I've been reading theses threads for a few weeks now and I've started my own ribbon speaker project. A few things were immediately evident.
Firstly, is that the knowledge that is needed to build a speakers isn't very transparent. That is it's nearly impossible to find any source about audio design and theory.(at least for me ) and secondly, that it's hard to find a source for the materials and equipment you need in order to build such devices.
It just seemed a shame to me, financial markets are pretty efficiently. knowledge is shared relatively easily through all participants, why not so with engineering. It just seems like we are hindering ourselves. but, i digress...
I built a simple prototype ribbon out of stuff i had laying around the house and 2 4"x1/2"x1/8" N42 Neodymium magnets.
The ribbon area is 4"x1/2" and is standard kitchen aluminum foil so I'm assuming it's 99% pure aluminum with a thickness of 12.5 microns.
Using the totally unvalidated formulas(eq. 1) I found on this page that gave me a theoretical performance of 84 db/watt @ 1 meter and I had no idea how to calculate the sensitivity. The number I got didn't make a bit of sense.
I don't understand why you would divide by log(10) when that just simplifies to 1 so why not instead wright (eq. 2) Also what happens when m goes to 0? does Efficiency go to infinity? If so then light would be the ideal sound source.
(eq. 1)
# SIMPLE RIBBON EFFICIENCY EQUATIONS
#
# A=ribbon area meter^2
# B=magnetic field in Telsa
# %= percent of ribbon which conducts current(100=100%)
# m= mass of ribbon in grams
# Efficiency = A^2 * B^2 * % / m
# spl 112.2 + 10*(LOG(wfficiency)/LOG(10))
# senS effeciency + LOG(8/resistance)/LOG(10)
(eq. 2)
# SIMPLE RIBBON EFFICIENCY EQUATIONS
#
# A=ribbon area meter^2
# B=magnetic field in Telsa
# %= percent of ribbon which conducts current(100=100%)
# m= mass of ribbon in grams
# Eff = A^2 * B^2 * % / m
# spl 112.2 + 10*LOG(Eff)
# senS Eff + LOG(8/resistance)
When I first hooked it up the foil was uncorrugated and rattled a lot. I first fixed this by tensioned the foil then using different size foil, latter I realized that corrugation increased horizontal rigidity and cut down on the rattling.
I have no way of checking the impedance on the speakers. DC resistance is well below an ohm. I don't have a transformer to present a descent load to the receiver so I just directly drive it with the audio turned down. The receiver also has a safety relay that trips when too much current is passed through. thankfully.
Latter in the stages in the prototype development I installed a high pass filter, and this improved performance a lot.
I have a load of pictures that I'll post later.
I plan on a few projects, first I want to complete a pair of full range ribbons, I plan on building my own subs and woofers for them from either a carbon fiber or Kevlar composite. Anyway I'll post something more in depth later.
Firstly, is that the knowledge that is needed to build a speakers isn't very transparent. That is it's nearly impossible to find any source about audio design and theory.(at least for me ) and secondly, that it's hard to find a source for the materials and equipment you need in order to build such devices.
It just seemed a shame to me, financial markets are pretty efficiently. knowledge is shared relatively easily through all participants, why not so with engineering. It just seems like we are hindering ourselves. but, i digress...
I built a simple prototype ribbon out of stuff i had laying around the house and 2 4"x1/2"x1/8" N42 Neodymium magnets.
The ribbon area is 4"x1/2" and is standard kitchen aluminum foil so I'm assuming it's 99% pure aluminum with a thickness of 12.5 microns.
Using the totally unvalidated formulas(eq. 1) I found on this page that gave me a theoretical performance of 84 db/watt @ 1 meter and I had no idea how to calculate the sensitivity. The number I got didn't make a bit of sense.
I don't understand why you would divide by log(10) when that just simplifies to 1 so why not instead wright (eq. 2) Also what happens when m goes to 0? does Efficiency go to infinity? If so then light would be the ideal sound source.
(eq. 1)
# SIMPLE RIBBON EFFICIENCY EQUATIONS
#
# A=ribbon area meter^2
# B=magnetic field in Telsa
# %= percent of ribbon which conducts current(100=100%)
# m= mass of ribbon in grams
# Efficiency = A^2 * B^2 * % / m
# spl 112.2 + 10*(LOG(wfficiency)/LOG(10))
# senS effeciency + LOG(8/resistance)/LOG(10)
(eq. 2)
# SIMPLE RIBBON EFFICIENCY EQUATIONS
#
# A=ribbon area meter^2
# B=magnetic field in Telsa
# %= percent of ribbon which conducts current(100=100%)
# m= mass of ribbon in grams
# Eff = A^2 * B^2 * % / m
# spl 112.2 + 10*LOG(Eff)
# senS Eff + LOG(8/resistance)
When I first hooked it up the foil was uncorrugated and rattled a lot. I first fixed this by tensioned the foil then using different size foil, latter I realized that corrugation increased horizontal rigidity and cut down on the rattling.
