Building better High sensitivity headphones
amdx
I'm interested in a discussion about building a high sensitivity set of
headphones.
The intended use would with a crystal radio.
The usual thing is to find an old set of Brandes headphones with 2000 ohms
DC resistance.
It seems like the collective intellect here could design or modify
existing headphones
to provide a better impedance match and increase the sensitivity of
headphones.
Maybe there are some better materials now then there were 80 years ago.
Any ideas about the ideal characteristics for a set of crystal radio
headphones?
--
MikeK
Scott Dorsey
You want an infinitely high input impedance and no shunt capacitance.
The old method was to use the finest possible wire, and as many turns of
it as possible. This results in some design issues; because your coil is
now extremely massy, you can't use a moving coil transducer design and
get good high frequency response. So most of those high-Z headphones were
variable reluctance types.
Another old method is to use a piezoelectric transducer. I think that is
still a viable method, maybe using some of the Motorola piezo horn elements.
The easiest way to do it today would be to use a modern low-z headphone
with a JFET follower in front of it. Gives you as high an input Z as you
would ever want. But, really, that would be cheating, wouldn't it?
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."
John Larkin
Or use a transformer.
John
Tim Wescott
> amdx<am...@knology.net> wrote:
>> Hi Guys,
>> I'm interested in a discussion about building a high sensitivity set of
>> headphones.
>> The intended use would with a crystal radio.
>> The usual thing is to find an old set of Brandes headphones with 2000 ohms
>> DC resistance.
>> It seems like the collective intellect here could design or modify
>> existing headphones
>> to provide a better impedance match and increase the sensitivity of
>> headphones.
>> Maybe there are some better materials now then there were 80 years ago.
>> Any ideas about the ideal characteristics for a set of crystal radio
>> headphones?
>
> You want an infinitely high input impedance and no shunt capacitance.
>
> The old method was to use the finest possible wire, and as many turns of
> it as possible. This results in some design issues; because your coil is
> now extremely massy, you can't use a moving coil transducer design and
> get good high frequency response. So most of those high-Z headphones were
> variable reluctance types.
I think some were magnetically biased, with an iron diaphragm, a
permanent magnet, and the variations in the coil acting to vary the
force on the diaphragm.
Come to think of it, a lot of the old amateur regenerative receivers
called out the head phone connection in the plate circuit of the final
audio amp (that's right! Wrap the plate supply around your head!).
This would have provided the necessary bias right there, albeit in a
manner that Ralph Nader probably would shake his head at.
> Another old method is to use a piezoelectric transducer. I think that is
> still a viable method, maybe using some of the Motorola piezo horn elements.
>
> The easiest way to do it today would be to use a modern low-z headphone
> with a JFET follower in front of it. Gives you as high an input Z as you
> would ever want. But, really, that would be cheating, wouldn't it?
Get Thee Behind Me, Satan!
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
Scott Dorsey
>On 07/22/2010 11:17 AM, Scott Dorsey wrote:
>> The old method was to use the finest possible wire, and as many turns of
>> it as possible. This results in some design issues; because your coil is
>> now extremely massy, you can't use a moving coil transducer design and
>> get good high frequency response. So most of those high-Z headphones were
>> variable reluctance types.
>
>I think some were magnetically biased, with an iron diaphragm, a
>permanent magnet, and the variations in the coil acting to vary the
>force on the diaphragm.
Yes, that's a variable reluctance headset.
>Come to think of it, a lot of the old amateur regenerative receivers
>called out the head phone connection in the plate circuit of the final
>audio amp (that's right! Wrap the plate supply around your head!).
>This would have provided the necessary bias right there, albeit in a
>manner that Ralph Nader probably would shake his head at.
Yes.
>> Another old method is to use a piezoelectric transducer. I think that is
>> still a viable method, maybe using some of the Motorola piezo horn elements.
>>
>> The easiest way to do it today would be to use a modern low-z headphone
>> with a JFET follower in front of it. Gives you as high an input Z as you
>> would ever want. But, really, that would be cheating, wouldn't it?
>
>Get Thee Behind Me, Satan!
Look for "infinite impedance detector" in the ARRL Handbook if you want to
go one step farther. You can always use a triode if you want.
Tim Williams
> Or use a transformer.
I have a few 50kohm microphone transformers, if you're interested. IIRC, the primary is 50-300-600 ohm.
Tim
--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Rich Grise
> amdx <am...@knology.net> wrote:
>> Hi Guys,
>> I'm interested in a discussion about building a high sensitivity set of
>>headphones.
>>The intended use would with a crystal radio.
> The easiest way to do it today would be to use a modern low-z headphone
> with a JFET follower in front of it. Gives you as high an input Z as you
> would ever want. But, really, that would be cheating, wouldn't it?
Well, there's cheating and then there's cheating. If there's a powerful
AM or (analog) TV station nearby, you could tune one Xtal set to that,
and just derive DC from the detected RF, then power the headphone for
your selected station by the DC that's generated by detecting ambient
RF.
Cheers!
Rich
Phil Allison
"amdx"
** Standard 32 to 100 ohm headphones can be used with a small "100 volt
line" matching transformer to increase the effective impedance to a much
higher number.
Eg: http://www.altronics.com.au/index.asp?area=item&id=M1109
Using the 0.5 watt input and with 32 ohms connected to the 8 ohm secondary
tap, the nominal input impedance is over 50 kohms.
.... Phil
amdx
--
MikeK
"Scott Dorsey" <klu...@panix.com> wrote in message
news:i2a205$8ou$1...@panix2.panix.com...
> amdx <am...@knology.net> wrote:
>> Hi Guys,
>> I'm interested in a discussion about building a high sensitivity set of
>>headphones.
>>The intended use would with a crystal radio.
>>The usual thing is to find an old set of Brandes headphones with 2000 ohms
>>DC resistance.
>> It seems like the collective intellect here could design or modify
>>existing headphones
>>to provide a better impedance match and increase the sensitivity of
>>headphones.
>>Maybe there are some better materials now then there were 80 years ago.
>> Any ideas about the ideal characteristics for a set of crystal radio
>>headphones?
>
> You want an infinitely high input impedance and no shunt capacitance.
>
> The old method was to use the finest possible wire, and as many turns of
> it as possible. This results in some design issues; because your coil is
> now extremely massy, you can't use a moving coil transducer design and
> get good high frequency response. So most of those high-Z headphones were
> variable reluctance types.
>
Variable Reluctance... my inspection of old headphones leads me to
believe
this would basically be an iron cored electromagnet with a ferous plate near
it
to vibrate. The electromagnet would have a magnet on the core to bias the
ferous plate.
I don't doubt that this is called a variable reluctance headphone (could be
wrong)
but where is the rulactance varying?
I guess the forous plate changing it's distance from the electromagnet would
change the reluctance.
Hmm... any advantage to completeing more of the magnetic path inside
the headphone? It's a long air path from the backside of the electromagnet
to the edges of the ferous plate
> Another old method is to use a piezoelectric transducer. I think that is
> still a viable method, maybe using some of the Motorola piezo horn
> elements.
>
I think that is driving a capacitor, any circuit changes to optimize for
the piezo.
> The easiest way to do it today would be to use a modern low-z headphone
> with a JFET follower in front of it. Gives you as high an input Z as you
> would ever want. But, really, that would be cheating, wouldn't it?
> --scott
Yup, it would.
MikeK
Graeme Zimmer
the "Sound Powered" Navy types. These are of "Rocking Armature" type
construction.
Google on "RCA Big Cans"
see
http://www.crystalradio.net/soundpowered/
and
http://www.crystalradio.net/soundpowered/repair/index.shtml
and
http://www.antiquewireless.org/otb/dxxtal.htm
............ Zim
Robert Baer
other super-strong magnet with space for more wire (say in 10K region)?
Robert Baer
drain..
Ian Bell
The sensitivity depends only on the electromagnetic circuit of the
phones so you will not be able to make one that is 'more sensitive' than
any currently available i.e. circa 110dB SPL per mW input. Your only
real option is a transformer to match current ones to the crystal set.
Cheers
Ian
amdx
"Ian Bell" <ruffr...@yahoo.com> wrote in message
news:i2cjl0$o33$1...@localhost.localdomain...
What is the cause of the losses?
Mike
GregS
Space.
GregS
I though crystal headphones were efficient.
k...@att.bizzzzzzzzzzzz
That makes sense. Professional headsets tend to spec 100dBA/mW.
Ian Bell
Show me some specs to back that up.
Cheers
Ian
Ian Bell
Yup, and the audiophile ones come in around 90dBA/mW
Cheers
Ian
GregS
Only that its pure capacitive, and small at that. Very
little energy required to drive.
greg
Ian Bell
That says nothing about efficiency. The only way real power can be
transferred is into the resistive part of the device. The fact that it
is 'largely capacitive' says nothing about the energy required to drive it.
Cheers
Ian
amdx
"amdx" <am...@knology.net> wrote in message
news:c99ca$4c488925$18ec6dd7$45...@KNOLOGY.NET...
KBT-33-RB-2CN.
Not a lot of info on the build but does seem to say these would be quite a
bit more sensitive than
even the best sound powered balanced-armature headphones.
http://wiki.waggy.org/dokuwiki/piezo_headphone
and the piezo element page.
http://www.avx.com/docs/masterpubs/piezo.pdf
MikeK
Angelo Campanella
> Here's a page with info using piezoelectric elements, specifically
> Kyocera KBT-33-RB-2CN.
The Kyocera data book is
http://www.avx.com/docs/masterpubs/piezo.pdf
KBT-33-RB-2CN data is found on page 20. It's about 35mm in diameter.
The impedance is capacitive, stated as 3k at 1 kHz, and less at higher
frequencies, being about 1 k at 3 kHz.
Where is the 107dB SPL measured? Perhaps not at a distance of 1m. Perhaps at
the transducer surface?
I have not seen reports of crystal receiver rectifier (CR) output
voltages... perhaps millivolts?
It's still practical to test CR circuits with the radiation from local AM
broadcast stations.
My last venture with a CR such worked OK on WOSU AM radiation, that
transmitter antenna being about two miles east of me. [I used an old 2k-ohm
magnetic headphone set, which also produces enough sound from my old B&K
2203 SLM ac-out port signal to show me whether that SLM was working OK in
the field...]
Ange
Scott Dorsey
>
>"amdx" <am...@knology.net> wrote in message
>news:4a781$4ca92512$18ec6dd7$56...@KNOLOGY.NET...
>> Here's a page with info using piezoelectric elements, specifically
>> Kyocera KBT-33-RB-2CN.
>
>The Kyocera data book is
>
>http://www.avx.com/docs/masterpubs/piezo.pdf
>
> KBT-33-RB-2CN data is found on page 20. It's about 35mm in diameter.
>
>The impedance is capacitive, stated as 3k at 1 kHz, and less at higher
>frequencies, being about 1 k at 3 kHz.
ALL of those devices have narrowband resonances all over the place. The
response looks like a hedgehog. And they have to be driven with a high-Z
source or it gets worse.
>Where is the 107dB SPL measured? Perhaps not at a distance of 1m. Perhaps at
>the transducer surface?
Probably, if the transducer is not necessarily intended to couple into air
anyway.
>It's still practical to test CR circuits with the radiation from local AM
>broadcast stations.
