audio recording on IC -help wanted

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[Eeyore]

Andre Majorel wrote:

Eeyore wrote:
Tman wrote:

First one, the type of connector is just like the traditional
IDE or floppy drive connector on motherboards. I need that
in 10 pin, 16 pin, etc. but that basic style. What's it
called?

It's called an IDC connector.

An "IDC connector" could just as well be a 50-way micro-ribbon
or a 3.96 mm MTA or god knows what else... The actual family
name of those .1" dual-row connectors is, as far as I know,
"HE10". That won't help you with Digi-Key but you could get
manufacturer order codes from RS Components' and Farnell's
sites.

You're kinf of making my point for me actually. Most people I know who
talkof an 'IDC connector' in the context of computers would mean one of
that type with 0.05" pitch cable. The fact that there are so many other
types merely illustrates the problem.

And then if it's PATA 66/? /100 or 133 then it needs a 0.025" cable with
interleaved ground wires.

Second type -- well you have to see a picture. Pls see
http://home.comcast.net/~tman9/connectors.jpg I need the
right thing to plug into J7, the red connector on this SBC.

Looks like it may be a Molex brand type KK.

Yep, looks like a .1" KK or MTA. The friction lock looks odd but
it may be the photograph...

I know what you mean. After looking carefully I reckoned KK. Most likely
2.54mm pitch, but you never can tell just by looking.

Graham

 

[GregS]

In article <Xns9979863DC3052zoodlewurdle@140.99.99.130>, Lostgallifreyan <no-one@nowhere.net> wrote:

Caspian <zero@nomail.com> wrote in
news:46a87ac5$0$19518$8f2e0ebb@news.shared-secrets.com:

Some domestic appliances come with rechargeable cells that need charging
for many hours before their first use.

For example, I have a Panasonic DECT phone with two AAA NiMH cells which
needs charging for 7 hours.

But maybe you can't wait that long for some reason and you simply
"must" use the appliance!

How much would cell life be reduced if such an appliance is used after,
say, only 30 to 60 minutes of charging instead of the 7 hours?


First, take care that NiMH and NiCd are different, Nicads have a 'memory
effect' that makes this unwise. You've got Nickel metal hydrides, so you
can do it. Most of the charge goes in during the first part of the charge,
the increase isn't linear.

That means you can grab a useful charge in the first hour of charge, but to
make sure the full charge is properly applied, use up that charge before
trying to recharge, then give it the full time. Don't try to guess how much
is needed if you charge from empty for an hour, use for half an hour, and
such, you'll probably overcharge the batteries.

In short, you can get a lot of capacity from a charge of as little as one
third of the time for full charge, but whatever you do, always start the
charge on a freshly discharged MiMH battery so you know where you are.

I have done the quick test of a device sometimes without even plugging it
in, or after a short charge. I don't try to drain the device, and I never
seemed to have a problem. Here is a fairly neat page on battery effects......

http://www.dansdata.com/gz011.htm

greg

 

[Engineer]

Could someone please help me. I need Hayt: Engineering
electromagnetics, 1981. 4th edition solution manual. It will help me a
lot if someone of you good people send it to me on my mail or post a
download link, torrent...
Tnx in advance!

 

[Geronimo Stempovski]

Hi folks,

I am planning to design a PCB featuring three Virtex-4 FX60 FPGAs and some
periphery. The three FPGAs are needed due to the data processing complexity
and the amount of high-speed IOs (MGTs). What I am most concerned about
right now is to find an appropriate clocking solution.

In my opinion, there are mainly three alternatives to design the clocking
scheme:

a) connection of the clock in a star-like topology, feeding each of the
three FPGAs with the same clock signal (which has to be possibly duplicated
by a clock buffer to generate three out of one clock reference signal,
thereby introducing additional jitter)

b) clock in daisy-chain, feeding each of the three FPGAs with the identical
clock signal which is routed from one device to another (in terms of jitter
this is also not an optimal solution)

c) each FPGA device is supplied with its own clock (which than can be
optimally routed to the device in short distances), but synchronization is a
major issue then

Does anyone have sufficient experience in designing clock trees and is
willing to share his experience, comments, hints and suggestions with me?

