How is SLOPE = -1 / Ra ??

[Lectroid]

I should start by saying I've had a fair amount of exposure to Cartesian coordinates up through first year calculus.  I came across this first in Blencowe's "Designing Tube Preamps...", page 24,  but I've read similar statements in other books:

"We can see from the figure that the DC load line has a slope of –1/Ra, but the addition of the AC load causes the working load line to rotate clockwise around the bias point, and it now has a steeper slope of –1/(Ra||Rl)."

Image attached below.

To me, slope is equal to "the rise over the run."  My question is that when I see "-1/Ra" I want to substitute "-1 / 100K." When I do that, I get 0.00001.  I have no idea how that translates in a slope.   

I'm sure I'm missing something obvious.  Can someone more learned please tell me where I'm confused?

My thanks in advance. 

[acheld]

Well, I'm not more learned, just older.

I think he is talking about proportionality in this case, though I may be wrong.   Those load lines are derived from empiric test data which is published (lifetimes ago!) by the manufacturers.   

This is one case where I think Rob Robinette has done a great job of consolidating information about load lines, see:  https://robrobinette.com/Drawing_Tube_Load_Lines.htm, better than Merlin, who is really talking about systems engineering.   I love Merlin's books, but they are aimed at a different audience.

You are correct: the derivative of the 100kOhm load line at the bias point is -1mA/100V.   

[shooter]

 
wait til you get to the part where -1/0 =i

[tubeswell]

Do you mean?

wait til you get to the part where -1/0 =ij

[shooter]

it's been 40yrs cut me some slack 

[Lectroid]

Do you mean?

wait til you get to the part where -1/0 =ij

Imaginary numbers don't faze me.  I don't bother them and they don't bother me.   


However, on a serious note, still looking for an help with my slope confusion if anyone can help me out.

...I think he is talking about proportionality in this case, though I may be wrong.   Those load lines are derived from empiric test data which is published (lifetimes ago!) by the manufacturers.   

This is one case where I think Rob Robinette has done a great job of consolidating information about load lines, see:  https://robrobinette.com/Drawing_Tube_Load_Lines.htm

I've been over RobRob page many times.  I don't think he addresses this question.  At least, not in a where that I understood what he was getting at.   

You are correct: the derivative of the 100kOhm load line at the bias point is -1mA/100V.   

That looks like tautology.     Seriously,  -0.5mA / 50V will give you the same result. (0.00001)  So will -2mA/200V.  I'm sure I'm may be missing something completely obvious, but I still don't get why the slope = -1/Ra



 

[PRR]

You are over-thinking.

A high slope is a low resistance. Maybe not on your side of the hill, but in most electronics (and particularly the Brits).

[Lectroid]

You are over-thinking.

A high slope is a low resistance. Maybe not on your side of the hill, but in most electronics (and particularly the Brits).

@PRR:  I could certainly be overthinking this...but--"a high slope is a low resistance?"  I wish I could say that clears everything up.  But no, I'm still in the dark about why "slope = -1/Ra."  I do understand that in my example from Merlin, the lower resistance (Ra||ra = 28K) line has a steeper slope than the (Ra = 100K) line. 

But I don't see where that dimension-less ( -1 ) comes from, or why dividing it by Ra yields the slope.  ached alluded to it actually being -1mA.  Is that right?  Then why doesn't Merlin say "-1mA?"  Thanks. 

[tubeswell]

You are over-thinking.

A high slope is a low resistance. Maybe not on your side of the hill, but in most electronics (and particularly the Brits).

@PRR:  I could certainly be overthinking this...but--"a high slope is a low resistance?"  I wish I could say that clears everything up.  But no, I'm still in the dark about why "slope = -1/Ra."  I do understand that in my example from Merlin, the lower resistance (Ra||ra = 28K) line has a steeper slope than the (Ra = 100K) line. 

