Plate Curves and Load Lines!
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- เผยแพร่เมื่อ 6 พ.ย. 2020
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Warning!
The circuits discussed in this
video contain High Voltage.
There is a risk of injury and death
when working with these circuits.
Please take precaution.
Don't attempt to build these circuits if you
are unsure of your ability to do so safely.
One of the best explanations of load lines and the effect on amplification I've seen in years! Thank you for posting this.
Incredible deep dive into load lines. Thank you!
Wow, Doug, the best explanation of loadlines ever... the white boards, blue-tack, and load line sticks are so much better than any computer-generated graphic!
I have been in the electronic repair business for almost 30 years, and have loved vacuum tubes through all the integrated circuits and solid-state stuff I've had to slog through!
I have been to college, have the Radiotron Designers Handbook, and too many others to list, and am getting into more tube design then repair work now that I have turned 65 (age has its benefits!).
The one thing I haven't wrapped my head around yet is the primary impedance change when one of the tubes shuts off in class AB. I'm going to dig into my books and see if I can figure this one out.
Your video wasn't too long at all, I plan on watching it at least two or three more times. Thank you for generously sharing of your knowledge!
Thanks for the kind words! It means a lot!
I've added a couple of questions and then deleted them because I've ultimately answered my own questions upon further study. I've watched this video about 5 times now and I'm still learning. This is just outstanding. Thanks so much. I've plotted load lines so many times but never really stopped to fully fully understand everything about them. I'm doing that now, with your help. It's people like you who make TH-cam worthwhile. Know that your time and and effort will not go unnoticed and unappreciated for years to come. Cheers.
Wow man, it’s definitely comments like this that make it all worth it! 🙂
Thank You ! Doug. Took me 2 year to found this video. 🙏
I couldn’t hit the “like” and “subscribe” fast enough. Very nicely done!
Many thanks for sharing this amazing and so precious video. One of the best/the best explanation about it.
Great video. Thanks
Very cool video. Thank you.
This is the best i have seen. Now i understand much better. Thank You :)
Brilliant!
great video, very good!
Good explanation, Thank!
Awesome lecture! Thank you very much. 😄
Great tutorial, thank you
Happy to help
This is the single most informative video about this topic, super useful!
Thanks for share . Great vídeo.
Holly cow!
This is the best of the best of everything I have seen on TH-cam.
Amazing work. Congratulations.
This video is, basically, what to know to build a tube/valve amplifier.
I was expecting to see a video on triode loadlines and you've covered everything.
I have read a lot about this, watched a million videos and, with this one, I have finally understood some aspects that were not clicking for me.
Thanks a million.
This makes me very happy!!
idem this video is amazing
52:27 brilliant explanation, always wondered how a push-pull Class A amplifier worked as far as the load line goes. Love it
My wife, and I, studied: "REACTANT PHYS -- UCSD -- CURRICULUM - there be -- none of this!"
Youngya -- + De -- + Doug
Very well explained. Chapeu bas!
I have come back to this video many times over to refresh my memory. I just want to compliment you again or what a fantastic video this is and thank you very much for uploading.
It really makes it all worth it when people leave a comment like this. Thanks!!
Thank you very much for this explanation. On of the best explanations i've seen around the web those days!
That’s nice of you to say! 🙂
Great video! I love it. I've got one question. What to do if I'd like to design a Push-Pull power stage with multiple parallel tubes? Let's say that I'd like to use specific tubes and match the output transformer to them. I assume that I should just calculate the needed transformer primary plate to plate impedance as with only two tubes, and divide it by 2 in case of 4 tubes, 3 in case of 6 tubes etc. Am I correct? But divide or multiply?
The plates will be in parallel so you need to do the parallel resistance calculation.
I think is a good idea to clarify .How those two points that define the load line are selected ..
1) Max plate Voltage at 0 current .. THS IS the tube in cutoff ,,, no current flows ..this happens when grid bias is very negative
here the Max Plate voltage = Power supply ..So you will chose this point equal to the power supply
2) MAx current at 0 plate volts ( yes is the PLATE voltage not power supply) ..This happens when tube is in saturation ,,and the plate is at ground level .This happens when grid bias is very positive ..
Max plate voltage equals the supply voltage in a resistive loaded stage. It swings higher than supply in a inductively loaded stage. Thanks for the comment.
Great video. Not to be too picky but I noticed you said around 19:05 “when you load the tube” and increase the steepness of the load line.
I’d think you mean to say as you load the power supply in the sense that the power supply is applying the fixed voltage to the circuit the tube is in. No?
Increasing the load on the power supply with less load resistance would steepen the load line resulting in more current thru the tube.
