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?
Thanks
Thanks
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.
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.
🙂
🙂
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.
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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Of course, as you can see from the picture you posted.
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).
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?
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?
2 double triodes for one channel or 4 double triodes for 2 channels with each channel using half of the triode maybe.
OK.
Yes.
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.
Um, you'll generally want some sort of load between B+ and the plates (except in cathode follower).
Man, you lost me with this one - above you said 1 resistor for four triodes, now you're saying 4 resistors.
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.
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.
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Mirroros are new to me. Something new to read about.
This was just an example of how to lower impedance.
I'm experimenting with low plate resistance tubes in parallel to create a headphone amp that can drive most headphones without the need for a transformer.
Just DIYing
Note that every one of those triodes should have a separate grid stopper mounted right at the socket. Duals in a single envelope should get one grid stopper per section.
That's a RIAA preamplifier for MM pickups. Paralleling a bunch of 12AX7s (or any tube) decreases the tube's noise and gives them higher gm and lower rp. A single 6922 does the same job😉
Indeed it looks like he connects the four grid together and a single grid resistor to that node. I second Kevin and would use seperate gridstoppers on each grid. Those resistors would then also function to wire all the grids together.
Better to have seperate cathode resistors as well, instead of one shared by all cathodes.
You can do the same with your 6080s.
Thxs guys.
So I'm thinking I would use 2 triodes for each channel.
one channel: each grid connected together with resistor, so I've got 4 grids together.
100V-120V at plate.
300-400 ohm cathode resistor
30-40V at cathode (or would that be 60-70v on cathode? plate v - cathode voltage = grid voltage).
80-100mA current
3-4 watts power through cathode resistor.
Maybe a CCS with some DN2540s at plate.
So I'm thinking I would use 2 triodes for each channel.
one channel: each grid connected together with resistor, so I've got 4 grids together.
100V-120V at plate.
300-400 ohm cathode resistor
30-40V at cathode (or would that be 60-70v on cathode? plate v - cathode voltage = grid voltage).
80-100mA current
3-4 watts power through cathode resistor.
Maybe a CCS with some DN2540s at plate.
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