Hello to everybody,
I have problem to figure out the exact meaning and mathematical construction of the dynamic loadline.
All the information that i found over internet are under payment (i mean i have to buy them).
The books that i have are not very clear over this subject and so i ask for your help.
hope somebody can give me some usefull link where to download informations.
Thanks in advance.
Regards,
Stefano.
I have problem to figure out the exact meaning and mathematical construction of the dynamic loadline.
All the information that i found over internet are under payment (i mean i have to buy them).
The books that i have are not very clear over this subject and so i ask for your help.
hope somebody can give me some usefull link where to download informations.
Thanks in advance.
Regards,
Stefano.
A resistive loadline is a nice straight line, and could be called static. A loudspeaker is very rarely resistive and will cause the loadline to depart into an ellipse. You could call that dynamic. There are exotic waveforms (bearing little relation to music) that can cause loudspeakers to draw even larger currents than would be suggested by their complex impedance. These would certainly be called dynamic.
thanks for the explanation.
Now i understand that is the deviation from the resistive behaviour that leads to distortion (if i haven't understood wrong).
What i need is the mathematic explanation e construction and some other detailed information.
I'm sure there are information but as i said, those info are not free.
Now i understand that is the deviation from the resistive behaviour that leads to distortion (if i haven't understood wrong).
What i need is the mathematic explanation e construction and some other detailed information.
I'm sure there are information but as i said, those info are not free.
With general regards to loadlines, a place to start might be to learn to draw them for the single ended common cathode amplifier stage. Here is one tutorial http://www.tubecad.com/articles_2003/Grounded_Cathode_Amplifier/Grounded_Cathode_Amplifier.pdf
if I have understood correctly 😉
if I have understood correctly 😉
You can construct a dynamic loadline by first plotting the resistive element in the standard way, then calculating the peak current drawn by the complex element of the impedance at your chosen frequency and output voltage, subtracting and adding this to your quiescent current to give two further points. Then draw an ellipse to go through these points and the peak voltages you chose. It's usually a bit of a waste of time because it would need to be done at a number of frequencies to show anything meaningful.
It is far better to use an active crossover and add a Zobel network across each loudspeaker to cancel the inductive component, then only drive each loudspeaker above its resonant frequency, thus seeing each loudspeaker as a resistor...
It is far better to use an active crossover and add a Zobel network across each loudspeaker to cancel the inductive component, then only drive each loudspeaker above its resonant frequency, thus seeing each loudspeaker as a resistor...
well i know how to draw a static load line, of course.
The point is that i don't understand the exact meaning of the dynamic loadline neather do the sense of that.
What can i see trhough a-c load line?
I'm interested to see how MATHEMATICALLY the a-c load line is built.
I can't find anything enough clear that helps!
I'm looosssttt.....help me please 🙂 !!!
The point is that i don't understand the exact meaning of the dynamic loadline neather do the sense of that.
What can i see trhough a-c load line?
I'm interested to see how MATHEMATICALLY the a-c load line is built.
I can't find anything enough clear that helps!
I'm looosssttt.....help me please 🙂 !!!
The use of an AC loadline is that it allows you to see how reactive a load you can drive. If your load is of the form Z = a + jb and V is your maximum peak output voltage, then peak reactive current = +/-V/jb. Plot these excursions from your quiescent current (no voltage swing at this point) and you now have four points of your ellipse, which is the AC loadline for your chosen frequency and output voltage. Now look at the ellipse and see if it crosses through anywhere unpleasant, like the negative resistance region of pentode curves, Vg = 0, or (worst) Ia = 0...
where did you get this information from?
Do you have any helpfull link with pictures and formulas?
Or, could you refere me at any good book that covers this issues?
Thanks
Do you have any helpfull link with pictures and formulas?
Or, could you refere me at any good book that covers this issues?
Thanks
If you are interested in the equation for an ellipse, then it is (x^2/a^2) + (y^2/b^2) = 1, where 2b is the minor axis, and 2a is the major axis.
Basically, if you fit a rectangle just outside the bounds of the elllipse, the height and width will be 2b and 2a.
May the Good Lord help you if you actually want to use this equation, or if you attempt to correlate a speaker to this equation (make it fit).