The beginning part of the video seems to explain a graphical method of or approach to solving a circuit (finding significant or operating point(s) of interest given consideration to dc voltage supplying circuit and load resistor and triode tube performance curves and voltage at control grid).
I just say this because it’s a subtle but important point.
I think sometimes you say when you load the circuit which is technically more correct given the triode behavior.
But for a pentode you could probable say load the tube because the pentode operates more like a current source or current driver.
But actually after thinking about it loading the pentode would be load line with decreasing slope.
But this is an excellent video and great reference source and I will study it.
I have Simulink and I’m going to build a model of circuit with tube behavior in a kind of black box model and plot out the performance curves.
I’m going to extend the operation to saturation beyond what the tube can dissipate as if the tube is ideal vacuum tube with unlimited heat rejection.
But you have answered the question about the part of the behavior where the curves end at operation point where tube operation is heat rejected limited and in the shape of constant power.
I keep coming to this vid haha, I'm in the process of designing a guitar amp with a push pull power stage using two PL504 and this is super helpful.
Thanks for the comment. It definitely feels good that people find this helpful!
@@Edgarbopp sure man. Hey, would you check my spice model of the design? I hope I’m not bothering. No probs if you can’t, the thing is I’m a rookie in this field and lacking feedback. Thanks anyway since you’ve already helped me a lot with these vids =)
@@leandrocarg if you show me the circuit I’ll have a look at it and let you know if I see any issues.
@@Edgarbopp thanks 🙏🏻! May I send you the files via email?
@@leandrocarg do you have Facebook or Discord?
32:59.
IF the load is infinity, meaning an open circuit, the ruler would be down at 0 amperes on the horizontal, indicating the ability for a full voltage swing over the open circuit, with zero current flowing through it as you would expect.
Conversely, when the load is a short circuit, or 0 Ohms, the load line will be on the ordinate along the vertical axis allowing for a full current swing but zero voltage ( can't swing a voltage across a closed switch).
Every single XY between the horizontal voltage axis and the vertical current axis represents some form of resistance or impedance, that is, between an open and a short.
An individual resistor would be represented by the slope of a line running in the same general direction as the plate voltage lines in a positive direction. Y=mx+b. Same as I=mV (b=0) so m (slope) = I/V. As such, I/V = 1/R so Y=1/R*V or I= V/R.
Now, knowing that individual resistors are lines with a slope running in the direction of the grid curves, the load line is a line of the opposite slope that intersects many individual resistor lines of positive slope (the grid curves), and is therefore is a line that represents a variable resistor, which is what a vacuum tube in effect is.
And on the load line once plotted, the vacuum tube becomes a variable resistor, and as you apply more negative voltage to the grid, you move on the load line to the right, and the vacuum tube becomes more resistive and less current flows through it and more voltage is dropped across it. Vice versa as you apply less negative voltage to the grid the opposite happens. Not so scary this load line thing. One could simulate what a vacuum tube is doing with a potentiometer and the negative voltages for the grid would be commands for given rotations of the potentiometer. That would be one way of thinking about it.
I think you’ve got it! Except the infinity load is AC only. It’s like having a infinite resistance with a infinite Voltage on top so you can maintain the DC operating point. You can actually approx this with active devices like a choke or ccs load. Have a look at a Mu follower circuit to see one.
35:41 If we need to choose an actual transformer, knowing that our primary impedance is say 4.5K, and our load is say 8 Ohms, then our windings ration is sqrt(4500/8) = 24:1. Mehopes.
Close. That will get you the impedance ratio. To get the turns ratio use sqrt(zpri*zsec). T be clear, multiply rather than divide.
if I wanted to calculate the output power that the amplifier provides before switching to class B, how can I do it? that is, I want to know the power only in class A. thanks
Can you make a same tutorial for biasing otl amps? cuz I just built one recently and I wanna tune it to sound good because it sounds very bad right now😂 my tubes are 6h8c and 6l6gc my setup is self biasing but I think I think I messed up with my resistor values on the cathode
nice video thanks, a doubt arises at minute 36, if the supply voltage is 360v how does the signal oscillate up to 520?
Good question! Because the output transformer is an inductor it has a reactance not a resistance. It supplies the voltage above supply. Inductors supply voltage to preserve current. They’re a type of current source.
@@Edgarbopp That's incredibile, so when the tube oscillates in class A, before the cut-off at the bottom, it always reaches that extra voltage? Thanks for these clarifications
hi sir..i saw your video you are perfect bravo... sir you explain about how we can bias power section with trans for impedance.. now i want to know can i use this design that you do for power section..can i use for otl systems ..pleas lead me and i want appreciate of you for make this video
Definitely! For OTL you just draw the load line for the load you want to drive. Most OTL are using parallel tubes so you need to calculate for the load sharing of the output tubes.
