I have a pair of output transformers burning a hole in my pocket, no rush to build with them but would prefer to come up with some ideas rather than selling.
They are Lundahl LL9202 gapped for 50mA. Can be wired for 6.5K, 11K, and 23K primaries. I breadboarded them in a 801A A2 design wired for 11K, but found the bandwidth due to parasitics unfavorable and have gone another direction.
More likely would use them wired for 6.5K, but sort of an oddball transformer in that case, 6.5K:8ohm at 50mA. Looking for some ideas on single-ended output tubes that might pair well with it. Despite the bandwidth issues they are good sounding transformers.
One that came to mind is the EL33 / EL3N / EL11 which I have seen used with this transformer before, although quite a wimpy amplifier even by SET standards. Another thought would be a triode-strapped directly-heated pentode, although I don't know which.
Any ideas welcome, thanks.
They are Lundahl LL9202 gapped for 50mA. Can be wired for 6.5K, 11K, and 23K primaries. I breadboarded them in a 801A A2 design wired for 11K, but found the bandwidth due to parasitics unfavorable and have gone another direction.
More likely would use them wired for 6.5K, but sort of an oddball transformer in that case, 6.5K:8ohm at 50mA. Looking for some ideas on single-ended output tubes that might pair well with it. Despite the bandwidth issues they are good sounding transformers.
One that came to mind is the EL33 / EL3N / EL11 which I have seen used with this transformer before, although quite a wimpy amplifier even by SET standards. Another thought would be a triode-strapped directly-heated pentode, although I don't know which.
Any ideas welcome, thanks.
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Looks like the 4 primary series windings could be connected in two series sets of 2 parallel windings to get 1/4 the primary Z. Half the voltage and twice the current (100 mA). That would fit trioded DHP tubes better with their lower screen voltages. An 8042 DHP might fit then.
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The 50 mA gap notation is really a 50 mA x turns gap for the 4 windings in series. So 50 mA thru the total wire. With the windings re-arranged into two sets of two parallel windings, each wire section still has 50 mA as before, but the tube sees two of those in parallel. Hence 100mA x 1/2 the turns is the same magnetization as far as the core is concerned.
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Thanks for the ideas everyone. Paralleling the primary windings is interesting, a lower primary Z at a higher current rating would open up many more possibilities. I may touch base with Lundahl / Kevin at K&K Audio and get their input.
The 50 mA gap notation is really a 50 mA x turns gap for the 4 windings in series. So 50 mA thru the total wire. With the windings re-arranged into two sets of two parallel windings, each wire section still has 50 mA as before, but the tube sees two of those in parallel. Hence 100mA x 1/2 the turns is the same magnetization as far as the core is concerned.
Aren't you talking AC magnetization?
I referred to DC magnetization as it is SE.
Rearranging primary sections will not change that.
I believe smoking-amp is referring to DC magnetization. With four primaries in series, it is rated for 50mA.
However with two sets of parallel windings in series and a tube biased at 100mA, each primary winding still only carries 50mA, so the DC magnetization is the same for the core. Correct me if I am wrong, smoking-amp.
However with two sets of parallel windings in series and a tube biased at 100mA, each primary winding still only carries 50mA, so the DC magnetization is the same for the core. Correct me if I am wrong, smoking-amp.
Just for illustration, versus the apparant disagreements in this thread . . .
Take 4 windings, 300 turns each.
And, all 4 windings have nearly the same DCR.
Magnetic Field is proportionate (~) to 'Amp x Turns'.
4 windings, 300 Turns each, all in series (1200 Turns).
Then 0.05A x 1200 Turns = 60 Amp Turns
2 windings in parallel, each winding gets 0.025 A (50mA divided by 2 into equal DCRs)
0.025A x 600 Turns = 15 Amp Turns
Or if you want to look at it this way . . .
0.050A x 300 Turns = 15 Amp Turns
So, now with 2 windings in a parallel set, and 2 more windings in a parallel set,
and with those two parallel sets in series, we get:
15 Amp Turns + 15 Amp Turns = 30 Amp turns.
30 Amp turns is 1/2 of 60 Amp turns.
You get 2 times the DC current capability with 2 parallel in series with another 2 parallel.
Inductance:
Inductance is proportionate to (Turns)Squared (~ Turns Squared)
4 in series = (4)Squared = 16
2 in parallel in series with 2 in parallel = (2)Squared = 4
Conclusion:
You will get 2 x the DC current capability, but with only 1/4 of the inductance.
And you will get 2 x the AC current capability before saturation, but at only 1/4 of the inductance.
Take 4 windings, 300 turns each.
And, all 4 windings have nearly the same DCR.
Magnetic Field is proportionate (~) to 'Amp x Turns'.
4 windings, 300 Turns each, all in series (1200 Turns).
