I'm trying to draw load lines manually instead of relying on an online calculator so I will fully understand how to do it. Since others may use this amplifier after me, possibly wanting/needing more gain, it seems that 6SL7 just makes more sense than 6SN7. A 6SL7 tube to try it isn't expensive, so no problem there.
Have I drawn the first two lines correctly for 1/2 6SL7?
Have I drawn the first two lines correctly for 1/2 6SL7?
Yes, that is the correct DC loadline for the anode load. Signal loadline is rotated clockwise from there, pivoting at the idle point. A cathode resistor line isn't really straight, but close enough for this type (6SL7 is very linear, as can be seen).
All good fortune,
Chris
All good fortune,
Chris
Ok, did I draw these lines correctly? Yellow = AC load line, which does in fact rotate clockwise.
Yes, but remember that the brown lines should end at the yellow line, to be useful.
All good fortune,
Chris
All good fortune,
Chris
the brown lines should end at the yellow line, to be useful.
Oh yes, I see that mistake now. I am using this detailed tutorial to understand it, and I do see that they drew their brown lines that way.
https://robrobinette.com/Drawing_Tube_Load_Lines.htm#Intro
I guess my next question is, what can be done (if anything) to improve this result, and when do we simply say "good enough" and not change anything? In a perfect world, we would not have 92V and 105V, correct?
Most schematics for this amplifier with 6SL7 use 220k for R9 instead of 470k.
That seems to result in this:
That seems to result in this:
This imperfection causes even order harmonics, so yes, but it's not a perfect world. Your goal in choosing an operating (idling) point is to be situated in the best possible/practical spot, on your loadline, that allows enough anode voltage swing to drive the following stage, with a little headroom.In a perfect world, we would not have 92V and 105V, correct?
You'll notice that for triodes, both gain and linearity improve monotonically as you move up and left on the loadline. And also that both improve as the loadline is rotated CCW.
There is a limitation approaching the zero VDC line; the grid begins to conduct approaching zero bias, giving a very non-linear load to the preceding stage (or even the simple source resistance), so you can't expect to swing all the way to zero bias. How close? I'd say a safe number for type 6SL7 is maybe 0.5 to 1.0 VDC.
You're in the deep weeds now!
Chris
This grid leak resistor is limited in size by the output valve's construction, and most folk tend to just believe the datasheet recommendations by manufacturers who are all long gone. The bigger the better, from the 6SL7's POV, but limited by output valve grid current and its sensitivity to heating. Tricky to guess about any current production.Most schematics for this amplifier with 6SL7 use 220k for R9 instead of 470k.
All good fortune,
Chris
enough anode voltage swing to drive the following stage, with a little headroom.
It seems that is what I need to figure out next. My next research topic.
... for triodes, both gain and linearity improve monotonically as you move up and left on the loadline. ...
There is a limitation approaching the zero VDC line ... very non-linear load to the preceding stage ... I'd say a safe number for type 6SL7 is maybe 0.5 to 1.0 VDC.
I was wondering how far it could be pushed in the real world. Thanks.
for triodes, both gain and linearity improve ... as the loadline is rotated CCW.
I can see the gain part now that I have done the drawings above. Not sure I understand the linearity part as loadline rotates though.
You're in the deep weeds now!
I'd say so. It's interesting. 👍
If you draw the extreme case, a horizontal loadline, infinite load impedance, you can see how linear a triode vacuum valve can be - pretty amazing. This is the best possible (but not attainable) condition, and is also how the valve's mu is defined. In contrast, a steeply vertical loadine must sacrifice both gain and linearity.the linearity part as loadline rotates
All good fortune,
Chris
If you draw the extreme case, a horizontal loadline
Oh, I can see that. Hadn't thought of that.
Linearity is shown graphically on the loadline. Perfect linearity means that any change of grid voltage causes the same change in anode voltage, at all levels - the same multiplication. In the perfect condition, all grid lines would be equally spaced at all anode voltages (and currents) but that's reserved for Heaven (is a place where nothing ever happens).Not sure I understand the linearity part as loadline rotates though.
It's actually possible, if you had accurate curves, to directly calculate all distortion products, even by pencil and paper. But eyeball is plenty good enough for designing a valve amplifier. That potential for understandability seems to me to be its own enticement into vacuum valve DIY.
All good fortune,
Chris
all grid lines would be equally spaced at all anode voltages (and currents)
Right. Horizontal line makes it easy to see.
Trying now to figure out just how much voltage swing I'm going to need out of the 6SL7 given pentode mode operation (possibly a 300V regulated screen supply) and the desire to use some GNFB.
