magnetic field

[Don Klipstein]

In article <daa69o02n34@drn.newsguy.com>, Winfield Hill wrote:

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.

What about neon sign transformers and oil burner ignition transformers?

Those that I see mostly have AC output of 6 to 15 KV, at least
usually center tapped - 3 to 7.5 KV from either end to ground. That's
a lot more than 250V.

Modern CRT flyback transformers put out pulses in one direction of a few
KV, and "triplers" double this, "quintuplers" triple this, and so on.
Even though two diode-capacitor "steps" add only one increment of peak
input voltage (add to this increment whatever the peak voltage in the
other direction is, presumably not zero), this may be practical in systems
requiring low weight, small size of the initial high voltage generator,
and/or battery power.

Get an older flyback transformer for color TV sets that did not use
multipliers, and you have pulses of a good 25 KV or so, possibly 30 (I
have achieved 30 KV from some of those).

How about use of an automotive ignition coil? Those generally produce
25 KV or more.

How about a Tesla coil? AC of voltage mostly from 40 KV to hundreds of
KV!

Now, let's worry about selecting suitable diodes for high AC (or
pulsating DC) voltages and AC (or pulsating DC) voltages for such diodes.

- Don Klipstein (don@misty.com)

 

[Jamie]

rgregoryclark@yahoo.com wrote:

Jamie 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 ?


--


The usual CW multipliers are *additive*:

Cockroft Walton Voltage Multipliers.
http://home.earthlink.net/~jimlux/hv/cw1.htm

So after one doubler the voltage would go from V to 2V. If you add
another stage accoding to the usual circuit, it becomes 4V. So after 10
stages it's only 20V.
I'm asking why can't you change the wiring so that it's 1024V.


Bob Clark

count on your fingers.
2,4,8,16,32,64,128,256,512,1024
lets see, our RDI vaults produce a little over 1 M volts
using a CW multiplier in an insulated pressured gas sourcing
from a 3000 Kv OSC via pie RF transformers.
i can tell you that we don't have a mile rectifier stacks
in there.




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

 

[Guy Macon]

Jamie wrote:

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

You, sir, are a Real Programmer!

 

[ehsjr]

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

Instead of restricting it to CW, lets call it a voltage

"exponenter" instead of "multiplier" that yields (mathmatically)
X^n times whatever the original voltage was. The ugly real world
steps in, current decreases, and each stage adds more and more
loss. I don't know what the practical limit of such voltage
multipliers is, but logically at stage ten, V will have increased
by less than the mathmatical X^10 times due to accumulated losses.

Specifically to the CW and your question:
Each stage in the CW rectifies and stores the input peak Vac by
charging 2 capacitors. One is charged to the negative peak, the
other is charged to the positive peak. The capacitors are charged
individually, but are discharged in series, so each stage *adds* the
voltage it has stored to the previous stage. The net effect is
Total_V = (Original_V * 1.414) * N where N is the number of stages.

If you examine the schematic of a CW you will see that it
can't multiply the voltage stored in the previous stage - it
can only add to it. If you don't examine the schematic, then
just recognize that the ac input is required by each stage.
In essence, all stages are charged in parallel and discharged
in series.

Ed

 

[Guy Macon]

Jamie wrote:

lets see, our RDI vaults produce a little over 1 M volts
using a CW multiplier in an insulated pressured gas sourcing
from a 3000 Kv OSC via pie RF transformers.

i can tell you that we don't have a mile rectifier stacks
in there.

That's roughly 333 stages. The breakdown voltage in air is
roughly 25kV to 75kV per inch (depending on electrode shape,
humidity, etc) so lets give each stage an an inch to be really
sure. That's a bit under 40 feet total, which will fit in your
garage with sepentine construction.

As for the output, here is a fellow making 4-foot sparks in air:
[ http://homepage.ntlworld.com/electricstuff/marxthree.html ]
So I can see why one would want to have an insulator better
than air.

 

[CWatters]

"Manny" <cktmanny@yahoo.com> wrote in message
news:1120436155.453380.294180@g43g2000cwa.googlegroups.com...

If adding cheap low pass filter to a square wave power
inverter can convert the square to sine wave. How come
square wave power inverter manufacturers don't use a
low pass filter to produce cheap sine wave??

Because of the high currents involved.

A simple low pass circuit is shown here..
http://upload.wikimedia.org/wikipedia/en/e/e7/Low_pass_filter.png

However the value or R would have to be very small to avoid voltage and
power losses. That means C would have to be huge or the cutoff frequency is
too high. You could improve things by replacing the R with an inductor but
Inductors also have resistance.

Anyway. For an inverter output of 220 volts and 60 hz.
What exact values of capacitors and inductors must I
use?

Well the corner frequency occurs when the impedance of the L is the same as
the impreance of the C.

Write the equation for each.
Put an equals sign between them
Rearrange to give L or C.

My place doesn't have high capacity capacitors.

Can I increase or decrease the inductor value to
compensate for it.

Yes but watch out for...

The resistance of the inductor (and the resulting power lost in it).
The ripple current rating of the capacitor and it's rated design life.

What's the actual formula.

Look it up.

Manny

Hope you're not Emma's brother!