>
>My last venture with a CR such worked OK on WOSU AM radiation, that
>transmitter antenna being about two miles east of me. [I used an old 2k-ohm
>magnetic headphone set, which also produces enough sound from my old B&K
>2203 SLM ac-out port signal to show me whether that SLM was working OK in
>the field...]
So what's wrong with using a JFET and a low-efficiency moving coil driver?
It's hard to get efficiency and flat response at the same time.
GregS
>Angelo Campanella <a.camp...@att.net> wrote:
>>
>>"amdx" <am...@knology.net> wrote in message
>>news:4a781$4ca92512$18ec6dd7$56...@KNOLOGY.NET...
>>> Here's a page with info using piezoelectric elements, specifically
>>> Kyocera KBT-33-RB-2CN.
>>
>>The Kyocera data book is
>>
>>http://www.avx.com/docs/masterpubs/piezo.pdf
>>
>> KBT-33-RB-2CN data is found on page 20. It's about 35mm in diameter.
>>
>>The impedance is capacitive, stated as 3k at 1 kHz, and less at higher
>>frequencies, being about 1 k at 3 kHz.
>
>ALL of those devices have narrowband resonances all over the place. The
>response looks like a hedgehog. And they have to be driven with a high-Z
>source or it gets worse.
>
>>Where is the 107dB SPL measured? Perhaps not at a distance of 1m. Perhaps at
>>the transducer surface?
Typically the output would be 86 dB at three feet 1 watt.. Add a horn and add 9-10 dB.
They are driving with equivalent 8 ohm .125 watt or 1 VRMS and who knows the distance.
or complete seal. 1 VRMS draws .00036 watt at 1 kHz roughly. At 300 Hz
very efficient headphone drawing .0001 watt.
>Probably, if the transducer is not necessarily intended to couple into air
>anyway.
>
>>It's still practical to test CR circuits with the radiation from local AM
>>broadcast stations.
>>
>>My last venture with a CR such worked OK on WOSU AM radiation, that
>>transmitter antenna being about two miles east of me. [I used an old 2k-ohm
>>magnetic headphone set, which also produces enough sound from my old B&K
>>2203 SLM ac-out port signal to show me whether that SLM was working OK in
>>the field...]
>
>So what's wrong with using a JFET and a low-efficiency moving coil driver?
>It's hard to get efficiency and flat response at the same time.
>--scott
I guess its not efficient. The orginal post I think is about crystal radios.
greg
GregS
>In article <i8hv0b$agt$1...@panix2.panix.com>, klu...@panix.com (Scott Dorsey)
> wrote:
I got a no SCI.PHYSICS.ACOUSTICS
So I changed it.
Angelo Campanella
> Angelo Campanella <a.camp...@att.net> wrote:
>> KBT-33-RB-2CN data is found on page 20. It's about 35mm in diameter.
>>The impedance is capacitive, stated as 3k at 1 kHz, and less at higher
>>frequencies, being about 1 k at 3 kHz.
>
> ALL of those devices have narrowband resonances all over the place. The
> response looks like a hedgehog. And they have to be driven with a high-Z
> source or it gets worse.
True.
BUT the critical point is to pay attention to the part of the audio
spectrum that purveys the maximum speech intelligibility. We find that it is
in the vicinity of 1,000-2,000 Hz. (The exact metric is to be found
somewhere in ANSI lore where the contribution to speech intelligibility of
each 1/3 octave band from maybe 100 Hz to 10kHz is catalogued).
Most practical audio piezoelectric resonators show a resonance in the
2,000-3,000 Hz range, which suits... more or less.
The original telephone systems resonances are to be found in the 1500 Hz
area, which is a crude balance between the the chief frequency range of the
sound of the vowels (500-800Hz) and the consonants (1,000-3,000Hz). Kryter
(also a ham), in a 1960's paper showed that if the dedicated speech
purveyance bandwidth had to be only 500 Hz, that the best place was that it
be centered at 1500 Hz (old telephones). If two 500 Hz bands were available,
then they should be centered at about 1,000 Hz and 2,000 Hz respectively. If
three such 500 Hz bands were available, he suggested about 800 Hz, 1500 Hz
and, I think 2500 Hz as the center frequencies. Such info was handy to
design crystal SSB filters in days-gone-by. So I think we should use these
criteria for selecting earpieces for CR...
>>Where is the 107dB SPL measured? Perhaps not at a distance of 1m. Perhaps
>>at
>>the transducer surface?
>
> Probably, if the transducer is not necessarily intended to couple into air
> anyway.
> So what's wrong with using a JFET and a low-efficiency moving coil driver?
> It's hard to get efficiency and flat response at the same time.
Absolutely, but that may require an external power supply, unless the DC
from the crystal rectifier is sufficient to feed the drain (or source) of a
JFET....
Ange
amdx
"GregS" <zekf...@zekfrivolous.com> wrote in message
news:i8i0hp$90j$2...@usenet01.srv.cis.pitt.edu...
The object is to use no external power, then hear as many stations as
possible
with a crystal radio. So we want the most sensitive headphones we can get,
also
want high impeadance to prevent loading the resonant circuit.
There are those that receive a strong signal and use that energy to power
an amplifier. The amp is then used to amplify the audio from a weaker
station.
MikeK
GregS
I wonder what the highest Z might be. I think 600 ohm nowdays. I don't
know if it would drive a transformer. ??
amdx
headphones.
I see the datasheet for the KBT-33-RB-2CN says 2.8 kohms at 1 khz. These are
often put
two in series (one for each ear) for 5.8 kohms.
MikeK
Scott Dorsey
>> response looks like a hedgehog. And they have to be driven with a high-Z
>> source or it gets worse.
>
>spectrum that purveys the maximum speech intelligibility. We find that it is
>in the vicinity of 1,000-2,000 Hz. (The exact metric is to be found
>somewhere in ANSI lore where the contribution to speech intelligibility of
>each 1/3 octave band from maybe 100 Hz to 10kHz is catalogued).
>
> Most practical audio piezoelectric resonators show a resonance in the
>2,000-3,000 Hz range, which suits... more or less.
That's true... the voice intelligibility from those things is actually pretty
good... if anything you could argue that some of the clanginess might even
improve the ability to make words out.
> The original telephone systems resonances are to be found in the 1500 Hz
>area, which is a crude balance between the the chief frequency range of the
>sound of the vowels (500-800Hz) and the consonants (1,000-3,000Hz). Kryter
>(also a ham), in a 1960's paper showed that if the dedicated speech
>purveyance bandwidth had to be only 500 Hz, that the best place was that it
>be centered at 1500 Hz (old telephones). If two 500 Hz bands were available,
>then they should be centered at about 1,000 Hz and 2,000 Hz respectively. If
>three such 500 Hz bands were available, he suggested about 800 Hz, 1500 Hz
>and, I think 2500 Hz as the center frequencies. Such info was handy to
>design crystal SSB filters in days-gone-by. So I think we should use these
>criteria for selecting earpieces for CR...
Maybe, unless your goal is the highest possible fidelity with that Canadian
big band station on 860 KC!
>> Probably, if the transducer is not necessarily intended to couple into air
>> anyway.
>> So what's wrong with using a JFET and a low-efficiency moving coil driver?
>> It's hard to get efficiency and flat response at the same time.
>
> Absolutely, but that may require an external power supply, unless the DC
>from the crystal rectifier is sufficient to feed the drain (or source) of a
>JFET....
So this is cheating? Even if I use a lemon and some zinc and copper plates
for power?
Angelo Campanella
>>> driver?
>>> It's hard to get efficiency and flat response at the same time.
>> Absolutely, but that may require an external power supply, unless the
>> DC
>>from the crystal rectifier is sufficient to feed the drain (or source) of
>>a
>>JFET....
> So this is cheating? Even if I use a lemon and some zinc and copper
> plates
> for power?
Well we'd have to define two classes; one with air RF only; the other
using a single 1.5v cell (AA, AAA, etc)
Ange
Answerman
news:i8hse4$an7$1...@news.eternal-september.org:
>
> "amdx" <am...@knology.net> wrote in message
> news:4a781$4ca92512$18ec6dd7$56...@KNOLOGY.NET...
>> Here's a page with info using piezoelectric elements, specifically
>> Kyocera KBT-33-RB-2CN.
>
> The Kyocera data book is
>
> http://www.avx.com/docs/masterpubs/piezo.pdf
>
> KBT-33-RB-2CN data is found on page 20. It's about 35mm in diameter.
>
> The impedance is capacitive, stated as 3k at 1 kHz, and less at higher
> frequencies, being about 1 k at 3 kHz.
>
> Where is the 107dB SPL measured? Perhaps not at a distance of 1m.
> Perhaps at the transducer surface?
Neither. The KBT-33-RB-2CN transducer is designed for use as a telephone
headset receiver. The specified response is measured by a microphone in a
IEC-318 coupler.
amdx
"Answerman" <answ...@home.com> wrote in message
news:GfKdnaQP9Yq9IjLR...@giganews.com...
I noted that test on the spec. page and the coupler used in the
measurement.
Now I wonder, to get the good output do I need there specific coupler to
mount the
transducer and then mount that assembly into the headset?
And if that is so, Why don't the sell it installed in the special designed
loading coupler?
I haven't found a seller that has them in stock with a price, only RFQ
pages.
MikeK
Answerman
news:6415d$4cafc194$18ec6dd7$22...@KNOLOGY.NET:
No. You install the transducer in a headset and measure the response of
the headset on an IEC-318 coupler.
The IEC-318 coupler is a device that simulates the average acoustic
characteristis of the human ear. The microphone in the IEC-318 coupler
measures the pressure that is expected to exist at the eardrum when the
transucer is tightly coupled to a real ear. In order to get the specified
response and acoustic output, the transducer needs to be coupled tightly to
either the ear simulator or to a real ear.
> And if that is so, Why don't the sell it installed in the special
> designed loading coupler?
Because everyone has one. It's your ear.
> I haven't found a seller that has them in stock with a price, only RFQ
> pages.
The IEC-318 coupler is a design/measurement tool. It's very expensive.
http://www.aes.org/technical/documentDownloads.cfm?docID=177
http://www.gras.dk/00012/00058/00166/00172/
MikeK
Answerman
news:4a781$4ca92512$18ec6dd7$56...@KNOLOGY.NET:
If you want to build a high sensitivity, high acoustic output headphone, a
piezoelectric transducer isn't the way to do it. The sensitivity of a
piezoelectric transducer is poor compared to that of a dynamic transducer,
and distortion is high, especially at high levels. A high sensitivity
dynamic transducer such as that which is used in the Sennheiser HDA-200
audiometric headphone is what is needed. The HDA-200 produces 114dB SPL at
the eardrum at 1kHz for 25mW electrical input (1 Vrms). At maximum 500mW
input, the acoustic output would be 127dB SPL. If you want even higher
sensitivity and associated acoustic output, there's the VeriPro headphone
that is used for testing the real-ear attenuation of insert hearing
protectors. The VeriPro headphone is capable of producing 120-125 dB SPL
for 25mW electrical input (1Vrms), At maximum 500mW input, the acoustic
output of the VeriPro is 133-138dB SPL as meeasured in an IEC-318 coupler.
That's some serious acoustic output.
amdx
"Answerman" <Answ...@home.com> wrote in message
news:V9ednb6tVY51TS3R...@giganews.com...
