daanve,
More laminations does not require double the wire diameter, for the same total turns.
Doubling the wire diameter will reduce the DCR to 1/4 for equal lengths.
The DCR/length is proportionate to the 'Diameter Squared'.
Increase the wire diameter by 1.414 times, and DCR Ohms per length goes down to 1/2.
Just an approximation:
I believe the 1.414 larger diameter wire, and two times the length, will fit into the doubled lamination size.
Therefore, the DCR will be the same, or vary similar.
More laminations does not require double the wire diameter, for the same total turns.
Doubling the wire diameter will reduce the DCR to 1/4 for equal lengths.
The DCR/length is proportionate to the 'Diameter Squared'.
Increase the wire diameter by 1.414 times, and DCR Ohms per length goes down to 1/2.
Just an approximation:
I believe the 1.414 larger diameter wire, and two times the length, will fit into the doubled lamination size.
Therefore, the DCR will be the same, or vary similar.
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The issue of a fixed air gap with a fixed lamination size, and fixed number of wire turns:
The magnetic field from the coil is dependent on the Amp x Turns.
Same turns, more amps, means more magnetic field.
More magnetic field means Saturation occurs earlier with the added current of the signal peaks.
(Quiescent current + peak current)
I am pretty sure that is how it works, and is exactly what makes it fall apart when you have too much current for a fixed lamination size, fixed number to turns, fixed low frequency, and fixed air gap size.
Reduce the air gap size, and Saturation, whether it is caused by 'quiescent current', or by 'quiescent current + peak currents' comes earlier.
And if the frequency is low enough, it starts to approach saturation at nearly the same rate as quiescent DC.
Is there another factor that I forgot?
Please let me know.
Thanks!
Just my opinions
The magnetic field from the coil is dependent on the Amp x Turns.
Same turns, more amps, means more magnetic field.
More magnetic field means Saturation occurs earlier with the added current of the signal peaks.
(Quiescent current + peak current)
I am pretty sure that is how it works, and is exactly what makes it fall apart when you have too much current for a fixed lamination size, fixed number to turns, fixed low frequency, and fixed air gap size.
Reduce the air gap size, and Saturation, whether it is caused by 'quiescent current', or by 'quiescent current + peak currents' comes earlier.
And if the frequency is low enough, it starts to approach saturation at nearly the same rate as quiescent DC.
Is there another factor that I forgot?
Please let me know.
Thanks!
Just my opinions
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Onetics 6K Primary Level 3 OPTs
Hi,
I found a pair of these on ebaY and bought them some months ago.
Planning to try them at Va-k=350V, Ia=70mA, Vg=-72.5V OP for:
Po=5.05W, H2=-35.5dB, H3=-58.5dB
We shall hear what we shall hear...
PM2As and/or 802s-Conical Horns/Petite Onkens with Altec 814s.
Cherro,
RC
Hi,
I found a pair of these on ebaY and bought them some months ago.
Planning to try them at Va-k=350V, Ia=70mA, Vg=-72.5V OP for:
Po=5.05W, H2=-35.5dB, H3=-58.5dB
We shall hear what we shall hear...
PM2As and/or 802s-Conical Horns/Petite Onkens with Altec 814s.
Cherro,
RC
If DCR can be a make or break specification, what would constitute an acceptable DCR figure?
Acceptable DCR will depend a bit on the intended primary impedance. The LL1664 is 148R for the primaries in series and the 1682 is a bit higher, but that's suitable for the design. These aren't the lowest DCRs ever, but they aren't crazy high for what they are.
I wouldn't want to see a 3K OT with a 300 ohm primary, that's for sure. A lot of the dinky vintage output transformers have this issue.
I wouldn't want to see a 3K OT with a 300 ohm primary, that's for sure. A lot of the dinky vintage output transformers have this issue.
Whatever the reason - the 5K into 5ohms primary or the 50mA gap which raises the inductance - the LL1682/50 just sounds great in my 300b amp. It's the best Lundahl I've used with a 300b. On paper it isn't the ideal match, but I love it and for a while it's replaced my O-Netics OPT 300b SE which I love equally.
Effects of DCR in the primary and secondary.
DCR causes a real loss of power:
Start with a 5000 Ohm primary with 500 Ohms DCR, and an 8 Ohm secondary with 0.8 Ohms DCR.