That's what they make curve traces, trial and error, and experience for.
Basically, if you fit a rectangle just outside the bounds of the elllipse, the height and width will be 2b and 2a.
May the Good Lord help you if you actually want to use this equation, or if you attempt to correlate a speaker to this equation (make it fit).
That's what they make curve traces, trial and error, and experience for.
well......actually right now i'm studying the theory of the tube's amp.
I read a 3-4 books by now and just yesterday i found this dynamic loadline.
honestly your word confused me even more.
I know what an elliplse is and its equation (i'm an electronic enineer).
I just would like to find a clear book that would explain as well as the static load line is usally explained everywhere.
The sourec of infometion that i got don't consider this subject very clear!
first i don't understand why the loadline has to pass trhoug the bias point.
I've just proved the following equation:
given a load zb=rb+jxb
sinusoidal plate current ip=Ipm*sin(wt)
the plate voltage: ep=-ip*zb
ep^2 + 2 * rb* ip * ep + mod(zb)^2 * ip^2= xb * Ipm^2
which is an ellipse with origin on (0,0) so i don't understand how it can pass for the bias point.
But then, from here, i don't know how to move over and match this with the static tube characteristics, what has to do with the distortion and i don't know, mainly, what it represents exacly in the contest of an amplifier stage ....and so on and so forth.
Hope to clearify this point.
Regards.
P.S. i see that you are from milwaukee...my wife is from wisconsin (oconomowok WI)...and i've been to milw last summer....i've been to a jazz festoval there....it was super nice.....the only thing...in wi...they pretend to imitate too much the italian food...and they are not able to get it all the time 😉
I read a 3-4 books by now and just yesterday i found this dynamic loadline.
honestly your word confused me even more.
I know what an elliplse is and its equation (i'm an electronic enineer).
I just would like to find a clear book that would explain as well as the static load line is usally explained everywhere.
The sourec of infometion that i got don't consider this subject very clear!
first i don't understand why the loadline has to pass trhoug the bias point.
I've just proved the following equation:
given a load zb=rb+jxb
sinusoidal plate current ip=Ipm*sin(wt)
the plate voltage: ep=-ip*zb
ep^2 + 2 * rb* ip * ep + mod(zb)^2 * ip^2= xb * Ipm^2
which is an ellipse with origin on (0,0) so i don't understand how it can pass for the bias point.
But then, from here, i don't know how to move over and match this with the static tube characteristics, what has to do with the distortion and i don't know, mainly, what it represents exacly in the contest of an amplifier stage ....and so on and so forth.
Hope to clearify this point.
Regards.
P.S. i see that you are from milwaukee...my wife is from wisconsin (oconomowok WI)...and i've been to milw last summer....i've been to a jazz festoval there....it was super nice.....the only thing...in wi...they pretend to imitate too much the italian food...and they are not able to get it all the time 😉
OK, I haven't actually drawn a dynamic loadline and have never felt the need. It does seem that the ellipse would grow larger dependent on level and so this cannot be as versatile as a loadline for a resistive load and you may want to draw several. I also thought the ellipse should revolve around the bias point but it must start there on the first cycle.
Maybe all you need to do is to plot the extremes of a max output, steady cycle. Or what some here do in practice and try to just mitigate the effect of a reactive load instead 😉.
Maybe all you need to do is to plot the extremes of a max output, steady cycle. Or what some here do in practice and try to just mitigate the effect of a reactive load instead 😉.
Stefanoo said:
The only things we do well are medium rare prime rib, cheese curds, and dumplings. You know, all the stuff that makes us as fat as we are.
From one EE to another, you are punishing yourself. You can strain your brain over these equations, and possibly even figure it all out. But then what do you do with it? What type of mods do you add/subtract from your amp to correct for reactance? It seems to me your design may stray into such complexity that the human ear may not appreciate it in the end.
My recommendation would be as others: assume a resistive load, correct with zobel if needed, and trust your negative feedback to do its job.
If you still simply want to know the theory, irrespective of building it, it may continue to be very difficult to find your answer. If for no other reason, not many consider it important to study to produce a great amp. As a result, there is little information on the subject.
But "well done" to anyone who can significantly add to the discussion.