The output from a crystal radio can be down in the microwatts,
and maybe even nanowatts if I'm understanding this page;
http://www.bentongue.com/xtalset/28SqLDtL/28SqLDtL.html
Which can be found as a link here.
http://www.bentongue.com/xtalset/xtalset.html
This is the Ben Tongue from Blonder and Tongue Labs,
he did quite a lot of fine research on crystal radio design.
MikeK
Scott Dorsey
>If you want to build a high sensitivity, high acoustic output headphone, a
>piezoelectric transducer isn't the way to do it. The sensitivity of a
>piezoelectric transducer is poor compared to that of a dynamic transducer,
>and distortion is high, especially at high levels.
They don't care about high acoustic output. They care about high efficiency
but they _mostly_ care about highest possible load impedance.
>A high sensitivity
>dynamic transducer such as that which is used in the Sennheiser HDA-200
>audiometric headphone is what is needed. The HDA-200 produces 114dB SPL at
>the eardrum at 1kHz for 25mW electrical input (1 Vrms). At maximum 500mW
>input, the acoustic output would be 127dB SPL.
That's nice, but now you need a substantial step-down in front of it, and
the consequent transformer losses. On top of which it's really hard to make
a transformer with much higher than 50k input impedance and it would be
_nice_ to have an order of magnitude higher.
Answerman
news:7efee$4cb125ed$18ec6dd7$88...@KNOLOGY.NET:
Then YOU need to be precise in defining your terms. If YOU define
transducer sensitivy as acoustic output per applied unit power, as
opposed to acoustic output per applied volt, then a piezo transducer will
almost always win because, except at its resonant frequency, its
impedance is almost purely reactive (capacitive) and virtually NO power
is consumed.
amdx
k...@att.bizzzzzzzzzzzz
If no power is consumed no work is done.
amdx
<k...@att.bizzzzzzzzzzzz> wrote in message
news:nlp4b65clf1vn9h6u...@4ax.com...
Ya, my thoughts ran hmm... "virtually NO power is consumed", virtually no
air is moved,
virtually no sound.
Anyway, I'm not finding a US source for the Kyocera KBT-33-RB-2CN
piezoelectric elements.
k...@att.bizzzzzzzzzzzz
Sorta, ya.
> Anyway, I'm not finding a US source for the Kyocera KBT-33-RB-2CN
>piezoelectric elements.
The Kyocera site doesn't recognize it either. Ate you sure of the P/N?
amdx
<k...@att.bizzzzzzzzzzzz> wrote in message
news:sb15b6ldv5u2541tl...@4ax.com...
the KBT-33-RB-2CN is the last page of the pdf file.
I don't know the relationship between AVX and Kyocera.
k...@att.bizzzzzzzzzzzz
Where?
Answerman
news:17275$4cb2f4e1$18ec6dd7$15...@KNOLOGY.NET:
In terms of piezo receiver sensitivity, you are not going to get much
better than either the Kyocera KBT-33-RB-2CN or the the MuRata PKD17EW-
01R/PKD22EW-01R. Unfortunately, they are all obsolete, and the likelihood
of finding any of them in small quantities is zip to none. The reason is
because the demand for high-sensitivity piezo receivers dried up about ten
years ago when manufacturers of cell phones realized their shortcomings for
that application (which was "the" major application). Consequently all
manufacturers of piezo receivers stopped pursuing the further development
of high sensitivity piezo receivers. In order to fabricate a high
sensitivity piezo receiver you need a paper thin ceramic element bonded to
a paper thin metal diaphragm. The fabrication of the paper thin ceramic
element is the problem and is not something that you can do in your
basement. Consequently, you can forget about building your own.
amdx
<k...@att.bizzzzzzzzzzzz> wrote in message
news:6537b6tvqni3u31on...@4ax.com...
MikeK
k...@att.bizzzzzzzzzzzz
Thanks. Looks like Kyocera bought AVX:
http://www.medibix.com/company.jsp?company_id=43673
"AVX Corp majority owned by Kyocera, ceramic capacitors, film capacitors,
tantalum, thermistors, thin film fuses, zinc oxide varistors, electronic
connector couplers, inductors, EMI Filters, thin film resistor, piezo
ceramics, clock oscillators"
Still don't see a supply anywhere.
Answerman
news:56140$4cb444d3$18ec6dd7$30...@KNOLOGY.NET:
I dug up my file on some work that I did 10-12 years ago involving piezo
receivers. Here's a short list of the piezo receivers that were available
at that time.
Manufacturer Model dia dB SPL @1Vrms
Taiyo Yuden CD15AARC 17mm 103
Taiyo Yuden CD22AARC 20mm 109
MuRata PKD33EP 33mm 110
Primo CR9-II 20mm 106
Panasonic WM70S 30mm 107
Proj.Unlim AT006 22mm 108
Kyocera KBT33RB 30mm 107
Doing a quick Google search, I came up with the following:
Panasonic WM-71A111M available at www.allelectronics.com
Panasonic WM-71A111M also available at www.surplustraders.net
Panasonic WM-71A108M available at www.surplustraders.net
I don't have data sheets for either of the WM-71xx receivers, but
surplustraders claims outputs of 106+/- 3db, impedance 2K ohm. 22mm
diameter. I suggest that you start experimenting with these. If you are
happy with the result and want to get a bit more acoustic output, I have
about ten MuRata PKD33EP receivers left over from the project work that I
did 10-12 years ago.
Answerman
news:i8spvs$70i$1...@panix2.panix.com:
> Answerman <Answ...@home.com> wrote:
>>If you want to build a high sensitivity, high acoustic output
>>headphone, a piezoelectric transducer isn't the way to do it. The
>>sensitivity of a piezoelectric transducer is poor compared to that of
>>a dynamic transducer, and distortion is high, especially at high
>>levels.
>
> They don't care about high acoustic output. They care about high
> efficiency but they _mostly_ care about highest possible load
> impedance.
Then they need to specify their source impedance beacuse 1) piezo receivers
are capacitive, and 2) increased receiver sensitivity is only achieved at
the expense of increased receiver capacitance. Consequently, in and of
itself, the sensitivity spec alone can be very misleading. If the source
impedance is very high, a low-capacitance, low-sensitivity receiver could
produce as much if not more acoustic output as a high-capacitance high-
sensitivity receiver.
Scott Dorsey
>klu...@panix.com (Scott Dorsey) wrote in
>news:i8spvs$70i$1...@panix2.panix.com:
>
>> Answerman <Answ...@home.com> wrote:
>>>If you want to build a high sensitivity, high acoustic output
>>>headphone, a piezoelectric transducer isn't the way to do it. The
>>>sensitivity of a piezoelectric transducer is poor compared to that of
>>>a dynamic transducer, and distortion is high, especially at high
>>>levels.
>>
>> They don't care about high acoustic output. They care about high
>> efficiency but they _mostly_ care about highest possible load
>> impedance.
>
>Then they need to specify their source impedance beacuse 1) piezo receivers
>are capacitive, and 2) increased receiver sensitivity is only achieved at
>the expense of increased receiver capacitance. Consequently, in and of
>itself, the sensitivity spec alone can be very misleading. If the source
>impedance is very high, a low-capacitance, low-sensitivity receiver could
>produce as much if not more acoustic output as a high-capacitance high-
>sensitivity receiver.
The source impedance depends on the station in question, but it's certainly
in the hundreds of megohms region if not higher. I know that building an
infinite impedance detector with a triode, there's a difference in performance
between a 100M leak resistor and a 680M resistor so I would expect the source
impedance to be effectively higher than that.
I believe the whole key to crystal radio performance is to get the load
Z as high as possible.
amdx
I think your estimate of hundreds of megaohms is a little high.
I think the formula for the impedance at resonance of the LC in a crystal
radio tank circuit is,
Rp=2pifLQ If we assume a very high Q of 1000 for the LC and a frequency of
1Mhz
and 240uh inductor then we get,
2 x 3.14 x 1,000,000 x .00024 x 1000 or 1,507,200 ohms of source impedance.
Please correct me if I'm wrong.
MikeK
Scott Dorsey
>
> I think your estimate of hundreds of megaohms is a little high.
>I think the formula for the impedance at resonance of the LC in a crystal
>radio tank circuit is,
>Rp=2pifLQ If we assume a very high Q of 1000 for the LC and a frequency of
>1Mhz
>and 240uh inductor then we get,
>2 x 3.14 x 1,000,000 x .00024 x 1000 or 1,507,200 ohms of source impedance.
> Please correct me if I'm wrong.
That makes sense, yes. So you're thinking about the total output impedance
as being the source impedance of the antenna in parallel with the impedance
of the tank circuit. I was thinking only about the source impedance of the
antenna itself.
amdx
And then comes the detector and a capacitor to act as a low pass filter.
I don't know how that affects the impedance, but I think it lowers it
further.
MikeK
Answerman
> amdx <am...@knology.net> wrote:
>>
>> I think your estimate of hundreds of megaohms is a little high.
>>I think the formula for the impedance at resonance of the LC in a
>>crystal radio tank circuit is,
>>Rp=2pifLQ If we assume a very high Q of 1000 for the LC and a
>>frequency of 1Mhz
>>and 240uh inductor then we get,
>>2 x 3.14 x 1,000,000 x .00024 x 1000 or 1,507,200 ohms of source
>>impedance.
>> Please correct me if I'm wrong.
>
>
> That makes sense, yes.
According to my reference books, Q=wL/R.
So, R=wL/Q, not wLQ.
amdx
> klu...@panix.com (Scott Dorsey) wrote in
> news:i972fe$fd7$1...@panix2.panix.com:
>
>> amdx <am...@knology.net> wrote:
>>>
>>> I think your estimate of hundreds of megaohms is a little high.
>>>I think the formula for the impedance at resonance of the LC in a
>>>crystal radio tank circuit is,
>>>Rp=2pifLQ If we assume a very high Q of 1000 for the LC and a
>>>frequency of 1Mhz
>>>and 240uh inductor then we get,
>>>2 x 3.14 x 1,000,000 x .00024 x 1000 or 1,507,200 ohms of source
>>>impedance.
>>> Please correct me if I'm wrong.
>>
>>
>> That makes sense, yes.
>
> According to my reference books, Q=wL/R.
> So, R=wL/Q, not wLQ.
I'm a little over my ability here but, we are calculating the impedance
at resonance of a parallel LC circuit. So the higher the Q of the circuit,
the higher the impedance is.
Your R = wL/Q is the loss in an inductor when you know Q.
MikeK
Answerman
news:d76a$4cb77d50$18ec6dd7$41...@KNOLOGY.NET:
>
> "Answerman" <Answ...@home.com> wrote in message
> news:i97tko$qtm$1...@news.eternal-september.org...
>> klu...@panix.com (Scott Dorsey) wrote in
>> news:i972fe$fd7$1...@panix2.panix.com:
>>
>>> amdx <am...@knology.net> wrote:
>>>>
>>>> I think your estimate of hundreds of megaohms is a little high.
>>>>I think the formula for the impedance at resonance of the LC in a
>>>>crystal radio tank circuit is,
>>>>Rp=2pifLQ If we assume a very high Q of 1000 for the LC and a
>>>>frequency of 1Mhz
>>>>and 240uh inductor then we get,
>>>>2 x 3.14 x 1,000,000 x .00024 x 1000 or 1,507,200 ohms of source
>>>>impedance.
>>>> Please correct me if I'm wrong.
>>>
>>>
>>> That makes sense, yes.
>>
>> According to my reference books, Q=wL/R.