I have no way of checking the impedance on the speakers. DC resistance is well below an ohm. I don't have a transformer to present a descent load to the receiver so I just directly drive it with the audio turned down. The receiver also has a safety relay that trips when too much current is passed through. thankfully.
Latter in the stages in the prototype development I installed a high pass filter, and this improved performance a lot.
I have a load of pictures that I'll post later.
I plan on a few projects, first I want to complete a pair of full range ribbons, I plan on building my own subs and woofers for them from either a carbon fiber or Kevlar composite. Anyway I'll post something more in depth later.
Hello ,
Kitchen foil has very poor memory and while it can sing it is not a good choice . then again a 4x .5 inch diaphragm is pretty small so you might get away with this .
84dB /W / M seems to be the sensitivity of the unit , not sure why you would state this then ask what is the sensitivity .. I would also place a 2 ohm resistor in series for testing , your amplifier will find it more palatable..
Regards ,
Kitchen foil has very poor memory and while it can sing it is not a good choice . then again a 4x .5 inch diaphragm is pretty small so you might get away with this .
84dB /W / M seems to be the sensitivity of the unit , not sure why you would state this then ask what is the sensitivity .. I would also place a 2 ohm resistor in series for testing , your amplifier will find it more palatable..
Regards ,
I was referring to the last two lines of the group of equations
1) spl 112.2 + 10*(LOG(wfficiency)/LOG(10))
2) senS effeciency + LOG(8/resistance)/LOG(10)
It seems like it yields two different variables SPL and SenS which i'm assuming is sensitivity.
also I remember a post by Linesource that read
and he was using the same formulas so thus the confusion about sensitivty. Unless I'm missing something?
What do you mean by poor memory? I wanted to put a resistor in series but I don't have one that below 100 ohms.
1) spl 112.2 + 10*(LOG(wfficiency)/LOG(10))
2) senS effeciency + LOG(8/resistance)/LOG(10)
It seems like it yields two different variables SPL and SenS which i'm assuming is sensitivity.
also I remember a post by Linesource that read
and he was using the same formulas so thus the confusion about sensitivty. Unless I'm missing something?
What do you mean by poor memory? I wanted to put a resistor in series but I don't have one that below 100 ohms.
This is the FEMM model for my ribbon motor. I'm using ferrite bars to shield the magnets and it gave me around .65 tesla flux in the gap I was hopping to get around 1. I ended up not having enough ferrite to cover it the way i wanted to so the gap flux is probably a good deal lower.
Attachments
I didn't want to waste any bell wire on making a resistor so I chose instead to use aluminum foil cut into very small strips. It has a resistance of around 3 ohms. Now my receiver doesn't kick off any more. I was testing the limits I could push it and The current vaporized a small piece of the resistor it was pretty freaking awesome. So It's a kinda fuse/ resistor but it also brought about a timely realization. when I create my main ribbons I need to keep in mind the maximum power I can deliver to them safely so this will limit me to the size I can make them as too small of ribbon will flash over too easy.
So I need a few thing when designing the next ones.
1) How much current does it take to vaporize a ribbon
2) What will be the maximum current that will be expected from a standard receiver.
I've decided on building my ribbons from a modular frame. I'm designing it now in Autodesk and I'll post more on it latter.
So I need a few thing when designing the next ones.
1) How much current does it take to vaporize a ribbon
2) What will be the maximum current that will be expected from a standard receiver.
I've decided on building my ribbons from a modular frame. I'm designing it now in Autodesk and I'll post more on it latter.
Attachments
FrankCashio said:
I have seen a Krell fry a ribbon
But it was the bearing film that crumbled
I the foil leads were ok still
But if you use very thin foil it may happen
A known problem to some of the best ribbons
There are a few ribbons for pro studio use
Though they are also with good sensitivity, and need less power
btw, your proto ribbon looks about the size Im working on
Well i have a hunch that the foil size I'll use will be thick enough to preform under pretty rigorous conditions. The ribbon of foil used on my 5 minute foil resistor preformed rather well, it failed only twice where the foil was especially narrow.
I did frequency sweeps on abnormally high volume the foil got warm but, being foil it was pretty effective at dissipating the heat. I think the amp would have failed before the foil as it was getting hot as well.
I just need to find the energy that will be converted to heat due to resistance and then at what point would the heat cause a critical failure.
Profanity filter picture. 😱
fuckfuckfuckfuckfuckfuckfuck
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****----------****----------****
****-----fuckfuckfuck -----****
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fuckfuckfuckfuckfuckfuckfuck
cheers!