But I don't see where that dimension-less ( -1 ) comes from, or why dividing it by Ra yields the slope.  ached alluded to it actually being -1mA.  Is that right?  Then why doesn't Merlin say "-1mA?"  Thanks.

The plate characteristics chart is in 2 different scales. The x-axis is in volts. The y axis is in milliamps. Looked at this way, mathematically, a gradient of 0.00001 isn't out of the ballpark

[PRR]

.... I don't see where that dimension-less ( -1 ) comes from, or why dividing it by Ra yields the slope.  ...

OIC.

We want to know what the TUBE sees. Yes, if we plotted the resistor we would get E/I. But the tube sees the part of battery voltage that is left-over after the resistor drops it. (V+)-(Vr)

[acheld]

I still don't get why the slope = -1/Ra

It doesn't.   It's an "expression" of conductance.   

As for imaginary numbers, they do not bother me either, and I haven't messed with them in 50 years.  Second order partial differential equations do bother me, even after 48 years . . . 

[Lectroid]

The plate characteristics chart is in 2 different scales. The x-axis is in volts. The y axis is in milliamps. Looked at this way, mathematically, a gradient of 0.00001 isn't out of the ballpark

Why not?  I understand that the characteristics graphs use different scales on their axes. I followed you that far.  You lost me making the jump to your conclusion.

I still don't get why the slope = -1/Ra

It doesn't.   It's an "expression" of conductance.   

That almost makes sense to me.  Except for that dimensionless -1/Ra.  I feel like you all know what that (-1) means, and so no one's made a point of actually explaining it.

I may just be terminally dense.

[PRR]

> -1/Ra

"..the part of battery voltage that is left-over after the resistor drops it. (V+)-(Vr)"

Simple subtraction. Not imaginary. Dimensional Analysis is important for proof but if you grok the physical system you may not need it.

[acheld]

I feel like you all know what that (-1) means, and so no one's made a point of actually explaining it.

I'll give it another shot.    In fact, my starting approach to load lines was much like yours -- here is a graph, with a line, and the line is visualizing an equation.

Well, not really.    Load line graphs are a tool to assist the designer understand what will happen when any one of the relevant variables changes.   Such as Ra, Rk, the following stage impedance, Vpp, and the input signal voltage . . .

So it's not really one equation you're looking at, but a method of visualizing how small signals are amplified by a tube.    Remember, load line graphs were invented a century ago (when engineers and school boys had slide rules), and designers needed a way of understanding how their circuits would work. 

If you think of the slope of the DC plate load line, it simply represents Ohm's law.   That's it.  The current that flows through a tube at a defined plate voltage.  By itself, you can infer gain, but the magic is when you find your operating point, and then start to see what happens with your AC load line, etc.

Seriously, go back to RobRob's "Drawing Load Lines" page and work through it.   He really explains it well.   Merlin assumes you know this already.

 

[tubeswell]

The plate characteristics chart is in 2 different scales. The x-axis is in volts. The y axis is in milliamps. Looked at this way, mathematically, a gradient of 0.00001 isn't out of the ballpark

Why not?  I understand that the characteristics graphs use different scales on their axes. I followed you that far.  You lost me making the jump to your conclusion.

1V/1I = a unit of impedance* (1R). With high enough impedance (like 100k or so), we don’t have to change I much (i.e. only a matter of a few mA) to get a relatively large swing in V (a couple of hundred V).

* I.e. we’re talking about resistance to AC.



Now if the chart was drawn using the same unit scales for V and I, the y-axis would be all squashed down and we wouldn’t be able to see the darn load line, and although we’d be theoretically able to plot a -0.00001 gradrient at scale for the AC load, it would visually look indistinguishable from the gradient for the DC loadline - unless you were able to squint really hard, both gradient lines would look horizontal. So we make the x and y axes a hundred-thousand scales different (i.e. thousandths of an amp one way and hundreds of volts the other way), so that we can see the loadline ‘gradient’ more easily.