The insertion loss due to the DCRs will be 1 dB in the primary, and 1 db in the secondary. That is a total 2 dB insertion loss.
10 Watts signal power from the output tube, will be 2 dB less than 10 Watts when it appears at the 8 Ohm load.
10 Watts from the output tube will be only 7.9 Watts to the 8 Ohm load.
Start with a 5000 Ohm primary with 250 Ohms DCR, and an 8 Ohm secondary with 0.4 Ohms DCR.
The insertion loss due to the DCRs will be 1/2 dB in the primary, and 1/2 db in the secondary. That is a total 1 dB insertion loss.
10 Watts signal power from the output tube, will be 1 dB less than 10 Watts when it appears at the 8 Ohm load.
10 Watts from the output tube will be only 8.9 watts to the 8 Ohm load.
DCR causes a loss of damping factor:
Start with a 5000 Ohm primary with 500 Ohm DCR, and an 8 Ohm secondary with 0.8 Ohm DCR.
Consider an output tube that has a plate impedance of 500 Ohms.
If the output transformer did not have any DCR in the primary, and did not have any DCR in the secondary, the damping factor would be 5000/500 = 10.
But, there is DCR in those windings . . .
Having DCRs of 10% of the primary and 10% secondary winding impedances (20% total), would have the same effect as increasing the output tube's plate impedance, rp, by 20%.
500 Ohms + 20% = 600 Ohms.
The "new" damping factor due to the DCRs of the windings becomes 5000/600 = 8.33.
That accounts for the loss of damping factor due to DCRs.
Now, if the output tube's plate impedance, rp is 2500 Ohms, and it drives a 5000 Ohm primary, the damping factor is already quite low (2), so a total of 20% of DCRs in the windings, is not going to affect the low damping factor by much (5000/(2500 + 500); 5000/3000 = 1.67
Always think of DCR in terms of a ratio of the winding impedances versus their DCRs.
DCR causes a real loss of power:
Start with a 5000 Ohm primary with 500 Ohms DCR, and an 8 Ohm secondary with 0.8 Ohms DCR.
The insertion loss due to the DCRs will be 1 dB in the primary, and 1 db in the secondary. That is a total 2 dB insertion loss.
10 Watts signal power from the output tube, will be 2 dB less than 10 Watts when it appears at the 8 Ohm load.
10 Watts from the output tube will be only 7.9 Watts to the 8 Ohm load.
Start with a 5000 Ohm primary with 250 Ohms DCR, and an 8 Ohm secondary with 0.4 Ohms DCR.
The insertion loss due to the DCRs will be 1/2 dB in the primary, and 1/2 db in the secondary. That is a total 1 dB insertion loss.
10 Watts signal power from the output tube, will be 1 dB less than 10 Watts when it appears at the 8 Ohm load.
10 Watts from the output tube will be only 8.9 watts to the 8 Ohm load.
DCR causes a loss of damping factor:
Start with a 5000 Ohm primary with 500 Ohm DCR, and an 8 Ohm secondary with 0.8 Ohm DCR.
Consider an output tube that has a plate impedance of 500 Ohms.
If the output transformer did not have any DCR in the primary, and did not have any DCR in the secondary, the damping factor would be 5000/500 = 10.
But, there is DCR in those windings . . .
Having DCRs of 10% of the primary and 10% secondary winding impedances (20% total), would have the same effect as increasing the output tube's plate impedance, rp, by 20%.
500 Ohms + 20% = 600 Ohms.
The "new" damping factor due to the DCRs of the windings becomes 5000/600 = 8.33.
That accounts for the loss of damping factor due to DCRs.
Now, if the output tube's plate impedance, rp is 2500 Ohms, and it drives a 5000 Ohm primary, the damping factor is already quite low (2), so a total of 20% of DCRs in the windings, is not going to affect the low damping factor by much (5000/(2500 + 500); 5000/3000 = 1.67
Always think of DCR in terms of a ratio of the winding impedances versus their DCRs.
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Very informative. So how would you evaluate the LL1682/50 SE as an example?
http://www.lundahl.se/wp-content/uploads/datasheets/1682.pdf
Primary is 210DCR and secondary 0.4DCR. That's like your second example, so insertion loss of 1db. Not too bad?
I'm not sure I follow the damping factor maths. Bit late in the evening for me here......
