magnetic field

[Jan Panteltje]

On a sunny day (30 Jun 2005 13:17:39 -0700) it happened "emma"
<mrandmrsrelativity@yahoo.com> wrote in
<1120162659.044721.130920@g47g2000cwa.googlegroups.com>:

I've got other idea. I'll get meters of high gauge wires and measure
3 ohms and use it as resistor. How do I calculate how much will it
heat up?
That depends on a lot of factors, for example when you wind it like a coil

it will heat up a lot more then when you leave it laying about.

I've got Resnick and Holliday 800 page book on
Electromagnetisms and still slowly going thru it but need the
information asap. Thanks.
Try it!

There is special resistance wire, that you can use to make your own
resistance. Google:
http://www.surplussales.com/Wire-Cable/Resistance.html
Then you can use shorter length.

Say. Is there a difference in performance if I have say a 20 meter
thin cooper wire measuring 3 ohms versus a thick 5 meter cooper
wire measuring 3 ohms also?
Well, since you want to measure magnetic fields, I would use a light bulb,

because that wire (especially when the 20 meter is wound) will be a coil
of its own, with its own magnetic field, interfering with your setup!
I presume you meant 20 meter thick versus 5 meter thin.

Thanks. It explains why I fried my power transistors 2 times
already. Well if I never go below 60 Hz but go higher to 10Khz.
Would it also fry the circuit?? I notice a high pitch sound in
the circuit when the frequency is increased linearly from 60 Hz,
to 10 kHz.
At higher [switching] frequencies the switch time of the transistors

becomes important (they will dissipate heat while not 100% on or off),
and also the losses in the core of the transform will increase.
Then there is 'skin effect' in the wire (electricity only flows in outside).
For this reason above say 3 kHz you will often see ferroxcube (ferrite)
cores.
And possibly litze wire (wire made up of many strands).

Some kind of non-linear detection system that is based on the
quantum potentials of the vectors. I can't explain it. I'll ask.
Squid!

Well that should be sensitive enough
But what about the liquid cooling and the totally magnetically screened room?
Well ask anyways

 

[St. John Smythe]

Jan Panteltje wrote:

On a sunny day (30 Jun 2005 13:17:39 -0700) it happened "emma"
mrandmrsrelativity@yahoo.com> wrote in
1120162659.044721.130920@g47g2000cwa.googlegroups.com>:

I've got other idea. I'll get meters of high gauge wires and measure
3 ohms and use it as resistor. How do I calculate how much will it
heat up?

That depends on a lot of factors, for example when you wind it like a coil
it will heat up a lot more then when you leave it laying about.

You fellows are being trolled big-time. Can't you see it?

--
St. John

 

[ehsjr]

John Fields wrote:

On 30 Jun 2005 12:46:58 -0700, "kell" <kellrobinson@billburg.com
wrote:



He's going to have to face some hard realities. At 100 uA,
if it can be achieved, physical construction becomes critical.
Assuming 5V Vcc, a 100 meg resistance draws 50 uA.


---
E 5V
I = --- = ------ = 5E-8a = 0.05ľa
R 1E8R
---

Sheesh! Usually, I hate it when I do that. But in
this case, I'm glad. I had dismissed the scheme based
on the OP's specs and my error. They are *way* less
stringent than I figured.

He can buy a chip that will provide the .1 mA current
and have at it: REF200AU-ND from Digikey for $5.20.

His spec
was .5% accuracy. 100uA * .005 = .5 uA
So if dirt/humiditity/whatever on his pc board yields a 100
meg or lower path for current to follow, he's blown away by
two orders of magnitude, or more.


---
No, at 100 megohms he's an order of magnitude better than the spec.
---


Frankly, I'm clueless as to how to meet his specs.

No more. Thanks John!

Ed

 

[Charles Jean]

On 30 Jun 2005 12:46:58 -0700, "kell" <kellrobinson@billburg.com>
wrote:

He's going to have to face some hard realities. At 100 uA,
if it can be achieved, physical construction becomes critical.
Assuming 5V Vcc, a 100 meg resistance draws 50 uA. His spec
was .5% accuracy. 100uA * .005 = .5 uA
So if dirt/humiditity/whatever on his pc board yields a 100
meg or lower path for current to follow, he's blown away by
two orders of magnitude, or more.

