OT Primaries

mcentee2

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Been doing a lot of reading on the JTM45 but can't get to the bottom of this one!

Just why is it that 2x el34 like an OT primary of about 3.6k, and 2x kt66 like one of about 8k ?

I think it's to do with things like transconductance values, anode resistance, max power transfer Vs distortion and similar, as well of choice of operating point but can't untangle it.

I know things like frequency dependant impedances not being linear, and grid curves behaviour under load, bias shifts etc ..... Also that the final choice of OT primary is just that, a choice that fits with the overall desired "tone" and behaviour, and that there is a practical range that is used in designs for each tube type, and they can overlap.

I have used vtadiy to see what common JTM45 type voltages look like, and then for jtm50/Plexi, and really can't see much speaking to me :(

The only obvious thing to me is that the el34 load line for 3.6k is really steep and goes way above the 25w limit for most of its travel, whereas the kt66 chart stays well within its watts limit.

No idea why one is "good" and the other still "good" but so different approach.

I have looked at whatever Merlin, Aiken, Rob say, as well as looking at data sheets for things line Ra of the tubes, and Rp-p, and power transfer curves (can only find one for kt66 not one for el34), worked through load line comparisons, but can't seem to understand the fundamental driving parameters behind that "standard" OT :(
 
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2L man

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A push pull loadline can go above tubes max power curve because tube operates there only 50% time and heat does not come too high.

EL34 often have 1k or bigger Screen resistors and it can be used to control max anode current. When Screen voltage on vtadiy-calculator is lowered, notice how Grid 0V curve gomes lower. Obviously this is major source for sound compression and sag?
 

andrewRneumann

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I have a few quick observations about this as it relates to the typical power pentode/beam tetrode in Class A or AB.

1. The optimal load line from an efficiency/power stand point goes through the knee of the 0Vgk curve. Why? Because at this point, you get the most current swing AND the most voltage swing. Remember power is a function of voltage AND current--you need as much of both as you can get. If the load impedance is too low, the tube reaches Vgk=0 to the right of the knee and voltage swing is limited. If the load impedance is too high, the tube reaches Vgk=0 below the knee--well before it reaches it maximum current.

2. So what sets the knee of the curve? Generally, that is a function of gm (transconductance) and screen voltage. A tube with higher transconductance will conduct more current for a given input signal--and therefore needs a load with less impedance for maximum power delivery.

So that explains a little of why the 2 tubes have different typical load impedances specified.
 

Ten Over

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1. The optimal load line from an efficiency/power stand point goes through the knee of the 0Vgk curve. Why? Because at this point, you get the most current swing AND the most voltage swing. Remember power is a function of voltage AND current--you need as much of both as you can get. If the load impedance is too low, the tube reaches Vgk=0 to the right of the knee and voltage swing is limited. If the load impedance is too high, the tube reaches Vgk=0 below the knee--well before it reaches it maximum current.
I think that what you mean is that the product of the current times the voltage maxes out around the knee of the 0Vgk curve yes/no? Neither the current nor the voltage are at their maximums at this point.
 

andrewRneumann

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I think that what you mean is that the product of the current times the voltage maxes out around the knee of the 0Vgk curve yes/no? Neither the current nor the voltage are at their maximums at this point.

details, details… maximums, minimums… good catch. :confused: What I should have said is that maximum voltage swing simultaneous with maximum current swing will occur when the load line goes through or near the knee. You can get more current swing by using less load impedance, but it comes at the expense of voltage swing. Conversely, you can get more voltage swing by using more load impedance but it comes at the expense of current swing. Is that better?
 

Ten Over

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What you don't have on the vtadiy.com graph is the screen grid current. In fact, the data sheets for EL34 and KT66 are pretty stingy with that info, but these tubes act just like other tubes when it comes to screen grid current. The screen grid current increases when the plate voltage decreases. The screen grid current skyrockets as the Vgk approaches zero and the plate voltage approaches its minimum. If the load impedance is too high, the load line will intersect the 0Vgk at a lower plate voltage. This small change in plate voltage from the knee voltage can have a dramatic effect on the screen current and it just might put the screen grid into over dissipation.
 

mcentee2

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Ok, I think I am getting closer.

It's more or less the choice of getting the load line first estimate near that knee more than anything ?

And that is driven by where the screen grid Vg lines fall, and by where you choose your quiescent bias point.


If I follow the above, whilst the el34 with 3.6k hits that knee just fine, the kt66 at 8k (or 7k) seems to hit the 0Vg curve well well below the knee, so that's where all that other info needs to be available to judge from :(

Edit: just caught this older thread on TGP, with some brilliant teaching from user hotblueplates

https://www.thegearpage.net/board/i...ce-with-6l6-5881-tubes.1925165/#post-32672532

Reading that and a few other posts by hotblueplates it looks like indeed that JTM45 OT isn't necessarily what you would "choose" as optimum at those voltages to get most power out.