Then 0.05A x 1200 Turns = 60 Amp Turns
2 windings in parallel, each winding gets 0.025 A (50mA divided by 2 into equal DCRs)
0.025A x 600 Turns = 15 Amp Turns
Or if you want to look at it this way . . .
0.050A x 300 Turns = 15 Amp Turns
So, now with 2 windings in a parallel set, and 2 more windings in a parallel set,
and with those two parallel sets in series, we get:
15 Amp Turns + 15 Amp Turns = 30 Amp turns.
30 Amp turns is 1/2 of 60 Amp turns.
You get 2 times the DC current capability with 2 parallel in series with another 2 parallel.
Inductance:
Inductance is proportionate to (Turns)Squared (~ Turns Squared)
4 in series = (4)Squared = 16
2 in parallel in series with 2 in parallel = (2)Squared = 4
Conclusion:
You will get 2 x the DC current capability, but with only 1/4 of the inductance.
And you will get 2 x the AC current capability before saturation, but at only 1/4 of the inductance.
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Late post, missed earlier posts:
AC or DC magnetization are measured in Ampere-turns. The core cannot tell if it has two turns with 50 mA each or 1 turn with 100 mA. Could twist and solder the wires together and call it 2 strand Litz wire.
DC Volts and AC Volts are completely different of course.
AC or DC magnetization are measured in Ampere-turns. The core cannot tell if it has two turns with 50 mA each or 1 turn with 100 mA. Could twist and solder the wires together and call it 2 strand Litz wire.
DC Volts and AC Volts are completely different of course.
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Like so?
For 6.5K:8ohm, all of the secondary windings are in series, which is how they are currently wired, so I will rewire the primary and measure the turns ratio.
View attachment 932831
Winding ratio for 6k5 : 8 ohm is 28.5 : 1.
With all primaries in series (DCR 600 ohm) there should be 4 secondaries in series.
As there are 8 secondaries, just parallel the pair of series connected secondaries (DCR 0,8 ohm).
In this case you have the maximum primary inductance (100H).
In your drawing you parallel connected a pair of series (two) connected primaries (DCR 150 ohm).
To get the same ratio connect two secondaries in series and parallel the four (DCR 0,2 ohm).
Now you have 25H of primary inductance.
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You can fix the turns ratio by changing the secondaries alone, but that will leave you with a 50 mA DC rating for the primary still.
To handle 100 mA DC from the tube, then the fix needs to be done on the primary side by the earlier paralleling up scheme. The primary L does drop to 1/4, but so does the primary Z. Same ratio of L to Z, so all is well. (This is the same process as designing the xfmr for 1/4 Zprimary and equal Watts in the 1st place.)
To handle 100 mA DC from the tube, then the fix needs to be done on the primary side by the earlier paralleling up scheme. The primary L does drop to 1/4, but so does the primary Z. Same ratio of L to Z, so all is well. (This is the same process as designing the xfmr for 1/4 Zprimary and equal Watts in the 1st place.)
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Thanks daanve and smoking-amp, I understand. I am not sure what tube to pair yet, will think it over, but a very usable ratio possible by re-wiring the primary would be 3.25K:8ohm with a 25H inductance at 100mA. Depending on the chosen tube, bias current may not even reach 100mA, so perhaps a more favorable ratio of L to Z. I think this is the route I will pursue.
Conclusion:
You will get 2 x the DC current capability, but with only 1/4 of the inductance.
And you will get 2 x the AC current capability before saturation, but at only 1/4 of the inductance.
And thanks 6A3 for the thorough explanation.
Just as a reminder, you can always just sell the transformer and go buy something better. Transformers do tend to move pretty quickly in the swap section of the forum. I don't want to dissuade you from building something, but at the same time, you seem to prefer sowter transformers based on your other builds.
If you feel like you must use these in an amp of some kind, I think a type 50 would be pretty much spot on perfect. 6.5K would be a very healthy load, and you might be able to get away with running the type 50s at 55ma on those transformers.
A good type 50 amp might knock your 6A5G amp off it's throne. ; )
If you feel like you must use these in an amp of some kind, I think a type 50 would be pretty much spot on perfect. 6.5K would be a very healthy load, and you might be able to get away with running the type 50s at 55ma on those transformers.
A good type 50 amp might knock your 6A5G amp off it's throne. ; )
450V, 50mA, 300B, about 9.5W output.
Euro21said it right at the start of the thread. Use it at 6.5K with a 300b. It will sound really nice. My daily 300b SE amp has LL1682 50mA OPTs. That's 5K into 5R so quite similar. I have a pair of LL1664/70ma which are 3K and though this might seem a better match on paper it sounds pretty average and nothing like the excellent sound I'm getting from the LL1682. Believe me, I've been there with this one. If you look at the WE data for the 300b, the lowest 3rd harmonic distortion figures are at 5K. I wouldn't hesitate. Drive it with a C3g or something like that and you're good to go.
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