Help me to understand this please

[VacuumVoodoo]

There is a common confusion/misunderstanding of what anode (plate) load line represents.
It is not a characteristic of how the anode (plate) behaves but how load ON the anode (plate) reacts i.e. it represents current drawn by the plate (anode) resistor from B+ supply as function of all B+ values available.
For the purpose of simplicity, one can assume the tube to be a switch. If ON the plate (anode) load resistor is connected between B+ and any actual "zero" reference potential. The current is thus (B+ -zeroreferference)resistance.
When switch is OFF no current flows in the resistor and voltage drop over it is 0V. You get a straight line.
Now you can draw a straight line originating in origo (0,0) point on the anode characteristics chart representing the cathode resistor.
Your first approximation of operating point is where these two straight lines cross.

[Rogue]

from "Design and Construction of Tube Guitar Amplifiers" by Robert Megantz, on page 85-86 for anyone that has the book.

It is discussing the reverb driver circuit of a typical fender type reverb. The parallel 12at7, transformer coupled (22.8k primary), 2.2k cathode resistor.

It has a picture of the characteristic curves with load line drawn from 415V (supply) 0 current, to 18.2 mA at 0 volts (415/22.8k).

Don't have the book (I think you can post the two relevant pages under the Fair Use Law, not the same as posting entire chapters or the full book) but if it says so, it's plain wrong.

When 12AT7 parallel plates receive no signal, idle current is not 0 by any means, since it's a single ended class A design it has the necessary idle current so that twice it will dynamically bring plate voltage to saturation (as close to 0 as possible) and on the opposite half cycle it will reach 0 mA ... but plate voltage will be about twice the idle value.

That's how a Class A transformer loaded power stage works and provides maximum symmetrical swing.

All that makes sense, but it goes on to say, "The stage as designed is not operated at maximum plate dissipation, so we must find the actual location of the load line using the cathode resistor value and plate current to determine the quiescent grid voltage. The operating point is located where the plate current at 415v times the cathode resistor equals the grid voltage.

Here he contradicts his earlier statement, where he said that at plate voltage=415V plate current was 0 !!!!!!
So which of 2 contradicting statements is true?

In this case, when the plate current is about 3.7 mA, the grid voltage is about -8V."

He's pulling this out of the blue.

To boot, grid voltage is already known: being referred to ground it's 0V !!!!!
What he is trying to (poorly) explain is called "grid to cathode voltage" , not unqualified "grid voltage"

And a new load line is drawn above the old load line that is shifted up 3.7mA.

Now we are getting closer, but the values come out of the blue ... he's just quoting tnhe Fender design but is not explaining how to design/calculate it out of a datasheet , pencil and paper and a calculator, which is what a real *designer* uses.
Definitely what Leo did., just forget the calculator and use pencil and paper or a slide rule.
Plus a human brain, of course.

I get that he's finding the voltage drop across the cathode resistor to determine the operating point, what I'm having trouble with is how did he come to conclusion there was 3.7mA flowing through the cathode resistor? He's not very clear how he made that determination.

See above.

So, what did he do here? I'm sure it's something simple and I'm just not seeing it, but I am puzzled.

You are rightly puzzled, because that's not the proper explanation.

I repeat, I write this based on your post, I think it would be reasonable and legal to post 1 or 2 relevant pages for analysis and discussion, maybe I'm being unfair and stuff is better explained than quoted here.[/i]

Thanks for taking the time to respond, JMF.

That's an exact quote. There isn't much said beyond that. He has a picture of the reverb driver (parallel 12at7, transformer, resistors, etc). Then the picture of the characteristic curve with a load line drawn from 415V 0A to 0V 18.2ma, but with that same line just above it but 3.7mA higher with "operating point" at the 3.7ma 415V point. Then the next paragraph is what I quoted.

I could take a picture I guess, but more than I was looking to do. I just wanted to know what in the world I was missing in how he got to his conclusion.