Damping factor = 5000/800 (300b) = 6.25
DCR of primary = 210/5000x100 = 4.2%
DCR of secondary = 0.4/5x100 = 8%
Total = 12.2%
800R + 12.2% = approx 900R
New damping factor = 5000/900 = 5.6
Does that make sense? Is 5.6 good or bad?
http://www.lundahl.se/wp-content/uploads/datasheets/1682.pdf
Primary is 210DCR and secondary 0.4DCR. That's like your second example, so insertion loss of 1db. Not too bad?
I'm not sure I follow the damping factor maths. Bit late in the evening for me here......
Damping factor = 5000/800 (300b) = 6.25
DCR of primary = 210/5000x100 = 4.2%
DCR of secondary = 0.4/5x100 = 8%
Total = 12.2%
800R + 12.2% = approx 900R
New damping factor = 5000/900 = 5.6
Does that make sense? Is 5.6 good or bad?
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andyjevans,
Personally, I would love to have a pair of those SE 50mA versions to try.
If I remember correctly, those are the exact versions you have.
I would not use the PP 100H version; and I would not use the 100mA 17H SE version.
Not so good numbers for PP and high current SE, versus my amplifier designs.
The primaries are 105 Ohms each, use them in series; 210 Ohms
The secondaries are wired in parallel, 0.4 Ohms
5500 Ohms and DCR 210 Ohms (3.8%)
5 Ohms and 0.4 Ohms (8%)
Total 11.8%
There is an 11.8% loss of voltage due to the DCRs.
1 - 0.118 = 0.882
10 x (Log 0.882) = -0.545 dB
That was an Un-Natural Log calculation, I used Log base 10, not Natural Log (Pun intended)
Power is proportionate to E squared
0.882 squared = 0.778
10 Watts from the output tube = 7.78 Watts to a 5 Ohm load.
See how easy that is?
If your speaker impedance is 8 Ohms instead of 5 Ohms, the insertion loss will be lower, and the damping factor will be higher.
If your speaker impedance is 4 Ohms instead of 5 Ohms, the insertion loss will be higher, and the damping factor will be lower.
But of course the different secondary load impedance, reflects an accordingly proportionate different primary load impedance.
The keys to proper application of that exact transformer are several:
1. The primaries are wound in 2 areas, over different air gaps, so they are not coupled primary to primary; to overcome that they are wired in series, the currents through them are identical, except for capacitive currents to secondaries, and capacitive currents to the core.
2. The 2 secondaries are also wound over 2 different air gaps, so the secondaries are not coupled to each other. But to overcome that, they are connected in parallel.
3. Do not use more than 50mA; or even use a little less current if you want low distortion at low frequencies. That will make for less power, a tradeoff.
There are not enough laminations to make lots of undistorted power at say 30Hz. Hey, most of us are not listening either to 32 foot pipe organs, or to thunder drums (at full volume anyway).
Good, you calculated it yourself.
You used 5000 Ohms, not the 5500 Ohm primary rating.
Your numbers are different because of that.
And for me a damping factor of 2.5 or better works OK for me on most all of my many different speaker brands, and different speaker models within a brand, although I prefer to get a damping factor of 3 or more, I want to be able to "drag" my amplifiers over to a friend, or to an amplifier "Shoot Out".
Worry, Worry, Worry. I will never enjoy the music. But not to Worry, I do enjoy the music.
Happy listening!
Personally, I would love to have a pair of those SE 50mA versions to try.
If I remember correctly, those are the exact versions you have.
I would not use the PP 100H version; and I would not use the 100mA 17H SE version.
Not so good numbers for PP and high current SE, versus my amplifier designs.
The primaries are 105 Ohms each, use them in series; 210 Ohms
The secondaries are wired in parallel, 0.4 Ohms
5500 Ohms and DCR 210 Ohms (3.8%)
5 Ohms and 0.4 Ohms (8%)
Total 11.8%
There is an 11.8% loss of voltage due to the DCRs.
1 - 0.118 = 0.882
10 x (Log 0.882) = -0.545 dB
That was an Un-Natural Log calculation, I used Log base 10, not Natural Log (Pun intended)
Power is proportionate to E squared
0.882 squared = 0.778
10 Watts from the output tube = 7.78 Watts to a 5 Ohm load.
See how easy that is?