Thanks in advance

Gero

 

[Phil Allison]

"Arny Krueger"

10-4 good buddy!

** Oh - how the once high and mighty have fallen.

Arny is now a grinning, dribbling, pathetic caricature of his once proud
self.

Dementia is most cruel.

But the Good Lord knows who to punish.

Arny's days are all numbered.





........ Phil

 

[Gaetan Mailloux]

Eeyore (rabbitsfriendsandrelations@hotmail.com) writes:

Gaetan Mailloux wrote:

Hello

Anybody have those cds photoresistors; NSL-7540 and NSL7550 ?

Those sound like Silonex part numbers. They're still around. In Canada
in fact.
http://www1.silonex.com/

Graham

Hello


I did send emails to Silonex but they never reply.


Thank

Gaetan

 

[Eeyore]

Lostgallifreyan wrote:

Eeyore <rabbitsfriendsandrelations@hotmail.com> wrote
Lostgallifreyan wrote:
"clicliclic@freenet.de" <clicliclic@freenet.de> wrote

the (national) postage is given as Euro 10.00 (this excludes
insurance). This is a bit on the high side, but within reason.

National. I asked about international. He never answered. If
answering that is too hard to do, but expecting me to go outside eBay
for all 3000 is preferable then I don't want to do business with the
man. As it is, none of his accepted payment methods give ANY
protection for eBayers. Curious, no?

Within the EU AIUI, postage is supposed to be the same as national
postage to any country within the Union.

OTOH if the guy won't answer questions, I'd have nothing to do with
him.

He finally responded. Took three tries over five days but it worked. He
wants 20 euros to send to the UK. Steep. Very. But not insane. He wants
Western Union though. While they're not as dodgy as eBay say they are, it's
still true that the seller ONLY accepts payment methods that offer NO
protection to the buyer! At least with insured banknotes sent by Airsure, I
get to claim them off the post office if they go awol, and so far far he
won't accept a method than gives me ANY cover, let alone cover against
problems caused by the seller directly.

It's not a great way to do international business. If he wants to restrict
that much, he ought to have appropriately restricted his market area to the
country he's in.

A demand to use Western Union is ridiculous within the EU.

Frankly he should take Paypal IMHO.

Graham

 

[Telespalla Bob]

Hi, someone ever it replaced the HP transistor 1853-0015 with standard one?

many thanks.


--
Prima di rispondermi devi togliere le dita dal naso!!!

 

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[jimma]

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ND420625 datasheet and PDF check from
http://www.chinaicmart.com/series-ND4/ND420625.html

Description:
Powerex Dual SCR/Diode Modules are designed for use in applications
requiring phase control and isolated packaging. The modules are
isolated for easy mounting with other components on a common heatsink.
POW-R-BLOKTM has been tested and recognized by the Underwriters
Laboratories.
Features:

 Electrically Isolated Heatsinking

 Aluminum Nitride Insulator

Compression Bonded Elements

Metal Baseplate

Low Thermal Impedance
for Improved Current Capability

Quick Connect Gate Terminal
with Provision for Keyed Mating
Plug

UL Recognized
Benefits:

 No Additional Insulation
Components Required

 Easy Installation

 No Clamping Components
Required

Reduce Engineering Time
Applications:

 Bridge Circuits

 AC & DC Motor Drives

 Battery Supplies

 Power Supplies

 Large IGBT Circuit Front Ends

 

[Joerg]

legg wrote:

On Fri, 18 Jul 2008 08:03:56 -0700, Joerg
notthisjoergsch@removethispacbell.net> wrote:

legg wrote:
On Thu, 17 Jul 2008 15:24:48 -0700, Joerg
notthisjoergsch@removethispacbell.net> wrote:

spin
They have all gone to lalaland. The MMBF4391 is still made by ON and
Fairchild but hard to get. And it's still <20mmhos.

