the smoking-amp examples (time to calculate)."
-
The formula I derived above (post # 879 ) assumes a constant Mu factor between plate and grid1 (& grid2) for the pentode, down to 0 current. It only considers grid1, since grid2 is 1/Mu less effective.
Vbias > 2 x B+ / (gm1 x Rp)
Plugging in the 6KG6/PL509 #'s
at 250 mA: Rp = 3600 and gm1 = 16666 uMhos
So for 500V B+
the formula gives Vbias > 1000/(0.01666 x 3600) = 16.66 Volts (minus)
However, looking at the 6KG6/PL509 datasheet curves, the Mu factor is changing with current as the 0.25 power. So its average value down to zero current is going to be something less, so grid1 will be slightly less effective toward cut-off. So grid1 will have to go further negative than the simple formula estimated. If I integrate the 0.25 power rule for Mu (down to zero current) it comes out with an average of 0.8 of the initial Mu. So the simple rule needs to be modified to: [note: 1/0.8 = 1.25]
Vbias > 1.25 x 2 x B+ / (gm1 x Rp)
giving Vbias > 20.8 Volts (minus) for the PL509 with 500V B+
but then if we take grid2 into consideration also (Mu = 4), then the combined grid1 and grid2 at -Vbias will be 1.25 x as effective. So we can reduce Vbias back to:
******* Vbias > 2 x B+ /(gm1 x Rp) ********** the original simple formula!
so Vbias is back to -16.66 V for the PL509 with +500V for B+
Another check on this is to look at page 2 of the PL509 datasheet for CUT-OFF VOLTAGE. Working those numbers down to Vg2 = 0 gives -115V on grid1 for 7000 V on the plate, as cut-off. So for 1000 V on the plate, that would give approx. 1/7 that, or -16.43 V on grid1 for cutoff. Agreeing well with the simple formula above.
However this does not take into consideration that grid2 is also going down to -Vbias. With a Mu of 4 for the above #'s, that would give an effectiveness of 1+1/Mu = 1 + 0.25. So we could drop Vbias (for grid1 and grid2 combined) to 16.42/1.25 = -13 Volts.
we now have Vbias (grid1 and grid2) at -13V, -16.66V, -16.43V
The truth may be out there somewhere!
Let us know what the PL509 actually needs on grid1 and grid2 combined to shut it off, with +1000 V on the plate. (usually spec'd at 1 mA)
https://frank.pocnet.net/sheets/010/p/PL509.pdf
https://frank.pocnet.net/sheets/084/6/6KG6.pdf
..
Last edited:
I finally got a 40KG6 tube to test here. (ie PL509) This is a Raytheon tube, made in Japan. Looks to be well made and measures well. Even has the RCA dark emitter heater.
All are curve traced at 50 mA/div Vert. and 50V/div Horiz.
1) normal g1 drive, -1.5V steps on g1, 78V on g2, top curve is 0V on g1
2) g2 drive, +5.5V steps on g2, g1 at 0V, bottom curve is 0V on g2
3) triode mode, -7.5V steps on g1, top curve is 0V on g1
4) Twin/Crazy drive (g2 and g1 with resistors), +3.5 V steps on g2, Rg2g1 = 888 Ohm, Rg1k = 2770 Ohm, bottom curve is 0V on g2 (and g1)
All are curve traced at 50 mA/div Vert. and 50V/div Horiz.
1) normal g1 drive, -1.5V steps on g1, 78V on g2, top curve is 0V on g1
2) g2 drive, +5.5V steps on g2, g1 at 0V, bottom curve is 0V on g2
3) triode mode, -7.5V steps on g1, top curve is 0V on g1
4) Twin/Crazy drive (g2 and g1 with resistors), +3.5 V steps on g2, Rg2g1 = 888 Ohm, Rg1k = 2770 Ohm, bottom curve is 0V on g2 (and g1)
Attachments
Last edited:
The curves I posted for the 40KG6/PL509 for Twin (Crazy) drive reached 1/2 an Amp at the max. What I have found is that if one goes to higher currents than 1/2 an Amp, the value of the Rg2g1 resistor needs to increase. Otherwise the grid 1 begins to suck up all the current at low plate voltage, causing the plate curve knees to move to higher voltage at the higher currents (see below). Not good for the grid 1 staying cool or for distortion correction. So I would linearly scale up the value of Rg2g1 (was 880 Ohms I think for the 40KG6 at 1/2 Amp max) with plate currents higher than 1/2 an Amp. Rg1k seems fairly insensitive, but could be made proportionately larger too, to reduce drive requirements on the previous stage. (One may be able to more closely optimize Rg2g1 with a curve tracer at the higher currents. I suspect the scaling might actually need to go up with the grid 2 drive voltage, which is not quite linear with current for grid 2 drive alone.)