I did frequency sweeps on abnormally high volume the foil got warm but, being foil it was pretty effective at dissipating the heat. I think the amp would have failed before the foil as it was getting hot as well.
I just need to find the energy that will be converted to heat due to resistance and then at what point would the heat cause a critical failure.
Profanity filter picture. 😱
fuckfuckfuckfuckfuckfuckfuck
****-------------------------****
****-----fuckfuckfuck-----****
****-----fuckfuckfuck -----****
****-----fuckfuckfuck -----****
****----------****----------****
****-----fuckfuckfuck -----****
****----------****----------****
****----------****----------****
****------****---****------****
****-------------------------****
fuckfuckfuckfuckfuckfuckfuck
cheers!
tinitus said:
I have blown 10 amp Output fuses driving my ribbons and have never burnt or damaged a driver ever ....
Frank is using Regular kitchen foil . it will burn easily ...
a.wayne said:
How will it burn easily? It's really hard to find any good light weight alloys to use. I decided I'd just cold roll some aluminum myself to around 5 or 6 microns. It would be considerably less expensive to build my own machine to do it for me that specially order it.
I would really love to find some sort of AL/Mg alloy to use by no surprise that it's impossible to find... anyone ever try smelting their own alloys? Wouldn't imagine it would be too hard. Build and inductive furnace, get the ingredients together and cast your own ingant. cold roll and your done.
Cheers!
a.wayne said:
Never had any problems with kitchen alufoil
Even with 5 feet ribbon crossed low
I wouldnt worry about it
If you need that high SPL maybe a ribbon is the wrong choise
Okay guys I crunched some numbers and this is what I came up with regarding aluminum ribbons.
I'm assuming a ribbon of 12.5 microns thick by 0.1016 meters long. (about 4 inches).
fig. 1 is a plot of width vs. current needed to melt the ribbon.
So with this particular length ribbon I would like the keep the width no thinner than 1cm to handle about a 5 amps load.
I figured that the melting point of aluminum is 660.25°C. I subtract room temperature from that (approx 25°C) and that gave me a temperature flux of 635.25°C
Specific heat of AL is 0.9J/g*°C thus 0.9*635.25°C = 571.73 J/g
So it takes 571.73 joules of heat to melt one gram of Aluminum.
Ohms law states that
so........... the square root of (571.73J / R)*ribbon weight should give me the current needed to melt a particular ribbon.
Then the only missing piece is R which can be found by.
Electrical resistivity of aluminum is (20 °C) 28.2 nΩ·m
Cheers!
I'm assuming a ribbon of 12.5 microns thick by 0.1016 meters long. (about 4 inches).
fig. 1 is a plot of width vs. current needed to melt the ribbon.
So with this particular length ribbon I would like the keep the width no thinner than 1cm to handle about a 5 amps load.
I figured that the melting point of aluminum is 660.25°C. I subtract room temperature from that (approx 25°C) and that gave me a temperature flux of 635.25°C
Specific heat of AL is 0.9J/g*°C thus 0.9*635.25°C = 571.73 J/g
So it takes 571.73 joules of heat to melt one gram of Aluminum.
Ohms law states that
so........... the square root of (571.73J / R)*ribbon weight should give me the current needed to melt a particular ribbon.
Then the only missing piece is R which can be found by.
Electrical resistivity of aluminum is (20 °C) 28.2 nΩ·m
Cheers!
Attachments
Excel spread sheet that calculates the resistivity and the current needed to melt a ribbon for you lazy and not mathematically inclined individuals 😀
Cheers!
Cheers!
tinitus said:
SPL ? You mean current , it's a 1 ohm ribbon ..............
Frank interesting graph ...
regards
Your estimate is probably low because you did not take into account that the ribbon will transfer heat to the air. For such a high surface to weight ratio I'm pretty sure this will be a significant factor.
arend-jan said:
May be, maybe not. It's hard to say. I usually take a conservative approach to design though I'll add it for completeness sake.
Cheers!
The heat lost to air due to convection is negligible.
The formula for Rate of Convective Heat Transfer:
q = hA(Ts − Tb)
h being the heat transfer coefficient of air @ 20C is 0.0257/W*m C
so......
q= 0.0257*ribbon area*635.25
which for my ribbon was 0.021 J which is nothing. translated to a drop in temperature of a thousandth of a degree.
unless I'm missing something??
The formula for Rate of Convective Heat Transfer:
q = hA(Ts − Tb)
h being the heat transfer coefficient of air @ 20C is 0.0257/W*m C
so......
q= 0.0257*ribbon area*635.25
which for my ribbon was 0.021 J which is nothing. translated to a drop in temperature of a thousandth of a degree.
unless I'm missing something??
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