Excuse the acronyms in the title but I wanted it to fit 
This is just a quick and I'm sure trivial question, but it's something I've yet (to butcher Douglas Adam's finest) to see written in clear, reassuringly simple language with nice, big, colourful pictures. So to speak. Thus to get my questions across this is actually rather long winded.
Anyway, here's the thing.
Imagine I have an output transformer labelled "10K Ohms into 8 Ohms".
Fairly straightforward - if I strap an 8 Ohm speaker on the secondary side, it is a 10k load to whatever I use to drive it.
Unless I have a push-pull (AB or B) circuit, in which case while the Anode-Anode impedance remains the same, each valve sees only 1/4 of the 10k => Each valve sees 2.5k, so will have a 2.5k ohm loadline.
I kind of know the answer to this but just want someone to say it: When designing this output stage, does one draw the 2.5k or 10k loadline? For completeness, is push-pull class A any different?
Another exception to the rule is if I have not only push pull but parallel push pull... and this is where I'm slightly less clear.
Question: With parallel output valves, assume in this case 2 per side, what load is presented to each valve, and what line do we then draw when planning the stage?
With our 10K:8 ohm example, does 4 valves (2 per side) in PPP appear as 5k, 2.5k, or 1.25k or some other value to each valve? What about with 6 valve (3 per side) and so on (i.e. general rule for impedance seen per valve, or if you look at it the other way around, OPT requirements).
I've read the ubiquitous MJ's Valve Amplifiers but this was not a point he has addressed directly, and infact the only references are on P.458 and in the (fairly hard to reduce to a generalisation) explanation of the Crystal Palace's output stage, on P.463.
It seems his method here was:
1) Loadline 625 ohms looks OK
2) Halve it (generalisation: multiply by 1/ValvesPerSide ?)
3) Multiply by 4 (generalisation: multiply by TotalValves ?)
Numerically:
1) 625 Ohms
2) 625 * 1/2 = 312.5
3) 312.5 * 4 = 1.25k Anode-Anode
Also: does the "ideal load is ~ 2ra" rule work at all with output transformers, or just straight up load resistances?
Thanks in advance for any help and sorry probably missing the point - I do think it helps to have things recorded clearly, though
Related Threads: Bevois Valley and Parallel Push Pull, modify parallel push pull EL84 to single push-pull, more output transformer questions (comes close).
Edit: Oh and just a warning in advance, if this thread goes as planned I will, unfortunately for you all, have another one in the pipeline
This is just a quick and I'm sure trivial question, but it's something I've yet (to butcher Douglas Adam's finest) to see written in clear, reassuringly simple language with nice, big, colourful pictures. So to speak. Thus to get my questions across this is actually rather long winded.
Anyway, here's the thing.
Imagine I have an output transformer labelled "10K Ohms into 8 Ohms".
Fairly straightforward - if I strap an 8 Ohm speaker on the secondary side, it is a 10k load to whatever I use to drive it.
Unless I have a push-pull (AB or B) circuit, in which case while the Anode-Anode impedance remains the same, each valve sees only 1/4 of the 10k => Each valve sees 2.5k, so will have a 2.5k ohm loadline.
I kind of know the answer to this but just want someone to say it: When designing this output stage, does one draw the 2.5k or 10k loadline? For completeness, is push-pull class A any different?
Another exception to the rule is if I have not only push pull but parallel push pull... and this is where I'm slightly less clear.
Question: With parallel output valves, assume in this case 2 per side, what load is presented to each valve, and what line do we then draw when planning the stage?
With our 10K:8 ohm example, does 4 valves (2 per side) in PPP appear as 5k, 2.5k, or 1.25k or some other value to each valve? What about with 6 valve (3 per side) and so on (i.e. general rule for impedance seen per valve, or if you look at it the other way around, OPT requirements).
I've read the ubiquitous MJ's Valve Amplifiers but this was not a point he has addressed directly, and infact the only references are on P.458 and in the (fairly hard to reduce to a generalisation) explanation of the Crystal Palace's output stage, on P.463.
It seems his method here was:
1) Loadline 625 ohms looks OK
2) Halve it (generalisation: multiply by 1/ValvesPerSide ?)
3) Multiply by 4 (generalisation: multiply by TotalValves ?)
Numerically:
1) 625 Ohms
2) 625 * 1/2 = 312.5
3) 312.5 * 4 = 1.25k Anode-Anode
Also: does the "ideal load is ~ 2ra" rule work at all with output transformers, or just straight up load resistances?