If your speaker impedance is 8 Ohms instead of 5 Ohms, the insertion loss will be lower, and the damping factor will be higher.
If your speaker impedance is 4 Ohms instead of 5 Ohms, the insertion loss will be higher, and the damping factor will be lower.
But of course the different secondary load impedance, reflects an accordingly proportionate different primary load impedance.
The keys to proper application of that exact transformer are several:
1. The primaries are wound in 2 areas, over different air gaps, so they are not coupled primary to primary; to overcome that they are wired in series, the currents through them are identical, except for capacitive currents to secondaries, and capacitive currents to the core.
2. The 2 secondaries are also wound over 2 different air gaps, so the secondaries are not coupled to each other. But to overcome that, they are connected in parallel.
3. Do not use more than 50mA; or even use a little less current if you want low distortion at low frequencies. That will make for less power, a tradeoff.
There are not enough laminations to make lots of undistorted power at say 30Hz. Hey, most of us are not listening either to 32 foot pipe organs, or to thunder drums (at full volume anyway).
Good, you calculated it yourself.
You used 5000 Ohms, not the 5500 Ohm primary rating.
Your numbers are different because of that.
And for me a damping factor of 2.5 or better works OK for me on most all of my many different speaker brands, and different speaker models within a brand, although I prefer to get a damping factor of 3 or more, I want to be able to "drag" my amplifiers over to a friend, or to an amplifier "Shoot Out".
Worry, Worry, Worry. I will never enjoy the music. But not to Worry, I do enjoy the music.
Happy listening!
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Andy, have you ever considered using a 16 Ohm speaker (on the 16 Ohm tap of the ouput transformer)? Everything else being equal this would draw half the currekt of an 8 Ohm speaker and 25 % of the current of a 4 Ohm speaker. A SET amplifier would be very grateful and reward you with much lower distortion and better sound. It's no accident that most of the speakers in the 1950'ies were 16 Ohm. when all amplifiers were of the tube variety. All the custom OPTs that I order have two 4 Ohm secondaries. I mostly use them in series for 16 Ohm, but I’m still hoping for some Western Electric 4 Ohm speakers.
boli46,
Why wait.
Two Western Electric 8 Ohm speakers in parallel = 4 Ohms
Two Western Electric 8 Ohm speakers in series = 16 Ohms
And the parallel or series DCRs of the voice coils is 1/2 or 2X respectively, versus the DCR of a single Western Electric speaker.
Just do not ask what the impedance of a loudspeaker driver is versus frequency.
That answer is often complex, and largely resists being purely resistive.
Why wait.
Two Western Electric 8 Ohm speakers in parallel = 4 Ohms
Two Western Electric 8 Ohm speakers in series = 16 Ohms
And the parallel or series DCRs of the voice coils is 1/2 or 2X respectively, versus the DCR of a single Western Electric speaker.
Just do not ask what the impedance of a loudspeaker driver is versus frequency.
That answer is often complex, and largely resists being purely resistive.
Your calculation of insertion loss is incorrect.
For your two examples insertion losses are 0,83 and 0,42 dB respectively.
When interested I can do the math.
Also, your calculation of damping factor is incorrect.
It is better to calculate the output resistance of the amplifier.
In your example (500 ohm plate impedance and 100 ohm primary DCR), for an output transformer with 5k primary impedance and 8 ohm secondary impedance the impedance ratio is 625 : 1.
As output resistance is ((plate impedance + primary DCR) / impedance ratio)) + secondary DCR, in your example the output resistance is ((600/625)+0.8)) is 1,76 ohm.
That is a damping factor of 4,55 @ 8 ohm.
Insertion loss of LL1682/50 SE is 0,5 dB.
Compared to for example a 5k quality OPT like Monolith Magnetics with insertion loss of 0,16 dB, the Lundahl is "bad"....
Let's say anode impedance of 300B is some 800 ohm (depending also on operating point), with the Lundahl OPT the output resistance of the amplifier will be around 1,3 ohm, so a damping factor of ~3,8 at the nominal 5 ohm load (or ~ 6 at 8 ohm load).
Whether this is good or bad depends on your loudspeakers (and ears).
OK-ish with well damped loudspeakers with benign impedance curve; with underdamped loudspeakers (most bass reflexes..) and irregular loudspeaker impedance curves your amplifier will behave like an effect box (which you might like..).
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daanve,
OK.