So replacing the anode resistor with a constant current source is said to produce a horizontal load line?If you draw the extreme case, a horizontal loadline, infinite load impedance, you can see how linear a triode vacuum valve can be - pretty amazing. This is the best possible (but not attainable) condition, and is also how the valve's mu is defined. In contrast, a steeply vertical loadine must sacrifice both gain and linearity.
All good fortune,
Chris
https://www.audioasylum.com/cgi/vt.mpl?f=tubediy&m=251856
"The output impedance of a stage built with a tube that's using a CCS plate load will be higher than the same tube, same operating point, with a plate load resistor. It can be almost as high as the plate impedance of the tube used, depending on the input impedance of the following stage.
The stage gain will be higher vs a plate resistor. It can be almost as high a the mu of the tube depending on the input impedance of the following stage.
The harmonic distortion will be lower as a horizontal load line avoids the non-linear part of the tube where the mu of the tube varies.
Because of the very high impedance of the CCS vs. a plate resistor the stage will have a higher PSRR.
Note, A tube stage can be built using a plate resistor of a very high value would be the same as a CCS but that isn't practical. The B+ supply voltage would have to be a very high and the resistor would have to dissipate a lot of heat.
A plate choke is a CCS load with high impedance and low DCR so a "lower than normal" B+ voltage is doable."
https://wtfamps.com/ccss-and-loadlines/
I used a 10m45 in my Chinese SE amp with transformative results.
Sorry if I have thrown a spanner in the works!
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"Trying now to figure out just how much voltage swing I'm going to need out of the 6SL7"
Sorry, having incomprehensible issues with attribution, quotations, I'm just not smart enough to work it yet.
So, the 6SL7's anode voltage is directly translated to the output valve's grid, exactly the same signal voltages. The output valve's grid can also not get more positive than zero volts without conducting, so that limits signal swing (it's always symmetrical because not DC coupled) to a peak of bias voltage and a peak-to-peak of twice that (of course). That's how much X axis you need, plus a small headroom.
Worst case, a "Schade" feedback resistor, which introduces an additional load of the resistor's value divided by gain of the output stage plus one. Clever folk get this plan to work, but it's not obvious.
All good fortune,
Chris
Sorry, having incomprehensible issues with attribution, quotations, I'm just not smart enough to work it yet.
So, the 6SL7's anode voltage is directly translated to the output valve's grid, exactly the same signal voltages. The output valve's grid can also not get more positive than zero volts without conducting, so that limits signal swing (it's always symmetrical because not DC coupled) to a peak of bias voltage and a peak-to-peak of twice that (of course). That's how much X axis you need, plus a small headroom.
In parallel with all of the residual parasitic impedances. Following grid leak resistor, Miller C, other strays.So replacing the anode resistor with a constant current source is said to produce a horizontal load line?
Worst case, a "Schade" feedback resistor, which introduces an additional load of the resistor's value divided by gain of the output stage plus one. Clever folk get this plan to work, but it's not obvious.
All good fortune,
Chris
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a peak of bias voltage and a peak-to-peak of twice that
I found that here as well in reply #3.
https://el34world.com/Forum/index.php?topic=12160.msg112757#msg112757
What is the amount of bias my output stage is using?
The value of the "grid 1 driving voltage" or "peak input voltage" will be equal to the absolute value of the bias voltage of the output tubes. You will get all the power output the stage is able to make ... when you drive the grid momentarily to 0v.
That's not so hard.
I used a 10m45 in my Chinese SE amp with transformative results.
I came across that in some preamplifier schematics like this one https://blog.audioworkshop.org/6sn7_preamp_ccs_dn2540n5_ixcp10m45s/ How did you implement it in the chassis? It's a little extra complexity but not much.
I soldered it and the resistor on a small piece of stripboard and then wired it in place of the anode resistor. My amp use a tag strip rather than a pcb so it was easy as there are only the two wires to connect in place of the anode resistor.
No heatsink was needed as I am using 3ma. I haven't yet experimented with higher current through the 6SN7GT so using a preset would help.
It's worth experimenting with even though a ccs has many detractors.
No heatsink was needed as I am using 3ma. I haven't yet experimented with higher current through the 6SN7GT so using a preset would help.
It's worth experimenting with even though a ccs has many detractors.
You select the text (as if to copy it) and as soon as you release the mouse button, you will see a small "quote" appear under the copied text, you click on it and it takes you to a new writing window new message.Sorry, having incomprehensible issues with attribution, quotations, I'm just not smart enough to work it yet.
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