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Old 20th April 2008, 08:56 PM   #1
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Default Direct-Coupling Question

I'm trying to design an input/splitter frontend for a 6F6 amplifier. I'm looking to use a 6EM7 tube per channel.

Here's what I've got so far. I'm biasing the high-gain section (using TubeCAD's 12AT7 numbers) at 1.2V on the cathode (787 Rk, 1.5mA Ia) , with a 49K plate resistor and B+ of 180V. I have 106V on the plate. Now, how the heck to I bias the bigger section? I've got the plate curves in front of me but I don't understand how to find a reasonable bias point while still keeping Ra and Rk exactly the same... Am I missing something incredibly simple here?

P.S. the maximum B+ I have to work with is 300V or so. I'm using an original PA774 going mostly choke input to a 193J (10H 200mA). I'm tuning it to put out 300V, as opposed to 400V or more, so I can expect a bit more than 180mA from it, right? Each 6F6 is at 40mA, so I have at least 20mA for both channels' frontends, quite possibly more..

Thanks for the help!
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Old 20th April 2008, 09:22 PM   #2
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The first section seems closer to a 6SL7 than a 12AT7. In any case, let's say you end up with 120V on the plate. That means you'll have 120V on the second section's grid.

The biasing of the second section is less critical because it's running at 100% degeneration. We'd want to run the current high enough to minimize the loading effects as the output tubes go out of class A (assuming your design does that and that there's any appreciable Miller effect from the output stage). The low mu of this section is a positive in this respect.

We have two approaches to determining the second stage current: we can look at it from a bandwidth standpoint or a load/distortion standpoint. From bandwidth, we calculate the output stage grid capacitance, then from out desired bandwidth, we determine how low the driver stage source impedance needs to be. We can apply rules of thumb to optimize things (e.g., standing current should be a decade higher than the AC current needed to drive the input stage capacitance).

The other way to do this is to note that the plate resistance is likely to be 2-3k, so a total load of 20k will result in low distortion. 10k in both plate and cathode leads would result in a standing current of something like 14mA, which ought to be more than adequate.

OK, let's see what we need for DC. Assume we're going to want at least 100-150V cathode-to-plate. The grid is at 120V, cathode will be at 140V (more or less), so the plate will be at 290V or somewhere slightly south of that. 140V drop across the plate resistor and the required B+ is about 430V. If that's too high, you can couple the two stages with a step network.
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Old 20th April 2008, 10:34 PM   #3
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Thanks for the info. I don't know what I was thinking, I mixed up the 12AT7 with the 6SL7. Unit 1 has 40K or something of plate resistance so I'll probably want to use 80-100K there.

Unfortunately I don't have more than 300V available. What is a step network?

By the way, the plate resistance of the second 6EM7 triode is listed as around 700 ohms. Does this mean I could get away with smaller plate and cathode resistors, and therefore a B+ lower than 430V?
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Old 20th April 2008, 10:42 PM   #4
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Take a look at the curves. The plate resistance will be higher at more reasonable currents. But whatever, it's already pretty low.

A good step network could consist of a voltage divider from the plate of the first tube to ground. The split-load's grid is connected to the tap of the divider, then the upper resistor is bypassed for AC. For example, if you wanted to knock down the grid voltage from 140V to 70V, you could have two 1M resistors in series from the first stage plate to ground, tap the grid at their junction, and bypass the upper resistor with 100nF or thereabouts.
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Old 21st April 2008, 07:44 AM   #5
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Default #4

Quote:
Originally #4 posted by SY
Take a look at the curves. The plate resistance will be higher at more reasonable currents. But whatever, it's already pretty low.