Thanks in advance for any help and sorry probably missing the point - I do think it helps to have things recorded clearly, though
Related Threads: Bevois Valley and Parallel Push Pull, modify parallel push pull EL84 to single push-pull, more output transformer questions (comes close).
Edit: Oh and just a warning in advance, if this thread goes as planned I will, unfortunately for you all, have another one in the pipeline
Sorry for not reading your post completely but it seems like you try to extrapolate based on your basic assumption of .25*Za-a. This is my understanding. For class A, you assume each tube in PP sees half the primary impedance, in this case, 5K. For PPP, this is doubled back up to 10K. For class B, each tube sees 2.5K, and 5K for PPP. Class AB is somewhere in between.
Ray
All depends upon what you find as a requirement for distortion Vs power, as in load line balance.
In PP class A1, let's for example use the 300B. The 300 B needs 800 ohms of reactive impedance to work into, at whatever frequency response interests you on the low end. Do remember this is a complex number that includes the reflected load. To remain in deep PP class A requires 6.6k Z ohms of reflected impedance, which does only provide 1.65 kZ ohms, from 1/2 of the primary.
But. transformers do not work 1/2 of the time, regardless of the connections in front of it. All antenna event coupling from primary to secondary is an all turns to all turns situation. So, your load line calculation is with 6.6k in mind, not 1.65 k.
Class AB is murkier, but the transformer still couples all turns to all turns so the entire primary is involved in presenting the tube of interest with it's load line.
Meanwhile only one half of the primary is involved in the voltage division ratio's. So, the impedance ratio is equivalent to the full AC wave form but the turns ratio is equivalent to 1/2 the primary to full secondary for voltage derivation.
When you have a CT primary and 450 VDC as plate to plate, the voltage used to calculate the B Sat of the transformers is only 48% of that total B+ VDC.
Does this make it any clearer?
Bud
All depends upon what you find as a requirement for distortion Vs power, as in load line balance.
In PP class A1, let's for example use the 300B. The 300 B needs 800 ohms of reactive impedance to work into, at whatever frequency response interests you on the low end. Do remember this is a complex number that includes the reflected load. To remain in deep PP class A requires 6.6k Z ohms of reflected impedance, which does only provide 1.65 kZ ohms, from 1/2 of the primary.
But. transformers do not work 1/2 of the time, regardless of the connections in front of it. All antenna event coupling from primary to secondary is an all turns to all turns situation. So, your load line calculation is with 6.6k in mind, not 1.65 k.
Class AB is murkier, but the transformer still couples all turns to all turns so the entire primary is involved in presenting the tube of interest with it's load line.
Meanwhile only one half of the primary is involved in the voltage division ratio's. So, the impedance ratio is equivalent to the full AC wave form but the turns ratio is equivalent to 1/2 the primary to full secondary for voltage derivation.
When you have a CT primary and 450 VDC as plate to plate, the voltage used to calculate the B Sat of the transformers is only 48% of that total B+ VDC.
Does this make it any clearer?
Bud
Thanks for the help
budp, unfortunately not a great deal easier, no, what with the introduction of additional variables : (
sorenj's post cleared classes A and B up, at least, so now the only remaining issue is how to proceed with AB(1).
If we were to look at the problem from the other direction: that is, I have a chosen output valve, of which there will be 2 on each side of the primary, and using a set of plate curves and have determined that, say, 5K is a nice looking loadline.
For class A PPP, then, it would require a 5K Za-a OPT, whereas just 'regular' PP would be 10k.
For B, PPP would require 10K and 'regular' PP ... 20k?
Would the "class B rule" above work OK for AB or will the impedance switching from X to Y as one side stops conducting make it all go drastically wrong and give distortion etc etc? This does seem to be the method MJ used on Crystal Palace (don't worry I'm not planning anything that big just yet)
budp, unfortunately not a great deal easier, no, what with the introduction of additional variables : (
sorenj's post cleared classes A and B up, at least, so now the only remaining issue is how to proceed with AB(1).
If we were to look at the problem from the other direction: that is, I have a chosen output valve, of which there will be 2 on each side of the primary, and using a set of plate curves and have determined that, say, 5K is a nice looking loadline.
For class A PPP, then, it would require a 5K Za-a OPT, whereas just 'regular' PP would be 10k.