Stefanoo, I think you have the basic idea. I haven't verified the math, but trust that you are at least as able to get it right as me. The real problem is that the dynamic loadline you end up with depends on the load; the speaker. Unfortunately, they don't seem interested in behaving simply, even as simple as resistive with a little L thrown in.
In any case, the dynamic loadline does not need to pass through the static (DC) bias point. In the case of a 'simple' R+L load the loadline happily orbits the DC bias point under AC operation. Perhaps that puts your mind at ease a little.
The only advice I can offer is to choose an operating point (even a tube) that allows the dynamic loadline to wander quite a bit from the static straight line loadline without getting ugly. If we accept the idea that our speakers are not going to behave perfectly, then giving them as much room as possible to misbehave will give us the best chance of ending up with something that sounds good. That's my approach, though I understand that it does not address your basic question.
-- Dave
In any case, the dynamic loadline does not need to pass through the static (DC) bias point. In the case of a 'simple' R+L load the loadline happily orbits the DC bias point under AC operation. Perhaps that puts your mind at ease a little.
The only advice I can offer is to choose an operating point (even a tube) that allows the dynamic loadline to wander quite a bit from the static straight line loadline without getting ugly. If we accept the idea that our speakers are not going to behave perfectly, then giving them as much room as possible to misbehave will give us the best chance of ending up with something that sounds good. That's my approach, though I understand that it does not address your basic question.
-- Dave
Hey ziqzaqflux,
What no Brats? Most of my family is from West Bend, WI. Can't visit without getting my fill ... and beer, of course.
What no Brats? Most of my family is from West Bend, WI. Can't visit without getting my fill ... and beer, of course.
CarlyBoy said:
LOL !
OK, you got me on the brats. That is definitely one of Sheboygan's specialties.
...i like WI ..... except for the cold and as i said some of the food!
Anyways...i heard that this year is not as cold as it use to be in the past.....good for you guys 🙂 !!
i would love to go back a little bit there and see a baseball game again!!
..going back to the 3d...more than this considerations...i'm looking for a theorical consideration of the dynamic loadline.
It's just a theory's issue.
i would like to understand deep down the theorical meaning of the ac loadline.
Then....after your post.. i have understood that it's not really employed in practice 🙂 !!
Hope to find some good information about this because the more i read the books that i have and the more i get confused
Anyways...i heard that this year is not as cold as it use to be in the past.....good for you guys 🙂 !!
i would love to go back a little bit there and see a baseball game again!!
..going back to the 3d...more than this considerations...i'm looking for a theorical consideration of the dynamic loadline.
It's just a theory's issue.
i would like to understand deep down the theorical meaning of the ac loadline.
Then....after your post.. i have understood that it's not really employed in practice 🙂 !!
Hope to find some good information about this because the more i read the books that i have and the more i get confused
It's just occurred to me that all this talk of ellipses orbiting the bias point might not be your fundamental problem. The other use of the the term "dynamic loadline" is to distinguish between the loadline produced by a resistive anode load that is valid at AC and DC and the AC loadline produced by an inductive anode load that passes through the DC operating point but has a gradient set purely by AC parameters.
....where did you get this information from?All this considerations come from the analysis of the meaning of the loadline (AC) .... and i would like first to understand how this line or ellipse in the case of a reactive load is drawn!
till i won't get the mathematical meaning of this curve i won't be able to catch all this concepts that you guys are kindly expressing to me!
hope somebody could source me or indicate me a text book or link on the net that covers these issues in a simple-clear and detailed way! (too demanding? 🙂 )
Thank you guys for all this posts
I started to write out all the equations for a treatment of load lines, but ran out of patience- trying to do all the subscripts, superscripts, and Greek letters was too time consuming. And my scanner isn't up at the moment, so I can't do sketches. So, how about if I give you a brief handwaving explanation and cite a reference which treats it more rigorously?
With a reactive load Z, defined as R + X, with X being an imaginary quantity which is a function of w (the angular frequency), the AC current is equal to the peak current times sin(wt). AC voltage by Ohm's Law must be the peak current times the impedance times sin(wt + theta), where theta is X/R. Total current and total voltage have the DC bits added to them. With a little bit of math, you can convince yourself that the equation for this load describes an ellipse on the I/V plate curves with the operating point at its center. All of this assumes that the plate curves are distortionless, that is, the magnitude peak current and the trough current with respect to the operating point are equal.
