magnetic field

[Gaetan Mailloux]

Marra (cresswellavenue@talktalk.net) writes:

On 15 Jun, 22:54, b...@FreeNet.Carleton.CA (Gaetan Mailloux) wrote:
"Joel Kolstad" (JKolstad71HatesS...@yahoo.com) writes:
"Marra" <cresswellave...@talktalk.net> wrote in message
news:1180824052.564962.142130@o5g2000hsb.googlegroups.com...
On 2 Jun, 17:46, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
I made my own for =A320 in components and a couple of hours to write the
software.

If it's anything like your PCB design software, it'd be fair to say tha=
t it's
lacking a lot of functionality that those =A3500 units provide, yes?

There are a *lot* of "toy" USB oscilloscopes out there...

Hi

Is there any digital to analog converter that would have a minimum
sampling rate of 500 khz and that can be directly connect to a Pc computer
USB input ?

Thank

Gaetan

No.

You would need to buy a USB scope and PC software.

The alternative is getting into microcontrollers, USB and PC software.

Hi

I found a web site using a Pic but it's seem hard to program it.

Here's the link;

http://www.semifluid.com/?p=24

Gaetan

 

[Charles]

REMEMBER, IT IS
100% LEGAL!

No, it is 100% STUPID.

 

[Michael A. Terrell]

Charles wrote:

REMEMBER, IT IS
100% LEGAL!

No, it is 100% STUPID.

So is your constant replies to spambots.

--
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

 

[thesak]

http://web.mit.edu/w1mx/www/swapfest/swapfest-2007.07.pdf
http://web.mit.edu/w1mx/
www/swapfest/swapfest-2007.07.pdf
http://www.google.com/search?hl=en&q=flea+mit

 

[JeffM]

matthew.molloy@ compsys.com.au wrote:
:[entire message is in Subject: line]

You'll have to be specific.
Oscilooscopes, Time Domain Reflectometers,
or Effective Series Resistance meters?
http://groups.google.com/group/sci.electronics/msg/13651a897337a7a9?q=zzz+Charters+sci.electronics.equipment-Test-lab-*-*-*-*+the.rec.hierarchy+consumer-electronics
..
..

mm007@ computersystems.net.au
http://groups.google.com/group/comp.lang.c++/msg/23b6f2c0225de6a2?q=If-you-haven't-got-the-time-*-*-to-read-the-newsgroup-*-*+zzz+*-benefit-other-readers+Do-not-ask-for-replies-by-email

 

[Charles]

Do they have hamfests in Australia.

I have picked up some good buys at hamfests in the U.S.

 

[CWatters]

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]

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

There is a third equation we can write..

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

Sorry that should be

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

 

[Tony Williams]

In article <39Wmi.19438$Rw1.18390@newssvr25.news.prodigy.net>,
Eng. Tech. <EngTech@noreply.net> wrote:

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. Total current through 90 ohm resistor Rs must be 10
mA with S1 open. Ra + Rb = 10K, but this isn't a requirement. The
schottky shouldn't conduct when S1 is open. I only have control
over Ra, Rb, and Rc. All else is pre-existing equipment. If Ra
and Rb can set the 10 mA current, then Rc is not required. S1
could be closed for both test cases, but when open the schottky
shouldnt conduct (not sure the +5V supply can "sink" current from
7 idle channels).

The requirement that is impossible to satisfy with
just resistors is that the no-load voltage at the
junction of Ra and Rb does not exceed 5V.

All specs can be met by going sideways though.....

Rs
+14.5V<---/\/\/--+
90 |
\
860 /Ra schottky clamping
\ minimised.
| Rc | /
+--/\/\--/\/\/---+---o o--/\/\/---+---0
| 23 180 S1 47 |
| \
\_|_ /500
5V band gap ref /_\ "Rb" \
| |
| +---0
-+- |
- \
Use a precision 5V shunt regulator instead of Rb. /500
\
|
-+-
Ra sets the req'd constant 10mA, S1 open or closed.
The 23 ohm Rc trims the output current to 4mA.
Schottky clamping is minimised.