@@Edgarbopp ...ok and thankyou sir
hi sir...sir when i saw your video on time 39 you told us (in order change the plate by 90volts) i couldn,t understand this voltage come from whe re ..i dont have bad mean... i just dont know what is this voltage..please lead me and thankyou
36:11 the Q point here is chosen to be more than half of the maximum plate voltage, which could allow for AC voltage swings to exceed the maximum plate voltage, it appears. On calculating the output power, the inductive load will allow for positive voltage swings to exceed the max plate voltage. Putting that aside, here how I did it: The voltage could swing from 360 down to 180 (can't go lower than that on this triode). The associated current swing estimating from the plate characteristics graph is from 115mA down to 65 mA. So, (360-180)(.707) x (.115-.065)(.707) = 8.2W. It would appear that the inductive load would allow that a maximum plate V swing be about 360 + (360-180) = 540, slightly over the max plate voltage. Should be shift the Q left a bit by about 40V or so to be safe?
44:24 the windings ratio of the transformer is sqrt(8000/8) = 32. So when both tubes are conducting, in class A, the windings ratio is 32:1. However, as soon as one tube stops conducting, the windings ratio changes because one half of the output transformer is taken out of the circuit, and therefore the windings ratio for the conducting tube becomes 16. Therefore, the reflected impedance that the conducting tube sees will be: 8x16x16= 2048, or stated differently one quarter instead of one half of the overall primary impedance. So this explains the increase in load when transitioning into class B. Correct?
The max plate voltage rating for tubes is generally the dc operating point and the designers expected the tube to potentially be inductively loaded. So the max AC spec is about double number in the data sheet.
@@stereopolice Not so much. The turns ratio cannot change.
Again you have the formula for Zra not Tra. Tra= sqrt(8000*8) = 252.98.
The Zra is 31.62 at 8 Ohms Zsec and 8000 Ohms Zpr. When the one tube operates alone it works against half the tranny or Zpri is 4000 and Zra becomes 22.23(sqrt(4000/8)), Tra is a physical construction so it remains constant unless you change connections on taps. The center tap is the feed point and why operation is only across one half of it when it runs into class B.
Think of it this way: one tube single ended amp 4000:8, same tube and all as a push pull amp is like 2 SE amps face to face so 8000:8 (measuring across the entire primary) the load we assume is not going to change, we still want the 8 ohm load. Now when you add 2 more tubes you have parallel effects halving the load, and series effects doubling back in to arrive at the spec of a single tube: 4000:8.
You can get Zpri by the load line which is B+ / Max Current (r=e/i),
you can also use the Power Out / IMAX Squared r=p/(I*I), and
(B+ * B+) / Power Out ((e*e)/p).
@@gregraynard781 A ratio is? A way of relating one thing to another and we do that with division. For instance Frequency is the number of cycles per second in an AC wave form. Cycles/time e.g. We noticed a wave repeat itself 180 times in a 3 second period of time --> 180 cycles divided by 3 seconds --> 180/3 = 60/1 we see 60 cycles in one second.
An impedance ratio for an output transformer is relating the design impedance of the primary to the designed load impedance on the secondary. The example above is 8000 ohms primary to 8 ohms secondary, or 8000/8 = 1000/1. Specifically he wanted to find the windings ratio which is the square root of the impedance ratio sqrt(8000/8) = sqrt(1000/1) = 31.62277 roughly 32 to 1 also known as the turns ratio, the windings ratio is 32 turns or winds primary to 1 wind or turn secondary.
Your information is incorrect, Greg. sqrt(8000*8) is a product not a ratio.
I study these them I go to a fender circuit in a guitar amplifier and the voltages are almost doubled
hi sir excusme ...when i saw your video in time 38:46 you pointed to a voltage at limit 90volt i need to know how you calculated this voltage??? because i want to try make a circuit vacuum tube and i need to know very accurate...please lead me and thankyou for make this video
You can divide any change in plate voltage by any corresponding change in plate current along the load line to find the plate impedance. Just pick ones that are convenient and around your operating point.
@@Edgarbopp...i love you and thankyou
@@mohammadmoezi3813 Thank you! Good luck!
At 33:45 you set an operating point. What determines the angle of the line from that point?
I picked a slope that looked like a swept through the most voltage and current without exceeding the max dissipation. You can see it’s roughly 700v and 150ma. So using ohms law that’s a load of 4.666k or so. So an output transformer presenting a 4 or 5k primary load would probably work well there.
