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Old 4th December 2009, 12:59 PM   #1
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Default Maximum gain with 6SN7 ?

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,

GŁnther
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Old 4th December 2009, 02:56 PM   #2
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Hi,

if you don't want to use a cathode bypass capacitor I'd say either:
- tie the cathode to ground and use a negative grid supply
- or use LED/diode/zener bias in the cathode

Also, to maximize gain, use the biggest plate resistor you can get away with, given your B+ limit.

Kenneth
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Old 4th December 2009, 03:00 PM   #3
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Furthermore a little bit of grid current is okay but that rules out capacitive coupling at the input. So you would need a transformer (driven from a low impedance source) or a cathode follower or mosfet source follower DC-coupled at the input.
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Old 4th December 2009, 03:38 PM   #4
kruesi is offline kruesi  United States
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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?
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Old 4th December 2009, 04:42 PM   #5
kruesi is offline kruesi  United States
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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!
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Old 4th December 2009, 06:05 PM   #6
kevinkr is offline kevinkr  United States
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Quote:
Originally Posted by kruesi View Post
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?

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.
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Last edited by kevinkr; 4th December 2009 at 06:08 PM.
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Old 4th December 2009, 06:09 PM   #7
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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.
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Old 4th December 2009, 07:21 PM   #8
kruesi is offline kruesi  United States
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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.
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Old 4th December 2009, 07:22 PM   #9
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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....

GŁnther
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Old 4th December 2009, 08:23 PM   #10
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Well indeed the requirements are a partly contradictory, aren't they.

@Guenther: maybe you could use a 6SL7 which has a much larger amplification factor...?
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