High Sensitivity HeadPhones, new direction

Hi guys, I ran across this electromechanicoustic device (speaker) called a Gallows headphone. This is used as the speaker for a crystal radio. It is pictured about 2/3 of the way down the page. You will understand it's operation by its picture.

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I have cross posted looking at this as having three disciplines, acoustics, electronics and magnetics. I would like to see ideas to design one with these objectives in mind. Objectives: Maximum audio from minimal (crystal radio) signal (high sensitivity)

High impedance, if we can get enough wire on the electromagnet, the matching transformer could be eliminated. (100k to 1 meg ohms, tapped) That's a wish :-)

Questions; What should the diaphragm look like? low mass? with ridges? needs to be magnetic at some point.

Can we put an electromagnet on both sides in push pull? Can needs to be sealed.

Earphone is connected with hollow tubing, what type of tube would be low loss? does length matter? I'm thinking stethoscope earphones till something better is found.

Do we want the center of the diaphragm to move or the whole circumference to breath? How do we focus the magnetic field, if we do.

Thanks for your input, MikeK

Reply to
amdx
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Electromagnetic phones will cause to much load on a crystal radio. The output and selectivity will become to low to be useful. Their impedance is only about between 10 ohms and ~2kohms(with extremely thin wire, and a few thousand windings). The crystal earphones are a much better choice.

Reply to
Sjouke Burry

I would point out that Electromagnetic phones (sound powered headphones) are very commonly used with crystal radios and yes, that's why I want this to be very high impedance. So maybe we need 10.000 turns and high (AsubL?) core.

Thanks, MikeK

Reply to
amdx

I could not make head or tail out of these pictures. A well thought out diagram would be much more helpful.

In general, you have a number of impedance matching problems. The crystal set provides the source impedance. That power has to be matched to a mechanical device like the stirrup in your ear. That in turn is attached to a diaphragm which must be matched to the air. Everywhere in this path, a mismatch will reduce the amount of power that can be transferred. That is the trick.

Bill

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An old man would be better off never having been born.
Reply to
Salmon Egg

Ok, first page down about 2/3 until you see a caption that says Gallows headphone. It is a sealed container with a diapragm mounted on top. The sealed contain has a tube sticking out, rubber hose is connected and this is connected to an air piece. Above the diaphragm is a a coil with an iron core (electromagnet). When an audio electrical signal drives the coil it causes the diaphragm to move and sound is emitted from the tubing to the ear. There are two pictured one has a large container the other is much smaller.

Yes matching is the trick, that's why I ask the experts. MikeK PS. a perfect match is never going to be made over the useful audio range, hopefully just better than what's available now.

Reply to
amdx

Salmon Egg wrote in news: snipped-for-privacy@news60.forteinc.com:

Impedance matching is only important when the objective is to maximize power ransfer. Since the ear is only interested in pressure, maximizing power transfer serves no useful purpose. In fact, The Designer actually uses an impedance mismatch to create a 15-dB amplification of pressure at the eardrum at about 3000Hz (quarter wave ear canal resonance). It is this resonance combined with the resonance of the middle ear at about

1000Hz that provides the low threshold of audibility in humans that exists between about 1000Hz and 3000Hz.
Reply to
Answerman

Indeed, that is the trick. I do not think that you will find an expert that can make an almost perfect match across the band. You probably won't find one who would try developing a good broadbamd match cheaply.

Good luck!

Bill

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An old man would be better off never having been born.
Reply to
Salmon Egg

Maximum power across the band is what you want. Exceptions would be if you are listening to a a narrow audio band signal like tone modulation of an rf signal detected by the crystal. The amplification will only occur if the acoustic network is the equivalent of a lumped circuit. Resonance at 3kHz is not going to help listen to an ordinary voice signal.

Bill

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An old man would be better off never having been born.
Reply to
Salmon Egg

Salmon Egg wrote in news: snipped-for-privacy@news60.forteinc.com:

No, maximum pressure at the eardrum across the band is what you want.

I never said that it would. It's the combined effect of system resonances at 1kHz (middle ear) and 3kH (ear canal) that together contribute to the sensitivity of the human ear in the frequency range between 500Hz and 6kHz. The point, which you apparently either missed or ignored, is that when it comes to hearing sensitivity, an impedance mismatch can be and is an advantage, and not, as you claim, something to be avoided.

Reply to
answerman

Sounds like one could "steal" that design for the electromechanicoustic device..

Reply to
Robert Baer

Well, a simple diaphragm device like that shown, is "tuned" and has its "Q" according to stiffness, etc. The photos indicate a low quality and rather non-uniform attachment to the ring. Those mechanical distortions will result in multiple resonances that are not related to each other. If one wants a broader band coupling, the de-Q-ing method is not robust as it decreases the sensitivity. Better yet may be to use multiple "resonator" converters,each at a different frequency, with sensitivity "tails" overlapping some. That may tend to get one to "steal" from the ear design (shrinking diameter tube with spaced sensors) - OR - use a number of energized pipes coupling to diaphragm(s) looking similar to a Gatling gun. That second design has been around for a while, and maybe available as a "distance mike".

Reply to
Robert Baer

Even if maximum pressure were wanted for specialize listening situations, you would also want matching. You would want to match impedance with the appropriate resonant structure's dissipation. Assume a high Q resonator. For a series (electrical) resonator that would typically mean using an acoustic matching transformer to a low impedance (force/velocity) for the vibrating structure. For a parallel resonator you would have to transform source impedance to a higher level.

One caveat is that a mechanical schematic diagram for a series resonance looks somewhat like an electrical parallel resonant structure and vice versa. Remember that the series resonance will have equal velocity for the mass and the spring. In the diagram the mass will resemble the symbol for a capacitor in parallel with a coil (inductor} which look like a spring.