>> So, R=wL/Q, not wLQ.
>
> I'm a little over my ability here but, we are calculating the
> impedance
> at resonance of a parallel LC circuit. So the higher the Q of the
> circuit, the higher the impedance is.
> Your R = wL/Q is the loss in an inductor when you know Q.
> MikeK
>
I am well aware of what you are attempting to calculate, and you are
clearly way over your ability here.
By definition, the Q of a resonant circuit is Q = X/R, where X is the
capacitive or inductive reactance at resonance.
http://www.allaboutcircuits.com/vol_2/chpt_6/6.html
Inductive reactance is wL and capacitive reactance is 1/wC. At resonance
wL=1/wC. Accordingly, for an LC resonant circuit, Q=wL/R, or equivalently
Q=1/wRC, where w is the resonant frequency of the LC circuit.
Answerman
news:VeKdnbt-3YNDOCrR...@giganews.com:
Addendum:
Your mistake is that you are equating/confusing the impedance at
resonance with resistance. The impedance at resonance is maximum when
the resistance is minumum (and vice versa). In other words, when the
resistance (loss) in the circuit is zero, there is no loss and the
impedance of the tank circuit at resonance is infinite. Conversely, when
the resistance (loss) in the circuit is high, there is considerable loss
and the impedance of the tank circuit at resonance is very low.
robert bristow-johnson
>
> > By definition, the Q of a resonant circuit is Q = X/R, where X is the
> > capacitive or inductive reactance at resonance.
> >http://www.allaboutcircuits.com/vol_2/chpt_6/6.html
i think the most fundamental definition is that Q is 2*pi times the
ratio of energy stored to energy dissipated per cycle (i think maybe
this has to be specified at the resonant frequency).
so whether it's series or parallel,
f0 = 1/(2*pi*sqrt(L*C))
for a series RLC circuit,
Q = 2*pi*f0*L/R = sqrt(L/C)/R
and for a parallel RLC circuit
Q = 2*pi*f0*C/(1/R) = sqrt(C/L)*R
>
> > Inductive reactance is wL and capacitive reactance is 1/wC. At
> > resonance wL=1/wC. Accordingly, for an LC resonant circuit, Q=wL/R,
> > or equivalently Q=1/wRC, where w is the resonant frequency of the LC
> > circuit.
>
> Addendum:
> Your mistake is that you are equating/confusing the impedance at
> resonance with resistance.
they're not the same conceptual thing, but how is it a mistake?
series: Z = R + i*(wL - 1/(wC))
parallel: 1/Z = 1/R + i*(wC - 1/(wL))
in either case, when resonance happens wL= 1/(wC) and the imaginary
term goes to zero and the remaining impedance (or admittance) is only
R (or 1/R)
so what's the mistake? conceptually, the impedance at resonance need
not be the resistance (or the reciprocals being equal in the parallel
case), but it turns out to be the case.
> The impedance at resonance is maximum when
> the resistance is minimum (and vice versa). In other words, when the
> resistance (loss) in the circuit is zero,
bad semantics going on here. when the *conductance* (1/R) in a
parallel circuit is zero (the resistance is infinite), then "there is
no loss and the impedance of the tank circuit at resonance is
infinite."
> Conversely, when
> the resistance (loss) in the circuit is high, there is considerable loss
> and the impedance of the tank circuit at resonance is very low.
better get the semantics correct, and then maybe the statement above
can make some sense.
r b-j
amdx
The only reason I care about Q is how it affects the impedance of the tank
at resonance.
If I'm clearly way over my ability here, why don't you tell me what the
impedance of
the tank is a resonance using the following information.
Frequency--1 Mhz
Q of the tank---- 1000
Inductance---240uh
and please show your work.
Thanks, MikeK
amdx
Nothing there new to me. This is why we work hard to maximize Q, by special
winding techniques, low loss materials, and using litz wire.
MikeK
Angelo Campanella
>> Your mistake is that you are equating/confusing the impedance at
>> resonance with resistance. The impedance at resonance is maximum when
>> resistance (loss) in the circuit is zero, there is no loss and the
>> impedance of the tank circuit at resonance is infinite. Conversely, when
>> the resistance (loss) in the circuit is high, there is considerable loss
>> and the impedance of the tank circuit at resonance is very low.
>
> Nothing there new to me. This is why we work hard to maximize Q, by
> special
> winding techniques, low loss materials, and using litz wire.
> MikeK
Hey Guys!
Great discussion!
Just a note; there are two "resistors" to consider...
1- The first is the "series" resistance in an LC tank circuit, which
represents the power dissipation in the circulating current, and as noted,
the grater this resistance, the grater the losses. In the case of a
piezo-resonator (call it a PZR here), the acoustic (sound) radiation is
virtually this resistor. So the more efficient this resonator is at
producing sound, the greater this "resistance", and inevitably the "lower"
the Q, and we would like that.
2a- The second is the input impedance of a device whose output terminal are
across a tank resonator. As far as the PZR is concerned, the circulating
"current" is the momentum of the vibrating PZR surface, and resonance indeed
occurs, As imagined, no dissipation, no series resistance, the impedance at
the drive terminal is infinite, and as often touted, purely resistive. I
can't find the relation between Q (set by the series resistor) and the
output impedance at resonance. But it has to be something like the series
resistance multiplied by the Q, or the like (homework for someone). In any
event, contrary to the thought that a capacitative element such as a PZR has
a purely capacitative input impedance, the input impedance at resonance can
be nearly resistive if the piezomechanical coupling efficiency is very high,
and such resistance can be relatively low if the aeromechanical coupling
efficiency to air is also very high. Perhaps there are but a few PZR devices
that can go this route.
So much for the acoustical section of a crystal receiver (CR).
2b.i-Since the only energy source there is in a CR is the RF captured by the
RF antenna, one would have to compute said energy from its free space
source, which I am not sure how to do at the moment. Microvolts per meter
field strength combined with the 377 ohms per square free space impedance
is a start. Anyway, getting back to Tanks, there are two here, the latter
being the audio tank (PZR or whatever); the former being the RF tank, the
wire coil and RF capacitance (often the stray capacitance of the coil
winding). Again, there is a series resistor and the output impedance entity
which I am getting to. First to dispose of another series resistor, this
time that of the RF coil winding. at a megahertz, the "skin effect" of the
coil wire is important. Several thinner wires bundled have less RF
resistance that a single thick wire of the same DC resistance, so "Litz"
wire was born to reduce the RF tank circuit dissipative losses. The H part
of the AM radio wave inspires RF circulating current, and we have an energy
source available to operate. We choose to impose a nonlinear device, a
diode across this tank to get some useful DC or audio energy out of the RF.
2b.ii-Again, the tank output impedance, (Q times something) is relatively
high if we want any RF selectivity, otherwise we get all AM stations in our
vicinity muddled together. But we take what we can get for now. The diode is
operated near cut-off so that its impedance if high, the nonlinear current
vs voltage characteristic produced a dc current which varies with the
amplitude of the RF H-field, and we have an audio signal thereby. The output
impedance of this "Audio" generator has to be somewhere less than the
parallel resonance shunt impedance of the RF tank, but greater than the DC
resistance of the diode at it operating point. That coves a wide range....
like somewhere between 10,000 ohms and 100 ohms...
2b.iii- To summarize, the general goal is to have on the one hand, an RF
tank of sufficient size to capture RF energy out of domestic space that can
be converted by a rectifier of conversion impedance similar to the resonant
impedance of the RF tank. At the same time, the audio or DC impedance of the
rectifier has to be similar to the input impedance of the PZR.
Finally, no one has proposed to try to use some of the available DC to
operate an audio amplifier (FET or the like).
Ange
amdx
"Angelo Campanella" <a.camp...@att.net> wrote in message
news:i99tku$lep$1...@news.eternal-september.org...
We are still working on the impedance of a parallel LC at resonance.
Yes. we do want minimize losses in both the capacitor and inductor.
I think I have a good approximation from another group.
Z(tank)~=(2*pi*F*L)^2 / RL/RC) were RL is all loss resistance in the
inductor
and RC is loss resistance in the capacitor.
I start off suggesting Q of 1500 for the capacitor, I have since found info
it
is much lower. See the last page of this pdf. http://w7zoi.net/coilq.pdf
So if we do everything very well we may get an inductor with a Q of 1000,
and a Q of 400 for our capacitor. Assuming a 240uh L and 100 pf C
with the above Qs, that calculates to;
Z(tank) = (6.28*1.027Mhz *.00024)^2 / (1.55/3.87) = Oops.
The formula looked good until I put in a more lossy capacitor.
It may be close but it not right, help!!
> 2a- The second is the input impedance of a device whose output terminal
> are across a tank resonator. As far as the PZR is concerned, the
> circulating "current" is the momentum of the vibrating PZR surface, and
> resonance indeed occurs, As imagined, no dissipation, no series
> resistance, the impedance at the drive terminal is infinite, and as often
> touted, purely resistive. I can't find the relation between Q (set by the
> series resistor) and the output impedance at resonance. But it has to be
> something like the series resistance multiplied by the Q, or the like
> (homework for someone). In any event, contrary to the thought that a
> capacitative element such as a PZR has a purely capacitative input
> impedance, the input impedance at resonance can be nearly resistive if the
> piezomechanical coupling efficiency is very high, and such resistance can
> be relatively low if the aeromechanical coupling efficiency to air is also
> very high. Perhaps there are but a few PZR devices that can go this route.
>
> So much for the acoustical section of a crystal receiver (CR).
>
Sorry, I didn't follow all that, but once we have minimized the losses in
our tank
then we start loading our tank with three things, the antenna, the detector
and the PZR
device to get audio. (often a electromechanical speaker).
I'd be curious to see a graph of reactance and resistance of a piezo
element used
as the headphone. We want high impedance to keep the bandwidth of our tank
high,
for good selectivity. We also want a very efficient device to convert
current to sound.
Oh there are those that receive a strong local and use that energy drive
an
amp to hear more distant stations. Some say it's cheating :-)
Also there is FET detector circuit that's suppossed to work well.
http://www.creative-science.org.uk/epadmosfets.html schematic at bottom
Answerman
news:b8b42$4cb83ca1$18ec6dd7$62...@KNOLOGY.NET:
First of all, let's be clear on the model, which is a series resistance
(Rs) and inductance (L) that is in parallel with a capacitance (C)to form
a parallel tank circuit. In other words, the resistive loss is assumed
to be entirely in the inductor.
For this situation Q=wL/Rs. Subsitituting your numbers for Q, frequency
and inductance yields Rs=1.5 ohms
From here there are several ways to proceed, but the simplest is to
perform a series L+Rs to parallel L//Rp transformation.
http://wcalc.sourceforge.net/parallel_rl.html
The reason for doing so is because the capaaitive and inductive
reactances vanish at resonance and all that is left is Rp, the impedance
at resonance.
For Qs>>>1, the transformation is Rp=Q^2*Rs. For Q=1000, the impedance
of the specified tank circuit at resonance is resistive and is 1.5Meg.
A p-spice simulation of the specified tank circuit confirms this result.