Did you confirm the gfs of an xx4391 sample? I see that the ON
datasheet shows it still climbing at 30mA, with the minimum Idss for
the part exceeding 50mA. Same might be said for the MMBFJ310, although
its minimum Idss is half that. Just curious.

I haven't, don't have one here. But in an oscillator scenario it would
become quite toasty and you'd have to load it down at the output.

Is it intended to work at VHF, or is this just an incidental
characteristic of the BF862?

Most of them are like that. For me it rarely matters, you just have to
make sure they don't "sing" up there. Else the FCC sends the goons out ;-)

Should have thought, that with cost being important, the .104 price
might have justified the effort. This would likely change, on serious
inquiry, for a multiply-sourced part that hasn't recently been
ordered. (BF862 is .16 from the same vendor's single source). The
PMBFJ310 is .095 from the same vendor, for reference.

Digikey? There they'll go to around 7c for 50k or larger qties.

--
Regards, Joerg

http://www.analogconsultants.com/

"gmail" domain blocked because of excessive spam.
Use another domain or send PM.

 

[Russell Hafter News]

In article <501887a418Spambin@argonet.co.uk>, Stuart
<Spambin@argonet.co.uk> wrote:

In article <pu12m45jj4hjudigir9vv4ih8agva479sk@4ax.com>,
SoothSayer <SaySooth@TheMonastery.org> wrote:

It would be nice to see your document posted up in
alt.binaries.schematics.electronic, if, that is, you
have access to it.

It's a 200 page book and I'm afraid I don't really have
the time. Copyright might also be an issue. One or two
pages I suppose I could manage but which ones?

My current project is going through hundreds of old
electronics magazines and scanning to PDF, stuff that I
find of interest.

Gosh, Stuart, why??

I do the exact opposite - if I get an *interesting* PDF,
then I print it and store the hard copy.

I just cannot cope with trying to read complex documents on
a screen, and the pink highlighter on the screen does not
scroll with the page, either
)

--
Russell
http://www.russell-hafter-holidays.co.uk
Russell Hafter Holidays E-mail to enquiries at our domain
Need a hotel? <http://www.hrs.de/?client=en__MT&customerId=416873103>

 

[Baron]

Andre Majorel wrote:

On 2008-07-12, Baron <baron.nospam@linuxmaniac.nospam.net> wrote:
Andre Majorel wrote:

Manufacturers tend not to give a figure for the number of
insertions their DIN connectors can take. I've never seen one
fail but those don't tend to be the type you plug/unplug ten
times a day. Does anyone have data or experience with that ?

Yes I do this far more often ! I've yet to have a plug fail,
but I have had the odd socket loose a leg from the mating
part. After that all you can do is replace the socket. Which
sadly I've had to do several times.

I see, thanks. I'm after an 8- or 9-way connector for a home
brewed polyphonic modular synth. Frequencies involved are DC to
20 kHz, 30 Vpp, input impedances on the order of 100 k.

The frequencies are not a problem for the mini DIN and I doubt that the
natural impedance is either !

Modular connectors (RJ45-style) have for them low cost and the
availability of pre-made patch cords. But when life expectancy
is specified, it is very low, on the order of dozens of cycles.

AFAIW the mini DIN is similar ! Just a few hundred !

High quality DE-9 are available but even the Canon ones are only
spec'd to a few hundred insertions at most. And there is the
problem of making a D-shaped hole in the panel.

Their expensive and the punch for the hole is around ÂŁ25 !

It better be a low-cost part because I'll need several hundreds.

Most if not all low cost connectors don't have high make/break
lifetimes !

In itself, replacing a panel connector is not a big deal if
there is a wire-to-board connector behind it. The problem is
that spotting one bad contact can take a while.

Wouldn't it be prudent to have a chat with somebody like AMP's tech
support people. Quite a while back I needed a high voltage connector
5.5Kv and they were very helpful.