Curves below for 6HJ5 in Twin drive with reducing value of Rg2g1 to observe effect on plate curve knees. (all 50 mA/div Vert., 50V/div Horiz.)
1) Rg2g1 = 1780 Ohms
2) Rg2g1 = 700 Ohms
3) Rg2g1 = 600 Ohms
Obviously, currents above 1/2 an Amp will require higher values of Rg2g1 to avoid overheating grid 1.
The effective tube gain (gm) for Twin drive will reduce with higher values of Rg2g1, eventually approaching that of grid 2 drive alone.
--
Curves below for 6HJ5 in Twin drive with reducing value of Rg2g1 to observe effect on plate curve knees. (all 50 mA/div Vert., 50V/div Horiz.)
1) Rg2g1 = 1780 Ohms
2) Rg2g1 = 700 Ohms
3) Rg2g1 = 600 Ohms
Obviously, currents above 1/2 an Amp will require higher values of Rg2g1 to avoid overheating grid 1.
The effective tube gain (gm) for Twin drive will reduce with higher values of Rg2g1, eventually approaching that of grid 2 drive alone.
--
Attachments
Last edited:
More research...
Some graphs from last try:
All resistor values in graphs are in Rg2-g1 and Rg1-gnd.
Note that my tracer don't output less than 2V in screen and anode output.
Interesting to note that one tube need strong Twin Drive to make gm (almost) constant, and the Matsushita sample seems to be easy to obtain.
For these measutements a analog tracer is a must, since even if this tracer is very precise, is very slow too, so I spend almost 2 hours to found a good point and make this graphs.
Other thing I've noted is that the flat gm is almost impossible to obtain for all anode voltages/conditions, and for being constant for most points a critical trimming is needed. Even so, considering that all devices have at least some deviating from ideal gm curve, the results will be very practical. ((considering that my tracer is 100% ok))
Some graphs from last try:
All resistor values in graphs are in Rg2-g1 and Rg1-gnd.
Note that my tracer don't output less than 2V in screen and anode output.
Interesting to note that one tube need strong Twin Drive to make gm (almost) constant, and the Matsushita sample seems to be easy to obtain.
For these measutements a analog tracer is a must, since even if this tracer is very precise, is very slow too, so I spend almost 2 hours to found a good point and make this graphs.
Other thing I've noted is that the flat gm is almost impossible to obtain for all anode voltages/conditions, and for being constant for most points a critical trimming is needed. Even so, considering that all devices have at least some deviating from ideal gm curve, the results will be very practical. ((considering that my tracer is 100% ok))
Attachments
-
PL509 Matsushita 1k2-1k4 Va-Ia result.png -
PL509 Matsushita 1k2-1k4 gm.png -
PL509 Matsushita 1k2-1k4 g2 current.png -
PL509 Matsushita 1k2-1k g2 current.png -
PL509 pentode 80Vg2.png -
PL509 unknow 2k4-880 gm.png -
PL509 unknow 2k4-880.png -
PL509 unknow 210-4k4 g2 current.png -
PL509 unknow 210-4k4 gm.png -
PL509 unknow g2.png
Is good to have a multi condition gm tracing to see the gm deviations.
The "unknow" brand tube seems to be Ultron or even RSD.