For B, PPP would require 10K and 'regular' PP ... 20k?
Would the "class B rule" above work OK for AB or will the impedance switching from X to Y as one side stops conducting make it all go drastically wrong and give distortion etc etc? This does seem to be the method MJ used on Crystal Palace (don't worry I'm not planning anything that big just yet
)
The general rule I belive you are looking for has nothing to do with the Class of operation...
I believe you just want to know how to change the plate loading to acomodate more or less valves...
For example: You have 2 tubes in P-P , meaning one tube on each side of the primary with respect to Center-Tap....
Lets say this Plate load is 8K ......
Now with 4 tubes P-P-P ( 2 per side) ....and everything else stays the same you would now needa 4K load......
With 6 tubes P-P-P ( 3 per side)..... you use 2.66K plate load...
This assumes you are not changing the operating conditions or Class of operation....
Also.....the load line that is on the graph is 1/4 of the plate load in most 1st order analysis.....
If you really want to do this properly...then you need to do composite load line....which shows a NON-LINEAR load-line.....
In Class B P-P , you truly have each side seeing 1/4 the reflected load.... In Class A P-P each side sees 1/4 the reflected load in parallel with the plate resistance of the opposite valve....
In Class AB it is a mix of the two....and the mix depends on where you bias the valves....
Since plate resistance is not a true constant and depending on where you bias in Class AB1 the Load-Line will need to be derived from composite curves.....
Also, keep in mind to get the analysis right you need to use an eliptical load based on the characteristics of the speaker yu plan on using....
The Bottom line is unless you have real tube data from the same tubes you plan on using, you are wasting your time...
Nothing wrong with a very acurate analysis......just go out and try finding tubes that are within 1% of the ones on the data sheet and you should have agreement with your lab measurements and your analysis.... The reality is you will be curve matching tubes till your blue in the face to try to match the analysis....
Your best bet is to just draw up the load-line and mutiply it by 4 to get the Plate load you need....
Chris
I believe you just want to know how to change the plate loading to acomodate more or less valves...
For example: You have 2 tubes in P-P , meaning one tube on each side of the primary with respect to Center-Tap....
Lets say this Plate load is 8K ......
Now with 4 tubes P-P-P ( 2 per side) ....and everything else stays the same you would now needa 4K load......
With 6 tubes P-P-P ( 3 per side)..... you use 2.66K plate load...
This assumes you are not changing the operating conditions or Class of operation....
Also.....the load line that is on the graph is 1/4 of the plate load in most 1st order analysis.....
If you really want to do this properly...then you need to do composite load line....which shows a NON-LINEAR load-line.....
In Class B P-P , you truly have each side seeing 1/4 the reflected load.... In Class A P-P each side sees 1/4 the reflected load in parallel with the plate resistance of the opposite valve....
In Class AB it is a mix of the two....and the mix depends on where you bias the valves....
Since plate resistance is not a true constant and depending on where you bias in Class AB1 the Load-Line will need to be derived from composite curves.....
Also, keep in mind to get the analysis right you need to use an eliptical load based on the characteristics of the speaker yu plan on using....
The Bottom line is unless you have real tube data from the same tubes you plan on using, you are wasting your time...
Nothing wrong with a very acurate analysis......just go out and try finding tubes that are within 1% of the ones on the data sheet and you should have agreement with your lab measurements and your analysis.... The reality is you will be curve matching tubes till your blue in the face to try to match the analysis....
Your best bet is to just draw up the load-line and mutiply it by 4 to get the Plate load you need....
Chris
OK, thanks again... I'll have another crack at that sbench composite loadline tutorial, I think (it all seemed a bit complex and now I think I see why that is the approach taken ).
The reason this came about is I'm trying to figure out the load I need for an OPT given that the tubes are in this case 6h6p and the HT is 250-280V. No doubt it will need to be a custom OPT but the value I was coming out with was like Za-a = 25k which seemed a bit extortionate and pricey!
Composite lines here I come!
).I'm not trying to get to the point where I can say OK I need a OPT with a Za-a of 13.450891k, just trying to understand how to come up with the required primary impedence given the number and type of tubes.
The reason this came about is I'm trying to figure out the load I need for an OPT given that the tubes are in this case 6h6p and the HT is 250-280V. No doubt it will need to be a custom OPT but the value I was coming out with was like Za-a = 25k which seemed a bit extortionate and pricey!