--
Best Regards:
Baron.

 

[John Fields]

On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ansari@gmail.com
wrote:

On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com   or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.  

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply.

---
That may be true once the capacitor has been charged and the charging
supply has been disconnected from the capacitor, but while the
capacitor is being charged the dielectric is acting like a spring,
allowing charge to accumulate until the pressure exerted by the spring
equals the pressure exerted by the charging supply.

All else being equal, the material the spring is made of will
determine the spring rate of the spring and the amount of energy which
can be stored for any given pressure exerted to compress the spring.

The same is true for the dielectric, with its energy storage
capability being determined by the material from which it's made.

It follows, then, that if the capacitance of a capacitor is determined
by the area of the plates, the distance between them, and the relative
static permittivity (dielectric constant) of the material the
dielectric is made from, changes in dielectric constant of the
dielectric will result in changes of capacitance in otherwise
identical capacitors.

In other words,

A
C = K e0 ---
d

Where, for a single parallel plate capacitor:

C is the capacitance in farads,

K is the dielectric constant of the dielectric,

8.85E-12 coulomb˛
e0 is -------------------, the permittivity of free space,
newton-meters˛

A is the area of the surface in contact with the dielectric of
one plate of the capacitor, in square meters, and,

d is the distance between the plates (the thickness of the
dielectric) in meters.

It's easy to see, therefore, that C goes with K, everything else being
equal.
---

On the other hand it is shown in the paper that capacitance of a
capacitor depends only on the configuration of its plats not also
on the existence of any dielectric.

---
Then what's "shown" in the paper is wrong, as is the implication that
capacitors have anything to do with scale maps showing the division of
land.

JF

 

On Jul 20, 6:25 pm, John Fields <jfie...@austininstruments.com> wrote:

On Sat, 19 Jul 2008 19:45:08 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:





On Jul 20, 2:44 am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com  or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease.  

The opposite, in real life, seems to be true.

JF

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply.

---
That may be true once the capacitor has been charged and the charging
supply has been disconnected from the capacitor, but while the
capacitor is being charged the dielectric is acting like a spring,

When a dielectricless capacitor is connected to a battery it gathers
some charge. When a dielectric is inserted between the plates of
this capacitor it causes gathering more charge. It seems that
you agree that in this state the dielectrics acts as an emf supply.
But why do you think that the situation is different when
instead of a battery we use an alternative votage? In the paper
it is analysed that here too, dielectric acts as an additional
emf supply.

allowing charge to accumulate until the pressure exerted by the spring
equals the pressure exerted by the charging supply.

All else being equal, the material the spring is made of will
determine the spring rate of the spring and the amount of energy which
can be stored for any given pressure exerted to compress the spring.

The same is true for the dielectric, with its energy storage
capability being determined by the material from which it's made.

It follows, then, that if the capacitance of a capacitor is determined
by the area of the plates, the distance between them, and the relative
static permittivity (dielectric constant) of the material the
dielectric is made from, changes in dielectric constant of the
dielectric will result in changes of capacitance in otherwise
identical capacitors.

In other words,

               A
     C = K e0 ---
               d

Where, for a single parallel plate capacitor:

      C is the capacitance in farads,

      K is the dielectric constant of the dielectric,

              8.85E-12 coulomb˛
      e0 is  -------------------, the permittivity of free space,
               newton-meters˛

      A is the area of the surface in contact with the dielectric of
        one plate of the capacitor, in square meters, and,

      d is the distance between the plates (the thickness of the
        dielectric) in meters.

It's easy to see, therefore, that C goes with K, everything else being
equal.
---

On the other hand it is shown in the paper that capacitance of a
capacitor depends only on the configuration of its plats not also
on the existence of any dielectric.

---
Then what's "shown" in the paper is wrong, as is the implication that
capacitors have anything to do with scale maps showing the division of
land.