Schottky clamping will happen whenever the new "Rb"
voltage is above the internal 5V. However, the
voltage differences are small and the resultant current
is limited by the 23+180 ohms connecting resistance.
eg, A 5% tolerance difference results in only 1.23mA of
clamping current.

This shows the first of 8 test cases. Next will be 3.5V and 15
mA, 3.0V and 20 mA, etc.

Simple changes to Ra and Rc?

--
Tony Williams.

 

[Tony Williams]

In article <wvWni.2911$Dx2.525@newssvr17.news.prodigy.net>,
Eng. Tech. <EngTech@noreply.net> wrote:
[snip]

And your equation 3 now becomes simply;

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

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

No, you also have to add in the current drawn by the
Schottky clamp when S1 is open.

A slightly better resistive circuit is to make
Rb= infinity and just work with Ra and Rc.
The calculated values for Ra and Rc below assumed
an exact 5V and Vdiode= 0.2V.

90
14.5V---/\/\---+----+
| |
Rc \ \Ra -+-5V
2119.2/ /2164 _|_
\ \ /_\ SD, 0.2V Fwd drop.
| | |
| | 180 | S1 47
| +---/\/\-+-+/+--/\/\--+-->Vout
| |
| [500+500]
| |
-------------+--------------------------+----

With S1 closed there is 4mA of output current and
about 10.4mA load current through the 90 ohm.

With S1 open there is 10mA of load current through
the 90 ohm, and about 3.58mA Schottky current.

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.

Whatever the simplification the sums end in a quadratic
equation with one sensible root. There are *probably*
values for Ra/b/c that gets the 10mA with S1 open or
closed, and the 4mA with S1 closed, but also probably
at a higher Schottky current.

Tony, the 5V reference seems like a good idea.

The low output resistance of a 5V reference is the thing
that removes the interaction between output current and
load current. It also produces minimum Schottky current.

--
Tony Williams.

 

[CWatters]

"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.

 

[Tony Williams]

In article <mefpi.17778$rL1.11688@newssvr19.news.prodigy.net>,
Eng. Tech. <EngTech@noreply.net> wrote:

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.)

Here goes, although the low output resistance band
gap reference is still the easier/better way to go.

Lucky I didn't throw the scribbles away. Please note
that I am also known as Tony (Typo) Williams.

-------------------------------------------------------

1. S1 closed, current in the 1227 must be 4mA.

90 Rx
14.5V---/\/\---+----+ Vx--/\/\----+
| | |
Rc \ \Ra \Ra
/ / /
\ \ \
| | -----> |
| | |
| \ \
| /1227 /1227
| \ \
| | |
-------------+----+ -------------+

The 90+Rc attentuator can be redrawn to an equivalent
Vx and Rx.

Vx = 14.5*Rc/(90+Rc), and Rx = 90*Rc/(90+Rc).

The three resistors in series must draw 4mA from Vx.

So Vx/(Rx + Ra + 1227) = 0.004 amps.

Do some algebra on that and you should get.....

Rc = (90.Ra + 110430)/(2308 - Ra) ....... eqn.1
-------------------------------------------------------

2. S1 open, current through the 90 ohm must be 10mA.

90
14.5V---/\/\---+----+ <--- 13.6V when I90= 10mA.
| |
Rc \ \Ra -+-5V
/ / _|_
\ \ /_\ SD, 0.2V Fwd drop.
| | |
| | 180 |
| +---/\/\-+ <-- 5.2V estimated.
|
|
|
-------------+

13.6/Rc + (13.6 - 5.2)/(Ra + 180) = 0.01 amps.

Expand and again reduce to Rc in terms of Ra.....

Rc = 1360*(Ra + 180)/(Ra - 660) ....... eqn.2
-------------------------------------------------------

Combine equations......

(90.Ra+110430)/(2308 - Ra) = 1360*(Ra+180)/(Ra-660)

Flog through yet more algebra to get.......

Ra^2 - 1960.7*Ra - 439919 = 0.

