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Old 14th April 2010, 09:18 PM   #1
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Default 6SL7 distortion

I've been trying to understand load lines and how to optimally set up a tube, but thus far I haven't been able to understand how to find out the lowest distortion operation point.

Say I have a 6SL7, with plate at 100 - 110 V (in a direct coupled circuit, so plate voltage has to be around these numbers) - how could I get the least amount of distortion from it?

Any help would be much appreciated. I'm still very new to hifi, my main tube experience is with guitar stuff (i.e. getting pretty much distortion out of tubes).
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Old 14th April 2010, 10:03 PM   #2
bigwill is offline bigwill  United Kingdom
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Distortion is low when the grid lines intersect the load line (whatever that happens to be) at even intervals
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Old 14th April 2010, 10:15 PM   #3
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Throw a pencil at the graph randomly. If it has a positive resistance slope, it will have low distortion.

On a related subject, you can use this technique to approximate pi.
Buffon's needle - Wikipedia, the free encyclopedia

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Old 14th April 2010, 10:20 PM   #4
bigwill is offline bigwill  United Kingdom
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Well it's not quite that simple obviously.
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Old 14th April 2010, 10:27 PM   #5
Merlinb is offline Merlinb  United Kingdom
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In general, the distortion reduces as you make the anode resistor larger, however, you have to avoid the 'bent' grid curves near the bottom of the graph. The modern solution is to use a constant-current source, as this will give you a horizontal load line, but you can 'slide it up' to any current you like.
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Old 14th April 2010, 10:29 PM   #6
bigwill is offline bigwill  United Kingdom
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A gyrator is a fun, modern solution

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Old 14th April 2010, 10:30 PM   #7
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Quote:
Originally Posted by MrCurwen View Post
Say I have a 6SL7, with plate at 100 - 110 V (in a direct coupled circuit, so plate voltage has to be around these numbers) - how could I get the least amount of distortion from it?
Use an active plate load. Triodes always perform with less distortion as the plate load is made very much larger than the dynamic plate resistance. An active load (constant current source) comes as close to the ideal of infinite resistance as you're likely to see in real world applications.
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Old 15th April 2010, 06:15 AM   #8
mach1 is offline mach1  Australia
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I think this question is in relation to a 6SL7 driving a 1626. If I remember correctly the 1626 is not a particular linear triode, and in Bob Danielak’s ‘Darling’ using an 8532 front end there is a fair amount of fortuitous H2 cancellation going on.

A flat loadline may not give the best audible results in this application. I would use the operating point shown in this schematic as a starting point and then EXPERIMENT.
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Old 15th April 2010, 09:17 AM   #9
Gordy is offline Gordy  United Kingdom
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At least you are starting with a nice tube!

Looking at the datasheet curves, my best guess says: 250V anode, 2.3mA and about -2V bias. Big internal impedance means it needs a really high resistance load to stop it distorting.

110V of anode is not looking good, as you will never get out of the -0.5V ~ -1 V region which restricts your input swing, and the curves are not as good in that region.

Best to forget the direct coupling and add in capacitor-resistor coupling. Then you can use the tube at an optimal bias point for low distortion.

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Old 15th April 2010, 01:32 PM   #10
Yvesm is offline Yvesm  France
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The real question you should answer first is: what do you want to drive ?
That is: wich level and in wich load impedance.
The second one is: how much gain do you need.

That done, you will be able to choose the tube and the surrounding circuit that can do the job.

BTW, as said above, 100 volts on a 6SL7 anode does not promise large and clean voltage swing and its current limitations does not make it a good choice to drive an trioded output stages.

Yves.
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