Putting the equiations together:
Q=wL/Rs
Q^2=Rp/Rs
Therefore Rp, the impedance of the tank circuit at resonance, is
Rp=Q^2*Rs = (wL)^2/Rs
Using your numbers
Rp=((6.28x10^6)(0.24x10^-3))^2 /1.5= 1.5Meg
Answerman
I disagree with many of your comments, but rather than going through them
one by one, let me simply state "my" bottom line. Specifically, when
dealing with a purely passive system, such as a crystal radio receiver,
one must consider the interacton among all of the components of the
system. Like it or not, acoustic transducers are specified in terms of
acoustic output per applied volt, with complete disregard for the
interaction between transducer impedance and its driving source
impedance. Consequently, a higher "pressure per applied volt"
sensitivity for a transducer does NOT necessarily translate to higher
acoustic output and higher OVERALL system sensitivity when the transducer
is installed in a passive system.
s of the system interact. can not
amdx
"Answerman" <Answerman@home> wrote in message
news:DL-dnWm6Md7HTCXR...@giganews.com...
It's been fun!
MikeK
Now about those piezos, I may ask for a quote just see what they are
thinking.
Answerman
news:da2aa$4cb8d7dd$4501214e$32...@KNOLOGY.NET:
It should be, because it's the same equation.
Rp=(wL)^2/Rs =wLQ (Q=wL/Rs)
That's the good news. The bad news is that the impedance of the tank
circuit at resonance is irrelevant bcause the impedance of the tank circuit
is not the impedance that is seen by the crystal earphone.
Szczepan Bialek
Uzytkownik "Angelo Campanella" <a.camp...@att.net> napisal w wiadomosci
news:i99tku$lep$1...@news.eternal-september.org...
> >
> 2b.i-Since the only energy source there is in a CR is the RF captured by
> the RF antenna, one would have to compute said energy from its free space
> source, which I am not sure how to do at the moment. Microvolts per meter
> field strength combined with the 377 ohms per square free space impedance
> is a start.
You also should know about the field emission of electrons from a
transmitting antenna. So the electrons are captured by RF antenna. Of course
they flow in form of pulsatile flow combined with the oscillations.
In physics are the two method: the field method and the charge method.
For the electronocs the charge method is better.
S*
Angelo Campanella
"Szczepan Bialek" <sz.b...@wp.pl> wrote in message
news:4cb95006$0$27041$6578...@news.neostrada.pl...
> You also should know about the field emission of electrons from a
> transmitting antenna.
Electrons are NOT enitted from a transmitting antenna. E/M waves re
launxhed from a transmitting antenna.
> So the electrons are captured by RF antenna. Of course they flow in form
> of pulsatile flow combined with the oscillations.
Waves are received by a reciever's Antenna.
> In physics are the two method: the field method and the charge method.
A temporary charge can be found on antenna during each voltage swing.
have nothing.
> For the electronocs the charge method is better.
Component interations are based on charge location ad intensiy.
Ange
Szczepan Bialek
>
> "Szczepan Bialek" <sz.b...@wp.pl> wrote in message
> news:4cb95006$0$27041$6578...@news.neostrada.pl...
>> You also should know about the field emission of electrons from a
>> transmitting antenna.
>
> Electrons are NOT enitted from a transmitting antenna. E/M waves re
> launxhed from a transmitting antenna.
E/M waves are in the papers. In antennas are electrons. They flow from (or
to) the ground (or chassis) and antennas.
In the cristal radio is a diode. Is in such radio the electrons flow?
>
>> So the electrons are captured by RF antenna. Of course they flow in form
>> of pulsatile flow combined with the oscillations.
>
> Waves are received by a reciever's Antenna.
But what the electrons do?
>
>> In physics are the two method: the field method and the charge method.
>
> A temporary charge can be found on antenna during each voltage swing.
> have nothing.
In transmitting antena the voltage is doubled at the end and the electrons
jump off.
>
>> For the electronocs the charge method is better.
> Component interations are based on charge location ad intensiy.
Charge is an excess or lack of electrons. The field emission/absorption is a
fact.
S*
Angelo Campanella
> But what the electrons do?
They jump into the tank to make waves.
> In transmitting antena the voltage is doubled at the end and the electrons
> jump off.
No, waves from the tank jump off the end of the antenna.
> Charge is an excess or lack of electrons. The field emission/absorption is
> a fact.
All electrons stay in the tank wire.
Ange
robert bristow-johnson
> "Szczepan Bialek" <sz.bia...@wp.pl> wrote in message
>
> news:4cbabda6$0$22803$6578...@news.neostrada.pl...
>
> > But what the electrons do?
>
> They jump into the tank to make waves.
essentially, charge sloshes back and forth in the transmitting antenna
element, which causes charge to slosh back-and-forth in the receiving
antenna element.
imagine you are holding a +1 coulomb charge and i am holding a -1
coulomb charge in my hand. we both allow movement of our charge up/
down and left/right, but we restrict the movement along the direction
of the line that connects the two of us.
since the charges are opposite sign, they attract, but we restrict
their motion so that they don't crash into each other.
now if i move my charge up, your charge will follow it up. if i move
mine down, yours will follow it down. if i move mine to my left, your
charge follows. if i move it up and down repeatedly, your charge
follows up and down repeatedly. *that* is an EM wave.
if i move it up and down a million times per second, you can tune in
the carrier on an AM radio. if i move it up and down 100 million
times per second, you can tune in it on an FM radio. if i move it up
and down 500 trillion times per second, you will see it as a blur of
orange light.
but electrons are not leaving my hand. i am just waving them around
and they have an attractive force on the positive charges in your hand
that vary as the relative position varies.
r b-j
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:a08dcbc5-1ac9-4bf0...@k22g2000yqh.googlegroups.com...
On Oct 17, 8:56 pm, "Angelo Campanella" <a.campane...@att.net> wrote:
> "Szczepan Bialek" <sz.bia...@wp.pl> wrote in message
>
>
>> They jump into the tank to make waves.
>essentially, charge sloshes back and forth in the transmitting antenna
element, which causes charge to slosh back-and-forth in the receiving
antenna element.
It take place in the original Hertz apparatus. So the receiver must be
parallel to transmitter.
>imagine you are holding a +1 coulomb charge and i am holding a -1
coulomb charge in my hand. we both allow movement of our charge up/
down and left/right, but we restrict the movement along the direction
of the line that connects the two of us.
>since the charges are opposite sign, they attract, but we restrict
their motion so that they don't crash into each other.
>now if i move my charge up, your charge will follow it up. if i move
mine down, yours will follow it down. if i move mine to my left, your
charge follows. if i move it up and down repeatedly, your charge
follows up and down repeatedly. *that* is an EM wave.
You described the electrons in the antennas. *that* is rather an electric
wave.
>if i move it up and down a million times per second, you can tune in
the carrier on an AM radio. if i move it up and down 100 million
times per second, you can tune in it on an FM radio. if i move it up
and down 500 trillion times per second, you will see it as a blur of
orange light.
>but electrons are not leaving my hand. i am just waving them around
and they have an attractive force on the positive charges in your hand
that vary as the relative position varies.
You have a magic force acting between the antennas.
In reality the electrons are leaving your hand. Electrons have mass. If your
waving (voltage) is intensive they jump off.
The father of the radio is Tesla. His radio waves are longitudinal.
According him the EM is a myth.
S*
Greegor
as good as the ancient "CANS" the military once used?
Most crystal radios nowadays are about
scientific demonstration or nostalgia, right?
Historically accurate ones are generally extremely durable.
A good modern crystal radio (AM) might be a good
thing to pack away in an emergency or survival kit
as a backup for extreme (Civil Defense)
emergencies where batteries have all run out.
To tune in FM Radio even on headphones
would inherently require some sort of
power source, correct?
For something like this, photovoltaic cells
might actually be worth their high cost, right?
(Though access to light might be a weakness.)
I actually had to sleep on a Red Cross
cot for two weeks in June of 2008 and
I can tell you that lots of people had
batteries that were dead when
needed, or ran out quickly.
amdx
We have antenna loading and detector/headphone loading.
Also to make the numbers worse I have found more realistic numbers
for the Qs.
I'm a little surprised how low the impedance at resonance is.
With a resonant frequency of 1 Mhz.
240uh inductor Q=800 XL= 1507 ohms R=1.884 ohms
105pf capacitor Q=400 XL=1507 ohms R=3.768 ohms
Z=2piFl^2/R1+R2
Z=2271652/5.652
Z=411,531 ohms
From what I can find out, the Qs I used are high but doable,
so this is a high tank impedance for a crystal radio.
MikeK
Michael A. Terrell
I had about 12 pounds of fresh alkaline cells with me when I spent a
couple weeks in a county hurricane shelter, along with two radios and
five flashlights. I hd over a month's supply of each medication, and a
basic set of tools in my truck in the parking lot. I even took my
rolling chair from my computer desk so i didn't have to sit on a folding
metal chair or emergency cot all the time. It doubled as a cart to haul
my stuff into the shelter, and out when my street was declared safe to
return to.
--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.
robert bristow-johnson
> Is there any currently produced off the shelf product
> as good as the ancient "CANS" the military once used?
>
> Most crystal radios nowadays are about
> scientific demonstration or nostalgia, right?
> Historically accurate ones are generally extremely durable.
>
> A good modern crystal radio (AM) might be a good
> thing to pack away in an emergency or survival kit
> as a backup for extreme (Civil Defense)
> emergencies where batteries have all run out.
>
> To tune in FM Radio even on headphones
> would inherently require some sort of
> power source, correct?
>
when i was really little (pre-1965) i had a crystal radio (from Knight-
Kit that was a division of Allied Radio) that worked okay for AM radio
around Fargo ND. i never did this with a crystal set, but later, when
i was a ham radio operator, we had this technique called "slope
detection" for FM that used an AM receiver and i did try it once. it
sorta worked sorta crappily.
what we did was position the received FM signal on one of the two
slopes (a.k.a. "transistion region" between passband and stopband) of
the selectivity filter (a piezo-crystal lattice network) where the
slope seemed linear. this means that the desired signal was quite
attenuated because it was on the slope and there was potentially an
interference signal that could end up right in the middle of the
passband of the selectivity filter that could potentially blow your
desired FM signal away.
do you understand how positioning the FM signal on the slope of the
selectivity filter of an AM receiver might work?
so an unpowered AM crystal receiver could conceivably pick up an FM
signal.
and one could design a crystal receiver to have two adjacent passbands
very close to each other that somehow went up and down with a two-
ganged variable capacitor. the right down-slope of the lower passband
would be aligned with the left up-slope of the higher passband. both
would have a crystal rectifier on the output and the headphones would
be wired in between the two.
r b-j
robert bristow-johnson
can you check your email/USENET client program to make sure it quotes
the poster you're responding to sensibly?
On Oct 18, 4:02 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> "robert bristow-johnson" <r...@audioimagination.com> wrotenews:a08dcbc5-1ac9-4bf0...@k22g2000yqh.googlegroups.com...
> On Oct 17, 8:56 pm, "Angelo Campanella" <a.campane...@att.net> wrote:
>
> > "Szczepan Bialek" <sz.bia...@wp.pl> wrote in message
>
> >> > But what the electrons do?
>
> > essentially, charge sloshes back and forth in the transmitting antenna
> > element, which causes charge to slosh back-and-forth in the receiving
> > antenna element.
>
> It take place in the original Hertz apparatus. So the receiver must be
> parallel to transmitter.
there are definitely issues regarding polarization of the EM wave and
the position of the antennae. for two simple dipole antennae, it's
best that they be parallel.
> > imagine you are holding a +1 coulomb charge and i am holding a
> > -1 coulomb charge in my hand. we both allow movement of our charge
> > up/down and left/right, but we restrict the movement along the
> > direction of the line that connects the two of us.