But mathematically, the AC loadline gradient is still -1/Ra (because Ohms Law is 1R x 1I = 1V)

[Lectroid]

Okay, we've made some progress. 

@acheld and @tubeswell,
I will go back and re-read Rob's page about this.   I can see that tube graphs are not the tool I thought they were.

I finally get tubeswell's point about the difference in scales between hundreds of volts and milliamps. You're right, a slope of .00001 is mathematically quite reasonable.  I've been looking through the wrong end of the telescope all this time.  Dang.

Thanks guy.  Sorry I had to have my nose rubbed in it to see it.   (that terminally dense thing.)   

[shooter]

once you're looking through the correct end AND know what you're looking at;
steal everything from others, tweak as needed using ears, friends, and random street people, the time savings from all the math and brain energy is music to the speakers 

[Lectroid]

1.  Completely down with the stealing.  Been doing that for decades.   

2. Tweaking underway after a slow start, once I realized that I could do this magic too.  "Bwa-ha-ha-ha!  I've got a capacitor and I know what to do with it!"

3 But all that math in school!  It's gotta be good for something, right...     And after my hair fell out, it keeps my head warm.   

[HotBluePlates]

Maybe folks are having a hard time articulating the "obvious" so here's a quick run through.

"We can see from the figure that the DC load line has a slope of –1/Ra, but the addition of the AC load causes the working load line to rotate clockwise around the bias point, and it now has a steeper slope of –1/(Ra||Rl)."
...
To me, slope is equal to "the rise over the run."  ...

You are correct that slope is "Rise over Run."

   "Rise" is Y-axis.  Moving left-to-right we see the Blue line move downward from 3mA to 0mA on Y-axis.  That's -0.003 Ampere

   "Run" is X-axis.  Moving left-to-right we see the Blue line span from 0v to 300v.  That's +300 Volt.

   Now Rise-over-Run:  -0.003 A / 300 V = -0.00001

   What is the inverse of this figure? 1 / -0.00001 = -100,000 ---> 100kΩ


This result matches Blencowe's statement of "-1/Ra."

   It has this form because this line represents the Plate Volts left on the plate (X-axis figure) after the plate current (Y-axis figure) creates a voltage drop across the plate load resistor.

   This resistor is subtracting from supply voltage to leave an amount on the plate.  Therefore the Rise-over-Run is negative.

Now pretend we just want Volts (X-axis) across a resistor for a given amount of current (Y-axis).

   Say this resistor is connected to ground, so the line (going left-to-right) will start at (0,0).

   Say the resistor is 50kΩ.  For 1mA there should be 50kΩ x 1mA = 50v across the resistor.  Put a dot at (50v, 1mA).

   For 2mA there should be 50kΩ x 2mA = 100v across the resistor.  Put a dot at (100v, 2mA).

   For 3mA there should be 50kΩ x 3mA = 150v across the resistor.  Put a dot at (150v, 3mA).


Now calculate the rise-over-run for this resistor:  Run is 150v, Rise is 3mA --> 0.003 A / 150v = 0.00002

   The inverse of this number is 1 / 0.00002 = 50,000 Ω

By the way:  you could figure out how 2 resistors will split an applied voltage graphically with this method.

   Pretend one resistor is connected to the voltage supply, and the other is connected to ground.

   Plot the line representing the resistor-to-ground starting from (0v, 0mA) and proceed as in the last example.

   Plot the line representing the resistor-to-supply starting from (Supply Volts, 0mA) and proceed as when drawing a DC loadline.

   The 2 lines will cross at the value of total current through the divider, and show how many volts are dropped across each resistor.  Count volts from "0v upward" for the resistor-to-ground" and count from "supply volts downward" for the resistor-to-supply.

Why do we keep having to take an inverse to get the resistance value?

   Because the data sheet curves draw Volts as the Independent Variable (X-axis) while Current is the Dependent Variable (Y-axis).

