... my DIY ribbon 
does this mean I can just mount it with a 1ohm resistor and have 4ohm ?
not sure I understand that
I thought Qts would rise
> does this mean I can just mount it with a 1ohm resistor and have 4ohm ?
You lost me on that one .
> not sure I understand that ......... I thought Qts would rise
Me neither .... I thought it would stay the same
Perhaps because the tester is reading the resistance
of the xfmr winding rather than Re .......
You lost me on that one .
> not sure I understand that ......... I thought Qts would rise
Me neither .... I thought it would stay the same
Perhaps because the tester is reading the resistance
of the xfmr winding rather than Re .......
because ribbons have zero impedance without transformer
and with a modest 2:1 autoformer trafo it needs 'something'
hence the 1ohm
btw, is this correct connection ?
Attachments
because ribbons have zero impedance without transformer
and with a modest 2:1 autoformer trafo it needs 'something'
hence the 1ohm
btw, is this correct connection ?
Your ribbon has significant resistance, which is impedance without any imaginary component.
To calculate how much resistance it has, see the post at:
http://www.diyaudio.com/forums/plan...epan-ribbon-tweeter-question.html#post3471872
Or here's the short version:
R = pL/A,
where R is in Ohms, and the conductivity of aluminum, p, is 2.6 x 10^(-8) Ohm-meters (where 2.6 x 10^(-8) = 0.000000026), and L is length in meters, and A is cross-sectional area in square meters.
From the equation above:
Ohms per Meter = .026 / (cross-sectional area in sq mm)
for aluminum.
OR, you could just measure the resistance with an ohmmeter.
As mentioned in the linked thread, if your ribbon is pleated, you can measure its resistance with an ohmmeter and then calculate the original un-pleated length (assuming you know the thickness and width).
Last edited:
ahh, you think of Apogee clones etc, with long foil lines glued to a film
yeah, I build a few long time ago
and I might do it again
they can sound very good
but this one is just a small and simple experiment with a piece of kitchen foil, 12mm wide and 150mm long
resistor alone was not good
attenuated too much
never got around winding a trafo
and actually dumped the whole thing
but this one I have to try
yeah, I build a few long time ago
and I might do it again
they can sound very good
but this one is just a small and simple experiment with a piece of kitchen foil, 12mm wide and 150mm long
resistor alone was not good
attenuated too much
never got around winding a trafo
and actually dumped the whole thing
but this one I have to try
Oh!
Well, standard kitchen foil is about 0.016256 mm thick. So maybe you could just add some length (and/or lower the width) to get a higher resistance.
At 12 mm wide, the cross-section is presumably 0.1951 sq. mm.
That gives 0.026 / 0.1951 = 0.1333 Ohms per meter!
So you really do have only 0.15 * 0.1333 = 0.02 Ohms; basically zero.
I'm guessing you don't want to make 15 2-meter ones in series, to get 4 Ohms.
Let's see, if you cut it down to 3 mm wide, you could get 0.5332 Ohms per meter. Then you'd "only" need 7.7 m to get 4 Ohms.
Nope. So let's see, to get 4 Ohms while keeping the same length, it would need to be 4/0.15 = 26.67 Ohms per meter. So the cross section would need to be 0.026 / 26.67 = 0.000975 sq mm. Wow. If the foil is 0.016256 mm thick, then the width would need to be 0.06 mm. Or make it 1 Ohm with a width of 0.24 mm. Ummm... No.
Never mind!
Well, standard kitchen foil is about 0.016256 mm thick. So maybe you could just add some length (and/or lower the width) to get a higher resistance.
At 12 mm wide, the cross-section is presumably 0.1951 sq. mm.
That gives 0.026 / 0.1951 = 0.1333 Ohms per meter!
So you really do have only 0.15 * 0.1333 = 0.02 Ohms; basically zero.
I'm guessing you don't want to make 15 2-meter ones in series, to get 4 Ohms.
Let's see, if you cut it down to 3 mm wide, you could get 0.5332 Ohms per meter. Then you'd "only" need 7.7 m to get 4 Ohms.
Nope. So let's see, to get 4 Ohms while keeping the same length, it would need to be 4/0.15 = 26.67 Ohms per meter. So the cross section would need to be 0.026 / 26.67 = 0.000975 sq mm. Wow. If the foil is 0.016256 mm thick, then the width would need to be 0.06 mm. Or make it 1 Ohm with a width of 0.24 mm. Ummm... No.
Never mind!
I don't know the theory ....
BUT !
I have tried passing audio through standard
power transformers ('EI' granted not 'C') and
it does not work. (Try a squarewave to see)
Toroids however pass a nice signal.
(which is why some don't like them for power
because they pass all the cr*p on the mains
into the power supply)
BUT !
I have tried passing audio through standard
power transformers ('EI' granted not 'C') and
it does not work. (Try a squarewave to see)
Toroids however pass a nice signal.
(which is why some don't like them for power
because they pass all the cr*p on the mains
into the power supply)
How much time do you spend listening to square waves? Personally I don't find them terribly musically satisfying.
se
I believe it did not work.
I also believe you have associated power E-I cores with all E-I cores.
There are several (if not more) types of material used in laminations. (whether E-I or C)
Power transformer manufacturers will not spend the extra cost on a material that runs low loss to 20kHz when they only need it to work at 60Hz.
When I spec inductors for the simulated load at work I have to specify what frequency they will be exposed to...kind of important.
I would have suggested to tinitus that he insure the C-cores he was about to try were audio cores and not power cores.
But that is just me.
Well, after Michael Fremer's gushing review of the Tara Labs Cyberlight cables, I don't believe in quantification anymore, crude or otherwise.
se
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.