There's the quadratic. (b^2 - 4ac) is positive.
Sensible root is Ra = 2164. Rc then also falls out.
-------------------------------------------------------

--
Tony Williams.

 

On Jul 30, 6:29 pm, Nutrition.t...@gmail.com wrote:

check out Future Mobile Phonehttp://mobilephone-guide.blogspot.com/2007/07/future-mobile-phone.html

Here are more links about mobile phones, that might be interesting:
http://www.hooqs.com
http://www.1888express.com
http://www.cell4you.com

 

[www.21cn-shoes.com]

http://www.21cn-shoes.com
21cn-shoes Co.,Ltd is the WORLD'S LARGEST SUPPLIER OF FAMOUS BRAND'S
GOODS.
we are one of the leading company that wholesales and retail brand
products,
We clome to be in 21cn-shoes.com, all of our products are
exclusive(top quality),
and we are an hoest and real companies offering such a wide variety of
Famous Brand Goods inc:
1)Brand Shoes:Nike Shox,AirMax,AirJordan,nike air force one,nike
dunk,nike
kobe,nike james,Nike sandle shoes and other brand shoes inc
Bape,Adidas,Puma,Gucci,
Timberland Prada,Lv,4us,Richmond,Ice cream,Diesel,Chanel
D&G,DSQUARED,etc.
2)Jeans of Diesel,Rock&Republic,Seven,red monkey,Evisu,lee,D&G,True
religion,bape,antik,
jack
jones,armani,kepasa,apple,bbc,levi's,guess,cocobongo,only,replay,on
line,MNG,von dutch
and cocolulu.
3)Brand T-
shirt olo,lacoste,burberry,boss,Tommy,BBC,Bape,Gucci,D&G,A&F,Versach,marlbolo.
4)brand handbag with LV,
Chanel,Gucci,Fendi,Chole,hermes,Dior,coach,balenciaga.
we also sell world brand watches and caps and ipod nano.All our
products are in best quality with lowest price.
welcome to contact with us. www.21cn-shoes.com
email: shoes21cn@yahoo_com
msn: shoes21cn@hotmail_com
wholesale cheap Dunk SB sneakers
wholesale cheap air force ones

wholesale cheap air jordan sneakers from china

wholesale cheap gucci sneakers from
china

cheap nike max 90 from china

china wholesale jordan sneakers
cheap nike sneakers from china
china wholesale jordan 11 sneakers [/
URL]
[URL=http://www.21cn-shoes.com]jordan sneakers from china
wholesale gucci sneakers from china [/
URL]
[URL=http://www.21cn-shoes.com]wholesale gucci sneakers from china [/
URL]
[URL=http://www.21cn-shoes.com]china wholesale gucci sneakers
cheap nike tn from china
china wholesale nike sneakers
cheap nike max 90 from china
china air nike 2004 shoes wholesale
price
cheap gucci shoes from china
china wholesale nike shox sneakers [/
URL]
[URL=http://www.21cn-shoes.com]china wholesale jordan sneakers
cheap jordan sneakers from china
china wholesale nikes sneakers
china wholesale jordans sneakers
cheap jordan sneakers

china wholesale nike sneakers
cheap lacoste shoes
jordans sneakers china wholesale


www.21cn-shoes.com wholesale D&G shoes,4us,dior,DSQUARED,and
chanel,prada gucci shoes

 

I need 2nd edition solution manual of elements of chemical reaction
engineering (not 3rd edition.)

Thank you.

 

I need 2nd edition solution manual of elements of chemical reaction
engineering (not 3rd edition.)

Thank you.

 

[Steven]

On Sep 24, 12:59 pm, "Michael A. Terrell" <mike.terr...@earthlink.net>
wrote:

" wrote:

"Green Xenon [Radium]" <gluceg...@excite.com> wrote in message
news:46f6d850$0$11068$4c368faf@roadrunner.com...
Hi:

One major reason the luminance signals of television are broadcasted on an
AM-carrier instead of FM is because FM requires large amounts of
bandwidth. Is there a way to use FM video without hogging so much
bandwidth?

