Hello,
How do I achieve maximum gain with one half of a 6SN7 ? (A must is conventional schematic, without CCS, without choke and without transformer. And kathode not bypassed by a capacitor !). Btw, current driving capability is in this case no issue.
Additional requirements are that B+ is exactly 355V and that bias should not be below -2,7V, because this is the maximum voltage to be fed to the grid. What about when the grid comes a little positive, is this very bad for the 6SN7 sound wise or not really a problem ? - So let's say, bias only 2V to achieve less NFB through kathode resistor and more gain because of higher current, but risk of positive grid voltage on peaks ?
Currently, I tried 68k for anode load and 820 Ohm for kathode, this brings up about 2,8V bias.
Which anode resistor and which bias voltage would you choose for...
1) maximum gain ?
2) high gain AND good distortion ?
Thanks,
Günther
How do I achieve maximum gain with one half of a 6SN7 ? (A must is conventional schematic, without CCS, without choke and without transformer. And kathode not bypassed by a capacitor !). Btw, current driving capability is in this case no issue.
Additional requirements are that B+ is exactly 355V and that bias should not be below -2,7V, because this is the maximum voltage to be fed to the grid. What about when the grid comes a little positive, is this very bad for the 6SN7 sound wise or not really a problem ? - So let's say, bias only 2V to achieve less NFB through kathode resistor and more gain because of higher current, but risk of positive grid voltage on peaks ?
Currently, I tried 68k for anode load and 820 Ohm for kathode, this brings up about 2,8V bias.
Which anode resistor and which bias voltage would you choose for...
1) maximum gain ?
2) high gain AND good distortion ?
Thanks,
Günther
Here's my first-order understanding of the solution:
The mu of a 6SN7 is 20, so that's the maximum gain possible. -so if you apply 2.7V P-P to the grid, you'll see 2.7 * 20 = 54 V P-P at the plate -BUT this gets divided down- the plate resistance and the plate load resistance act as a voltage divider. The higher you make the plate load (the resistor from plate to B+) the greater the voltage that you'll see at the output. This means that the plate current will be very low, and the stage following must have a very high input impedance.
In addition, the Cathode bias voltage simply subtracts from the max possible plate swing, so for every volt bias, you lose a volt of plate swing. So this is a secondary effect- the effort of maximizing the plate load is more important.
Also, the mu of a triode is dependent on the plate current, and the 6SN7's mu begins to fall off below about 4 mA. So we construct a load line with one end of the line at 355 V and the line passing through the intersection of -2 grid volts and 4 mA plate current.
At this point, there will be 80V across the tube and 4mA through it.
Since the B+ is 355V, this leaves 355-80=275V across the plate load resistor, and that resistor has close to 4mA flowing through it. Therefore R=275/.004 = 68.75k ohms -VERY close to the value you settled on(!)
At this point the 6SN7 has 80V across it and 4mA through it, so its plate resistance is 80/.004=20k ohms. So you've got a 54V (P-P) source feeding into a voltage divider consisting of 20k and 68 k, The output voltage is then 54 * (68/88) = 41.7 V P-P. Oh yeah... the cathode is unbiased, so we still need to subtact our 2V bais... the output is then 39.7V P-P.
I *think* I've got most of this right... Anyone else care to comment?
The mu of a 6SN7 is 20, so that's the maximum gain possible. -so if you apply 2.7V P-P to the grid, you'll see 2.7 * 20 = 54 V P-P at the plate -BUT this gets divided down- the plate resistance and the plate load resistance act as a voltage divider. The higher you make the plate load (the resistor from plate to B+) the greater the voltage that you'll see at the output. This means that the plate current will be very low, and the stage following must have a very high input impedance.
In addition, the Cathode bias voltage simply subtracts from the max possible plate swing, so for every volt bias, you lose a volt of plate swing. So this is a secondary effect- the effort of maximizing the plate load is more important.
Also, the mu of a triode is dependent on the plate current, and the 6SN7's mu begins to fall off below about 4 mA. So we construct a load line with one end of the line at 355 V and the line passing through the intersection of -2 grid volts and 4 mA plate current.
At this point, there will be 80V across the tube and 4mA through it.
Since the B+ is 355V, this leaves 355-80=275V across the plate load resistor, and that resistor has close to 4mA flowing through it. Therefore R=275/.004 = 68.75k ohms -VERY close to the value you settled on(!)
At this point the 6SN7 has 80V across it and 4mA through it, so its plate resistance is 80/.004=20k ohms. So you've got a 54V (P-P) source feeding into a voltage divider consisting of 20k and 68 k, The output voltage is then 54 * (68/88) = 41.7 V P-P. Oh yeah... the cathode is unbiased, so we still need to subtact our 2V bais... the output is then 39.7V P-P.
I *think* I've got most of this right... Anyone else care to comment?
Oops...
"Cathode is unbiased" --> "cathode is un bypassed"
And I note you're using 2.7V bias, not 2.0V as my example indicates.
The cathode resistor is 2.7V / 4 mA = 675 ohms, again rather close to the 820 ohms you're using.
I'd say you're doing about as well as can be done by using the values you're using. I wouldn't change a thing!
"Cathode is unbiased" --> "cathode is un bypassed"
And I note you're using 2.7V bias, not 2.0V as my example indicates.
The cathode resistor is 2.7V / 4 mA = 675 ohms, again rather close to the 820 ohms you're using.