Composite lines here I come!
OK...I see...
If you want a method for determining the proper plate load to use in P-P amps...here is the quick and dirty method....which happens to be the same method used by almost every classic Hi-Fi amp...
The key is finding the correct IV curves with pentodes.... Most impotantly you must make sure the IV curves you want are for the SCREEN VOLTAGE you are running....
The most important curve is the 0v bias curve, when your not going into class 2....ie, no grid current....
You draw the load-line starting at the intended Plate voltage on X-axis and then using a straight edge draw a line intersecting the 0v bias curve just to the right of the knee, so that you stay in saturation region to some degree...
Continue this line till you intersect the Y axis....
Then to find load-line you divide plate voltage at X axis by Plate current on the Y-axis intersection....
Now to figure the plate load, multiply the load-line value by 4 ....
This plate load value is for 2 tubes P-P ........
Each time you add more tubes in parallel you need to adjust the plate load by paralleling it to itself the same number of times you add additional pairs of tubes...
For example you have a 8K plate load for 2 tubes in P-P ...... 20W
You want to now run 4 tubes in P-P-P ....so it is 8K//8K = 4K...40W
If you want to run 6 tubes in P-P-P......8K//8K//8K = 2.66k ...60W
If you want to run 8 tubes in P-P-P ....8K//8K//8K//8K = 2k...80W
If you want to run 10 tubes in P-P-P..8K//8K//8K//8K//8K= 1.6K..100W
One additional thing to keep in mind....
The voltages you draw your load-line to should be the voltages your power supply droops to under full power.... i.e., if your idle voltage is 400V on the plate and you power supply droops to 345V at full power output, you would need to draw the load-line with the 345V.... that would also include the SCREEN voltage droop as well if it applies...
Hope this helps....
Chris
If you want a method for determining the proper plate load to use in P-P amps...here is the quick and dirty method....which happens to be the same method used by almost every classic Hi-Fi amp...
The key is finding the correct IV curves with pentodes.... Most impotantly you must make sure the IV curves you want are for the SCREEN VOLTAGE you are running....
The most important curve is the 0v bias curve, when your not going into class 2....ie, no grid current....
You draw the load-line starting at the intended Plate voltage on X-axis and then using a straight edge draw a line intersecting the 0v bias curve just to the right of the knee, so that you stay in saturation region to some degree...
Continue this line till you intersect the Y axis....
Then to find load-line you divide plate voltage at X axis by Plate current on the Y-axis intersection....
Now to figure the plate load, multiply the load-line value by 4 ....
This plate load value is for 2 tubes P-P ........
Each time you add more tubes in parallel you need to adjust the plate load by paralleling it to itself the same number of times you add additional pairs of tubes...
For example you have a 8K plate load for 2 tubes in P-P ...... 20W
You want to now run 4 tubes in P-P-P ....so it is 8K//8K = 4K...40W
If you want to run 6 tubes in P-P-P......8K//8K//8K = 2.66k ...60W
If you want to run 8 tubes in P-P-P ....8K//8K//8K//8K = 2k...80W
If you want to run 10 tubes in P-P-P..8K//8K//8K//8K//8K= 1.6K..100W
One additional thing to keep in mind....
The voltages you draw your load-line to should be the voltages your power supply droops to under full power.... i.e., if your idle voltage is 400V on the plate and you power supply droops to 345V at full power output, you would need to draw the load-line with the 345V.... that would also include the SCREEN voltage droop as well if it applies...
Hope this helps....
Chris
Peter....
Is the 6h6p a TRIODE ??????
If it is then the choice of plate load is a bit different then figuring it for a pentode......
If it is a TRIODE your load-line limit would be tangent to the MAX Plate-Dissipation curve....which would be to the right side of the load-line.... This would provide the smallest value of plate load, also highest power output for the P-P triode pair....
So the game with triode P-P is to start at that load-line limit and increase it in increments and re-calculating the power output and the distortions.... Eventually you choose a Plate-Load based on the Power Output vs. Distortions.....
As the plate load value gets bigger the distortion drops as well as the power output...not propotionately is the good news...
So you choose a happy spot....don't get too big with the plate load, otherwise controling the output transformer parasistics will be next to impossible...ie, leakage inductance and capacitance will become a bloody nightmare to contain..
Chris
Is the 6h6p a TRIODE ??????