Frankly, I'm clueless as to how to meet his specs. You
could use an op amp to get even better current regulation,
but I don't know how to accomodate that on a PC board with
those specs. Thompson could design it inside a chip,
combining the thermistor function with regulation and
whatever else, and yielding an output that is not critical.
I don't think it can be done with discretes.

I can't find the original post. What purpose does the OP have in mind
for this circuit... maybe he'll find a solution off the shelf, like a
temperature sensor chip.
___

Couldn' t find the original post, but the OP should check out the
Dalis DS18B20. It's an addressable 1-wire 12-bit serial device with
an accuracy of +/- 0.5 deg. C, and a readability of 0.0625 deg. C.
Maxim-Dalis has it for $2.57 one-off, but the lead time is 7 weeks.

 

[Strange Indeed]

"emma" <mrandmrsrelativity@yahoo.com> wrote in
news:1120172827.422569.115130@g14g2000cwa.googlegroups.com:

St. John Smythe wrote:
You fellows are being trolled big-time. Can't you see it?

--
St. John

Clearly matches one model but has been amateurish in execution when
compared to the other trolls who are regulars here.

You thought that way because I asked too many questions. Well.
I have gotten the necessary and important information already
so I'll carry on myself with the help of Resnick thick book on
electromagnetism and not ask too many questions (that is starting
to irritate some folks).

Resnick is too advanced for you.

Sorry and many thanks.


emma

Nice, this, but futile. It is clear you didn't actually understand
any of the answers.

 

[me]

Can anyone think of another solution to my problem? I always try my
best to hook into as many circuits as possible, and distribute my
equipment evenly, but I still have trouble.

I find your power requirements unlikely at best. But if true, buy a
portable 5kW+ generator...

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

 

[Vidar Lřkken]

DJ Craig wrote:

Forgive me if I've posted this message in the wrong group. I'm not
really sure where I should post this.

I'm a mobile DJ. I often DJ outside people's houses, or at hotels,
camp grounds, appartment blocks, churches, etc. These places usually
will let you draw about 2000 watts from one of their circuits before
you blow the circuit breaker. My equipment requires between 4000 and
5000 watts. The lighting effects, fog machine, etc, requires about
2000-2500 watts, and the audio equipment requires about 2000-2500
watts. I have about 1000 watts-worth of equipment that *must* remain
on througout the event, such as CD players, mixer, amp and speakers.
It is a nightmare trying to distribute my equipment across multiple
circuits so that I dont blow fuses, which I often do. It is not too
bad if the lights and fog machine turn off during an event, but it is
important that my audio equipment keeps running thoughout an event.

I thought that it may be possible to build a gadget that would hook
into 2 circuits, where i would plug my audio equipment into the
"reliable" plug, and the lighting into the "unreliable" plug. If
either of the two circuits blow, then the box would switch the audio
equipment over to the remaining circuit, and turn off the lighting.
Does such a gadget already exist, or would it be possible, and
practicle to build?

One issue is that if my CD players are turned off, even for an instant,
the music will stop and I have to re-cue the music. So a box that just
uses a relay to switch the circuits may not work. I'm not sure, but
the CD players may not be able to handle the delay caused by the relay.
Especially if the voltage from the circuit that is blowing fades
before it goes right out, creating even more delay and causing the CD
player's condensers to empty. Could it be done with a triac instead?
Would a triac be able to handle the high wattage?

Can anyone think of another solution to my problem? I always try my
best to hook into as many circuits as possible, and distribute my
equipment evenly, but I still have trouble.

If you where to design anything, I'd suggest ensuring that multiple
outlets was right regarding to hot/neutral, and connect all of them to
the input of a UPS. Some of APC's UPSes accept multiple inputs,but they
weigh above 100kg, and costs accordingly, but provides power for a while...

--
MVH,
Vidar

www.bitsex.net

 

[Old Man]

"John Fields" <jfields@austininstruments.com> wrote in message
news:l738c1tspg5lbfuncv52av7r3dukhdv4ju@4ax.com...

On Wed, 29 Jun 2005 13:18:41 -0500, "Old Man" <nomail@nomail.net
wrote:


"Tm" <smiller615@nospamcast.net> wrote in message
newsI6dnfn9gcjPvV_fRVn-iw@comcast.com...