> >
> > since the charges are opposite sign, they attract, but we restrict
> > their motion so that they don't crash into each other.
> >
> > now if i move my charge up, your charge will follow it up. if i move
> > mine down, yours will follow it down. if i move mine to my left, your
> > charge follows. if i move it up and down repeatedly, your charge
> > follows up and down repeatedly. *that* is an EM wave.
>
> You described the electrons in the antennas. *that* is rather an electric
> wave.
sure, in the antenna element (of either transmitter or receiver),
which is of metal or some conductive material, there is wave motion of
electrons back and forth.
> > if i move it up and down a million times per second, you can tune in
> > the carrier on an AM radio. if i move it up and down 100 million
> > times per second, you can tune in it on an FM radio. if i move it up
> > and down 500 trillion times per second, you will see it as a blur of
> > orange light.
> >
> > but electrons are not leaving my hand. i am just waving them around
> > and they have an attractive force on the positive charges in your hand
> > that varies as the relative position varies.
>
> You have a magic force acting between the antennas.
i have a *fundamental* physical force (one of the four fundamental
forces, which, if you choose, you may call "magic" since we really
don't know the cause of the force).
> In reality the electrons are leaving your hand.
oh, a few electrons are leaving my hand now as i type. in fact,
because of quantum mechanics, i s'pose there is some non-zero
probability that some electron "orbiting" some atom in my hand is
hanging out somewhere around Mars, but i would not say that reality is
likely.
> Electrons have mass.
consider the charge-to-mass ratio of electrons.
> If your waving (voltage) is intensive they jump off.
maybe a few do (but the waving would have to be just unreal to
overcome their attachment to the object they are attached to).
nonetheless. this has nothing to do with what is causing the charge
that you are holding in your hand to move around. what is causing the
charge in your hand to move when the charge in my hand moves is the
fundamental EM force that opposite charges attract (as well as like
charges repel).
> The father of the radio is Tesla. His radio waves are longitudinal.
> According him the EM is a myth.
well, reality could be a myth. maybe none of you are real and this
experience i am having typing into a computer is just an illusion and
my brain (such as it is) is in some mad scientist's laboratory (in a
reality where there is no EM) and this mad scientist is just providing
stimulus to my consciousness that i am reacting to.
but if reality is what it appears to be, i think Coulomb (who preceded
Tesla by a little bit) established that, in reality, these charged
objects influence these other charged objects, even though we cannot
see anything of substance between the objects.
r b-j
Graeme Zimmer
signal.
If you google on "FM crystal set" you'll find umpteen using Slope Detection,
and a bunch more using Ratio and Foster-Sealy detectors.
.............. Zim
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:2164a4ab-8c6b-4060...@j18g2000yqd.googlegroups.com...
>can you check your email/USENET client program to make sure it quotes
the poster you're responding to sensibly?
I am not fluent with PC.
On Oct 18, 4:02 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>>
> >> > But what the electrons do?
>
>> > essentially, charge sloshes back and forth in the transmitting antenna
> > element, which causes charge to slosh back-and-forth in the receiving
> > antenna element.
>
>> It take place in the original Hertz apparatus. So the receiver must be
> parallel to transmitter.
>there are definitely issues regarding polarization of the EM wave and
the position of the antennae. for two simple dipole antennae, it's
best that they be parallel.
Electric waves are not polarized. The dipole is like the two loudspeakers.
The two waves are "coupled".
> > imagine you are holding a +1 coulomb charge and i am holding a
> > -1 coulomb charge in my hand. we both allow movement of our charge
> > up/down and left/right, but we restrict the movement along the
> > direction of the line that connects the two of us.
> >
>> > since the charges are opposite sign, they attract, but we restrict
> > their motion so that they don't crash into each other.
> >
>> > now if i move my charge up, your charge will follow it up. if i move
> > mine down, yours will follow it down. if i move mine to my left, your
> > charge follows. if i move it up and down repeatedly, your charge
> > follows up and down repeatedly. *that* is an EM wave.
>
>> You described the electrons in the antennas. *that* is rather an electric
> wave.
>sure, in the antenna element (of either transmitter or receiver),
which is of metal or some conductive material, there is wave motion of
electrons back and forth.
In each real wave must be the drift. "" More generally, the Stokes drift
velocity is the difference between the
average Lagrangian flow velocity of a fluid parcel, and the average
Eulerian flow velocity of the fluid at a fixed position. This nonlinear
phenomenon is named after George Gabriel Stokes, who derived expressions
for this drift in his 1847 study of water waves."
>> > if i move it up and down a million times per second, you can tune in
> > the carrier on an AM radio. if i move it up and down 100 million
> > times per second, you can tune in it on an FM radio. if i move it up
> > and down 500 trillion times per second, you will see it as a blur of
> > orange light.
> >
>> > but electrons are not leaving my hand. i am just waving them around
> > and they have an attractive force on the positive charges in your hand
> > that varies as the relative position varies.
>
>> You have a magic force acting between the antennas.
>i have a *fundamental* physical force (one of the four fundamental
forces, which, if you choose, you may call "magic" since we really
don't know the cause of the force).
Step by step the cause will be know.
>> In reality the electrons are leaving your hand.
>oh, a few electrons are leaving my hand now as i type. in fact,
because of quantum mechanics, i s'pose there is some non-zero
probability that some electron "orbiting" some atom in my hand is
hanging out somewhere around Mars, but i would not say that reality is
likely.
The field emission is not visible. But if the voltage is high enough the
sparks are reality.
>> Electrons have mass.
>consider the charge-to-mass ratio of electrons.
It is very high. For this reason the "c" is so high.
>> If your waving (voltage) is intensive they jump off.
>maybe a few do (but the waving would have to be just unreal to
overcome their attachment to the object they are attached to).
nonetheless. this has nothing to do with what is causing the charge
that you are holding in your hand to move around. what is causing the
charge in your hand to move when the charge in my hand moves is the
fundamental EM force that opposite charges attract (as well as like
charges repel).
>> The father of the radio is Tesla. His radio waves are longitudinal.
> According him the EM is a myth.
>well, reality could be a myth. maybe none of you are real and this
experience i am having typing into a computer is just an illusion and
my brain (such as it is) is in some mad scientist's laboratory (in a
reality where there is no EM) and this mad scientist is just providing
stimulus to my consciousness that i am reacting to.
>but if reality is what it appears to be, i think Coulomb (who preceded
Tesla by a little bit) established that, in reality, these charged
objects influence these other charged objects, even though we cannot
see anything of substance between the objects.
In space is ISM. Rare plazma plus dust. Ions are medium for acoustic waves
and electrons for electric.
The same is in conductor. The are also acoustic and electric.
S*
r b-j
Paul Keinanen
<gree...@gmail.com> wrote:
>Is there any currently produced off the shelf product
>as good as the ancient "CANS" the military once used?
>
>Most crystal radios nowadays are about
>scientific demonstration or nostalgia, right?
>Historically accurate ones are generally extremely durable.
>
>A good modern crystal radio (AM) might be a good
>thing to pack away in an emergency or survival kit
>as a backup for extreme (Civil Defense)
>emergencies where batteries have all run out.
>
>To tune in FM Radio even on headphones
>would inherently require some sort of
>power source, correct?
I assume this refers to some US situation in which "AM" refers to
amplitude modulation in the 0.5-1.6 MHz band and "FM" refers to
frequency modulation in the 88-108 MHz band.
The signal power from a resonant dipole is proportional to the square
of the wavelength. Thus, a dipole tuned for 1 MHz will deliver 10000
times (+40 dB) more power to the receiver compared to a 100 MHz dipole
at the same field strength.
So if you intend to use a slope (or a proper ratio) detector at the RF
frequency, be prepared to add 40 dB more audio gain in the 100 MHz
case.
robert bristow-johnson
> "robert bristow-johnson" <r...@audioimagination.com> wrotenews:2164a4ab-8c6b-4060...@j18g2000yqd.googlegroups.com...
>
> >can you check your email/USENET client program to make sure it quotes
> >the poster you're responding to sensibly?
>
> I am not fluent with PC.
>
> On Oct 18, 4:02 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
>
> > >> > But what the electrons do?
>
> >> > essentially, charge sloshes back and forth in the transmitting antenna
> >> > element, which causes charge to slosh back-and-forth in the receiving
> >> > antenna element.
>
> >> It take place in the original Hertz apparatus. So the receiver must be
> >> parallel to transmitter.
> >
> >there are definitely issues regarding polarization of the EM wave and
> >the position of the antennae. for two simple dipole antennae, it's
> >best that they be parallel.
>
> Electric waves are not polarized.
you're evidently not fluent with classical physics either. that's
okay, but then you need to listen. we're done explaining anything to
you.
r b-j
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:a070c25d-4372-4ed6...@k22g2000yqh.googlegroups.com...
In physics is also the classical optics. There is wrote that the light and
the sound are fully analogous.
Maxwell assumed that light is polarized and did proper math for it. There
the magnetic substance had mass but electricity is massles and
incompressible.
But later the electrons were discovered. They have mass and the electron gas
is compressible.
You are using electrons. It is not classical electromagnetism. Now the
antennas are polarized not waves.
One dipole is sending pulses from the two ends. Monopoles from one. Phaze
radar from many. If are more sources than one the interference works.
Do not mix the electrons with the classical EM (it was wrote by Heaviside -
Maxwell model was quite different)).
S*
Angelo Campanella
"Paul Keinanen" <kein...@sci.fi> wrote in message
news:6e1rb6d7lv6fqkrju...@4ax.com...
> On Mon, 18 Oct 2010 12:19:53 -0700 (PDT), Greegor
> <gree...@gmail.com> wrote:
>>A good modern crystal radio (AM) might be a good
>>thing to pack away in an emergency or survival kit
>>as a backup for extreme (Civil Defense)
>>emergencies where batteries have all run out.
>>
>>To tune in FM Radio even on headphones
>>would inherently require some sort of
>>power source, correct?
One thing to consider the population of nearby or powerful AM (or FM)
stations in an all-out emergency.
If it's pragmatism you want vs purity, then a solar cell or hand
(shaker) generator is the answer to get the flea-power needed to power the
average crystal radio set.
> I assume this refers to some US situation in which "AM" refers to
> amplitude modulation in the 0.5-1.6 MHz band and "FM" refers to
> frequency modulation in the 88-108 MHz band.
Correct. The RF tank for FM will need different wire coil and tuning
capacitor (both smaller) and of a greater Q for "slope detection" to work. A
relatively high RF tank Q is needed since the FM swing is about 100 kHz
while the RF fundamental is about 100 MHz. The Q should be in the
hundreds... Whether slope tuning is practical remains to be seen.
> So if you intend to use a slope (or a proper ratio) detector at the RF
> frequency, be prepared to add 40 dB more audio gain in the 100 MHz
> case.
That remains to be determined.
Ange
amdx
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:a070c25d-4372-4ed6...@k22g2000yqh.googlegroups.com...
On Oct 19, 4:16 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>>
> >can you check your email/USENET client program to make sure it quotes
> >the poster you're responding to sensibly?
>
> I am not fluent with PC.
I did some correction. Is it better now?
What do you think on this?
"Alejandro Lieber" <alej...@Use-Author-Supplied-Address.invalid> wrote
news:iac90k$57m$1...@tioat.net...
