What does "Linear" mean?

[dbishopbliss]

The term "linear" is used frequently when referring to plate curves and load lines, but I don't really know what it means.  I have an idea, but I'm never quite sure if my idea is correct.

I've attached a copy of the plate curves for a 5670 tube.  I've marked the "typical" operating point of 150V and 8.2mA and drawn a load line of 100K. 

Please answer the following questions:

• How do I know if this operating point/load is linear?
• How can I determine if there is a more linear operating point/load?

[Paul Joppa]

A good question! Naturally, a complete answer won't fit in a forum post - in fact, it could make a small book, which as far as I know nobody has ever written ...

You can calculate the distortion from the curves - most old tube books (Terman, Seely, RDH4, etc) have the detailed procedure. Do this for a few load resistances and you will learn a lot...  :^)

The lowest distortion will occur with a flat load line, i.e. a current source load. If the grid curves are equally spaced, this will be zero distortion. When I talk of a triode being "linear", I usually mean in this sense. Even a perfect triode will have distortion if the load line is not flat, so my terminology is a simplification.

The distortion will change for different values of quiescent current and plate-cathode voltage. The wider the range over which the horizontal-load-line distortion is low, the more nearly perfect and linear the triode is - again, that's just the way I use the term.

Incidentally, note that 8.2mA through a 100K plate resistor drops 820 volts, so you need a power supply of 970 volts to run this circuit.

[dbishopbliss]

The lowest distortion will occur with a flat load line, i.e. a current source load.

So, a 240K load would be more linear because it is flatter (I know this probably depends upon operating point but for this example). 

Incidentally, note that 8.2mA through a 100K plate resistor drops 820 volts, so you need a power supply of 970 volts to run this circuit.

I based the operating point and load line on the "typical operation" and "class a resistance coupled amplifier" sections of the data sheet.  Sounds like I am interpreting those values incorrectly.  What would be a more reasonable operating point?

http://tdsl.duncanamps.com/pdf/5670.pdf

[Paul Joppa]

...
I based the operating point and load line on the "typical operation" and "class a resistance coupled amplifier" sections of the data sheet.  Sounds like I am interpreting those values incorrectly.  What would be a more reasonable operating point?

Your only mistake was in thinking these two sections of the data sheet had anything to do with each other  :^)  - they don't. They each say only what they say, nothing more. Which means neither of them provides an answer to your question.

A very crude answer would be my rule of thumb for resistor loaded triode amplifier stages - the plate voltage should be about half the power supply voltage. So the power supply would be 300v, and the plate load resistor 150v/8.2mA or about 18K.

I would argue that there is actually no such thing as a "reasonable operating point". For any set of constraints and goals, there are regions of operation which come close to the goals without violating the constraints; there may even be an optimum. Those would be reasonable under those assumptions. The science of getting from the constraints and goals to an actual design is called engineering. Good design includes coming up with good constraints and goals. I think you already have to tools to do these things, it's just work from here on out.

If this effort delights you (and I suspect it does, given the kind of questions you are asking) then you have the makings of an engineer. If it doesn't, then you have find things that someone else has designed so you can do the things that do delight you - for example, actually building the thing. If you want a 5670 circuit that I've designed, then it's the soft-start upgrade kit circuit. The goal was low distortion and adequate current driving the grid of a 300B or 2A3; the constraints were 375v maximum transistor voltage , near zero grid current, reliable and long life operation, adjustable plate voltage, and moderate cost - as perceived by Doc B, of course!

[dbishopbliss]

A very crude answer would be my rule of thumb for resistor loaded triode amplifier stages - the plate voltage should be about half the power supply voltage. So the power supply would be 300v, and the plate load resistor 150v/8.2mA or about 18K.

I'm all about crude answers.  :-)  Enough information to be dangerous (figure of speech, I'm actually very safety conscious while building). 

If this effort delights you (and I suspect it does, given the kind of questions you are asking) then you have the makings of an engineer. If it doesn't, then you have find things that someone else has designed so you can do the things that do delight you - for example, actually building the thing.

I've got the building part down.  Now I'm trying to figure out the "engineer" part.  I flunked physics in college but now I think that has more to with being 18 in Florida and taking a 9:00AM physics class.  Unfortunately for me, my learning style is ask, clarify, ask, restate, ask, do.  Reading a book just doesn't work very well for me.  These threads have been very useful.  Now that I have some "crude answers" to start with, I can move on to start refining my understanding.