Low Pass will do it. But you will need more than a 110 volt square wave
to
start with if you expect to get a 110 volt sine.

No loss in RMS power.

In principle (R_ series = 0, R__parallel = infinite), an LC
low pass filter is lossless.

---
1. There is no such thing as "RMS power"

Semantics. Write the equation.

2. Since, for a square wave, RMS and peak voltage are the same and
since for a sine wave they're not, a lowpass filtered 120V 60Hz
square wave will yield a 120V _peak_ 60Hz sine wave. That's about
an 85VRMS sine wave.

No. Simple addition doesn't cut it. In a series LC circuit,
voltages are added in quadrature. The voltages across L
and C (+ R_load) aren't in phase.

3. Of course there's a loss in power.

No. R_ series = 0, R__parallel = infinite. Give the equation.
All of the power goes into the resistive load. None elsewhere.

Where do you think all the
energy in the harmonics went, into the fundamental?

Exactly. Energy and power are conserved.

John Fields

Professional Circuit Designer

Professional ? Fields displays delusions of competence.

[[Old Man]

 

[Vidar Lřkken]

John wrote:

John
(I also speak X.25)

Hope that is not serious...Talks x25? Feel free...

--
MVH,
Vidar

www.bitsex.net

 

[Mark VB]

cheian07@yahoo.com wrote:

i see. thanks. my next questions are, even if they are all connected to
the same hub, it can happen that they will be in a different blocks of
addresses?

Sure, it can.

How can i make them on same block of address when the router
is off because most of the time i turn off the router and dsl line but
i want them to see each other eventhough the router is off.

Use fixed IP's on your computers and turn off DHCP.



HTH,
Mark Van Borm

 

[Bob Monsen]

redbelly wrote:

kell wrote:


I can't find the original post. What purpose does the OP have in mind
for this circuit... maybe he'll find a solution off the shelf, like a
temperature sensor chip.


Here is the OP, courtesy of the OP. For some odd reason, it is showing
up as the third message in the thread, at least when read at Google
Groups:

I'm looking for a basic, easy-to-build constant current source to drive
thermistors, nominally at 0.1 mA. The thermistors would be the common
10 k-ohm @ 25 C variety.

I spent time Googling this topic without success. One site mentioned
that an LM317 can be used to generate a current, but the current would
have to be a minimum of 5 or 10 mA. I need a 0.1 mA source.

I'd like to build something, preferably for under $10.00, with better
than 0.5% accuracy (that should correspond to temperature readings
within 0.1 or 0.2 C). My plan is to have several thermistors in series
driven with the circuit, and then I can read off the voltage of each
thermistor with an A/D hooked up to a PC, and be able to monitor
temperature readings over several hours or up to a day.

TIA,

Mark

One trivial way to do this is using a TL431.

The TL431 uses almost no current in through the adj terminal, so by
putting a 25k resistor from adj to ground, and putting the thermistor
in the same branch, you can measure the thermistor resistance by noting
the voltage at the cathode, as below:

5V
|
.-.
| | 1k or so
| |
'-'
|
.----o------ output V = 2.5 + Rt*100u
.---o-. |
| | |
Thermistor | | |
'---o-' |
| |
| |
| -
o--- ^ TL431
| |
| |
.-. |
25k | | |
| | |
'-' |
| |
| |
| |
'----'
GND

(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Slightly cheaper than a current source.

--
Regards,
Bob Monsen

If a little knowledge is dangerous, where is the man who has
so much as to be out of danger?
Thomas Henry Huxley, 1877

 

[Fred Abse]

On Thu, 30 Jun 2005 09:32:03 -0500, John Fields wrote:

I have a 50 watt 12V halogen lamp I just checked, and it has a cold
resistance of 0.22 ohms, so I suspect your hundred watt lamp should
have a cold resistance of about half that, or around 0.1 ohms.

Be careful, John. A few years ago, in the course of some work, I had
occasion to measure the cold resistance of several hundred 100W halogen
lamps. About half of them actually appeared open circuit on a low voltage
DMM, (300mV open circuit) but nonetheless lit normally at rated voltage.

Looking at samples using a curve tracer proved interesting. Most exhibited
rectifying properties in both directions, with a barrier potential of
about half a volt, which turned out to be due to the method of fixing the
filaments to the seal wires using a crimp technique, rather than
spot welding.