It could easily use the same 8k OT and get similar output power to the later el34 jtm50/Plexis.

Marshall's choice was prob made on what was available, and also poss not necessarily reading all the technicals at the time just to get the amp built and out to market :)

Benefit of doubt is that they knew exactly what they were doing and chose that OT as it gave the amp some OT distortion characteristics that were preferable to raw power....
 
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printer2

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I have not read it for a while but this might help.

http://www.turneraudio.com.au/loadmatch-4C-PP-tetrodes-pentodes.html

Also a bunch of random posts that I thought might be of interest.
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I built one, but used a 6600 ohm OPT since that is what I had. I got more power output (about 30 WPC) since I was using up more of the tubes current capability with the lower impedance load. Not satisfied, I turned up the B+ voltage. I found the limit for the 6JN6's. With 520 volts of B+ and a 6600 ohm load I got 70 WPC from the board. Any attempt to lower the load or increase the B+ further was met with higher distortion, red glow from the tubes, or both. I had reached the dissipation AND peak current carrying capability of the tubes. All tubes with a 17 watt dissipation rating seemed limited to about 70 WPC.
I swapped the output tubes for 6HF5's which have a higher dissipation and peak current carrying capability. Resuming the testing with the SAME conditions, I got the same 70 WPC. I could increase the B+ voltage to get more power but I was already pushing the limit of the capacitors on the board, so I reduced the load impedance. At 3300 ohms I could get 125 WPC. So using bigger output tubes allows me to use a lower load impedance to get more power. Attempts to use a 2500 ohm load were met with glow and distortion. I tried several tubes in the 24 to 28 watt dissipation class and found 110 to 125 watts to be the limit.
Again, I reached for bigger tubes. This time I used 6KD6's and I could get to 2500 ohms, still at 520 volts and got 150 WPC. I believe that none of the well used 6KD6's were in good condition limiting the power somewhat. I tried other tubes in the 30 to 35 watt range and got 150 to 175 WPC.

I modified the board to handle more voltage and installed some NOS 35LR6's. With 650 volts and a 2500 ohm load I could get 250 WPC. These were 35 watt tubes, but operating over spec. This was from the same board that Pete designed for 17 WPC.
I currently have E130L output tubes in the amp (27 watt dissipation rating). These have a transconductance rating of 27500. THe higher transconductance does not allow for more output power, it does however bring the input sensitivity down where I can hit clipping (120 watts) with 1/2 a volt of drive.
I have stuffed at least 20 different output tubes into this board and I am not done yet. What have I learned?
Given a good driver design the output tube determines the maximum output power CAPABILITY of a given amp. The operating conditions applied to those output tubes may use all or part of that capability. This is a trade off between reliability, power output, distortion, and transient handling.
Within reason the transconductance of the output tubes only affects the amps input sensitivity. If negative feedback is used around the output tubes, the transconductance will affect the feedback level. Swapping to a tube with a different transconductance will require tweaking the feedback components.
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Actually, computing mu and gm from curves is fairly simple for triodes, and is derived directly from the definitions of the terms.

mu is the change of Vpk per unit of change of Vgk, at constant plate current.

Therefore, pick an operating poit Vp, Ip on the curves. Then draw a horizontal line through this point. Take a convenient change in Vg from the one at the operating point, and move along your horizontal line to this different (delta) Vg, then check on the X axis (Vp) how much the Vp changes with your chosen Vg change. Divide the two (delta Vp / delta Vg) and you have mu. Typically, you chose an operating point on one of the Vg curves. This is because it makes it simple to see what the change of Vp is for the adjecent Vg curves on the left and right, along a horizontal line through the operating point.

gm is the shange in plate current Ip with change in grid voltage Vg, at constant plate voltage Vp.

Therefore, chose an operating point on the curves. This time draw a vertical line through the operating point, epresenting a constant Vp. Again, move up and down on this line for a convenient change in Vg, checking on the Y axis of the curves (Ip) how much Ip changes. Divide the change in Ip with Vg and you have gm. Again, chosing an operating point on one of the Vg lines in the curve diagram siplifies things considerably.
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also, if you have 2 in the sheet, you can work the other one out.
its oddly a little similar to ohms law

u = Gm rP

rp = u / gM

ignore cases
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Average Plate Characteristics, page 3, top diagram. This is the one you need for both mu and gm. Each curve in the diagram shows the relatinship between plate voltage (X axis) and plate current (y axis) for a given constant grid voltage.

Here is an example. Let's say we chose an operating point at Iplate = 1.2mA and Vplate = 200V. You will find that this point sits exactly on the -1.5V grid voltage line, which is why I chose it - simple convenience.