OK, that's where the mysterious ellipse comes from.
Now, let's look at a dynamic situation. Again, we'll assume distortionless plate curves. Suppose we load the tube with an ideal transformer having a low DC resistance but a high AC impedance (e.g., an output transformer with a few ohms of DCR but relecting back the load at several thousand ohms). Now, to determine the operating point (which is at DC), we consider that low DC resistance and draw the usual static load line. It will be fairly steep since we know that the DC resistance is low. We set the static operating point somewhere on that line.
What happens when we start wiggling the voltage? Clearly the AC component of the plate current will have to change, but since the plate load at AC is a high impedance in series with the low DC resistance, it will not move along the low impedance DC line, it will move along that AC line, but with its "zero" at the operating point- that is, we need to draw a line with the slope corresponding to the reflected impedance, but passing through the operating point. Is that clear so far?
Assuming yes, we now have some grip on the difference between a dynamic and static load line. Now here's the twist: if we now consider that the tube has distortion, that is, the swing down from the operating point is not equal to the swing up, we see that when we construct the dynamic load line, the effective dynamic operating point will actually move away from the DC operating point, since there effectively is an offset voltage due to the uneven swing- a transformer can't transform DC, so the effective operating point as far as AC is concerned has to shift to a point which removes the DC component. So we have the static operating point and now a dynamic operating point.
For a rigorous treatment, try H. Reich, "Theory and Application of Electron Tubes." He does a very nice analysis for those willing to slog through the math. In somewhat simpler form, Reich does a graphical treatment in "Principles of Electron Tubes." The latter book has, I think, been reissued in paperback and is very worthwhile to own.
I hope this was helpful.
With a reactive load Z, defined as R + X, with X being an imaginary quantity which is a function of w (the angular frequency), the AC current is equal to the peak current times sin(wt). AC voltage by Ohm's Law must be the peak current times the impedance times sin(wt + theta), where theta is X/R. Total current and total voltage have the DC bits added to them. With a little bit of math, you can convince yourself that the equation for this load describes an ellipse on the I/V plate curves with the operating point at its center. All of this assumes that the plate curves are distortionless, that is, the magnitude peak current and the trough current with respect to the operating point are equal.
OK, that's where the mysterious ellipse comes from.
Now, let's look at a dynamic situation. Again, we'll assume distortionless plate curves. Suppose we load the tube with an ideal transformer having a low DC resistance but a high AC impedance (e.g., an output transformer with a few ohms of DCR but relecting back the load at several thousand ohms). Now, to determine the operating point (which is at DC), we consider that low DC resistance and draw the usual static load line. It will be fairly steep since we know that the DC resistance is low. We set the static operating point somewhere on that line.
What happens when we start wiggling the voltage? Clearly the AC component of the plate current will have to change, but since the plate load at AC is a high impedance in series with the low DC resistance, it will not move along the low impedance DC line, it will move along that AC line, but with its "zero" at the operating point- that is, we need to draw a line with the slope corresponding to the reflected impedance, but passing through the operating point. Is that clear so far?
Assuming yes, we now have some grip on the difference between a dynamic and static load line. Now here's the twist: if we now consider that the tube has distortion, that is, the swing down from the operating point is not equal to the swing up, we see that when we construct the dynamic load line, the effective dynamic operating point will actually move away from the DC operating point, since there effectively is an offset voltage due to the uneven swing- a transformer can't transform DC, so the effective operating point as far as AC is concerned has to shift to a point which removes the DC component. So we have the static operating point and now a dynamic operating point.
For a rigorous treatment, try H. Reich, "Theory and Application of Electron Tubes." He does a very nice analysis for those willing to slog through the math. In somewhat simpler form, Reich does a graphical treatment in "Principles of Electron Tubes." The latter book has, I think, been reissued in paperback and is very worthwhile to own.
I hope this was helpful.
One other way to look at the same thing: The total instantaneous plate current equals a constant current plus an ac term. The reactance only operates on the ac part.
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