--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)

 

[CWatters]

<cheian07@yahoo.com> wrote in message
news:1120270378.210228.4320@g43g2000cwa.googlegroups.com...

i see. thanks. my next questions are, even if they are all connected to
the same hub, it can happen that they will be in a different blocks of
addresses?

yes I believe it's called subnetting but I might be wrong.

How can i make them on same block of address when the router

is off because most of the time i turn off the router and dsl line but
i want them to see each other eventhough the router is off.
like this unit i am using. i can browse on the net but it cannot be
seen by other computers.but its connected to the internet anyway...
thanks in advance.

Normally you would set your network up so that all the PC get IP addresses
from the router vis DHCP. If you want to turn the router on and off (why?)
you could allocate each PC, the router and anything else (printer) static
IPs in the range 192.168.0.x and all with the subnet mask 255.255.255.0

 

[Vidar Lřkken]

CWatters wrote:

cheian07@yahoo.com> wrote in message
news:1120270378.210228.4320@g43g2000cwa.googlegroups.com...

i see. thanks. my next questions are, even if they are all connected to
the same hub, it can happen that they will be in a different blocks of
addresses?


yes I believe it's called subnetting but I might be wrong.

Yes, it is subnetting. You assign subnets via the netmask:

10.100.8./255.255.255.64
which means you could use from 10.100.8.1 to 10.100.8.63 (lowest and
higest address goes away...)
That'd be a 7-bit wide subnet, so it could be written 10.100.8.0/31.
You can divide your net into any portion that is dividible by a power of
two (i.e 2-4-8-16-32-64-128 leaving respectively 126-62-30-14-6-2-2 IP's
availvable.)
I seldomly see any reason for subnetting, unless you have more than a
/24 bit (255.255.255.0) subnet, in which cases windows uses ages for
discovering its network neighbourhood. If you have under 254 machines, I
see no point in subnetting. For security, use VLANs.

How can i make them on same block of address when the router

is off because most of the time i turn off the router and dsl line but
i want them to see each other eventhough the router is off.
like this unit i am using. i can browse on the net but it cannot be
seen by other computers.but its connected to the internet anyway...
thanks in advance.


Normally you would set your network up so that all the PC get IP addresses
from the router vis DHCP. If you want to turn the router on and off (why?)
you could allocate each PC, the router and anything else (printer) static
IPs in the range 192.168.0.x and all with the subnet mask 255.255.255.0

Yes, and I suggest yo turn off dhcp in the modem/router if you turn it
off. Enable a dhcp server somewhere else.
Here is my dhcp setup:
ddns-update-style none;
default-lease-time 14400;
option domain-name "vidarlo.net";
option domain-name-servers 10.0.0.10;
option routers 10.0.0.10;
subnet 10.0.0.0 netmask 255.255.255.0 {
range 10.0.0.20 10.0.0.200;
default-lease-time 14400;
max-lease-time 172800;
}

DHCP can be found on http://freshmeat.net/dhcp


--
MVH,
Vidar

www.bitsex.net

 

[Jamie]

rgregoryclark@yahoo.com wrote:

Why couldn't you have the output of a CW voltage doubler lead into the
input of a another doubler? It seems to me that instead of the voltages
being additive with additional stages as done now, with this method you
could double the voltage each time.
So with 10 repetitions you could multiply the voltage by 2^10 = 1024.


Bob Clark

to add to that, if you look you can repeat the CW stages.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

 

[Jim Thompson]

On Sun, 03 Jul 2005 18:31:49 -0700, Jamie
<jamie_5_not_valid_after_5_Please@charter.net> wrote:

rgregoryclark@yahoo.com wrote:

Why couldn't you have the output of a CW voltage doubler lead into the
input of a another doubler? It seems to me that instead of the voltages
being additive with additional stages as done now, with this method you
could double the voltage each time.
So with 10 repetitions you could multiply the voltage by 2^10 = 1024.


Bob Clark

Hmmm
You answered your own question.
what do you think it does now ?

Are "Real Programmers" sure about that ?

