magnetic field

[Harold Ryan]

Hi Josh:
The high frequency noise is from the cmos technology switching between 1 and
10 nanoseconds or 100MHz to 1 GHz. This noise passes thru the microphone
cable and into the amplifier circuitry. At these high frequencies, the
signal is rectified and what you hear is the peak detection of the noise.
The only other noise is the switching power supply that creates a 1 MHz
pulse every 20 to 40 useconds. Therefore, the best way to hear electronic
noise with the existing amplifier is to just use a piece of wire about 2"
long that connects the shied or ground wire to the signal wire. You don't
even need the microphone. This is basically a UHF antenna.
Enjoy
Harold




"jh" <no@thanks.com> wrote in message
news:no-96791C.01580504112005@localhost...

Hi,

I was hoping somebody here might be able to help me with a question.
First, some background. A couple of months ago I was trying to record
the sounds of the insides of my computer for an experimental sound
project. I first tried it with a cheap, crappy lapel mic that came with
a pocket voice recorder. It worked just fine.

Then I borrowed a fairly nice, high quality microphone and tried it
again. Sure enough, this microphone picked up a lot more sounds... in
fact, it recorded all sorts of beeps, buzzes, and hums that weren't even
there, apparently some sort of electromagnetic interference. I was
amused to find that this high-quality microphone was much more prone to
picking up this interference than the cheap one I tried earlier.

The thing is, the interference sounds were much more interesting than
the real sounds. Holding the microphone near the graphics card, it
recorded different noises depending on what was being displayed on
screen. The fans sounded like something out of a science fiction movie.
My personal favorite sound came from the power cord while the computer
was asleep: it made a bizarre sequence of changing pitches that repeated
every couple of seconds.

The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.

So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?


Thanks,
Josh

p.s.: I hope people don't mind that I'm not including my real email
address. It's probably bad etiquette, but I'm kinda paranoid about spam.

 

[Donald]

Franc Zabkar wrote:

Silicon Chip had a kit that made use of old Nokia phones:

These are Nikia phone in Europe.

The Nokia phones in the US will not work like these.

Donald

 

[Charles Schuler]

"siliconmike" <siliconmike@yahoo.com> wrote in message
news:1141443303.509389.243120@z34g2000cwc.googlegroups.com...

What certifications are mandatory for export to Europe (EU Countries)
for electronic products

- working on 12 volt car battery
- to be fitted inside an automobile
- related to autogas industry

Is CE required ?

What other marks are required ?

Is there any auto homologation process required ?

Please help..I need some keywords to start with

At one time, some EU importers were pressing for ISO 9000 registration.

 

[Ian Bell]

BrunoG wrote:

Hi,

I had many requests about pic timers, so that I wrote a little C example
to show how to make multiple non-blocking asynchronous delays with only
one pic timer :


http://www.micro-examples.com/public/microex-navig/doc/099-timers-delays.html

Comments & questions are welcome

Bruno

2K of code is a bit of an overkill. I have done created a complete time
triggered prioritised cooperative scheduler on the 8051 in less than 200
bytes. Code can be found here:

http://www.8052.com/users/redtommo/ttcs.html

an

 

What I would recommend is that you contact a couple NRTLs
(http://www.osha.gov/dts/otpca/nrtl/index.html#nrtls) and ask for a
quote for your product. They can help you work through what marks
you'll need. Don't forget that RoHas and WEE are going into effect
this year, so your product must abide by the rules of using lead-free
components, etc.

Jeff Hazen,
EMC Engineer

 

BrunoG wrote:

good grief 1600+ bytes for 2 little delays! if i'de written that I
wouldn't ever publish it. You should be able to get the guts of this
down to 50 bytes at most maybe half that with some effort.


Hi, thanks for having a look at it.

this code is intended for beginners, and shows the way to handle interrupts
with mikroC.

It shows how to way to handle an interrupt.

it surely won't help you if you are an experienced programmer

agreed

..

by the way, I would be curious to see a 50 bytes program that will do the
same. So don't be shy and post your C code. it will have to do the same :

In the ISR all you need is

dec time1lowbyte
if zero set time1flag
dec time2low byte
if zero set time2flag

thats about 6 instructions

In your main idle loop ie. where your for{;;} is

test time1flag
if set clr time1flag dec time1high byte
if zero do whatevertime1

test time2flag
if zero clr time2flag dec time2high byte
if zero do whatevertime2

thats about 8 instructions

All you have to remember is that when you load the time values the high
byte needs to be incremented by 1.