Quotes from
http://groups.google.com/group/sci.electronics.basics/msg/0c013cf5371...:

Multiple-level quadrature modulation,
"constellation modulation",
is most common for packing
lots of bits per Hz of bandwidth.
The more you pack,
the better the s/n ratio has to be.

http://en.wikipedia.org/wiki/Quadrature_amplitude_modulation

http://en.wikipedia.org/wiki/Constellation_diagram

Does this mean that Quadrature Modulation and Constellation Modulation
can -- at least in theory -- be applied to FM video so that excessive
bandwidth is not needed? If so, then what would be the minimum
radio-frequency required to transmit the video signal?

No.

These are like special modified forms of AM.

You can only use these to reduce bandwidth of an FM signal in the same way
that you can use AM to reduce the bandwidth of an FM signal. It just won't
be FM any more.

radium is a troll. he tries this crap on a lot ot different
newsgroups, and asks stupid questions to entertain himself. If you
don't believe me. look at his posting history on Google Groups, then
kill file him.

--
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

We're all trolls, damn you, least the exceptional you.

 

[Caine]

After looking through the available data sheets for Model 127 on
http://www.simpsonelectric.com, I think I may have found something...

While the specs may have changed slightly or even significantly over
the years, I did notice something of interest...

The unit currently described in the
http://www.simpsonelectric.com/uploads/Files/datasheets/rectangular_datasheet.pdf
describes the following:

DC Millivoltmeters, Self Shielding Movement
Option 1:
Range: 0 - 50
Approx. Impedance (Ohnms) @ 60 Hz: 10
Model/Size and Catalog Number - 2-1/2" Model 127: 06990

Option 2:
Range: 0 - 100
Approx. Impedance (Ohnms) @ 60 Hz: 20
Model/Size and Catalog Number - 2-1/2" Model 127: Available on special order.

With regards to the unit I recently procured, it would seem to have
some unique design features that may not necessarily be available on
current models offered by Simpson Electric, Co.

The meter in my possession is of the 2-1/2" variety and is accompanied
by four (hardwired) test leads attached to four connecting posts on the
back of the meter.

According to the accompanying detail within the lid of the unit's case...

There is a plain (common) wire. This would have been connected to the
inner thermocouple lead or white powerpile lead.

There is a black (0-50 MV) lead. This would have been the outer
thermocouple lead.

There is a red (0-500 MV lead. This, in conjunction with the yellow
(meter) lead would be connected to the red powerpile lead.

There is a yellow (0-1000MV) lead. This, in conjunction with the red
(meter) lead would be connected to the red powerpile lead.

From the detail listed on the underside of the lid, this particular
unit seemed to have been used for diagnostics with regards to the
thermocouple and/or powerpile of a boiler (based on my google
research). Other part numbers in the printed on the detail seem to
reference various (deprecated) Honeywell thermocouples and powerpiles.

Once again, I would be grateful for any additional information that
anyone may have regarding this unit.


On 2008-02-20 14:54:21 -0800, Caine <caine@sonic.net> said:

I found additional information on the meter itself.

Patent Numbers: 2,051,399 & 2,433,165, Simpson Electric Co. Chicago, IL
USA Model 127

Any information you can provide would be greatly appreciated.

On 2008-02-20 11:51:44 -0800, Caine <caine@sonic.net> said:


I have recently acquired a working, antique Simpson /
Minneapolis-Honeywell millivoltmeter, model W129A2X1.
This unit is extremely old and I have had a difficult time tracking
down any information pertaining to the unit.

On the inside of the case is a condensed chart with reference to
thermocouple and powerpile readings -- I would presume this unit was
designed to work with a boiler.

I am hoping someone here may have or know of somewhere/someone that I
can obtain an operator's manual for this unit -- if there is such a
thing.

I have recently contacted both Simpson Electric and Honeywell, but have
yet to hear back from either regarding this interesting device.