If it is then the choice of plate load is a bit different then figuring it for a pentode......
If it is a TRIODE your load-line limit would be tangent to the MAX Plate-Dissipation curve....which would be to the right side of the load-line.... This would provide the smallest value of plate load, also highest power output for the P-P triode pair....
So the game with triode P-P is to start at that load-line limit and increase it in increments and re-calculating the power output and the distortions.... Eventually you choose a Plate-Load based on the Power Output vs. Distortions.....
As the plate load value gets bigger the distortion drops as well as the power output...not propotionately is the good news...
So you choose a happy spot....don't get too big with the plate load, otherwise controling the output transformer parasistics will be next to impossible...ie, leakage inductance and capacitance will become a bloody nightmare to contain..
Chris
PP Loadline, Pt. I
How about an illustration to make this clearer? Actual Design PP Loadline.
As for OPTs, I used a 4K4 : 8R PP OPT. As for the impedances involved, this xfmr matches 8 ohms to an 1100 ohm load. 1100R : 8R corresponds to a voltage transformation of 11.73 : 1. If you want to make a push-pull OPT, then tou need two symmetrical halves for the primary, or a voltage ratio of 23.46. Doubling the voltage ratio increases the impedance transformation by a factor of four, therefore: 8(23.46)^2= 4403, or ~4400 : 8 across the whole primary. Each tube connected to the primary still sees the 1100R as its load since the primary center tap is at AC ground. Thus, when doing the loadline, use 1100R as the load resistance.
The DC loadline is red. This is used to determine the static Q-point. Since there's 350V at the plates, and the 6BQ6 has a Pd= 12W, 25mA of static DC gives a Pd= 8.75W -- comfortably under the max plate dissipation rating. This loadline intersects on the -30V grid line, so this establishes the Q-point.
The dynamic loadline (blue line) starts at the (350, 0) point. The justification for this being that the OPT primary responds to differential current only. With no signal, each VT draws the same current, and there is zero differential current, therefore no output, and no DC core magnetization, which is why PP OPTs don't require an air gap to avoid core saturation. This line ends at the (0, 318.2) point since: I= 350/1100= 318.2mA. Since this line starts at (350, 0) it's considered to be the "Class B" loadline. If the 6BQ6s were actually to be biased to Ip= 0 at no signal, then they'd be operating in Class B.
Having drawn the loadline, the max current is read where it intersects the Vgk= 0 line, since this design is to be a Class AB1 that doesn't pull grid current. This is 262.2mA, and this is a peak current. Since one 6BQ6 will pull that current at max no-clip signal while the other one is way into plate current cutoff, that will also be the total primary current, and so the output power becomes:
Irms= Ip / sqrt(2)
Irms= 262.2E-3 / sqrt(2)= 0.185Arms
Po= Rl * Irms^2
Po= (1100)(0.185^2)= 37.65W (Not including the ineviatble core and copper losses in the OPT)
Since the DC rail supplies two VTs, the total rail current is twice the peak plate current, or 524.4mAp. The average DC current then becomes:
Iave= Ip / pi
Iave= (0.5244)(3.414)= 0.167A
The DC input power: (350)(0.167)= 58.45W
Plate dissipation: Pd= (Pdc - Po) / 2
Pd= (58.45 - 37.65) / 2= 10.4W per plate, and we're still comfortably within specs.
To see just how deep into Class AB we are, add the Class A loadline. Since Class A means a duty cycle of 360 deg, the resulting primary load will be 2200R since we basically have two "generators" in series feeding the primary. Start at the intersection of the DC current and voltage lines. Since that means dropping 350V, the change in current for a change of 350V becomes: 350 / 2200= 159.1mA. This delta current is added to the Q-point current of 25mA, so the Class A line hits the current axis at (0, 184.1).
The intersection of the Class A and Class B loadlines is the transition point. Read the volatge there as: 297.14V. The Class A power becomes:
Vp= 350 - 297.14= 52.86Vp
Vrms= Vp / sqrt(2)= 52.86 / sqrt(2)= 37.28Vrms
Po= V^2 / R= (37.28^2) / 1100= 1.26W per VT. Since two VTs contribute to this, the total Class A power is: 2.52W.