   But when we plotted our points we had to start with current (Independent Variable) and find the Volts by multiplication (Dependent Variable).  The axes used by data sheet curves (Current / Volts) is inverse of what Ohm's Law says (Volts / Current), which is why we wind up having to find the inverse of the caulcated slope to find the resistance value.

EDIT: adding Lectroid's original graph, now with an Orange 50kΩ line.

[AlNewman]

You have a way with words, HBP.
I'm not sure what you said, but it was certainly said well.

[shooter]

not sure what you said

turn the telescope around 

[HotBluePlates]

By the way:  you could figure out how 2 resistors will split an applied voltage graphically with this method.

Now that Lectroid's original graph has been modified in my post to show the newly-plotted 50kΩ line in Orange, we can observe this ourselves.

   Orange is the new 50kΩ resistor connected to ground.
   Blue is the original 100kΩ resistor connected to a 300v supply.
   If these two resistors were in series, their intersection now shows how they will split that supply voltage.

Let's observe how each line has a Positive or Negative slope:

   We will count from the intersection of the two lines.

   The Orange Line moves Up, and to the Right.
   Rise (current change) is in a positive direction.
   Run (voltage change) is in a positive direction.
   Rise / Run is positive:  2mA / 100v = 0.00002

   The Blue Line moves Down, and to the Right.
   Rise (current change) is in a negative direction.
   Run (voltage change) is in a positive direction.
   Rise / Run is negative:  -2mA / 200v = -0.00001

Peeping the graph, we see these lines cross at 2mA, and 100v.

   The graphical intersection is what Kirchoff would call a "node."
   Kirchoff's Current Law ("KCL") tells us the sum of currents entering and exiting a node equals 0 (what goes in comes out).
   If we move consistently from Left to Right on the X-axis, the graphed current adheres to KCL:
     The Orange Line rises from 0mA to 2mA, so represents 2mA entering the node.
     The Blue Line falls from 2mA to 0mA, so represents 2mA leaving the node; numerically -2mA.

     Sum of Currents entering/exiting the node:  2mA + -2mA = 0  ----> Kirchoff is Happy

Now graphically analyze the Voltage Division:

   Count from "Ground up" to the intersection to observe the Volts across the 50kΩ (Orange):  0v to 100v = +100v
   Count from "intersection up" to the Supply Volts to observe the Volts across the 100kΩ (Blue):  100v to 300v = +200v
   Check that Volts dropped across each part of the divider equals total voltage applied:  100v + 200v = 300v  √

Let's mathematically analyze this voltage divider to see we arrive at the same conclusions:

   Total Resistance = 100kΩ + 50kΩ = 150kΩ
   Total Voltage: 300v
   Current through series-connected resistance:  300v / 150kΩ = 2mA
     The resistors are series-connected, so current through any resistor equals the current through the circuit.
     This practical use of KCL means current though each resistor is 2mA.

   50kΩ Voltage Drop:  2mA x 50kΩ = 100v
   100kΩ  Voltage Drop:  2mA x 100kΩ = 200v

   Check that Volts dropped across each part of the divider equals total voltage applied:  100v + 200v = 300v  √

Let's mathematically analyze this voltage divider with "ratio of resistors" to see we arrive at the same conclusions:

   R1 = 100kΩ
   R2 = 50kΩ
   V_in = 300v
   V_out = Voltage across R2, and also the "Node Volts" with respect to reference.

   V_out = V_in x [R2 / (R1 + R2)] = 300v x (50kΩ / 150kΩ) = 100v   √

We have now checked & proven several different ways that:

   We can graphically inspect how 2 resistors divide voltage.
   One Resistor-Line will have a positive slope, because current is flowing into the node (intersection of resistors).
   One Resistor-Line will have a negative slope, because current is flowing out of the node (intersection of resistors).
   The Resistor-Lines will intersect graphically at the Node Voltage, and the Circuit Current.
   Resistor-Line slope is 1/R because the graph Rise/Run = Current/Volts, but Ohm's Law is R = Volts/Current.