>
> I haven't read all the thread.
> I always see the diode in series between the antenna and the headphones.
Like here: http://scitoys.com/scitoys/scitoys/radio/homemade_radio.html
>
> You will get a much louder audio signal if you connect a germanium diode
> ( 1N34 ) in parallel with the phones.
> \/ antenna
> |
> |________
> | |
> -- O
> diode \/ ) phones
> -- O
> | |
> |_______|
> |
> ____
> __ ground
> -
Will it be work with the diode in reversed direction?
S*
robert bristow-johnson
...
>
> > You will get a much louder audio signal if you connect a germanium diode
> > ( 1N34 ) in parallel with the phones.
> > \/ antenna
> > |
> > |________
> > | |
> > -- O
> > diode \/ ) phones
> > -- O
> > | |
> > |_______|
> > |
> > ____
> > __ ground
> > -
>
> Will it be work with the diode in reversed direction?
as well as it works with the diode in the direction shown.
dunno how you tune a particular station with such a circuit.
r b-j
Szczepan Bialek
Uzytkownik "robert bristow-johnson" <r...@audioimagination.com> napisal w
wiadomosci
news:308d1a90-6808-432d...@l8g2000yql.googlegroups.com...
It is a simplst radio. All is wrote here: :
http://scitoys.com/scitoys/scitoys/radio/homemade_radio.html
But there the diode is always in series.
It seems that in series " electrons are leaving my hand". DC consists of
electrons.
In parallel also?
S*
P.S. Does the corrections works?
robert bristow-johnson
> Uzytkownik "robert bristow-johnson" <r...@audioimagination.com> napisal w
> On Oct 29, 3:38 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> ...
>
>
>
>
>
> > > You will get a much louder audio signal if you connect a germanium diode
> > > ( 1N34 ) in parallel with the phones.
> > > \/ antenna
> > > |
> > > |________
> > > | |
> > > -- O
> > > diode \/ ) phones
> > > -- O
> > > | |
> > > |_______|
> > > |
> > > ____
> > > __ ground
> > > -
>
> >> Will it be work with the diode in reversed direction?
> >as well as it works with the diode in the direction shown.
> >dunno how you tune a particular station with such a circuit.
>
> It is a simplst radio. All is wrote here: :http://scitoys.com/scitoys/scitoys/radio/homemade_radio.html
>
> But there the diode is always in series.
in a practical crystal radio, the diode is in series with the
headphones and behind the diode is some sorta "tank circuit" with
reactive elements. the crystal set i built when i was a kid in the
1960s had both a hand-wound coil (with a tap where the crystal diode
was attached) and a variable capacitor for tuning. i don't remember
the exact circuit, it was 45 years ago.
the antenna and tank circuit formed an output with sufficiently low
series impedance and the headphones had sufficiently high impedance
that it mattered little if the diode was conducting or not.
> It seems that in series " electrons are leaving my hand".
i cannot account for what "seems" to you, but if electrons are leaving
your hand at a sufficient rate, you would feel it. if they never
returned by a return path, charge would build up in very little time.
touch a metal doorknob or ground and you would feel it.
> DC consists of electrons.
DC is *current*. in normal conduction, it is the movement of free
electrons around a *circuit* (and i think that "circuit" means in a
circular path of some sort).
it's not DC, but has a DC component. it also has an AC component at
audio frequencies and another AC component at the carrier frequency
and its harmonics.
> P.S. Does the corrections works?
dunno what corrections. but i can read your post using Google.
what you need to correct is your misconception that the mechanics of
radio transmission and reception somehow involves electrons leaving
the transmitting antenna and traveling to the receiving antenna.
that's not how it works.
r b-j
Szczepan Bialek
On Oct 29, 12:21 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>>
>> >> Will it be work with the diode in reversed direction?
>> >as well as it works with the diode in the direction shown.
> >dunno how you tune a particular station with such a circuit.
>
>> It is a simplst radio. All is wrote here:
>> :http://scitoys.com/scitoys/scitoys/radio/homemade_radio.html
>> But there the diode is always in series.
>in a practical crystal radio, the diode is in series with the
headphones and behind the diode is some sorta "tank circuit" with
reactive elements. the crystal set i built when i was a kid in the
1960s had both a hand-wound coil (with a tap where the crystal diode
was attached) and a variable capacitor for tuning. i don't remember
the exact circuit, it was 45 years ago.
My older brother built the same 60 years ago.
>the antenna and tank circuit formed an output with sufficiently low
series impedance and the headphones had sufficiently high impedance
that it mattered little if the diode was conducting or not.
>> It seems that in series " electrons are leaving my hand".
>i cannot account for what "seems" to you, but if electrons are leaving
your hand at a sufficient rate, you would feel it. if they never
returned by a return path, charge would build up in very little time.
touch a metal doorknob or ground and you would feel it.
He feels it: " I put up a 200 foot antenna between two trees over my house,
and tuned to a 50,000 watt station about 30 miles away, and now I get 175
microamps of current through my meter."
I am interesting in direction only.
>> DC consists of electrons.
>DC is *current*. in normal conduction, it is the movement of free
electrons around a *circuit* (and i think that "circuit" means in a
circular path of some sort).
>it's not DC, but has a DC component. it also has an AC component at
audio frequencies and another AC component at the carrier frequency
and its harmonics.
Oscillations plus pulsatille flow.
>what you need to correct is your misconception that the mechanics of
radio transmission and reception somehow involves electrons leaving
the transmitting antenna and traveling to the receiving antenna.
that's not how it works.
And what about this: "The simplest radio is like a photocell. There
are the two copper plates with the semiconductor between. Electrons always
flow from the side exposed to light to side in shadow. Never in opposite.
One plate is working as an antenna and the other as the chassis."
The all is not my conception. For Helmoholtz and Tesla the radio waves are
longitudinal. Such waves exhibit the Stokes drift.
This names are more famous than Heaviside.
S*
robert bristow-johnson
well, again, you or your client software doesn't bother to quote
properly so that people reading what you post can tell who says what.
i fixed in manually in the past, i am not bothering to do it again.
On Oct 30, 3:37 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> Uzytkownik "robert bristow-johnson" <r...@audioimagination.com> napisal w
> wiadomoscinews:8143ca2a-777b-4e48...@l8g2000yql.googlegroups.com...
> On Oct 29, 12:21 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
...
> >> DC consists of electrons.
> >DC is *current*. in normal conduction, it is the movement of free
>
> electrons around a *circuit* (and i think that "circuit" means in a
> circular path of some sort).
>
> >it's not DC, but has a DC component. it also has an AC component at
>
> audio frequencies and another AC component at the carrier frequency
> and its harmonics.
>
> Oscillations plus pulsatille flow.
>
"pulsatille" is not an English word. dunno precisely what it means in
French. i get a lotta flowers when i Google the word.
> >what you need to correct is your misconception that the mechanics of
>
> radio transmission and reception somehow involves electrons leaving
> the transmitting antenna and traveling to the receiving antenna.
> that's not how it works.
>
> And what about this: "The simplest radio is like a photocell.
a radio is not only about energy conversion. it is about the wireless
conveyance of information. yes, you can use a photocell for a
receiver if the transmitter is a keyed (or modulated) lamp of some
sort.
it still has nothing to do with clearing up your misconception.
> There
> are the two copper plates with the semiconductor between. Electrons always
> flow from the side exposed to light to side in shadow. Never in opposite.
> One plate is working as an antenna and the other as the chassis."
>
> The all is not my conception.
you are fully screwed up. you know little or nothing about the simple
physics involved in radio transmission and reception at frequencies
less than, say, 300 MHz.
you think that electrons are somehow leaving the transmitting antenna
and moving to the receiving antenna. they don't. the cause and
effect that provide for the transfer of information is the movement of
electrons in the transmitting antenna element which, because of the
fundamental interaction of E&M (one of the four fundamental
interactions), cause movement of electrons in the element of the
receiving antenna.
in a semiconductor photocell, there is a totally different physics
going on where a photon sorta dislodges an electron from atoms and a
current is formed when that electron (due to the electrostatic
potential that results) tries to find its way back home.
> For Helmoholtz and Tesla the radio waves are longitudinal.
i never implied that radio waves were not transversal.
>Such waves exhibit the Stokes drift.
fluff. this has nothing to do with anything regarding a crystal
radio.
> This names are more famous than Heaviside.
more name dropping. you really have little idea of what you are
writing about.
r b-j
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:193a184b-9a5a-4f92...@j25g2000yqa.googlegroups.com...
>well, again, you or your client software doesn't bother to quote
properly so that people reading what you post can tell who says what.
i fixed in manually in the past, i am not bothering to do it again.
I must add ">" manually to your posts.
On Oct 30, 3:37 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> Uzytkownik "robert bristow-johnson" <r...@audioimagination.com> napisal w
> wiadomoscinews:8143ca2a-777b-4e48...@l8g2000yql.googlegroups.com...
>
>> Oscillations plus pulsatille flow.
>
>"pulsatille" is not an English word. dunno precisely what it means in
French. i get a lotta flowers when i Google the word.
""Unsteady flow is present in man, machine and nature. The flow of blood in
arteries and capillaries in the human body is pulsatile- composed of a mean
flow superposed with an oscillating component. The tides that wash in and
out of rivers, harbors and estuaries are unsteady flows with very long
periods of oscillation. Many engineering devices employ pulsatile and
oscillating flow. Pulsating flow is defined here as a periodic flow with a
net displacement of fluid over each flow cycle. Oscillating flow is defined
as a periodic flow with a zero mean over each cycle." From:
http://gltrs.grc.nasa.gov/cgi-bin/GLTRS/browse.pl?1995/CR-198416.html
S*
Should be pulsatory (probably). The "net displacement of fluid over each
flow cycle" is the Stokes drift.
> >what you need to correct is your misconception that the mechanics of
>
>>> radio transmission and reception somehow involves electrons leaving
> the transmitting antenna and traveling to the receiving antenna.
> that's not how it works.
>
>> And what about this: "The simplest radio is like a photocell.
>a radio is not only about energy conversion. it is about the wireless
conveyance of information. yes, you can use a photocell for a
receiver if the transmitter is a keyed (or modulated) lamp of some
sort.
>it still has nothing to do with clearing up your misconception.
>> There
> are the two copper plates with the semiconductor between. Electrons always
> flow from the side exposed to light to side in shadow. Never in opposite.
> One plate is working as an antenna and the other as the chassis."
>
>> The all is not my conception.
>you are fully screwed up. you know little or nothing about the simple
physics involved in radio transmission and reception at frequencies
less than, say, 300 MHz.
>you think that electrons are somehow leaving the transmitting antenna
and moving to the receiving antenna. they don't.
An experiment decides.
> the cause and
effect that provide for the transfer of information is the movement of
electrons in the transmitting antenna element which, because of the
fundamental interaction of E&M (one of the four fundamental
interactions), cause movement of electrons in the element of the
receiving antenna.
>in a semiconductor photocell, there is a totally different physics
going on where a photon sorta dislodges an electron from atoms and a
current is formed when that electron (due to the electrostatic
potential that results) tries to find its way back home.
>> For Helmoholtz and Tesla the radio waves are longitudinal.
>i never implied that radio waves were not transversal.
>>Such waves exhibit the Stokes drift.
>fluff. this has nothing to do with anything regarding a crystal
radio.
A cristal radio is the best tool to measure the ground current (net flow of
electrons).