A good step network could consist of a voltage divider from the plate of the first tube to ground. The split-load's grid is connected to the tap of the divider, then the upper resistor is bypassed for AC. For example, if you wanted to knock down the grid voltage from 140V to 70V, you could have two 1M resistors in series from the first stage plate to ground, tap the grid at their junction, and bypass the upper resistor with 100nF or thereabouts.
Hi SY,
have a look at post #1. Please note that the author asked
for a direct coupled solution.
Your bypass cap is a coupling cap. Direct coupling is made to
eliminate this coupling cap. At very low frequencies you
get the divided signal voltage and at high frequencies you
get full level. Your solution is like a sound regulator in
fixed position.
Step network coupling is possible without a coupling cap.
This schematic shows an example link .
Comes from this website link .
The split network is a voltage divider.
Note that the values of resistors: 68KΩ 22KΩ
You'll get a loss in gain and this loss is desired.
It provides you a better S/N ratio in this application.
The frequency curve is linear, no loss at low frequencies
caused by the step network. You don't loose the advantages
of direct coupling.

Kind regards,
Darius
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Old 21st April 2008, 09:20 PM   #6
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If chosen sensibly, the capacitor in the step network does not cause any attenuation at audible frequencies. The values that SY mentioned will have a -3dB point at ~1.6 Hz. Also, it acts like direct coupling if within a feedback loop, providing stability right down to DC.

If you miss out the capacitor, it's no longer a step network, just a potential divider. How can loss of gain be desirable? How can a resistive attenuator improve signal/noise?
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Old 22nd April 2008, 09:00 AM   #7
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Default #6

Quote:
Originally #6 posted by ray_moth
If chosen sensibly, the capacitor in the step network does not cause any attenuation at audible frequencies. The values that SY mentioned will have a -3dB point at ~1.6 Hz. Also, it acts like direct coupling if within a feedback loop, providing stability right down to DC.

If you miss out the capacitor, it's no longer a step network, just a potential divider. How can loss of gain be desirable? How can a resistive attenuator improve signal/noise?
Hello ray_moth, thanks for your question.

The two sections of the 6SN7 are making too much gain.
This causes two problems:

First: You have to attenuate the input signal
Second: You can hear noise from the input section
of the 6SN7 (schschsch...)

Making an attenuator at the input does not solve
the noise problem. An attenuator at the input does
not attenuate the noise generated by the first section
of the 6SN7.
Negative voltage feedback and current feedback
eliminates the triode characteristic and is not desired.
Reducing the gain by a low anode resistance of the
input section causes transconductance distortion.
Means the stage becomes more pentode-transistor like.
Thus I decided to put the attenuator after the input stage.
In this configuration the step resistors attenuates the
signal voltage and the noise voltage.

The ratio of Signal voltage to noise voltage is increased
by the attenuator at the input of the second stage
because you need more signal voltage at the anode of the
input stage to get the same output level. See attachment.

I agree that the low frequency cut and the time constant
R x C caused by your coupling cap is not a disaster since there
are two cathode decoupling caps in this 6SN7 driver stage.

This is one reason why I prefer another topology.

BTW: There is no feedback loop.

Kind regards,
Darius
Attached Files
File Type: zip conventional_se_signal_to_noise_22.april2008.zip (13.1 KB, 67 views)
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Old 24th April 2008, 10:42 AM   #8
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Default #7

Hello ray_moth,
did you read it?
Kind regards,
Darius

(There is a mistake in the schematic. I'll correct it soon. )
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Old 24th April 2008, 11:26 PM   #9
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Hi Darius,

Yes, I read it. Thanks for answering my questions. In your example I can see your point, although I would have approached it differently.

I don't see how it would help sorenj07, though, unless he specifically needs an attenuated signal, which I doubt. SY's step network suggestion would avoid that problem. It's a perfectly valid approach and, as I said, would work just as well as direct coupling if it happens to be inside a feedback loop, which sorenj07 might want for all we know.
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Old 25th April 2008, 03:02 AM   #10
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Heh. At the moment I'm leaning more towards a more conventional setup involving an LTP with 6SN7's driving the output tubes through cap coupling, fed by an input tube which might or might not be direct-coupled. I'm leaning between half a 6SN7 or half a 12AX7, like an Eico HF87 except in "mini" for perhaps 10-12 watts of output.

The problem is that I don't know what kind of swing I'm going to need in order to drive UL-connected 6F6's to full power...
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