JF- Hide quoted text -

- Show quoted text -

 

[MooseFET]

On Jul 19, 11:06 pm, "cliclic...@freenet.de" <cliclic...@freenet.de>
wrote:

Hamid.V.Ans...@gmail.com schrieb:



On Jul 20, 2:44?am, John Fields <jfie...@austininstruments.com> wrote:
On Sat, 19 Jul 2008 03:56:11 -0700 (PDT), Hamid.V.Ans...@gmail.com
wrote:

Only visit:
http://hvansari.googlepages.com??or
http://www.geocities.com/hamid_vasigh

---
Your premise seems to be that if the dielectric constant of material
interposed between parallel conductors is greater than that of the
vacuum, then the energy storage in that system should decrease. ?

The opposite, in real life, seems to be true.

Dielectric attracts more charge onto the plates of the capacitor,
so it plays the role of an emf supply. On the other hand it is
shown in the paper that capacitance of a capacitor depends
only on the configuration of its plats not also on the existence
of any dielectric.

In any case, you should try to verify this revolutionary theory before
you publish it!

You might for example take one of those old AM radio tuning capacitors
with air dielectric, connect it to a capacitance meter, and observe
what happens to the reading when you immerse the capacitor in
destillated water, or in turpentine, etc. You will then discover that
your theory is contradicted by experiment, so there must be something
very wrong with it.

Calculating the capacitance of an actual ceramic capacitor according
to your theory, and comparing the result with the specification or a
meter reading might also help you see the problem. I suggest you check
the capacitance of an old single-layer disk capacitor first, and break
it open then. (For modern multilayer models you would need to study
the cross-section under a microscope.)

Even for the old single layer ones you need the microscope. The shape
of the plates of the capacitor is often very rough making the surface
area much higher than you would see with the unaided eye.

Martin.

 

[Joerg]

John Larkin wrote:

On Fri, 05 Sep 2008 12:55:09 -0700, Joerg
notthisjoergsch@removethispacbell.net> wrote:

John Larkin wrote:
On Fri, 05 Sep 2008 10:18:59 -0700, Joerg
notthisjoergsch@removethispacbell.net> wrote:

So here I am, needing more than the usual 32V opamp. Chose ye olde
MC33174 because it can take 44V but was surprised when I read that the
usual SO14 package is only rated at 154C/W. Since I am burning 500mW
that would become too toasty, ouch. Or I'll have to idle two amps each
in there and double the number of chips.

So, is 500mW in a SO14 really too much? What do thee say?
Epoxy a heat sink on top. I have a board right here on my desk that
has a heatsink on an FPGA, and two more on some dual dacs.

ftp://66.117.156.8/DSC01786.JPG

A heatsink glued onto a BGA? Wow, that takes guts.

Guts? Why? It's soldered to the board in 456 places, which should be
pretty stiff.

Yeah but if that number is reduced to 455 places some grief could set in

Or put a power pad below on the pcb layout, some thermal vias to a
plane, and use a dab of epoxy between the part and the board.

Nah, I'll go to singles but right now I am trying to knock down the
power via some inductive kicker tricks. It's a HV switch bank where
these opamps have to hold programmable biases. If I can shave off
another 500uA/channel I might just be back to a quad package. Unless
there is a really compelling reason like I am sure there is in many of
your cases I try to keep everything cool, to the point where you can
operate without heatsinks yet touch everything with screaming. Well, ok,
one should not touch this high voltage stuff ...

I've got used to touching everything to see how hot it is, or to
finagle suspected RF oscillations. And get occasionally
blistered/shocked for the trouble.

So do I. But if I get a blister I want to know why and what can be done
about the circuit to chill it.

Reminds me of a sad case. A very experienced power engineer guided a
tour for some bigshots. He was a neat freak. The whole place looked
spiffy, not a speck of dust. Then, a fly landed on a distribution rail.
Instinctively he wanted to flick it away. That was the last millisecond
of his life :-(

--
Regards, Joerg

http://www.analogconsultants.com/

"gmail" domain blocked because of excessive spam.
Use another domain or send PM.