@@Edgarbopp thank you! It's taken me so long to understand, but I'm finally getting their
@Edgarbopp it's all making sense now. Thank you again. So much. In regard to the perfect current source/ perfect voltage source. What helped me understand was to look at it another way. If the tube went dead short to ground the voltage would drop to zero, and the current would be limited by the plate load resistor dropping the full voltage of b+ across it. That is where the load line hits the graph on the left side. If the tube shuts completely off, the current drops to zero, and the voltage rises all the way to b+. That is where the load line hits the graph on the bottom. We are biasing, choosing a plate load resistor, and setting a dc region to rest in so that the grid voltage can swing either way, undistorted, within the watt limitations.
I think your presentation is very good and am sure you know this and in a presentation I know it is easy to speak and overlook small things but just after 44:00 where you say the plate-to-plate Z is 8K and half of that is 4K is incorrect. It is actually 1/4 the L from one side of the OPT to CT and would be 2K instead of 4K.
I just reviewed that portion of the video and I believe what I’ve said there is correct. A 8k primary will look like a 4k for each plate of the output pair and will transition to 2k when the stage enters class B.
In a transformer winding like the primary of a vacuum tube output transformer, we have the series inductance of the coils from the CT which is L1 + L2 and we also have the mutual inductance between the coils as they are wound on the same core and share the same magnetic flux. Neglecting leakage inductance which is usually in the mH range, and assuming the coupling factor to be 1, the mutual inductance doubles the series inductance. A common value for vintage output transformers like the UTC or Acrosound transformers is 20H plate to plate and 5H plate-side to CT. Direct measurements with an LCR meter will confirm this.
@@ElPasoTubeAmps I don’t think I talk about inductance anywhere in this video. This is about the impedance of the primary and the resulting voltage and current swings at the plate. In this video the inductance of the primary is assumed to be sufficient to load all audio frequencies. Of course in real life not every transformer has sufficiently high inductance to load the lowest frequencies but that was beyond the scope of this video.
@@ElPasoTubeAmps I’ve just reviewed several articles to be completely sure on this and they all agree with my previous understanding. Have a quick read here as a example www.valvewizard.co.uk/pp.html
I commented because the inductance (L) of 1/2 the P-P winding (one anode side to CT) is 1/4 the P-P winding L, and therefore the Z would be 1/4 also. Maybe inductance should be dropped from the discussion. I realize you were only speaking of Z but L has to go along with it and your video is about load lines and not about transformers. L and Z is easily measured with an LCR and Rs meter on the bench in a static condition.
The article you linked to is interesting and I can see that both of you have more experience in guitar/HiFi amplifiers than I do although I did do repair on such amplifiers for a couple of years here in El Paso after retiring. Quite honestly, I took up the repair of guitar amplifiers up so I could study the amplifier builds and designs... Anyway, I see where the article says that once the amplifier/transformer goes into class-B operation (which I am actually not buying that concept from just a signal drive perspective) the L and therefore the Z of the transformer winding from anode to CT changes from 1/2 to 1/4 the L and Z of the P-P winding. My point is, I do not believe the Z (or L) of half of the winding from anode to CT is ever at 1/2 the P-P impedance and remains 1/4 the P-P Z regardless of how the amplifier/OPT is operated. Maybe it changes under dynamic conditions I do not understand yet but I am open to learn and understand. My question would be how would we verify the claim in the linked article by measurement? I can't see how it would be done with plate characteristic curves as the load Z is a derived parameter from the characteristic curves and not an input parameter. Maybe I am missing something here. Best regards.
If my b+ is at 300v, and a cathode resistor is dropping 100v does that move the bias point to 200v on a reactive load?
The voltage is the plate to cathode voltage. So you don’t count the voltage across the cathode resistance in a cathode biased stage.
@@Edgarbopp thanks again! Much appreciated
Hi sir thank you for make this video..but I have some problem..I couldn't understand how we can calculate output impedance..sir can you help me about it I saw some other TH-camrs but they didn't say anything about impedance..and I need to know about this section please lead me ..and thank you again
Asking what the impedance is, is asking the question “when I change the voltage at this point how much current will it require?” So once you’ve drawn your load line pick a convenient place to measure the change in voltage along it and then measure the current change for that same part of the line. Then V/I=R. The R in this case can be thought of as the impedance. Though impedance is also time dependent so will change somewhat with frequency. But for now don’t worry about that.
@Edgarbopp again hi sir. And i am full of thanks for answer my question...sir I design my circuit...I choose my load line and everything is done ...but I need to know how I must calculate output impedance..I don't know could I gave you my intention...
@@Edgarbopp I mean about impedance is inner impedance
Sir if you need more information I can give you my email or my telegram or Instagram...you just tell me sir and thank you
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