How this circuit works (Baby Huey type)
 

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!

What value of grid current did you use?

Hm? Not sure what you mean. What would that affect?

Thanks!

Well... some amount of DC grid current must be flowing to arrive at a 0.3 grid voltage...

Yeah, I get that for sure, but what does that affect in terms of tube conduction?

Cheers! :)

The tubes conduct primarily based on the voltage difference between the grid and the cathode.

Yeah, I know. I'm just confused about what you said about grid current, and how that's relevant here. If one grid is at .3 volts at quiescent state, and one is at 0 volts, and the cathode voltage is the same, how can conduction be equal?

You are saying that there is 0.3 Volts at one of the grids... that implies a current flow does it not?

Where does the 0.3 come from... and how did you arrive at that?

:)

This amplifier has been discussed before -- here i think.

What i understand is that the price for using the output pair as the phase splitter is that it is only capable of delivering half the power of a more conventionally driven push-pull pr.

dave

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