Help me to understand this please

[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).

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

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

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.

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

[Phil_S]

Just apply Ohm's Law. V=I * R.
We know I and R. I=.0037. R= 2200
.0037 - 2200 = 8.14V
This has nothing to do with the grid voltage.
So, is there 8V at the cathode? According the AB763 schematic, the answer is 8.6V. So there you have it.

Actually, backing into the schematic, I'd say we should solve for I:
8.6/200 = 3.9mA. We really can't quibble about 0.2mA can we?

[Rogue]

Thanks Phil

How did he come to the conclusion 3.7mA was what was flowing through the tube (and thus the cathode resistor)? This is where I'm confused. He just states it and didn't go into detail how he arrived at that conclusion.

[roberto]

8.6 volts on the cathode.
2200 Ohms as cathode resistance.
V = i x R
i = V / R
8.6 / 2200 = 0.0039 A = 3.9 mA

[tubeswell]

8.6 volts on the cathode.
2200 Ohms as cathode resistance.
V = i x R
i = V / R
8.6 / 2200 = 0.0039 A = 3.9 mA

...all up for both triodes. (1.95mA per triode)

[Rogue]

Sorry guys, I guess I've not been clear.

The 3.7mA and -8V are unknown values.

The known quantities are 415V supply, the 2.2k cathode resistor, and the 22.8k primary.

He draws a load line at 0 mA 415V and 18.2 mA 0V. He then states the tube isn't operated a max dissipation so has to find the "actual location of the load line".

"The operating point is located where the plate current at 415v times the cathode resistor equals the grid voltage. In this case, when the plate current is about 3.7 mA, the grid voltage is about -8V."

He then draws another load line exactly like the first but 3.7mA higher.

So how is he coming up with 3.7 mA? If he's just using a value he's either measured or knows, that's fine, but it's a design book. You'd think it would be predetermined. If he did use previously known values, how does a designer then determine where the load line really exists?

[Stevem]

It's interesting how Fender over runs the plate voltage on these by well over 100 above there design spec!
Well I guess back then in the 60s if I recall right 12At7s ćost some 4 bucks and 6v6's ran about 5.50 !

[martin manning]

"The operating point is located where the plate current at 415v times the cathode resistor equals the grid voltage."

To begin with, this statement is incorrect. The DC operating point is not at 415V because the DC load line is not vertical. It's slope is determined by the resistive load, which is the sum of the transformer primary DCR and Rk.

Hammond lists DCR for a Fender replacement reverb driver transformer at 1065 ohms, so we can assume the resistive load to be 2*(2200+1065) = 6530 (I prefer to double impedances rather than double the current on the anode curves for paralleled triodes). Plot this line from 0 mA at 415V to (pick a convenient voltage) 115/6530 = 17.6 mA at 300V.

The DC operating point lies on the load line where Ia*Rk = Vg-k. To find it, assume Vg-k = -7V, and use Rk to find Ia: 7/(2*2k2) = 1.59 mA. Plot that point on the -7V Vg-k line. Then, assume Vg-k = -8V, and find Ia = 8/(2*2k2) = 1.82 mA. Plot that point on the -8V Vg-k line.

Connect the points on the -7V and -8V Vg-k lines. The intersection of this line and the DC load line is the DC operating point. Using RCA 12AT7 anode curves, I get 403.7V and 1.73 mA for each triode.

[Rogue]

Thanks Martin. That makes a little more sense and that method for finding the operating point is the one I'm familiar with from the Merlin book.

The final result from the Megantz book has the second load line ending at 415V at 3.7mA, with the operating point being there at -8V.

I couldn't figure out the means in which he arrived at that conclusion with the known data.

One question regarding your example....the 17.6mA at 300V. If I draw out that load line, it goes plumb off the chart before it gets down to 100v. Is that a reasonable result for what we're doing here?

[martin manning]

One question regarding your example....the 17.6mA at 300V. If I draw out that load line, it goes plumb off the chart before it gets down to 100v. Is that a reasonable result for what we're doing here?

Absolutely. It's the AC load line that determines the dynamic current vs. voltage relationship.

[Rogue]

Ah, right. The working load line wouldn't be as steep a slope because the AC impedance would be greater. Thanks!

So what is Megantz's reasoning for just shifting the original load line straight up by 3.7mA and keeping operating point right at the supply voltage?

I wished he would have went into more detail here, because it has caused some confusion for me.

[martin manning]

I don't know, trying to keep it simple? In his common cathode stage discussion he again ignores the contribution of the cathode resistor to the load resistance, and the concept of an AC load line is not introduced at all. In my opinion Blencowe does a much better job of explaining the theory, in a more complete and accurate fashion.

[Rogue]

Yeah, Blencowe's book has so far been the standard to which others are compared. I really wish he'd do a power amp book. I'd buy two.

I'm just reading all I can get my hands on now. Megantz is the current one. It's not bad, but not "complete". There's been a few times I've had to stop and say, "wait, what just happened".

Thanks again for your contribution!

[JMFahey]

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]

[Reeltarded]

It might be in google books... in whole or nearly whole..

Because they are copyright bandits and the most criminal institution in the history of mankind, with friends and all benefits, right up there with Pandora, Spotify, and PB(redacted because the mention of the name sends me into a frothing spew)S.


Ooops.. sorry.