Building better High sensitivity headphones

"find even one exception" to "analogous technologies regarding EM and sound"?

Light and sound.

EM is in the two version. In Maxwell's model the lines of force rotate and nothing is flowing along them. In Heaviside's model the lines not rotate but something is flowing along them.

In the both the electricity is incompressible and massles. Your electrons qute oppsite.

In the result in Maxwell's solenoid something rotate and for this reason the polarisation plane also rotate. In Heaviside's solenoid the flux flows but nobody know what rotate the polarisation plane.

Of course no ""analogous technologies regarding EM and sound".

But such are with the radio waves, light and sound.

Now is possibility to read originals. In another texbooks is wtote: " What is a pity that to teaching was chosen Heaviside instead Ampere". Ampere's "current elements" are like electrons. See:

S*

One is using Google Groups, the other is Using Outhouse Express. Both are complaining about each other's mis-quoted posts. It doesn't get any better than this, folks! (I can't tell who is saying what...)

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Blah Blah Blah...

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1- THIS (sci.physics.acoustics) IS AN *ACOUSTICS* GROUP!

2- I'm suspecting that Stephan is an Anglo-Spoofer, as no one educated in common science can be that completely misiinformed.

3- If your arguments can't be specifically about *high-efficiency headphones*, then please take this rondo private.

Ange

Uzytkownik "Angelo Campanella" napisal w wiadomosci news:iamaom$ffo$ snipped-for-privacy@news.eternal-september.org...

I have found:

"If you don't have a strong station nearby, you can add a battery in series with the LED (a small 1.5 volt battery works fine). The LED will light up, and the radio will play much louder than without the battery (if the LED doesn't light up, try connecting the battery the other way around). This arrangement is the best detector I have used so far, and is louder than the

1N34A germanium diode."

Acoustic waves, radio waves and light are very simmilar. Helmholtz was informed in the all. You also should. S*

GY.NET:

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mW

PL

c

. =A0

The last thing you want on a xtal set is headphones capable of that kind of output. Imagine a nice storm crackle coming in... the make it a very nearby lightning strike. All with no electronics to limit it.

NT

when i was a kid and i had a crystal radio with a 40 meter long-wire antenna outside, besides having a lightning arrestor on it, i was pretty much forbidden to use it or even connect it up when there was anything like an electical storm around. this cheap Knight-Kit crystal radio had no limiting electronics in it, as best as i can recall.

besides have as much of an aperture as possible (a long wire) for the antenna, you wanted as sensitive or efficient headphones as possible. the way it was, i only got 2 or 3 local AM stations. my dad had some old WW2-vintage headphones that worked pretty good. i wish i knew what happened to them. they would be a nice collector's item now.

r b-j

I put a neon bulb between my long-wire antenna and my ground, and it lit up nicely when there was a storm anywhere in the general area.

That is how this thread started.

--scott

--
"C'est un Nagra. C'est suisse, et tres, tres precis."

When I was a kid I had an 80m dipole. During electrical storms the PL-259 on the end would arc across.

In college one of RF profs built an antenna powered radio that drove a speaker directly (it played in the lab 24/7). It used an LED as the detector. It really didn't matter much what the diode was but the LED was more impressive. ;-)

the longwire antenna we had to our high scholl electronics shop ran from the school building to the football stadium's press box. Six inch sparks to an outlet were common.

Light Rock?

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

Nah, he was a '40s kind of guy (a friend of my father's).

eaker

=A0It

ive.

i would think that, in the unpowered and unamplified context, you would want the turn-on voltage of the diode to be as small as possible. isn't this why germanium diodes, like the 1N34, (about 0.3 v) were normally used, rather than a silicon diode (0.7 v)? if the peak-to-peak amplitude never exceeded about 1.4 volts, you would hear nothing with Si, but something with Ge.

r b-j

on

speaker

impressive.

Woosh!!! Maybe I should have said: 'Light' Rock? ;-)

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

Guys:\ Have we settled yet the most sensitive headphones (WattSoundPower/WattsRF)?

Ange

--
Those were called "2000 ohm headphones", back in the day, and I
happened across a set when I was about 12, and they blew me away.

Lost them somewhere along the way. :-(

on

speaker

impressive.

I got it, just wanted to go another direction.

I thought we did. 1mW/100dBA is normal for high-quality headsets. Anything appreciably more than that will be difficult.

"Angelo Campanella" wrote in news:iapsuq$46d$1 @news.eternal-september.org:

I certainly hope not, because the question doesn't appear to have a definitive answer within the context of a passive crystal radio receiver. This is so because the elements of the receiver interact. There must be a bleed resistor following the detector, or no demodulation of the AM carrier will take place. In the absence of a bleed resistor, the capacitance of the piezo earphone will simply charge to the maximum detected voltage and stay there. The problem is that increasing the value of the bleed resisstor has opposing effects. On the one hand it reduces the load on the tank circuit thereby increasing Q and RF voltage applied to the demodulator. On the other hand, as just stated, it also decreases the output of the demodulator. Because the sensitivity of a piezo earphone (eardrum pressure per applied volt) is inversely proportional to its capacitancee, the overall sensitivity of a crystal radio may well be independent of the sensitivity of the piezo earphone. This is so because a higher sensitivity earphone implies higher load capacitance on the demodulator which in turn requires a lower value for the bleed resistor to provide efficient demodulation. But, a lower value for the bleed resistor increases the load and decreases the Q and RF output of the tank circuit.

The sensitivity of a crystal earphone is defined as eardrum pressure per applied volt. If you want maximum sensitivity, you need to drive the earphone from a voltage source whose impedance is small relative to the impedance of the earphone. Under these drive conditions, the most sensitive piezo earphone will be the one that has the highest capacitance. This is so because the largest diaphragm deflection is achieved when diaphragm stiffness is the smallest, which is when the piezo element and its metal backing are as thin as possible. The latter is limited by the ability to fabricate very thin discs of piezo ceramic material. At the present time this appears to be about 0.005".

PL-259 on

speaker

It

impressive.

Pun killer! ;-)

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

--
if only you'd have AM detected the flickering light from the neon lamp
with, say, a photoresistor, you might have heard God talking. ;)