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#1 |
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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When paralleling both sides of a dual triode and using a transformer plate load, I understand that the cathode bias resistor should be halved, but does the load line stay the same? Is there something else to consider?
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#2 |
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diyAudio Member
Join Date: Jul 2005
Location: South Florida
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Generally the load impedance should be halved also. The current will be doubled, so the power supply must be up to the task. In some experiments, I have simply wired in a second tube, using a second (identical) set of cathode, grid stopper, and screen stopper (even though it is connected to the plate) resistors , then connected my 8 ohm speakers to the 16 ohm tap. This will find the weak link in your power supply and driver capability, but if they are up to the task, you will get double the power out.
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Too much power is almost enough! Turn it up till it explodes - then back up just a little. |
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#3 |
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diyAudio Member
Join Date: Nov 2005
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If you wish to do your graphical work on the same actual set of curves; simply double the values on the current scale of the graph. Then proceed with your design normally.
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#4 |
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diyAudio Member
Join Date: Jul 2008
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Bringin up an old thread as I have a qq related to triodes in parallel.
I'm not sure how to wire them. Looking at Broskies pic here he has the grids connected together and only 1 grid stopper at the beginning and showing connection to one plate and one cathode. Now does he connect the 2 grids in a double triode directly together and B+ to both plates? Is each second tube here the second half of each double triode? I'm confused with this diagram. I'm trying to parallel 6080 in the same fashion but really confused. Please help. |
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#5 | ||
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diyAudio Member
Join Date: Feb 2009
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Quote:
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mod verb, transitive /mod/ to state that one is utterly clueless about the operation of device to be "modded" and into "fixing" things that are not broken; "My new amplifier sounds great so I want to mod it." |
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#6 |
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diyAudio Member
Join Date: Jul 2008
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Thxs.
So 4 double triodes in parallel for example. Each grid connected directly together. I can couple the cathodes in each triode with 1 resistor. And B+ to both plates. So 4 cathode resistors in total for the 4 triodes. Is this what you are referring to? |
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#7 |
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diyAudio Member
Join Date: Jul 2008
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2 double triodes for one channel or 4 double triodes for 2 channels with each channel using half of the triode maybe.
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#8 | ||||
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diyAudio Member
Join Date: Feb 2009
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OK.
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__________________
mod verb, transitive /mod/ to state that one is utterly clueless about the operation of device to be "modded" and into "fixing" things that are not broken; "My new amplifier sounds great so I want to mod it." |
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#9 |
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diyAudio Member
Join Date: Mar 2008
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If you are considering doing this you might also consider how critical to it is to you that, for instance, two tubes in parallel track each other. Though it often not that noticable I think it would be good, just on principle, to get them to track each other and act like one combined tube. It will be an equivalent tube with the same mu but twice the gm and half the ra of just one of the tubes.
Since these parallel driven tubes are presumably being driven in phase then you can use a sandy device, such as a current mirror in the cathodes, to make them one big powerful tube. No need to bypass them them with any kind of capacitor since they will be operating in phase. In other words you can have them track in "absolute" terms rather than average the combined current with capacitors, as you would if the tubes were being driven out of phase, as in push pull. You might have to go with a more sophisticated current mirror than just a plain jane one to get rid of the Early effect though in all cases. Current mirrors can be constructed for virtually any number of tubes being driven in parallel. Transistors are a good thing, as Martha Stewart would say, in these situations. Last edited by exeric; 29th August 2010 at 08:34 PM. |
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#10 |
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diyAudio Member
Join Date: Aug 2004
Location: Cambridge, England.
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