Foreplay I Circuit Questions

[dbishopbliss]

Since I started looking at the guts of my Foreplay I the other day, I started wondering how it works.  I reverse engineered the design and came up with the attached schematic (I hope Doc, doesn't mind.  I didn't include any values and since no circuit is really new I think it should be ok).

I got this from the Foreplay 2 description:

One half of the 12AU7 operates as a traditional voltage amplifier, directly coupled to the other half of the 12AU7 which operates as a cathode follower.

The triode on the left appears to be a grounded cathode (voltage amplfier), so I assume the triode on the right is the cathode follower.  

Why doesn't the cathode follower have a load resistor?

Does the cathode follower contribute to the total current of the circuit (I'm assuming it does, but not sure).  How would I go about calculating these values?

[Paul Joppa]

The cathode resistor is the load resistor in a CF.

I think it's time for you to get more systematic. Go to this site:

http://www.tubebooks.org/technical_books_online.htm

and get a book or two. The military basic training books listed under "Vacuum Tube theory & circuit design" are widely admired; for hifi stuff specifically Crowhurst from the fifties and sixties is pretty readable.

You asked "How would I go about calculating these values?" - the answer to this and many other questions is usually "Ohm's Law". The first chapter or two of any good beginning circuits book will give you the mindset as well as the equation - well worth the minor effort!

[dbishopbliss]

I will go download the Crowhurst books and re-read again. I have the Morgan Jones book and the RDH4 as well.  I just can't seem to make the leap from reading the books to creating a circuit. That said, now that I drew the Foreplay I circuit for myself and isolated the parts, the filaments (tube humor) are coming on.

Although, there are some practical things that I don't recall being pointed out in any books I've read.  Like considering the B+ voltage when choosing the load resistor and the current draw.  I have blindly picked an operating point and drawn a load line without thinking about it (e.g., 8.2mA and 100K load in another thread).  You pointed out how I would need a 970V power supply.  While I would have figured that out eventually, its one of those things that if you know beforehand guides your choices.

Thanks for your help so far.  I think I'm getting it (although, I think I have said that before).

[Paul Joppa]

On that load line thing, here's the trick: Continue the load line to the left until it intersects the voltage axis. That will be the power supply voltage.