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Old 14th September 2006, 01:53 AM   #1
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Default Parallel Triodes and Load Lines

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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Old 14th September 2006, 02:18 AM   #2
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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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Old 14th September 2006, 02:27 AM   #3
poobah is offline poobah  United States
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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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Old 29th August 2010, 04:09 PM   #4
Brit01 is offline Brit01  United Kingdom
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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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Old 29th August 2010, 04:17 PM   #5
Arnulf is offline Arnulf  Europe
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Quote:
Originally Posted by Brit01 View Post
Now does he connect the 2 grids in a double triode directly together and B+ to both plates?
Of course, as you can see from the picture you posted.

Quote:
Is each second tube here the second half of each double triode?
Tubes with multiple active elements are always depicted as distinct elements in schematic (possibly with incomplete outline or designation indicationg that they are only a portion of larger whole).
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Old 29th August 2010, 04:34 PM   #6
Brit01 is offline Brit01  United Kingdom
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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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Old 29th August 2010, 05:08 PM   #7
Brit01 is offline Brit01  United Kingdom
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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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Old 29th August 2010, 06:01 PM   #8
Arnulf is offline Arnulf  Europe
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Quote:
Originally Posted by Brit01 View Post
So 4 double triodes in parallel for example.
OK.

Quote:
Each grid connected directly together.
Yes.

Quote:
I can couple the cathodes in each triode with 1 resistor.
Yes, but 4 separate resistors will allow for better control of operation. When one triode fails the combined current through common resistor will drop to 3/4 so the bias voltage will drop to 3/4 as well which will bias the remaining three triodes much hotter (possibly above the maximum suggested dissipation curve), again possibly leading to to untimely demise of another triode of the remaining three and reduction of current and bias voltage to 2/4 etc. in a spiral until all four triodes fail.

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And B+ to both plates.
Um, you'll generally want some sort of load between B+ and the plates (except in cathode follower).

Quote:
So 4 cathode resistors in total for the 4 triodes.
Man, you lost me with this one - above you said 1 resistor for four triodes, now you're saying 4 resistors.
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Old 29th August 2010, 08:19 PM   #9
exeric is offline exeric  United States
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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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Old 29th August 2010, 08:37 PM   #10
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Quote:
Originally Posted by Brit01 View Post
Looking at Broskies pic here
What is that circuit for?!
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