[HotBluePlates]

By the way:  you could figure out how 2 resistors will split an applied voltage graphically with this method.

        Turns out, this is exactly what we do with graphical loadlines for tube circuits.

Notice how the Orange 50kΩ line I added to the graph is very close to the 0v grid-line, and quite nearly parallel to the 0v grid-line where the Blue 100kΩ line crosses each.

   The grid-line shows what happens when we hold grid voltage fixed, then increase plate current.
   The 0v grid-line shows that when plate volts increase from 0v to 100v, plate current increases from 0mA to a bit over 2mA.
   We can infer that this grid-line represents "a Resistance of a little less than 50kΩ."

Turns out, Internal Plate Resistance ("r_p") of a tube is defined as:

   Change of Plate Current / Change of Plate Volts
   This exactly matches our graph's axes, and Rise/Run!

From this we take away that the slope of grid-lines show us the internal plate resistance of the tube.

   If our tube had an r_p of exactly 50kΩ when grid volts = 0v and plate volts = 100v, then that grid-line would intersect the Blue 100kΩ line at (100v, 2mA).
   We see the slope of the 0v grid-line is close to 50kΩ, but is a little less.
   We know the r_p of the tube is less than 50kΩ because current (100v/r_p) is slightly higher than 2mA, and volts across r_p is slightly less than 100v.

So now we see that the grid-lines represent the tube's r_p: so the intersection of a grid-line and the loadline for R_L tells us how the tube & plate load will split the supply voltage.

   Further, the tube is the part between Ground and the "Node" so the intersection is the voltage across the tube: Plate Volts.

We could now make other observations about the grid-lines:

   The grid-lines are curves:  their slope changes, so they represent a changing resistance.
   The grid-lines are shallower-slope at low current:  this represents higher r_p.
   The grid-lines are not same-arc:  the change of r_p varies with different grid-voltages.

We could then note that every place our loadline intersects a grid-line, there is a different slope for the particular grid-line.

   The curved grid-lines mean the r_p is the "tangent to the curve."
   A challenge is that if we need to know r_p at a point, we cannot simply calculate it from Plate Volts / Plate Current.
      This would represent a line from (0v, 0mA) to the operating point, and be very much flatter-sloped than the grid-line at the operating point.
      The actual r_p has a steeper slope, and is therefore a smaller resistance.
   We can draw a "tangent to the curve" at the operating point to define a "r_p line" from which we can find r_p.

I've added a Red line for r tangent to the -1v grid-line where that grid-line crosses the Blue 100kΩ loadline.

   Swag the intersection:  about 165v and 1.2mA ---> 165v / 0.0012A = 137.5kΩ (not r_p!)
   Rise / Run of the Red r_p line:  Rise about 3.7mA, Run from about 85v to 300v --> 300v-85v = 215v
   Rise / Run = 1 / r_p ---> Run/Rise = r_p = 215v / 0.0037A = 58.1kΩ




There are mathematical methods to calculate stage gain that rely on knowing rp and seeing how ideal Amplification Factor is divided between R_L and r_p.  However, we generally need to find r_p graphically using the method above to apply the calculation.  Add the complication that r_p is changing over the signal cycle.

[Lectroid]

@HBP,

Thank you for explaining that in so much detail.  I came in here with (what I thought was) a simple question.  Not for the first time, you answered the question I asked clearly and concisely, but then also went on to hand me a big chunk of well-explained conceptual framework to hang it on. I know that took some work, not a two minute throwaway post.  I really appreciate it.  I keep a few interesting threads I find here into an offline doc.  This post will go in there. I'll be digesting everything in this for a while yet.      

And to everyone else who answered: you have all helped me straighten out a bunch of bad noob assumptions.  Any remaining backwards knowledge is all due to me and my mis-wired telescope.  Thanks to all of you.   

Response like this is exactly why I spend my time here.