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[legg]

On 3 Jul 2005 15:13:08 -0700, rgregoryclark@yahoo.com wrote:

Why couldn't you have the output of a CW voltage doubler lead into the
input of a another doubler? It seems to me that instead of the voltages
being additive with additional stages as done now, with this method you
could double the voltage each time.
So with 10 repetitions you could multiply the voltage by 2^10 = 1024.

The amplitude of the single working source is limited and the energy
storage elements are all capacitive, in this topology. Disregarding
possible resonant effects or intentional use of inductive elements,
the source voltage amplitude is the only active voltage available to
charge or discharge capacitances in any stage of the multiplier
string.

If more sources or switches are present, different voltage
amplifications are possible in each and any specific stage, but then
it's not a Cockroft-Walton circuit any more.

RL

 

[John]

On Sat, 02 Jul 2005 13:23:40 GMT, Vidar Lřkken <njus@vidarlo.net>
wrote:

John wrote:

John
(I also speak X.25)

Hope that is not serious...Talks x25? Feel free...

In the sense that I've been successful in implementing Cisco's XOT
(X.25 over TCP/IP) and writing programs for available battery-backed
PC's to run constant traffic over a test network (4 Telematics nodes).
The test network ran and collected performance statistics for almost a
year. X.25 isn't that common in the US (other than the modified BX.25
used by some telephone companies and specific sections of the US
government), although it has seen much usage elsewhere in the world.

Network diagram: http://www.jecarter.com/images/x25cfg.gif

John

 

[Guy Macon]

rgregoryclark@yahoo.com wrote:

Why couldn't you have the output of a CW voltage doubler lead into the
input of a another doubler? It seems to me that instead of the voltages
being additive with additional stages as done now, with this method you
could double the voltage each time.
So with 10 repetitions you could multiply the voltage by 2^10 = 1024.

It's certainly possible, but of course you can't just feed the output
of one CW into another; you need an inverter that will turn the DC[1]
output into AC (a square wave is best for the CW and easiest to make).

That's a problem, because it throws away the biggest advantage of the
CW - the fact that no component has to withstand the entire output
voltage.

Try designing one on paper paying attention to the voltage across
each component and you will see that the stage1-stage2 inverter is
expensive, the stage2-stage3 inverter is really, really, expensive,
and you can't build the stage2-stage3 inverter at all.

The reason you don't even see a two stage system in practice is
that there is a much better way to do it; use a step-up transformer
to drive your first stage with a higher AC voltage.

Excellent thinking, though. It's always good to think "out of
the box" and to look for something that everybody missed before.


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

(Those who are interested in a technical discussion only are
encouraged to move on to the next post without reading the OT
material below.)

[Note 1] Ignore the brain-dead flamers who think that they have
discovered a weakness in their technical superiors when actually
all they have done is to invent a nonexistent difference between
DC and a pulse that starts when you turn on the DC power supply
and ends when you turn off the DC power supply. True, realizing
that the two are simply different ways of describing the same
signal is a rudimentary mental skill that many of us "normal"
people take for granted, but we sometimes forget that there are
"challenged" persons in this world who find these things to be
difficult to understand. The sad part is that they have the
raw mental capability to arrive at a rational conclusion, but
are so controlled by their emotions and fears that they grasp
at any straw to attack the target of their schoolyard bullying.
It's best to just killfile them and to let them flame away and
shout into an empty hall.

--
Guy Macon <http://www.guymacon.com>

 

[Winfield Hill]

Guy Macon wrote...

rgregoryclark@yahoo.com wrote:

Why couldn't you have the output of a CW voltage doubler lead into
the input of a another doubler? It seems to me that instead of the
voltages being additive with additional stages as done now, with
this method you could double the voltage each time.
So with 10 repetitions you could multiply the voltage by 2^10 = 1024.

The reason you don't even see a two stage system in practice is
that there is a much better way to do it; use a step-up transformer
to drive your first stage with a higher AC voltage.

That's silly, and it's wrong. You often see them, and I often
use them, for example. Frequently it's inconvenient to get a
higher-voltage transformer. For example, check the meager HV
offerings from Signal Transformer. Now that tubes are not so
commonly used in industry, the available selection of ac-line
transformers with outputs above say 250V is very meager indeed.

(Those who are interested in a technical discussion only are
encouraged to move on to the next post without reading the OT
material below.)

[ snip @)$*&^#$@&$ ]


--
Thanks,
- Win