So for the first two-and-a-half watts out of ~37W, we're in Class A, so this is operating deep into Class AB. Crossover distortion could be a concern here, but it turned out to not be much of a problem. Since the 6BQ6 is a TV horizontal deflection type, it can be biased quite a bit "hotter" in operation since TV HDs were rated very conservatively, due to the demanding nature of the intended operation. I've run them as hot as 65mA of plate current per tube without red-plating them, so they are a good deal tougher than comparable audio finals, such as a 6V6.
If you were going to parallel up, you'd have the same voltages, but twice the current. This would give you a load of 1100 / n. For two 6BQ6s per phase, Rl= 550R, and you'd need a 2200 : 8 OPT.
How about an illustration to make this clearer? Actual Design PP Loadline.
As for OPTs, I used a 4K4 : 8R PP OPT. As for the impedances involved, this xfmr matches 8 ohms to an 1100 ohm load. 1100R : 8R corresponds to a voltage transformation of 11.73 : 1. If you want to make a push-pull OPT, then tou need two symmetrical halves for the primary, or a voltage ratio of 23.46. Doubling the voltage ratio increases the impedance transformation by a factor of four, therefore: 8(23.46)^2= 4403, or ~4400 : 8 across the whole primary. Each tube connected to the primary still sees the 1100R as its load since the primary center tap is at AC ground. Thus, when doing the loadline, use 1100R as the load resistance.
The DC loadline is red. This is used to determine the static Q-point. Since there's 350V at the plates, and the 6BQ6 has a Pd= 12W, 25mA of static DC gives a Pd= 8.75W -- comfortably under the max plate dissipation rating. This loadline intersects on the -30V grid line, so this establishes the Q-point.
The dynamic loadline (blue line) starts at the (350, 0) point. The justification for this being that the OPT primary responds to differential current only. With no signal, each VT draws the same current, and there is zero differential current, therefore no output, and no DC core magnetization, which is why PP OPTs don't require an air gap to avoid core saturation. This line ends at the (0, 318.2) point since: I= 350/1100= 318.2mA. Since this line starts at (350, 0) it's considered to be the "Class B" loadline. If the 6BQ6s were actually to be biased to Ip= 0 at no signal, then they'd be operating in Class B.
Having drawn the loadline, the max current is read where it intersects the Vgk= 0 line, since this design is to be a Class AB1 that doesn't pull grid current. This is 262.2mA, and this is a peak current. Since one 6BQ6 will pull that current at max no-clip signal while the other one is way into plate current cutoff, that will also be the total primary current, and so the output power becomes:
Irms= Ip / sqrt(2)
Irms= 262.2E-3 / sqrt(2)= 0.185Arms
Po= Rl * Irms^2
Po= (1100)(0.185^2)= 37.65W (Not including the ineviatble core and copper losses in the OPT)
Since the DC rail supplies two VTs, the total rail current is twice the peak plate current, or 524.4mAp. The average DC current then becomes:
Iave= Ip / pi
Iave= (0.5244)(3.414)= 0.167A
The DC input power: (350)(0.167)= 58.45W
Plate dissipation: Pd= (Pdc - Po) / 2
Pd= (58.45 - 37.65) / 2= 10.4W per plate, and we're still comfortably within specs.
To see just how deep into Class AB we are, add the Class A loadline. Since Class A means a duty cycle of 360 deg, the resulting primary load will be 2200R since we basically have two "generators" in series feeding the primary. Start at the intersection of the DC current and voltage lines. Since that means dropping 350V, the change in current for a change of 350V becomes: 350 / 2200= 159.1mA. This delta current is added to the Q-point current of 25mA, so the Class A line hits the current axis at (0, 184.1).
The intersection of the Class A and Class B loadlines is the transition point. Read the volatge there as: 297.14V. The Class A power becomes:
Vp= 350 - 297.14= 52.86Vp
Vrms= Vp / sqrt(2)= 52.86 / sqrt(2)= 37.28Vrms
Po= V^2 / R= (37.28^2) / 1100= 1.26W per VT. Since two VTs contribute to this, the total Class A power is: 2.52W.
So for the first two-and-a-half watts out of ~37W, we're in Class A, so this is operating deep into Class AB. Crossover distortion could be a concern here, but it turned out to not be much of a problem. Since the 6BQ6 is a TV horizontal deflection type, it can be biased quite a bit "hotter" in operation since TV HDs were rated very conservatively, due to the demanding nature of the intended operation. I've run them as hot as 65mA of plate current per tube without red-plating them, so they are a good deal tougher than comparable audio finals, such as a 6V6.