Bill

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An old man would be better off never having been born.
Reply to
Salmon Egg

"amdx" wrote in news:298bd$4cd0a8fc$18ec6dd7$ snipped-for-privacy@KNOLOGY.NET:

For whatever reason, you are still not listening. You can not talk about building a high-sensitivity headphone when the headphone is to be used in a passive circuit such as a crystal radio. This is so because the load impedance of the headphone affects the terminal voltage of the circuit to which it is connected and because the source impedance of the circuit to which the headphone is connected affects the acoustic output of the headphone. It is because of this interaction that the sensitivity of a headphone is specified in terms of the pressure that it produces in specified aoustic load (eg IEC-318 coupler) per applied volt when driven from a low-impedance voltage source.

Reply to
Answerman

Taken to the extreme, you're saying no headphone is any better than any other when connected to a crystal radio. Bullshit! I understand impedance matching, I understand loading the tank to an unusable Q. Go look at this 4 Megohm to 8 ohm audio transformer.

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If you don't like my semantics get over it, everyone else understands I want maximum audio volume for the input power available. You are welcome to ignore this thread. MikeK

Reply to
amdx

I am glad you have made a response. I thought mine was the only voice in the wilderness. When I was a kid before I truly understood conservation of energy and impedance matching, I thought that the higher the impedance of headphones, the better. I did not realize that all headphones of the same design would have essentially the same efficiency. I did not understand why you would not want to get the largest possible number of turns on the phones.

It took me a long time to understand why a 5-tube superheterodyne receiver could be as sensitive as it was. Now, I realized that you tried to match the source of rf power to the losses of the parallel tuned intermediate frequency transformers. The loss of power to the grid of a pentode was minuscule compared to that in the resonant circuit. Matching gave the highest grid voltage to a tube amplifier that responded in proportion to the applied voltage. Loss in the tuned circuit would have to be great enough to allow adequate signal bandwidth.

The same principles apply to acoustic and mechanical devices.

Bill

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An old man would be better off never having been born.
Reply to
Salmon Egg

"amdx" wrote in news:73e6f$4cd55d03$18ec6dd7$ snipped-for-privacy@KNOLOGY.NET:

I agree with your conclusion. The problem is that what you say I said isn't what I said or even implied.

That's a good start. .

Interesting but irrelevant.

Your semantics aren't the problem.

I'm not going to ignore this thread because I'm curious about what valuable information others may provide regarding the subject matter. But I am going to ignore you.

Reply to
Answerman

Not irrelevant if you want to connect a low impedance headphone to a detector with a high output impedance. At this point I don't know the output impedance of high Q tank driving a diode that matches the tank, but I do think 200Kohm to 2.5megohm are not unreasonable numbers*. MikeK I like factors greater than 10, almost easier the horseshoes to get close. ;-) (it does give a range and tank Qs vary a lot in crystal radio building)

Reply to
amdx

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=A0Thanks, MikeK

Yeah, the old electromagnetic headphones were high impedance; typically two coils interacting with a light magnetic plate. My father gave me a pair when I was young and they did work in crystal sets. I don't recall what their impedance was, but I want to say 10kOhms.

What a great link amdx. Thanks. I hope to try to make my own diode one day and have just started playing around with metal oxides in my woodstove, simply by putting in a tin can top and cranking the heat. Plenty of resistance and material that cracks off when bent, which can sort of be scratched through to the metal- must be oxides. I don't mind primitive and poor results when starting out... no diodes yet, but possible resistors?

Good luck with the headphone. I believe you will get it, but yes, lots and lots of fine windings. Some printer stepper motors have pretty fine wire windings, and the wire is in rolls, so is easy to reuse. It's not ultrafine, but it doesn't look like you'll need it with that contraption. Looks like a good prototype design. Don't forget, DC resistance and AC impedance are two different things.

- Tim

Reply to
Tim Golden BandTech.com

Agreed. The main difficulty in our understanding of the CR system is that at this time no model including the transformation of the electrical power derived from the radio wave into acoustical power radiated from a speaker transducer has been advanced. The "Gallows" phone is an example of effort to improve the conversion of electrical audio power (demodulated RF power, no amplification) into sound one can hear. When speaking of reproduced sound pressure level, the distance at which that is heard is not yet in the model as far as I can see.

..... which is a human ear simulator... It's a start. The "distance" at which the sound is heard is about an inch.

(My mnemonic for conversion of sound power data to SPL in dB is that the sound pressure found one foot from a point source (small speaker) is numerically equally to the sound power expressed in picowatts (10^-12 W)). So the sound dB sound level created in that coupler's geometry is numerically about 20 dB more than its sound power value... A +10dB sound power level produced by the transducer under test would then produce a +30 dB SPL in that coupler.... (ad-hoc WAGs by me).

(FYI, a vacuum cleaner emits about +85 to +90 dBA sound power).

Ange

Reply to
Angelo Campanella

Here are some numbers thrown out on one of the crystal radio groups.

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"The big question here is: Do phones that produce 125 dB SPL when 1 mW is applied to them . . . produce 35 dB SPL when 1 pW is applied ?

1 pW is 90 dB below 1 mW. I believe 35 dB SPL should be very comfortable listening. 1 pW audio might be what a crystal radio would deliver from a weak DX signal.

Much debate has transpired in the past over this issue, without any definite conclusions. Anecdotally speaking: some Sennheiser "120 dB/mw" earbud-type phones failed miserably on weak DX when compared to a good set of sound-powereds." MikeK

Reply to
amdx

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