[dbishopbliss]

Two more questions:

• when determining the voltage of the operating point, do you measure at point A (before the load resistor) or at point B (after the load resistor)?
• Assuming grounded cathode as in this ciruit, How do you determine the power rating of R4?

[JC]

Paul will probably be along with a more informative answer than this, but, in the meantime, is "Point A" B+, or the positive pole of the power supply for that circuit?

Power dissipation for the cathode resistor would be calculated, I imagine, using Ohm's Law and the power corollary to it, so once you know the resistor's value and the current you expect to be traveling through it, you should be good to go.

[dbishopbliss]

... is "Point A" B+, or the positive pole of the power supply for that circuit?

Language can be a funny thing...

Are you asking if Point A is either "B+" or "the positive pole of the power supply" (two different things)?
Or, are you asking if Point A is the positive pole of the power supply, also known as B+?  I have always thought they were the same, but then the wording confused me and I started thinking I am missing something. 

That said, Point A is connected to the last resistor/capacitor of the power supply.  If my operating point is 150V/8.2ma, should I have 150V at point A or Point B or does it not work that way?

Power dissipation for the cathode resistor would be calculated, I imagine, using Ohm's Law and the power corollary to it, so once you know the resistor's value and the current you expect to be traveling through it, you should be good to go.

Using the values above:

P=V*I
P=150*0.0082
P=1.23W

Then using the rule of thumb that the power resistor should be double the power rating I would need a 3W resistor.  Does this sound correct?

[JC]

You're operating point Voltage will be at the plate of the tube if I understand your question correctly. 

If it is the power dissipation of the cathode resistor you are calculating, you will need to use the Voltage drop across the cathode resistor as the "V" in your equation, which involves knowing the value of the resistance as well as the current through it.  Unless I'm misunderstanding you, this would not be the 150V you used in your example, since that is the plate Voltage.  So, we either need to know the Voltage from cathode to ground from a direct reading, or we need to know the value of the cathode resistance so that we can calculate the Voltage drop across it given your specified current.

[dbishopbliss]

... So, we either need to know the Voltage from cathode to ground from a direct reading, or we need to know the value of the cathode resistance so that we can calculate the Voltage drop across it given your specified current.

Is this the correct way to calculate the value for R4 resistor and or predict the voltage from cathode to ground?

• Draw the operating point of 150V/8.2mA on the grid curves. (see diagram above)
• Determine the grid voltage by either interpolating a value.  My example appears to be 2.1V
• Calculate cathode resistor using R=V/I, R=2.1/8.2, R=256 Ohms

Therefore the power rating would be, P=V*I, P=2.1*0.0082, P=0.017W. 

This seems low to me.  I think I 'm still missing something.

[Paul Joppa]

You're correct, even though you don't think so.  :^)

"Plate voltage" in tube data sheets is the voltage at the plate. Most often, the unwritten assumption is that the cathode is grounded, and a "grid voltage" would be the voltage between grid and cathode.

Occasionally you will see some variant of "plate supply voltage" together with a specified plate resistor - that would be the B+ power supply voltage. That table of operating points uses this terminology. In that table, Ebb is the plate supply voltage; Eb would be the plate voltage (but is not included in the table). Ib would be the plate current (also not mentioned in the table). These are standard abbreviations; capital E stands for voltage, I is current, and subscripts are b for plate, c for grid, and k for cathode. In recent decades people are using V in place of E. The origins of these letters are historical and complex - that would fill up several other posts!

In the case of the referenced 5670 data sheet, the grid is biased by the "cathode bias resistor" which is your R4. Your calculation of 256 ohms is quite close to the specified 240 ohms. In this case, the plate voltage is still the voltage at the plate, 150v, and the plate to cathode voltage is about 148 volts.

In both cases, the plate curves on which you draw the load line uses the plate to cathode voltage as the horizontal axis.

[dbishopbliss]

For some reason I always thought cathode resistors had to be higher wattage.  Although, I've been looking at power amplifiers mostly where the current and voltages are higher, so I guess this makes sense.

So, back to my rule of thumb (the power resistor should be double the power rating), a 0.25W resistor would be more than adequate.