Now, draw a horizontal line through this point, these are all points in the diagram where plate current is 1.2mA. Moving to the next grid voltage line left and right of the -1.5V along this line, and reading what the plate voltage is for a +-0.5V grid voltage change will give you the plate voltage change for that constant plate current, 1.2mA, now divide the plate voltage change with grid voltage change and you get mu. In this example, you get Vplate = 155V @ Vg = -1V and Vplate = 252V @ Vg = -2V, both following the constant current line of 1.2mA. (252 - 155)/(-1 -(-2))=97, and this is the mu of the 12AX7 at the chosen operating point.

For gm, we do a similar thing, but drawing a vertical line through the chosen operating point, to represent all points in the graph at which Vplate = 200V. Again, we move to the next grid voltage line this time up and down following the 200V line, and read the currents we get: Iplate = 2.13mA @ Vg=-1V, and Iplate = 0.53mA @ Vg=-2V. (2.13 - 0.53)/(-1-(-2)) = 1.6 mA/V, and this is our gm.

In reality, the figures are approximate - if you look carefully, the error is readily apparent for the gm calculation - the distance between the -1 and -1.5V grid voltage line and -1.5V and -2V grid voltage line along the Vplate = 200V line is not the same, and the calculation assumes that it is. This can often be the case as it is impractical to draw many grid voltage lines on the plate characteristics diagram because the places where they are close would become too cluttered to make any usefull reading. Still, unless you are looking at gross asymetry as you move to the next grid voltage line over, you will get very usable results, the error will likely be smaller than actual tube tolerances, anyway.

If you also look carefully, you will notice that for the given operating current of 1.2mA, the grid voltage line intersectios with the 1.2mA line are very evenly spaced. This very graphically shows the linearity of a tube used as a pure voltage (plate current is constant as in the mu calculation) vs pure current (plate voltage is constant as in the gm calculation) amplifier. The 12AX7 was made for linear voltage amplification, and this clearly shows. The procedure for deriving mu also shows you why a triode with a current source in the plate circuit has an amplification factor equal to mu - because such a connection drives the triode exactly the way mu is defined - with a constant plate current.
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look at page 4 of that datasheet. You will see that mu remains fairly constant with varying plate current but both Gm and Rp vary a lot. The lower the plate current, the lower the Gm and the higher the Rp. Page 4 shows you the relationships. With a triode, you cannot simply compute Gm from mu and Rp, you have to take into account the operating current.

When Gm and Rp are quoted in datasheets, they are quoted for specific operating conditions. You will see, at the foot of Page 1 under CHARACTERISTICS AND TYPICAL OPERATION, that different values are given for both Gm (1,250 and 1,600 umho) and Rp (80,000 and 62,500 k) at two different plate currents (0.5 and 1.2 mA.) Mu, however, is 100 at both currents.
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mcentee2

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Many thanks, and I actually ended up understanding most of those points, remarkable :) well, maybe not completely but certainly to a better level than I expected!

The posts on TGP by your good self and hotblueplates amongst others gave me the eye opening words around all this, especially wrt that Marshall "decision" to use 8k.

That was where my question came from as to "why", and it turns out not really to do with any "because that's the engineering best practice due to rules X, y, z for kt66" answer I was looking for, but more a production and availability expedience answer :)

Kt66 can run just as happily on a 3.6k OT, give out more power etc etc just like an el34 or, within whatever limits are chosen/desired.
 

andrewRneumann

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Many thanks, and I actually ended up understanding most of those points, remarkable :) well, maybe not completely but certainly to a better level than I expected!

The posts on TGP by your good self and hotblueplates amongst others gave me the eye opening words around all this, especially wrt that Marshall "decision" to use 8k.

That was where my question came from as to "why", and it turns out not really to do with any "because that's the engineering best practice due to rules X, y, z for kt66" answer I was looking for, but more a production and availability expedience answer :)

Kt66 can run just as happily on a 3.6k OT, give out more power etc etc just like an el34 or, within whatever limits are chosen/desired.

Right on.

Also remember the tube characteristics w/load line are a snap shot at one and only one plate voltage and screen voltage. When the amp is driven to it's maximum power, those voltages can sag dramatically. Unless you have a perfectly stiff power supply on the plates and screens, what the actual current and voltage will be at Vgk=0 is a bit of guess work. Once the amp is built, you can observe the sag and make a better estimate.

So now you have to go back and look at the power supply and see what they have going on there. Whenever I have time to sit down to design a power amp from scratch, I find I am really designing the power tubes/OT and the power supply simultaneously. Since the power tubes pull 95% of the current, they cannot be examined in isolation of the power supply.
 

printer2

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Came across this, makes you appreciate the extra elements in the tubes.

HoOSuD7.jpeg


9HOdQ1T.jpeg


7z0Z2rN.jpeg


XI57dJ8.jpeg
 




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