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Old 7th September 2012, 01:25 AM   #1
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Exclamation Linearity of Red RCA 5692

I'm trying to design a driver stage with a concertina phase splitter using a 6SN7. I found what a lot of people say are great sounding tubes (RCA 6SN7GT, Kenrad 6SN7GT, etc) but couldn't find a datasheet for them. I'm trying to carefully design this so I would like to buy a pair of tubes with their corresponding datasheet.

HERE IS MY QUESTION:
I found the RCA 5692 and their tube specs (curves and everything) but as I got my quiescent point I realized I was below 4[mA] for Anode Current. I've read a lot of times that you get bad distortion below 8[mA] for 6SN7 tubes. Is this the case for the 5692?

Also how close can I get to the minimum negative grid voltage? The datasheet says -1[V] but should I try have my peaks stay, let's say, 0.5[V] away from that?

I would really appreciate a fast response!! Thank you very much in advance!
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Old 7th September 2012, 04:03 AM   #2
SY is offline SY  United States
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5692 is just a 6SN7 with a lot of mechanical support. Electrically, it's identical. great tube, on the low end of distortion for the 6SN7 family.

Remember a few things about the concertina. First, you have nearly 100% feedback so even at low currents, the distortion will be low. Higher currents are better, though, since the transconductance will be higher, meaning source impedance is lower. Second, the cathode follows the grid, so Vgk will never get anywhere near 1V.
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Old 7th September 2012, 04:21 AM   #3
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I'm following the Valve Amplifier book by Morgan Jones to design the Concertina, and the I had to set the Q point for the driver at Vgk = -2 [V] to be able to get a current above 5[mA] (Va at 80[V]). I don't know if it is too low for this tube, more so after I keep reading that 6SN7 distort heavily below 8[mA].

Since my input has a peak voltage of 1[V], having my bias for the driver at -2[V] has me a bit worried. But if I move it to -2.5[V] my plate current drops from 5.4[mA] to 3[mA].

Is there something I'm doing wrong? Or is it OK to run this tube with that current? And swing my input so close to the minimum negative grid voltage?
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Old 7th September 2012, 04:22 AM   #4
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I guess I should add I'm not having problem with the concertina half of the tube, but the driver one.
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Old 7th September 2012, 09:17 AM   #5
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If you find that the grid voltage is too small at your chosen anode current then this could be a sign that your HT/B+ supply rail voltage is too low. Don't attach too much weight to rules of thumb about currents ('greater than 8mA'), but plot load lines. A load line will give you the 2nd order distortion, and perhaps even a rough estimate of 3rd order. Remember, however, that the datasheet is an average of several valves then plotted by hand so it won't be 100% accurate for any particular valve.
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Old 7th September 2012, 11:02 AM   #6
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Yes, 80V is much too low for the plate of a 6SN7. 180V is more like it.

I would look at the Williamson amp as an example of how to correctly use a 6SN7 for voltage amp/concertina input stage.
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Old 7th September 2012, 02:23 PM   #7
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DF96: I'd like to increase HT but according to the datasheet, it can't go above 330[V] for the RCA 5692.

SY: The Williamson was my "inspiration" for this design ! I'm also following how the book constructs the Bevois Valley since he gives a detailed explanation on how to design a driver/concertina with a dual triode. Using this method and the Max Ratings from the datasheet, it would seem to be like there is no way I can build this using these values, but I'm hoping I'm wrong.

The page with the Williamson doesn't have any voltage labeled so it makes it a little bit difficult for me to read it effectively.

I'm attaching the loadline for the 5692 and everything I did to try to get these values. I changed the resistors but maybe I'm still doing something wrong (and I hope I am, because this just isn't giving me any "normal" results)

imgur: the simple image sharer

Also, thank you so much for helping me out!

Last edited by EEApprentice; 7th September 2012 at 02:47 PM.
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Old 7th September 2012, 08:14 PM   #8
DF96 is offline DF96  England
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Somewhere around the middle of the load line looks OK to me. Bear in mind that people who design 'by ear' sometimes include a little distortion without realising it.
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Old 8th September 2012, 01:00 AM   #9
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Ok I'm posting the schematic for the driver/concertina (dual triode with RCA 5692) to better explain myself.

imgur: the simple image sharer

Here is the loadline.

http://i.imgur.com/Laijh.png

The black loadline is for the concertina, the blue is for the driver. The red line is the Q point for the concertina (at Vg = -2.5). The voltage at the Q point for the driver is at V = 100.


My question is, I'm inputting a signal that is 0.6Vrms and I'm worried this swing in the loadline is too close to Vg = -1 which is the limit for the tube. Am I in a danger zone with these number? Would this damage the tube?

Also, how does the negative feedback affect the bias to the driver? I'm worried that this will push my Vg from -2.5 CLOSER to the -1 [V] and push the peak input signals into that -1 [V] and damage the tube.

Once again THANK YOU to anyone who can help me out here! This is my first push-pull so I'm still trying to learn the ropes and I don't want to make the amp and have it fail on me because my values are pushing the components to the limit.
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Old 8th September 2012, 01:30 AM   #10
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Remember what the feedback does here- if you run open loop, you'll overload at something near the bias point (but the signal swing at the plate will be 15 times higher!), but with feedback, there's a signal injected at the cathode which "follows" the input signal. What the tube actually handles is the difference between the input signal and the feedback signal. That's a much smaller number than just the input signal alone.
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