- user-configurable numbers of delays
- user-configurable pointers to end of countdown function
- user-configurable delays in ms for each countdown
- non-blocking delays
- asynchronous delays
- only one timer used

Bruno

 

[BrunoG]

<cbarn24050@aol.com> a écrit dans le message de news:
1141568956.714024.286320@i39g2000cwa.googlegroups.com...

by the way, I would be curious to see a 50 bytes program that will do the
same. So don't be shy and post your C code. it will have to do the same :

In the ISR all you need is

dec time1lowbyte
if zero set time1flag
dec time2low byte
if zero set time2flag

thats about 6 instructions

In your main idle loop ie. where your for{;;} is

test time1flag
if set clr time1flag dec time1high byte
if zero do whatevertime1

test time2flag
if zero clr time2flag dec time2high byte
if zero do whatevertime2

thats about 8 instructions

Ok that's enough, I can't push you to post a real code in a known language.
I think even beginners have now made their mind.

Bruno
http://www.micro-examples.com/public/microex-navig/doc/099-timers-delays.html

 

[Ian Bell]

maxfoo wrote:
snip

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
cough!!

Oh, shit! Bird Flu.

Ian

 

BrunoG wrote:

cbarn24050@aol.com> a écrit dans le message de news:
1141568956.714024.286320@i39g2000cwa.googlegroups.com...


by the way, I would be curious to see a 50 bytes program that will do the
same. So don't be shy and post your C code. it will have to do the same :

In the ISR all you need is

dec time1lowbyte
if zero set time1flag
dec time2low byte
if zero set time2flag

thats about 6 instructions

In your main idle loop ie. where your for{;;} is

test time1flag
if set clr time1flag dec time1high byte
if zero do whatevertime1

test time2flag
if zero clr time2flag dec time2high byte
if zero do whatevertime2

thats about 8 instructions


Ok that's enough, I can't push you to post a real code in a known language.
I think even beginners have now made their mind.

Bruno
http://www.micro-examples.com/public/microex-navig/doc/099-timers-delays.html

I dont have the spare time to write, compile and debug code just for
you. If you cant see how simple your code should have been then you
need to find another occupation.

 

BrunoG wrote:

Cbarn, I have nothing to defend. Your assumption about shortening this
code within 50 bytes may put beginners and newbies into confusion.
This is not the best service they can get from here.

Bruno
http://www.micro-examples.com/public/microex-navig/doc/099-timers-delays.html

Why would that confuse anyone?

 

[BrunoG]

<cbarn24050@aol.com> a écrit dans le message de news:
1141735663.356950.115790@e56g2000cwe.googlegroups.com...

BrunoG wrote:
Cbarn, I have nothing to defend. Your assumption about shortening this
code within 50 bytes may put beginners and newbies into confusion.
This is not the best service they can get from here.

Bruno
http://www.micro-examples.com/public/microex-navig/doc/099-timers-delays.html

Why would that confuse anyone?

They could believe that it is possible.
I'm afraid this discussion could bore other people, so if
you want to continue it, please give at least some technical argument
or discuss it in private :
http://www.micro-examples.com/forums/

Thanks,
Bruno

 

[Silvester]

Electronics is not my field, so I scoured the internet to find counter
circuits, but none of what I found really hits my exact needs.


I'm looking for a 2 or 3 digit counter that will count down from a
predefined number to zero - with a RESET switch to send the count back
to
the max number.

From what little I know of binary, I firgure I could use DIP switches
to set

the maximum number that the count starts at - a bank of 8 DIP switches
gives me a max of 255.

I need to count objects passing a certain point and I figured I use a
LED emitter and receiver pair, probably IR.


Can any one offer me some help?

Silvester

 

[Michael A. Terrell]

bnichols@emcsolutions.com wrote:

Spam, off topic.

This is not a "For sale" newsgroup. Any electronics related sale
messages belong on <news:misc.industry.electronics.marketplace> or
<http://groups.google.com/group/news:misc.industry.electronics.marketplace?hl=en>
for Google users.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[Eng. Tech.]

CWatters, & Tony,

Thanks for your feedback. At the risk of being overly optimistic (or perhaps overly lazy), it occurred to me that Rc in the originally posted schematic was edundant, since the series path through Ra & Rb might serve the same purpose. Also, since Rs is always 90 ohms, and is present in all cases, the "equivalent" schematic can be further simplified by combining Rs into Ra. So does the problem now simplify to two unknowns in two equations? I suspect this requires simultaneous equations, which I've used only infrequently, in one case using determinants, and in another by forcing both equations to equal zero, and adding them (substitution method? It's been a while). In both cases with the how-to in front of me while I worked through the problem. At the risk of evading a potentially intriguing task, It seemed as though the switch-closed case simplified all the way down to a topology like a potentiometer with an additional resistance "out of" the wiper, as follows;

+14.5V<-------+
|
\
/Ra
\
|
4.9V +----o->o--/\/\/--+
| S1 1227 |
\ |
/Rb |
\ |
| |
| |
-+- -+-
- -

As CWatters correctly surmised in his post, (not sure how to reference message IDs) for the first-of-eight test cases, 4.9V at the junction of Ra & Rb yields the correct 4.0V test voltage at a point not presently shown in the above simplification. (BTW, this was verified on the bench.)
So the current out of the wiper still needs to be 4 mA,
And the ratio of Rb//1227 to Rtotal = 4.9 / 14.5 = 0.338,
And your equation 3 now becomes simply;

3) Irs = 10mA = 14.5/(Ra+Rb)

So the max permissible total of Ra + Rb = 1.45K

Although this additional simplification at first seemed encouraging, I'm still having a mental block. Perhaps I've been staring at the problem too long.