If you were going to parallel up, you'd have the same voltages, but twice the current. This would give you a load of 1100 / n. For two 6BQ6s per phase, Rl= 550R, and you'd need a 2200 : 8 OPT.
PP Loadline, Pt. II
One other indicator of performance is estimating harmonic distortion. From the Fourier series, to a first approximation, the h3 content is given by determining the plate current at the "half way" point. For a Vi= 30Vp, this would be at Vi= 15Vp. If one 6BQ6 has an instantaneous Vgk= -15V, the other one is well into plate current cutoff at Vgk= -45V. This makes it easier since there is no need to construct a composite characteristic for the Vgk= -15V line. The Class B loadline hits the Vgk= -15V line at 109.29mA (green line). The third order harmonic current is given by: I0.5= (Imax - 2I0.5) / 3
I0.5= (262.2 - 2*109.29) / 3= 14.54mA
The fundamental current is given by: If= 2(Imax + I0.5) / 3
If= 2(262.2 + 109.29) / 3= 247.66mA
The estimated THD= 100(I0.5 / If)
THD= 100(14.54 / 247.66)= 5.87%
Not real good, but not real bad either. Of course, it's a rough estimate since there may be higher order harmonics, and the assumption is perfect balance that perfectly cancels even order harmonics (not likely!) and all the higher orders of the Fourier series are ignored in favour of a simple graphical solution. Also, different VTs have different harmonic "personalities", and a low estimate doesn't mean good sound if the harmonics produced are all high order ones. The third isn't so bad since it's musically related, producing "edge" or "brightness". Besides, applying gNFB will improve linearity, damping, and reduce distortion arising from the front end and OPT nonlinearities.
In actual open loop operation, the third measured more like 3.0% with just a trace of a fifth and higher. Didn't sound bad running open loop, and gNFB improves the sonics quite a bit in practice.
One other indicator of performance is estimating harmonic distortion. From the Fourier series, to a first approximation, the h3 content is given by determining the plate current at the "half way" point. For a Vi= 30Vp, this would be at Vi= 15Vp. If one 6BQ6 has an instantaneous Vgk= -15V, the other one is well into plate current cutoff at Vgk= -45V. This makes it easier since there is no need to construct a composite characteristic for the Vgk= -15V line. The Class B loadline hits the Vgk= -15V line at 109.29mA (green line). The third order harmonic current is given by: I0.5= (Imax - 2I0.5) / 3
I0.5= (262.2 - 2*109.29) / 3= 14.54mA
The fundamental current is given by: If= 2(Imax + I0.5) / 3
If= 2(262.2 + 109.29) / 3= 247.66mA
The estimated THD= 100(I0.5 / If)
THD= 100(14.54 / 247.66)= 5.87%
Not real good, but not real bad either. Of course, it's a rough estimate since there may be higher order harmonics, and the assumption is perfect balance that perfectly cancels even order harmonics (not likely!) and all the higher orders of the Fourier series are ignored in favour of a simple graphical solution. Also, different VTs have different harmonic "personalities", and a low estimate doesn't mean good sound if the harmonics produced are all high order ones. The third isn't so bad since it's musically related, producing "edge" or "brightness". Besides, applying gNFB will improve linearity, damping, and reduce distortion arising from the front end and OPT nonlinearities.
In actual open loop operation, the third measured more like 3.0% with just a trace of a fifth and higher. Didn't sound bad running open loop, and gNFB improves the sonics quite a bit in practice.
Chris: Yes, it's a (dual) triode, so I was operating under the assumption that instead of aiming for the 'knee', one goes for a tangent to the Pa max as you said, though thanks for your continued clarifications!
Miles: wow! Took a while to read, but in the process I took a handwritten copy as well as saving the page verbatim *taps head* Thankyou so much, can't have been a quick thing to type up.
Finally some formulas/rigidly defined methods! Yessssss! This is more like it
Thanks again, I owe you all a beer if you ever come to Cardiff. (offer valid at the students union only though, I need to start saving for transformers...)
Miles: wow! Took a while to read, but in the process I took a handwritten copy as well as saving the page verbatim *taps head*
Thankyou so much, can't have been a quick thing to type up.Finally some formulas/rigidly defined methods! Yessssss! This is more like it
Thanks again, I owe you all a beer if you ever come to Cardiff. (offer valid at the students union only though, I need to start saving for transformers...)
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