LATE EDIT: the unknow tube is like a bipolar transistor with very low beta. Needs 100mA in g2-g1 assembly to obtain 200mA in anode and flatten the gm curve. Unsuitable... But is good in plain g2 drive, with good "current gain"... and is this tube I've tried in November. If I tried this Matsushita first.....
The "unknow" brand tube seems to be Ultron or even RSD.
LATE EDIT: the unknow tube is like a bipolar transistor with very low beta. Needs 100mA in g2-g1 assembly to obtain 200mA in anode and flatten the gm curve. Unsuitable... But is good in plain g2 drive, with good "current gain"... and is this tube I've tried in November. If I tried this Matsushita first.....
Last edited:
Only for fun...
...and for comparing with our beloved tubes , I show some semiconductor curves (ohhh no... heresy....!!!!)
Note the good beta linearity for a ancient Ge japanese transistor, 2SB481, and the resulting curves. Note the bunching at high currents for the ye olde RCA2955, corresponding with beta decay. The Japanese 2SC1226 is good, unlike the RCA, and better than this exists.
Note the MOSFETs (ahhh.... one thing we use with our tubes ) and their strong difficulty to turn-off, like an tube with semi-remote cut-off, but have high transconductance even at this point. The gm curve I saved only in the tracer format (.cuv), so I need to re-open and plot in png. But is a boring more-than-square-law curve (instead of being straigh tilted, is bowed like from PL509 in pentode mode.
And strange things occurs with overloading a signal BJT, see BC109 curves.
LATE EDIT: this HV MOSFET will be a excellent CCS
...and for comparing with our beloved tubes
, I show some semiconductor curves (ohhh no... heresy....!!!!)Note the good beta linearity for a ancient Ge japanese transistor, 2SB481, and the resulting curves. Note the bunching at high currents for the ye olde RCA2955, corresponding with beta decay. The Japanese 2SC1226 is good, unlike the RCA, and better than this exists.
Note the MOSFETs (ahhh.... one thing we use with our tubes
) and their strong difficulty to turn-off, like an tube with semi-remote cut-off, but have high transconductance even at this point. The gm curve I saved only in the tracer format (.cuv), so I need to re-open and plot in png. But is a boring more-than-square-law curve (instead of being straigh tilted, is bowed like from PL509 in pentode mode.And strange things occurs with overloading a signal BJT, see BC109 curves.
LATE EDIT: this HV MOSFET will be a excellent CCS
Attachments
Last edited:
On the g2 and Twin drive for PL519/509 tube, I just remembered this set of g2 drive curves online (left/below).
This set has a -33V permanently on grid1, which is causing the big kinks in the g2 curves. For some reason, a minus Vbias on grid1 has often been considered necessary for grid 2 drive.
I think the problem lies in the fact that 0V on grid1 and 0V on grid2 will NOT keep the tube turned off when the P-P OT swings to +1000V or more during the opposite phase (2 x B+). But this issue is solved by having the grid2 (and grid1 in Twin drive) swing sufficiently negative during the turned-off phase. No need then for the permanent -Vg1 (causing kinks).
Minimum -Vg2 is approx.:
-Vg2 turnoff = (2 x B+ x Mu)/(Rp x gm1) where Mu is for g2/g1 internal.
right/below is 40KG6 in g2 drive with 0V on grid1 (no kinks) [ 50 mA/div Vert., 50V/div. Horiz.]
This set has a -33V permanently on grid1, which is causing the big kinks in the g2 curves. For some reason, a minus Vbias on grid1 has often been considered necessary for grid 2 drive.
I think the problem lies in the fact that 0V on grid1 and 0V on grid2 will NOT keep the tube turned off when the P-P OT swings to +1000V or more during the opposite phase (2 x B+). But this issue is solved by having the grid2 (and grid1 in Twin drive) swing sufficiently negative during the turned-off phase. No need then for the permanent -Vg1 (causing kinks).
Minimum -Vg2 is approx.:
-Vg2 turnoff = (2 x B+ x Mu)/(Rp x gm1) where Mu is for g2/g1 internal.
right/below is 40KG6 in g2 drive with 0V on grid1 (no kinks) [ 50 mA/div Vert., 50V/div. Horiz.]
Attachments
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.