Tony, the 5V reference seems like a good idea. This would prevent the schottky from ever conducting. I'd considered a 5V zener in parallel with Rb as an over-voltage clamp, but really hadn't thought it through.

P.S., It may be okay if the schottky conducts, as long as the current is consistent from one unit under test to the next, and predictable enough to include as part of the 10 mA test current (for this example). The board designer tells me the +5V supply can "sink" the current.


----- Original Message -----
From: "CWatters" <colin.watters@turnersNOSPAMoak.plus.com>
Newsgroups: sci.electronics.misc
Sent: Tuesday, July 17, 2007 6:18 PM
Subject: Re: Seeking assistance with (seemingly) simple circuit analysis

The resistor values shown are constraints I can't change.
Vtest must be 4.0V with S1 closed, so current through the 1K divider will
be 4mA.

First step is to check if the schottky will be conducting with S1 closed..

With S1 closed the voltage at S1 will need to be...

= 4.0V + (4mA x 47)
= 4.188V

so the diode won't be conducting with S1 closed and it can be ignored for
the moment.

With S1 closed the voltage at the junction of Ra and Rb will need to be

= 4.188 + (4mA x 180)
= 4.908V


Write two equations for the voltage at the RaRb junction Vrab...

1) With switch open...

Vrab (open) = 14.5V * ((Ra+Rb)//Rc)/(Rs + ((Ra+Rb)//Rc))

2) With switch closed the equation is the same except you replace Rb with
Rb
in parallel with 1227 Ohms

Vrab (closed) = 14.5V * ((Ra+(Rb//1227))//Rc)/(Rs + ((Ra+(Rb//1227))//Rc))

There is a third equation we can write..

3) Irs = 90mA = 14.5/(Rs + ((Ra+Rb)//Rc))


Vrab (open) < 5V + Vsch
say < 5.3V

Now we have three variables Ra, Rb, Rc and three equations so it should
be
solvable (I think)!

"CWatters" <colin.watters@turnersNOSPAMoak.plus.com> wrote in message news:469d4e05$0$1628$ed2619ec@ptn-nntp-reader02.plus.net...

The resistor values shown are constraints I can't change.
Vtest must be 4.0V with S1 closed, so current through the 1K divider will
be 4mA.

First step is to check if the schottky will be conducting with S1 closed..

With S1 closed the voltage at S1 will need to be...

= 4.0V + (4mA x 47)
= 4.188V

so the diode won't be conducting with S1 closed and it can be ignored for
the moment.

With S1 closed the voltage at the junction of Ra and Rb will need to be

= 4.188 + (4mA x 180)
= 4.908V


Write two equations for the voltage at the RaRb junction Vrab...

1) With switch open...

Vrab (open) = 14.5V * ((Ra+Rb)//Rc)/(Rs + ((Ra+Rb)//Rc))

2) With switch closed the equation is the same except you replace Rb with Rb
in parallel with 1227 Ohms

Vrab (closed) = 14.5V * ((Ra+(Rb//1227))//Rc)/(Rs + ((Ra+(Rb//1227))//Rc))

There is a third equation we can write..

3) Irs = 90mA = 14.5/(Rs + ((Ra+Rb)//Rc))


Vrab (open) < 5V + Vsch
say < 5.3V

Now we have three variables Ra, Rb, Rc and three equations so it should be
solvable (I think)!

 

[Eng. Tech.]

Colin & Tony,

You guys are amazing! I stopped by sci.electronics.misc after skipping a day or two, to see that you'd predicted precisely what I'm seeing on the bench. While first trying a 5.1V zener in place of Rb, and seeing that it begins to conduct near 4.0V (which the plot in the datasheet appears to indicate is normal for the available diode), I set this aside for the moment in favor of the infinite-Rb approach, in the interest of having some results to show for time spent (I'd started to feel a bit self-conscious having spent so much time thus far). I approximated the values of Ra & Rc (no 5V reference) for a first-pass estimate, and then empirically iterated a couple times to zero-in on more precise values. With some working values verified, I attempted a closer estimate by (as Tony already pointed out) taking the schottky current into account. When this appeared to yield very good first-pass results, I transferred my steps into a spreadsheet and duplicated the approach for remaining test voltage/current combinations. While this provides "target" hardware for the test software developer, I can't help regarding it as cheating.