I'd say you're doing about as well as can be done by using the values you're using. I wouldn't change a thing!
I don't think the method of rp calculation is correct (at 4mA rp would be around 10 - 12K IIRC) and it also neglects the effect of the unbypassed cathode resistor on rp. I think rp is going to be a lot closer to 50 - 60K in this scenario.. (This is a SWAG so I could be more than a bit off the mark.)
This isn't really a good way to use the 6SN7 IMO I would run it at 8 - 9 mA with perhaps a 1K cathode resistor and a 27K load resistor on something close to a 400V supply.. This will give you about 21dB of gain.
Should you want more gain stack 4 red leds in series in place of the cathode resistor. This should get you to around 23 - 24dB of gain.
Last edited:
With an unbypassed cathode resistor, the gain of the stage will be below maximum. It acts as local feedback. In a nutshell: suppose the grid signal swings up towards 0V (decreasing Vgk), cathode current rises, voltage drop across the cathode resistor rises, increasing Vgk. So the result counteracts the original stimulus, in other words, negative feedback.
I agree with kevin as far as running the tube with a higher plate voltage and biasing it closer to a symmetrical output swing. I believe the constraints given were that the B+ had to be 355V and that the cathode resistor had to be unbypassed. (I took this to further imply that the stage would indeed be biased by a cathode resistor).
There are certainly ways to get more out of a 6SN7.
There are certainly ways to get more out of a 6SN7.
Hi,
I have only 355V, and try to get the best out of it. This is work on an _existing_ amplifier !
I just found that some 6SN7 (for example the russian 6h8C) get non-linear when the grid voltage gets below -1V . So I might stick to bias at ca. -3.5V instead, and at the same time bring up the current with a lower anode resistor than 68k.
This might not be such a drawback on ampliciation factor because according to my datasheets, the mu of the 6SN7 does not fall considerably down below 4ma current as mentioned, but instead falls down already below 6mA ! - That means more gain at 6mA instead of 4mA. - But if I would need to get 6mA trough the tube, and at minimum -3,5V bias, I would need a much lower anode resistor, which brings back to the more common 39k or 47k....
Günther
I have only 355V, and try to get the best out of it. This is work on an _existing_ amplifier !
I just found that some 6SN7 (for example the russian 6h8C) get non-linear when the grid voltage gets below -1V . So I might stick to bias at ca. -3.5V instead, and at the same time bring up the current with a lower anode resistor than 68k.
This might not be such a drawback on ampliciation factor because according to my datasheets, the mu of the 6SN7 does not fall considerably down below 4ma current as mentioned, but instead falls down already below 6mA ! - That means more gain at 6mA instead of 4mA. - But if I would need to get 6mA trough the tube, and at minimum -3,5V bias, I would need a much lower anode resistor, which brings back to the more common 39k or 47k....
Günther
The most you'll get is 20 (26dB). And that's with a CCS load and the cathode tied firmly to AC ground. If you specify "no CCS," your gain will go down and distortion will go up.
Gain = mu (Rload/(rp + Rload) for a grounded cathode. With a cathode resistor, gain drops further, rp then being effectively increased by (mu + 1)Rk.
Gain = mu (Rload/(rp + Rload) for a grounded cathode. With a cathode resistor, gain drops further, rp then being effectively increased by (mu + 1)Rk.
The 4mA figure I came up with for mu Vs. Ib was from a Tung-Sol datasheet. I suppose they're all different, never bothered to check.
If plate current increases from 4mA to 6 mA, mu might go up a bit, plate resistance goes down a bit, and the plate load resistor goes down (probably proportionally more than the plate resistance decrease).
Lots going on here! -It'd be interesting to solve this and plot a "usable stage gain" Vs. Ib curve.
If plate current increases from 4mA to 6 mA, mu might go up a bit, plate resistance goes down a bit, and the plate load resistor goes down (probably proportionally more than the plate resistance decrease).
Lots going on here! -It'd be interesting to solve this and plot a "usable stage gain" Vs. Ib curve.
Seriously just try it with a cathode resistor of 820 - 1K and 27K load resistance, despite the less than optimum implementation it will actually perform pretty well. A significant further improvement could be realized with led bias. note that you will be much happier with the sound if you bias to at least -6Vgc (effectively cathode 6V positive relative to the grid) - you do not want to draw grid current.
Hi,
I've seen often biasing people the input tubes quite low with very low currents. But I don't understand why (is it sound wise or because the tubes should last long ?)
Why should the sound be better with -6V bias instead of -3,5V bias ? - With -6V, there is much less current flow and less linear, especially when I have such a high anode resistor like 68k !
I think there will be never ever grid current with -3,5V bias, because the highest voltage ever on "my" grid will be 1,9V RMS. That is +-2,7V and that means 3,5-2,7= -0,8V is the lowest it will get on the grid, never positive.
Günther
I've seen often biasing people the input tubes quite low with very low currents. But I don't understand why (is it sound wise or because the tubes should last long ?)
Why should the sound be better with -6V bias instead of -3,5V bias ? - With -6V, there is much less current flow and less linear, especially when I have such a high anode resistor like 68k !
I think there will be never ever grid current with -3,5V bias, because the highest voltage ever on "my" grid will be 1,9V RMS. That is +-2,7V and that means 3,5-2,7= -0,8V is the lowest it will get on the grid, never positive.
Günther
Ignorance. The distortion of a 6SN7 is notably higher at currents less than 8mA.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.