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How this circuit works (Baby Huey type)

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Hi,
I found this during some browsing about the Baby Huey circuit, and it's pretty interesting.
http://www.triodeel.com/compact.html

I understand how the inverse of the signal is present on the cathode, but I don't understand why this would produce the same amplitude as the other tube. Does that make sense?

Using some math, and totally made up numbers:

Quiescent state:
Grid voltage = .3 volts
Cathode voltage = 1.3 volts
Plate current = 10 mA
Cathode resistor = 130 Ω

Alright, that's easy enough. Since it's in quiescent state shouldn't the other tube be conducting equally?

Grid voltage = 0 volts
Cathode voltage = 1.3 volts
Already we can see that since the grid to cathode voltage is not the same, different amounts of current will flow...

What's wrong here?

Thanks!
 
Having a tail resistor, instead of a CCS, will cause some assymetry between the output tube drives. The cathode driven tube won't get quite as much drive signal as the grid driven tube due to some AC current loss to the tail resistor. The grid input signal to the 1st tube will generally have to be twice as large to drive both output tubes (compared to normal dual inverted drives). But should be capable of about the same power output as normal (with doubled drive) if the assymetry is not so large as to cut down the power significantly from the second tube.

I think this scheme also increases the output impedance (higher damping factor then) due to effective resistance in the cathode circuit. Each tube sees approx. 1/gm resistance in its cathode circuit from the other tube and the tail R in parallel. Normal configuration would have the cathodes grounded.

Don
 
Ex-Moderator
Joined 2004
Poobah,

The cathode voltage is higher than the grid voltage in rockgardenlove's made-up figures. This will not cause grid current.

rockgardenlove,

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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