>> This names are more famous than Heaviside.
>more name dropping. you really have little idea of what you are
writing about.
I was trying to ask for "net displacement of electrons over each cycle".
Without succes.
S*
robert bristow-johnson
> "robert bristow-johnson" <r...@audioimagination.com> wrotenews:193a184b-9a5a-4f92...@j25g2000yqa.googlegroups.com...
>
>
>>>> what you need to correct is your misconception that the mechanics of
>>>> radio transmission and reception somehow involves electrons leaving
>>>> the transmitting antenna and traveling to the receiving antenna.
>>>> that's not how it works.
>
> > physics involved in radio transmission and reception at frequencies
> > less than, say, 300 MHz.
> > you think that electrons are somehow leaving the transmitting antenna
> > and moving to the receiving antenna. they don't.
>
> An experiment decides.
the correct tense of the word is "decided". settled long ago. air is
an insulator. to get electrons to move through it requires ionization
of the molecules which will manifest as a very hot arch. if you ever
visit Boston, try visiting the Museum of Science and the big
Vandergraf generator there. that's when you'll see an experiment of
electrons moving through the air.
if you're thinking of becoming a physics prof or even a high school
teacher of physics, i wouldn't quit the present job, if i were you.
r b-j
robert bristow-johnson
> "robert bristow-johnson" <r...@audioimagination.com> wrotenews:a070c25d-4372-4ed6...@k22g2000yqh.googlegroups.com......
> >> Electric waves are not polarized.
> >
> >you're evidently not fluent with classical physics either. that's
> >okay, but then you need to listen.
>
> the sound are fully analogous.
oh, i missed this early misconception.
would you be willing to divulge the author and title of an optics
textbook that has written inside it that light and sound are fully
analogous? i'll put in on a list of textbooks to avoid using for any
purpose. (sound waves are longitudinal, but EM radiation are
transverse waves.)
light and radio waves are not merely analogous, they are the same
thing. the only difference is quantitative and because of that
difference, which is significant, the transmitting and receiving
apparati is qualitatively different.
there *are* some analogous technologies regarding EM and sound. some
of the techniques i had once learned in a class on antenna theory and
design (like phased arrays) can also be used for sound (wavefront
synthesis for either loudspeaker or microphone arrays).
r b-j
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:70970195-ce8f-4b05...@v20g2000yqb.googlegroups.com...
On Oct 20, 12:56 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>>
...
> >> Electric waves are not polarized.
> >
>> >you're evidently not fluent with classical physics either. that's
> >okay, but then you need to listen.
>
>> In physics is also the classical optics. There is wrote that the light
>> and
> the sound are fully analogous.
>oh, i missed this early misconception.
>would you be willing to divulge the author and title of an optics
textbook that has written inside it that light and sound are fully
analogous?
Walter Weizel, Lerhbuch der theoretischen physik.
You wrote "there *are* some analogous technologies regarding EM and sound".
I bet that you do not find even one exception.
> i'll put in on a list of textbooks to avoid using for any
purpose. (sound waves are longitudinal, but EM radiation are
transverse waves.)
For Tesla and many others the electric waves were and are longitudinal.
>light and radio waves are not merely analogous, they are the same
thing. the only difference is quantitative and because of that
difference, which is significant, the transmitting and receiving
apparati is qualitatively different.
See "free electron laser". Apparati are the same.
>there *are* some analogous technologies regarding EM and sound. some
of the techniques i had once learned in a class on antenna theory and
design (like phased arrays) can also be used for sound (wavefront
synthesis for either loudspeaker or microphone arrays).
You are close to right way.
Start with the oryginal Hertz experiment:
http://people.seas.harvard.edu/~jones/cscie129/nu_lectures/lecture6/hertz/Hertz_exp.html
"According to theory, if electromagnetic waves were spreading from the
oscillator sparks"
EM starts there from current (the place where the sparks are). But Oliver
Lodge discowered the standing wave. At the end the voltage is doubled. From
the ends the alternate electric field is emitted.
Is Hertz dipole analogous to the two Kundt's tube (loudspeaker array).
Is Weizel right?
Weizel presents the all theories and evidences (also Michelson-Gale
experiment). But without comments.
Student should think.
S*
Szczepan Bialek
>the correct tense of the word is "decided". settled long ago. air is
an insulator. to get electrons to move through it requires ionization
of the molecules which will manifest as a very hot arch.
But now we know that in space is the Interstellar Medium (IMS). Rare plasma
+dust. Plasma is a conductor
> if you ever
visit Boston, try visiting the Museum of Science and the big
Vandergraf generator there. that's when you'll see an experiment of
electrons moving through the air.
But in "vacuum" they travel freely.
I only do not know why they do not like vacuum. The prefer body. They jump
from one to the other. Why they do not stay in vacuum?
>if you're thinking of becoming a physics prof or even a high school
teacher of physics, i wouldn't quit the present job, if i were you.
I am a retired engineer. I only want to know the simplest things on
electricity. But not in the school version. There the gravity, electricity
and magnetism must be seperate. In reality the mechanism must be the same.
It is not my opinion.
S*
robert bristow-johnson
> "robert bristow-johnson" <r...@audioimagination.com> wrotenews:70970195-ce8f-4b05...@v20g2000yqb.googlegroups.com...
> On Oct 20, 12:56 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
>
>
> ...
> > >> Electric waves are not polarized.
>
> >> >you're evidently not fluent with classical physics either. that's
> >> >okay, but then you need to listen.
>
> >> In physics is also the classical optics. There is wrote that the light
> >> and the sound are fully analogous.
> >
> >oh, i missed this early misconception.
> >would you be willing to divulge the author and title of an optics
>
> textbook that has written inside it that light and sound are fully
> analogous?
>
> Walter Weizel, Lerhbuch der theoretischen physik.
if that's true (and it's likely it's not), the textbook is not to be
depended upon for factuality.
>
> You wrote "there *are* some analogous technologies regarding EM and sound".
>
> I bet that you do not find even one exception.
>
how much do you bet? i'll bet US$10000 or any lessor amount above US
$500 (the minimum is just to justify the bother). i'm serious. i can
get a credit line on our home for at least that amount (in fact, i can
just get it from a couple of credit cards). can we find a neutral
party where we can both deposit our wager? perhaps Angelo?
you know what? i'll even grant you a 2-to-1 handicap. (i will
deposit twice what you do, if you win the bet, you would get twice
what i would if i win. think of it, S, you earn an easy US$10000 and
all you have to risk is US$5000 which will be returned to you along
with what used to be my $10000, if you are so sure of what you say.)
> > i'll put in on a list of textbooks to avoid using for any
>
> purpose. (sound waves are longitudinal, but EM radiation are
> transverse waves.)
>
> For Tesla and many others the electric waves were and are longitudinal.
i won't represent (or misrepresent) Tesla, but EM radiation is
transversal. every physicist knows that (besides a few engineers).
maybe you want to check out the first sentence here:
http://en.wikipedia.org/wiki/EM_radiation#Wave_model
S, you can name-drop all you want, but you don't know shit. you are
truly what we call here in English-speaking countries a "crackpot" or
"crank". your foolishness (and naivete) is manifest for all to see.
if you had *any* idea, you would be embarrassed, but you don't even
know enough to be embarrassed about the silly things that you continue
to say.
unless you're willing to "put your money where your mouth is", i am
not willing to continue this foolishness. (but don't quit your day-
job if you're thinking of becoming a physicist or any other scientist,
or engineer for that matter.)
r b-j
robert bristow-johnson
> Uzytkownik "robert bristow-johnson" <r...@audioimagination.com> napisal w
> On Oct 30, 1:37 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
>
>
> >> An experiment decides.
> >the correct tense of the word is "decided". settled long ago. air is
> > an insulator. to get electrons to move through it requires ionization
> >of the molecules which will manifest as a very hot arch.
>
> But now we know that in space is the Interstellar Medium (IMS). Rare plasma
> +dust. Plasma is a conductor
well, let's see how much conduction you get in space. or in a vacuum
created in the lab.
>
> > if you ever
> > visit Boston, try visiting the Museum of Science and the big
> > Vandergraf generator there. that's when you'll see an experiment of
> > electrons moving through the air.
>
> But in "vacuum" they travel freely.
only if they are first freed from the matter the electrons were
originally part of (usually thermal energy is used to do that). and
when that happens, they will "freely" travel right back to that matter
(because it will be left positively charged after the electrons are
knocked off of the surface) unless there is an even greater electric
field that draws them elsewhere.
do you have any idea how a vacuum tube (what the Brits call a "valve")
works? it's a rhetorical question. of course you don't.
> I only do not know why they do not like vacuum. The prefer body. They jump
> from one to the other. Why they do not stay in vacuum?
>
> >if you're thinking of becoming a physics prof or even a high school
> > teacher of physics, i wouldn't quit the present job, if i were you.
>
> I am a retired engineer.
i hope no bridges you may have designed have fallen
( http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_%281940%29 ).
> I only want to know the simplest things on electricity.
no you don't. you want your simple misconceptions about electricity
to be reinforced. you clearly are not interested in learning
*anything* factual that contradicts what you already believe.
> But not in the school version.
so you would rather not have your understanding of reality stand up to
scrutiny?
> There the gravity, electricity
> and magnetism must be separate. In reality the mechanism must be the same.
so which is it? like-signed charges attract (as do like-signed
masses)? or do like-signed masses repel (as do like-signed charges)?
even if you're willing to accept the sign change between EM and
gravity, have you discovered any negative masses yet?
> It is not my opinion.
it is your (and some other crackpots') opinion. it has nothing to do
with reality.
r b-j
Szczepan Bialek
"robert bristow-johnson" <r...@audioimagination.com> wrote
news:df9a239d-a3b9-4971...@u10g2000yqk.googlegroups.com...
On Oct 31, 3:29 am, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
> >
>> For Tesla and many others the electric waves were and are longitudinal.
>i won't represent (or misrepresent) Tesla, but EM radiation is
transversal. every physicist knows that (besides a few engineers).
maybe you want to check out the first sentence here:
http://en.wikipedia.org/wiki/EM_radiation#Wave_model
Here is the Maxwell's model:
http://en.wikisource.org/wiki/On_Physical_Lines_of_Force
And here the Heaviside's:
http://en.wikisource.org/wiki/Electromagnetic_effects_of_a_moving_charge
The both are transversal but they are quite different. In textbooks is
Heaviside's.
The must be the all theories.
>S, you can name-drop all you want, but you don't know shit. you are
truly what we call here in English-speaking countries a "crackpot" or
"crank". your foolishness (and naivete) is manifest for all to see.
if you had *any* idea, you would be embarrassed, but you don't even
know enough to be embarrassed about the silly things that you continue
to say.
>unless you're willing to "put your money where your mouth is", i am
not willing to continue this foolishness. (but don't quit your day-
job if you're thinking of becoming a physicist or any other scientist,
or engineer for that matter.)
Is the pulsatory flow of electrons in a ground cable?
S*
Szczepan Bialek
>i hope no bridges you may have designed have fallen
( http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_%281940%29 ).
It was aircraft engines.
> I only want to know the simplest things on electricity.
>no you don't. you want your simple misconceptions about electricity
to be reinforced. you clearly are not interested in learning
*anything* factual that contradicts what you already believe.
You believe in Heaviside' model. I in Stokes, Helmholtz and Tesla.
Take a glance at them.
S*