A 5000 Ohm primary that has no DCR, versus a primary (5000 + 500 Ohms) that has 500 Ohms DCR.
1 Volt across 5000 Ohms = 0.2 mA
1 Volt appears across the 5000 Ohms primary (that has no DCR).
The resultant power is transferred to the output (if there are no other losses).
1 Volt across 5500 Ohms = 0.18 mA
0.18 mA across 5000 Ohms = 0.909V
Where did the other 0.09V go?
0.909V + 0.09V = 0.999V
0.09V is across the 500 Ohm DCR.
0.18mA (0.09V) across the 500 Ohm DCR does not appear in the output signal power.
0.18mA (0.909 Volts) appears across the 5000 Ohm primary, and is transferred as signal power to the output (if there are no other losses).
Power = (E squared)/R
1V squared = 1
0.909V squared = 0.826.
OK, so it is only a -0.8287 dB loss with DCR, versus the power without DCR loss.
And so 10 Watts from the output tube, becomes 8.26 Watts
Now, does it matter if the only DCR is in the primary, and it is 10% of the primary impedance . . . versus a primary with DCR of 5% of its impedance plus the effect of a secondary with DCR that is 5% of its secondary impedance.
The total loss either way is due to a total of 10% DCR.
Right?
I need to stop doing those sleepy time calculations.
All of us make mistakes, I just make more of them.
And I love slop jar slide rule calculations. Early rocket motors did not have precision thrust and angular direction. That was a real good reason to have
mid-course corrections on the way to the moon (it is not fun to shoot by the moon, and not fun to plow directly into it either).
OK.
A 5000 Ohm primary that has no DCR, versus a primary (5000 + 500 Ohms) that has 500 Ohms DCR.
1 Volt across 5000 Ohms = 0.2 mA
1 Volt appears across the 5000 Ohms primary (that has no DCR).
The resultant power is transferred to the output (if there are no other losses).
1 Volt across 5500 Ohms = 0.18 mA
0.18 mA across 5000 Ohms = 0.909V
Where did the other 0.09V go?
0.909V + 0.09V = 0.999V
0.09V is across the 500 Ohm DCR.
0.18mA (0.09V) across the 500 Ohm DCR does not appear in the output signal power.
0.18mA (0.909 Volts) appears across the 5000 Ohm primary, and is transferred as signal power to the output (if there are no other losses).
Power = (E squared)/R
1V squared = 1
0.909V squared = 0.826.
OK, so it is only a -0.8287 dB loss with DCR, versus the power without DCR loss.
And so 10 Watts from the output tube, becomes 8.26 Watts
Now, does it matter if the only DCR is in the primary, and it is 10% of the primary impedance . . . versus a primary with DCR of 5% of its impedance plus the effect of a secondary with DCR that is 5% of its secondary impedance.
The total loss either way is due to a total of 10% DCR.
Right?
I need to stop doing those sleepy time calculations.
All of us make mistakes, I just make more of them.
And I love slop jar slide rule calculations. Early rocket motors did not have precision thrust and angular direction. That was a real good reason to have
mid-course corrections on the way to the moon (it is not fun to shoot by the moon, and not fun to plow directly into it either).
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I don’t need to ask what the impedance of my loudspeaker is versus frequency. In my three way system I have a Zobel network on the woofer that keeps the impedance at 16 Ohms with rising frequency. On the mid horn I use an autoformer volume control from Dave Slagle with a swamping resistor that keeps the impedance looking back into the crossover at a rock solid 16 Ohms. The same with the tweeter, except that the autoformer is a Crites 3636.
A little OT but it does relate to the OPT..... Where does the damping factor matter most? Is it just the bass response that needs damping? If so, the critical mids and highs might not be affected and the system might still sound good. Especially where the speaker is a bookshelf 2-way with limited deep bass.
to Boli
If you bass way is a bass reflex you have two peaks with a two curves about module that vary a lot.
If you put a Zobel maybe you can limit the peak but when the Z of loudspeaker goes down 16 ohm?
Have you measured the shape of modulus of your speaker system? ( and also phase)
Walter
If you bass way is a bass reflex you have two peaks with a two curves about module that vary a lot.
If you put a Zobel maybe you can limit the peak but when the Z of loudspeaker goes down 16 ohm?
Have you measured the shape of modulus of your speaker system? ( and also phase)
Walter