Another DIY Ribbon thread

[dhenryp]

Back on page page 2, Linesource posted a chart that showed 42" of 6 micron foil X .5 inch = .353 ohms. quarter inch should be double this or .705 ohms. Joules shows 6 micron X ~42 " X .25" = 4 ohms. Any idea of why the difference? Different alloy? Joule's number would get me pretty close to no-transformer range.

[dhenryp]

Here is the steel frame for my first full size ribbon. It's 47" long and wikk contain seven 5.25" ribbon segments, with 1.5" crossbars between each. This thing is HEAVY.

[LineSource]

Hi dhenryp,

My resistance calculations were for pure aluminum <1111> which has 2.65E-8 ohm-m of resistance. My handbook of metals shows a few non-ferric alloys with higher resistance, as well as many alloys that "allow" a trace of ferric materials. I have experimented with some alloys with 2x the resistance of pure aluminum to test corrugation annealing. Most alloys cannot be rolled as thin as pure AL, but provide superior shear strength. My ears think the harder alloys "sizzle" more than the softer pure aluminum.

[dhenryp]

All things being equal (e.g. mass, mechanical properties), I guess the best ribbon is the one with the least resistance. Energy lost as heat in the ribbon is just as bad as putting the equivalent resistor in series with the ribbon. In fact, ribbon resistance is worse because it can destroy the ribbon.

[Joules]

Energy lost to resistance in Watts = I^2*R

[Keld]

Originally posted by dhenryp
All things being equal (e.g. mass, mechanical properties), I guess the best ribbon is the one with the least resistance. Energy lost as heat in the ribbon is just as bad as putting the equivalent resistor in series with the ribbon. In fact, ribbon resistance is worse because it can destroy the ribbon.

OK this is the way I understand it: It is the current that flow trough the ribbon that makes it move, so the best ribbon would be zero ohm (to keep it cool), of course it will be difficult to find an amp that will drive that load. but what if you place a 4 ohm resistor with good cooling in series with the ribbon would that work?? I know ther must be something wrong with my thinking but what?.

[dhenryp]

The current flowing in the perfect ribbon, sitting in the magnetic field would make it move in the air doing work, which takes power. P = I * E . I think a back-emf would be generated in the ribbon * rhe current will equal that power. At least that's what I THINK will happen.

[Joules]

If you have 13 watts going into a transformer, you will have 13 watts, coming out of the transformer, minus some small amount of transformer loses,

If you have a corragated ribbon length of 6 inches x .5 wide made out of 1100 alum (99.0% pure) @ 2.9 x 10^-6 ohm centimeter, you wil have a ribbon resistance of about .288 ohms.

A transformer wil transform voltage and current by factors of the turns ratio. A transformer with 13/1 turns ratio with 13 volts on the primary will have 1 volt on the secondary. This same transformer with 1 amp on it's primary wil have 13 amps on the secondary. 13 watts in - 13 watts out. However for ohms law to work out the the impedence must be scale by a factor of turns ratio squared! Therefore the impedence ratio for this transformer will be 169/1!!! In this example primary R = 13v/1a = 13 ohms. 13 ohms /169, means the secondary resistance must be .077 ohms,
The numbers are consistant do the math V^2 / r = watts ----- I^2 * R = watts -----V * I = watts, for both primary and secondary circuits.

Now back to the ribbon example above with a resistance of .288 ohms x turns ratio squared (169) = 48.672 ohms tranformed impedence in to the primary circuit.

[dhenryp]

Hi Joules,

I found this interesting site that will calculate the resistance of aluminium for a given length and a given cross sectional area.

http://www.allmeasures.com/Formulae/...l%5D&formula=3

It corresponds reasonably closely with linesource's numbers and significantly lower than your numbers.

If your example is pure aluminium that is 5 micron (.000005 m) thick and 1/2 inch (.0127 m) wide = .0000000635 m^2 cross sectional area and 6 inches long; the site comes up with .0636 ohms instead of the .288 ohms you identified.

I follow the calculations you made for the 13:1 transformer example and it looks right to me. I'm not sure how it relates, or if it was intended to relate, to my "perfect ribbon" thought exercise. Would you clarify?

Denis

[LineSource]

dhenryp....if you want to estimate the SPL and sensitivity of a dipole ribbon you can run through these simple equations. The SPL is at 1 meter.

A 42" x 0.7" ribbon using 5.8 micron aluminum in a 0.5 T magnetic field would have
SPL ~ 97 db/watt @ 1m
sensitivity ~ 98 db/2.83 V

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)