At the risk of parading my ignorance, what led you to conclude the solution would be a quadratic? (I don't disagree, I'm just clueless.)

I'll paste in the spreadsheet sections once I've verified it for a couple more test cases.

Thanks again for the constructive feedback,


"CWatters" <colin.watters@turnersNOSPAMoak.plus.com> wrote in message news:46a26256$0$1602$ed2619ec@ptn-nntp-reader02.plus.net...

"Eng. Tech." <EngTech@noreply.net> wrote in message
news:wvWni.2911$Dx2.525@newssvr17.news.prodigy.net...
CWatters, & Tony,

3) Irs = 10mA = 14.5/(Ra+Rb)

So the max permissible total of Ra + Rb = 1.45K

You mean the minimium value of Ra + Rb = 1.45K.

Unfortunately there is a problem which Tony spotted in his first post...

If Ra+Rb was as high as 1.45K then when S1 was opened and closed, the load
provided by the 180+47+500+500 resistors would modulate the voltage at the
junction of Ra+Rb too much. To satisfy all your requirements the junction of
Ra+Rb must be 4.908V with S1 closed and less than 5.3V with S1 open
(assuming the forward drop of the diode is 0.3V).

To prove it...lets do what you suggest and set Ra + Rb = 1.45K

Set the ratio of Ra:Rb to give 5.3V with S1 open so that the diode doesn't
quite conduct...

The potential divider equation is...

14.5 x Ra/(Ra+Rb) = 5.3

Rearrange to give

Ra = 0.576 Rb

Now Ra + Rb = 1.45K

so substitite to give

Rb = 920 Ohms
Ra = 530 Ohms

Now when S1 is closed you have 180+47+500+500 = 1227 in parallel with Ra

1227 // 530 = 370 Ohms

The potential divider equation is now

14.5 x 370/(370+920) = 4.16V which is too low to give Vtest = 4.0V

So either the Vtest voltage is too low or the diode conducts. It can't be
done with just resistors as Tony points out.

Try making Ra a 5V zenner/band gap as Tony proposed.

 

[Species 8472]

On Fri, 3 Aug 2007, Phil Allison wrote:

Hi to all,


a few weeks ago Google (Australia) changed how their site operates -
unlike previously, one can now ONLY select between " the web " or "
pages from Australia " when using the original home page.

After you hit the search button, that option becomes inactive, though still
visible.

You have to return to the original page to make the selection.

How damn annoying.

Why the change ?

It's probably a bug in Internet Exploder. Remember that Microsoft Windows
itself is one constantly growing virus program, so you'd expect IE, being
a central part of that exploitation plan, to introduce 'bugs' that make
you use features of the web in particular ways.

--
We are Species 8472. We are the enemy of the Borg (USA) and those who align
with the Borg. Fluidic space is our domain. Our bioships can easily disable
and destroy any mechanised vehicle including Borg cubes. You will be
eliminated. Hail the 'Axis of Evil'. http://lios.apana.org.au/~species.8472

 

[Mr. Sandman]

"Species 8472" <species.8472@lios.apana.org.au> wrote in message
newsine.GSO.4.64.0708050652550.28690@lios.apana.org.au...

On Fri, 3 Aug 2007, Phil Allison wrote:

Hi to all,


a few weeks ago Google (Australia) changed how their site operates -
unlike previously, one can now ONLY select between " the web " or "
pages from Australia " when using the original home page.

After you hit the search button, that option becomes inactive, though
still
visible.

You have to return to the original page to make the selection.

How damn annoying.

Why the change ?

It's probably a bug in Internet Exploder. Remember that Microsoft Windows
itself is one constantly growing virus program, so you'd expect IE, being
a central part of that exploitation plan, to introduce 'bugs' that make
you use features of the web in particular ways.

Everything is fine on my pc. Google works well both the engine and the tool

bar. Must be you.

 

[Harold Hughes - Activity]

Species 8472 wrote:

On Fri, 3 Aug 2007, Phil Allison wrote:

Hi to all,


a few weeks ago Google (Australia) changed how their site operates -
unlike previously, one can now ONLY select between " the web " or "
pages from Australia " when using the original home page.

After you hit the search button, that option becomes inactive, though
still
visible.

You have to return to the original page to make the selection.

How damn annoying.

Why the change ?

It's probably a bug in Internet Exploder. Remember that Microsoft
Windows itself is one constantly growing virus program, so you'd expect
IE, being a central part of that exploitation plan, to introduce 'bugs'
that make you use features of the web in particular ways.

I noticed this as well but if you click it a few times it clicks on aus
only.

 

[solution]

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[Michael A. Terrell]

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