As you can see from the attached photos, the casing is somewhat
delapidated, but is still in a fairly sturdy condition. I took the
liberty of dismantling the unit in order to replace the four lead wires
as the originals were basically crumbling apart. I do not know whether
or not the supplied aligator clips are the originals or not. On the
bottom of each, they sport the following imprinting:

Mueller Elec. Co.
Cleveland Ohio U.S.A.
60 Series

Any information provided will be greatly appreciated.

Thanks in advance!

image

--
Caine
caine@sonic.net

 

[Caine]

I found additional information on the meter itself.

Patent Numbers: 2,051,399 & 2,433,165, Simpson Electric Co. Chicago, IL
USA Model 127

Any information you can provide would be greatly appreciated.

On 2008-02-20 11:51:44 -0800, Caine <caine@sonic.net> said:

I have recently acquired a working, antique Simpson /
Minneapolis-Honeywell millivoltmeter, model W129A2X1.
This unit is extremely old and I have had a difficult time tracking
down any information pertaining to the unit.

On the inside of the case is a condensed chart with reference to
thermocouple and powerpile readings -- I would presume this unit was
designed to work with a boiler.

I am hoping someone here may have or know of somewhere/someone that I
can obtain an operator's manual for this unit -- if there is such a
thing.

I have recently contacted both Simpson Electric and Honeywell, but have
yet to hear back from either regarding this interesting device.

As you can see from the attached photos, the casing is somewhat
delapidated, but is still in a fairly sturdy condition. I took the
liberty of dismantling the unit in order to replace the four lead wires
as the originals were basically crumbling apart. I do not know whether
or not the supplied aligator clips are the originals or not. On the
bottom of each, they sport the following imprinting:

Mueller Elec. Co.
Cleveland Ohio U.S.A.
60 Series

Any information provided will be greatly appreciated.

Thanks in advance!

image

--
Caine
caine@sonic.net

 

[J. P. Gilliver]

Brian Gregory [UK] wrote:

" <glucegen1@excite.com> wrote in message
news:46f73fc6$0$32520$4c368faf@roadrunner.com...
[]
I did a Google Search for "Quadrature Frequency Modulation" on
http://www.google.com/search?hl=en&lr=&q=%22Quadrature+Frequency+Modulation+%22&btnG=Search

None of those pages made any sense to me.

I had a quick look at a few of them.

Some guy seems to be trying to patent something he calls Quadrature
Frequency Modulation. I see this as similar to the many perpetual
motion machines that get patented.

If it works at all I can't see it offering anything different from
QPSK (aka QAM4) in performance or bandwidth requirements.

When I looked into various modulation schemes (for digital signals, or at
least conveying bits) about twenty years ago, I found many wonderful
techniques claimed for bandwidth reduction. They all boiled down to
filtering - either the individual bit stream(s) before modulation, or the
modulated (subcarrier) signal afterwards, or combinations of both.
Abbreviations often included an N or an M for "near" or "minimum".

Quite apart from the theoretical Shannon limits someone has already
mentioned, there was (and still is) the matter of clock recovery: many of
these papers often had "assuming perfect timing" hidden away somewhere (or
even just implied). In a real practical system, where you're trying to
convey information from one place to a distinct other, with no cheat wire
round the back of the bench carrying the clock signal, you have to recover
your clock timing - i. e. where the symbols start, end, and are to be
sampled - from the actual received signal. With many of these filtered
wonders, this is not practical!

Of course, digital signal processing has come on enormously since then (even
twenty years ago I had to just about admit, though it pained me to do so,
that it was possible to do better than Morse code, with _very_ complex
algorithms!), such that it _is_ possible to do a lot better at clock
recovery than it used to be, but the basic problem does still remain.
--
J. P. Gilliver | Tel: VNET 791 3298 (+44 1634 203298)
BAE SYSTEMS Electronics and | Fax: VNET 791 4831 (+44 1634 204831)
Integrated Solutions, Airport | Email: john.gilliver#baesystems.com
Works, Rochester, Kent, ME1 2XX, UK| (replace "#" with "@")
[]
(But in those days at what was then Marconi Research Centre.)