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Old 2nd May 2010, 03:43 AM   #1
jfitz57 is offline jfitz57  United States
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Default Low mu triode for phase splitter

So I have read that the split load phase splitter is not that bad compaird to the others. I'm thinking of using a disssimaler dual triode as an input VAS and phase spliter to drive a push pull output. So does it make a big differenc what the amplification factor of the tube that is used as the phase splitter is? Looks to me like it has a gain of a little less than 1 whatever tube is used.

Jim
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Old 2nd May 2010, 04:29 AM   #2
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Originally Posted by jfitz57 View Post
So does it make a big differenc what the amplification factor of the tube that is used as the phase splitter is? Looks to me like it has a gain of a little less than 1 whatever tube is used.
Not really. What's important is having adequate headroom for the cathodyne, and adequate grid drive for the finals. Triodes with the lower u-factors are more likely to provide that than the ones with higher u-factors.
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Old 2nd May 2010, 04:35 AM   #3
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(Alert: Me noobie, only used cathodyne splitters so far) Rule of thumb seems to be that a low mu tube/section is used as the splitter as the gain is going to waste. Use the high mu tube/section to drive it.
Some more experienced people will probably be able to explain why in detail.
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Old 2nd May 2010, 07:10 AM   #4
jfitz57 is offline jfitz57  United States
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The tube is 13DR7. I can heat it with a PC power supply. Huw would I figure out the cathode and plate resistors? I'm guessing about 2K ohms each (and I can figure out the wattage from there) B+ 300v 50mA. It's late no can think.

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Old 2nd May 2010, 07:35 AM   #5
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Originally Posted by jfitz57 View Post
The tube is 13DR7. I can heat it with a PC power supply. Huw would I figure out the cathode and plate resistors? I'm guessing about 2K ohms each (and I can figure out the wattage from there) B+ 300v 50mA. It's late no can think.
Donn't guess, use Loadlines. Scroll down a bit to the first table, there's a five part article that tells all about this.
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Old 2nd May 2010, 07:59 AM   #6
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Originally Posted by Miles Prower View Post
Donn't guess, use Loadlines. Scroll down a bit to the first table, there's a five part article that tells all about this.
Miles, thanks for the link (I had seen but not bookmarked Steve's page), and I'm poaching jfitz's thread but he will probably benefit from the answer to a question about cathodyne load lines:
Say I have an example schematic that shows 47k plate and 47k cathode resistors; in terms of drawing a load line what do I treat the load as - 47k or 94k? Each time I think I understand the answer, I have a niggling doubt.
For ac I think it's 47k because each half cycle of the amplified output sees 47k (in parallel with the load presented by the following stage), but what about the DC condition?
I'll deal with the confusion about the effect of DC coupling the previous stage later.
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Old 2nd May 2010, 08:32 AM   #7
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Originally Posted by jfitz57 View Post
The tube is 13DR7. I can heat it with a PC power supply. Huw would I figure out the cathode and plate resistors? I'm guessing about 2K ohms each (and I can figure out the wattage from there) B+ 300v 50mA. It's late no can think.

Jim
I had to look up a data sheet for a 6DR7 to get curves for this one.

Wow! When you said dissimilar triode, I was expecting one hi and one low mu small-signal triode. This thing is more like an all triode version of a 6BM8/ECL82. Interesting tube.
Is your B+ power supply limited to 50mA, or was that what plate current you were hoping to run through the big part of the tube? What is the stage you are trying to drive? If it's not a brutal load to drive (like class A2/B2), your plate/cathode resistors are probably going to be much higher than 2k.

You probably know more about this than I do, so I'll shut up now and hit the sack. Sorry to hijack your thread.
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Old 2nd May 2010, 09:50 AM   #8
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Originally Posted by rotaspec View Post
Miles, thanks for the link (I had seen but not bookmarked Steve's page), and I'm poaching jfitz's thread but he will probably benefit from the answer to a question about cathodyne load lines:
Say I have an example schematic that shows 47k plate and 47k cathode resistors; in terms of drawing a load line what do I treat the load as - 47k or 94k?
When you do the loadline, you draw a 94K loadline. The tube itself neither knows nor cares how the load is connected: between the plate and the positive rail, the cathode and DC ground, or whether it's split between the two. Once you have a loadline and a suitable Q-Point, you simply read off the maximum and minimum plate currents and use those currents to calculate the voltages at the plate and cathode, and to determine what the actual grid swing and gain will be.

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what about the DC condition?
I'll deal with the confusion about the effect of DC coupling the previous stage later.
What about it? You simply draw your loadlines on the basis of a reduced Vpp to account for the higher than normal DC voltage at the cathode. The voltages that count are those which are referenced to the cathode, not DC ground. It's a bit confusing since the characteristic charts always seem to refer to "plate voltage" when it's more properly plate-to-cathode voltage.

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I had to look up a data sheet for a 6DR7 to get curves for this one.

Wow! When you said dissimilar triode, I was expecting one hi and one low mu small-signal triode. This thing is more like an all triode version of a 6BM8/ECL82. Interesting tube.
Yeah, it was originally intended for use as a vertical deflection oscillator/vertical final. I have a project in the works that uses the 6DR7. The linearity of both the small signal and large signal sections is excellent. Furthermore, the large signal triode can stand up to some serious voltages, if you have the need for that. What I like about it for this project is the lower than usual r(p). This, and the use of fixed bias (gets r(p) increasing resistance out of the cathode, and avoids the resonance problem that can cause nasty distortion when bypassed cathode bias resistors are used -- capacitive currents in the cathode circuit are capacitive currents in the plate circuit when using triodes) should help considerably the bass performance with IST coupling to the grid of the final, as the r(p) will be a smaller percentage of the primary Xl. It should have the extra "moxie" to really slap those grids around. Inadequate grid drive seems to be a major factor in the less than satisfying sonics of SETs.

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Last edited by Miles Prower; 2nd May 2010 at 10:12 AM. Reason: Break to look up second post index no.
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Old 2nd May 2010, 10:47 AM   #9
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Originally Posted by rotaspec View Post
Say I have an example schematic that shows 47k plate and 47k cathode resistors; in terms of drawing a load line what do I treat the load as - 47k or 94k?
The triode can only swing somewhere between a short and an open circuit - so I plot those two points and draw the line.

Short = current at Va = 0
Open = voltage at Ia = 0
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Old 2nd May 2010, 01:45 PM   #10
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Some more experienced people will probably be able to explain why in detail.
The advantage of a low mu tube in this application is that if the load becomes unbalanced (for example, if the stage it is driving transitions from class A to class B), the splitter does not unbalance as much. If the loads are kept equal (e.g., if the driven stage is strictly class A as in Williamson circuits), the balance of the phase splitter will be essentially perfect, regardless of the mu. The figure of merit is really the transconductance, much as with a cathode follower.
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