Building better High sensitivity headphones

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.

Sorta, ya.

The Kyocera site doesn't recognize it either. Ate you sure of the P/N?

Here's a pdf from AVX that has Kyocera piezoelectric acoustic generators, the KBT-33-RB-2CN is the last page of the pdf file. I don't know the relationship between AVX and Kyocera.

Where?

"amdx" wrote in news:17275$4cb2f4e1$18ec6dd7$ snipped-for-privacy@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.

Sorry,

MikeK

Thanks. Looks like Kyocera bought AVX:

"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.

"amdx" wrote in news:56140$4cb444d3$18ec6dd7$ snipped-for-privacy@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

Panasonic WM-71A111M also available at Panasonic WM-71A108M available at

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.

snipped-for-privacy@panix.com (Scott Dorsey) wrote in news:i8spvs$70i$ snipped-for-privacy@panix2.panix.com:

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.

--scott

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

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

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.

--scott

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

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

snipped-for-privacy@panix.com (Scott Dorsey) wrote in news:i972fe$fd7$ snipped-for-privacy@panix2.panix.com:

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

"amdx" wrote in news:d76a$4cb77d50$18ec6dd7$ snipped-for-privacy@KNOLOGY.NET:

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.

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 wrote in news: snipped-for-privacy@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.

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 =3D 1/(2*pi*sqrt(L*C))

for a series RLC circuit,

Q =3D 2*pi*f0*L/R =3D sqrt(L/C)/R

and for a parallel RLC circuit

Q =3D 2*pi*f0*C/(1/R) =3D sqrt(C/L)*R

/R,

they're not the same conceptual thing, but how is it a mistake?

series: Z =3D R + i*(wL - 1/(wC))

parallel: 1/Z =3D 1/R + i*(wC - 1/(wL))

in either case, when resonance happens wL=3D 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.

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."

better get the semantics correct, and then maybe the statement above can make some sense.

r b-j

I'm not attempting to calculate Q. 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