How this circuit works (Baby Huey type) - Page 2 - diyAudio
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Old 14th November 2008, 09:57 PM   #11
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The cathode voltage is higher than the grid voltage in rockgardenlove's made-up figures. This will not cause grid current.


I think you have confused yourself with your "totally made up grid voltage = .3 volts". Where did that come from? Both grids will be at ground potential under no-signal conditions.

This is not a "Baby Huey" type of design. BTW, there is an incorrect statement in the article:
Only one tube is driven in this circuit as opposed to two in the normal type of output stage. This means that the required driving voltage is halved which, in this case, now amounts to 11 volts
That is not the case, because the voltage gain of the OP stage is halved in this arrangement. If you apply the drive signal to only one grid, it needs to be twice as high.

I dislike this design, personally, because it's suboptimal and it seems pointless. If the input stage used a double triode instead of a single triode, then a grounded cathode stage followed by a concertina splitter could be used and the OP stage could then follow conventional design. This would give much better balance. A concertina splitter would be fine to drive EL84s directly.
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Old 15th November 2008, 08:41 AM   #12
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Default go here...

lots of people using this tehnique now. Heres some...

and here

simple - low drive requirement, low parts count. All associated advantages and disadvantages...
"It may not be easy for some to not hear differences, even if they are not there." - Vacuphile,
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Old 15th November 2008, 08:53 AM   #13
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I think you were reading my mind Ray...
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Old 15th November 2008, 03:08 PM   #14
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Ahhh.., the half power output issue comes up because it has to run class A to do the inversion in the cathodes, while the normal inverted drives can be run class AB.
If it isn't broken, take it apart and mod it.
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