 

[Tomi Holger Engdahl]

"Manny" <cktmanny@yahoo.com> writes:

If adding cheap low pass filter to a square wave power
inverter can convert the square to sine wave. How come
square wave power inverter manufacturers don't use a
low pass filter to produce cheap sine wave??

The reason for this is that no simple cheap low pass
filter can do that. You need pretty much filtering
and components rated for considerable power to
do the needed filteting at the power range that
the typical square wave inverters use.
If it would be cheap and easy, there would be pretty
much more of those sinewave inverters out there.


--
Tomi Engdahl (http://www.iki.fi/then/)
Take a look at my electronics web links and documents at
http://www.epanorama.net/

 

[CWatters]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message >

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.

Thanks for posting that Fred. Nice example of how the real world is full of
surprises.

 

[John]

On 29 Jun 2005 22:07:23 -0700, cheian07@yahoo.com wrote:

good day!!!

when i test my newly installed lan, i have a problem setting up its
connections.I thought i can view all the stations or they will
recognize each other. but when i open network neighborhood, what
happened was there were groups of computer that recognized each others.
say S-1, S-2, S-3,S-5. The other groups of computer recognized
themselves. ex. S-4,S-6,S7.They are all connected to a single ethernet
hub but why is it that they seemed to group into 2?
We have 3 hubs here interconnected with the 3rd hub connected to a
router (connected to a DSL line).When i try typing IPCONFIG at the
command prompt, the first group of computers has no default gate way in
them so it did'nt recognize other computers connected to other hubs.
The second group of computers has a DEFAULT GATEWAY IN IT so it
recognizes other computers connected to other hubs including the hub
which is connected to the DSL line.
Any help would be gladly apprciated. thank you
ian

-- very simplified explanation --
When PC's with a network card are powered up, they try to get an IP
address from the network. Failing that, newer operating systems
provide a default IP address, such as 169.254.3.1 or 169.254.1.11 (or
some other default that's set by the user or an admin). It's possible
that some of your PC's are in one block of addresses (169.254.3.x) and
some in another (169.254.1.x). The PC's in a given network block will
see the other PC's in that block, but not the PC's in other address
blocks.

When the PC's are connected to the router, they all get an IP address
assigned by the DHCP server in the router. All the addresses will be
in the same block, so all the PC's will see each other.

John
(I also speak X.25)

 

[Guy Macon]

Winfield Hill wrote:

That's silly, and it's wrong.

silly and wrong.

Why must you behave in such a boorish and rude fashion? "Wrong"
is a fair comment. "Silly" is an insult. I have been around for
a while and I have never seen the specific configuration we were
discussing (A Cockroft-Walton multiplier that drives an inverter
that drives a second Cockroft-Walton multiplier). I have seen
many different kinds of voltage multipliers and have never seen
that particular configuration ever used.You claim that that
particular configuration is used, which was a surprise to me.
There is nothing "silly" about not knowing about a particular
rarely-used configuration.

People sometimes make mistakes. I appear to have done so out of
lack of knowledge in this case, and I acknowledged my error as
soon as I saw your post claiming that the configuration in
question is indeed used, without commenting on your use of the
term "silly." Seeing that I didn't rise to the bait, you repeated
it after I said that I was wrong. What is the point in being a
flaming asshole, Winfield? Why turn what could be a learning
experience into an adversarial relationship? Do you really think
that flaming a well-meaning person who made an error and then
acknowledged it is acceptable behavior?

I think you are a better person than the one I am seeing in your
posts.

 

[Fred Bloggs]

rgregoryclark@yahoo.com wrote:

The usual CW multipliers are *additive*:

Cockroft Walton Voltage Multipliers.
http://home.earthlink.net/~jimlux/hv/cw1.htm

Multiplication by an integer is equivalent to adding an integer number
of times. The "multiplier" terminology refers to adding n times where n
is the number of CW stages- the output is n*Vin.

So after one doubler the voltage would go from V to 2V. If you add
another stage accoding to the usual circuit, it becomes 4V. So after 10
stages it's only 20V.

Nope- the second stage produces 3V and not 4V. The 10 stages gets 11V
and not 20V- each stage increments by 1V.

I'm asking why can't you change the wiring so that it's 1024V.

Doesn't seem that anyone has hit on that method yet.

 

[Chip Anderson]

Guy Macon <http://www.guymacon.com/> wrote in
news:11chho723iuj50@corp.supernews.com:

Jamie wrote:

lets see, our RDI vaults produce a little over 1 M volts
using a CW multiplier in an insulated pressured gas sourcing
from a 3000 Kv OSC via pie RF transformers.

i can tell you that we don't have a mile rectifier stacks
in there.

That's roughly 333 stages. The breakdown voltage in air is
roughly 25kV to 75kV per inch (depending on electrode shape,
humidity, etc) so lets give each stage an an inch to be really
sure. That's a bit under 40 feet total, which will fit in your
garage with sepentine construction.

As for the output, here is a fellow making 4-foot sparks in air:
[ http://homepage.ntlworld.com/electricstuff/marxthree.html ]
So I can see why one would want to have an insulator better
than air.

This fellow is going to accidentally kill himself one day.


--
---
Chip

"Oderint dum metuant."
- Lucius Accius

 

[CWatters]

Anyone got time to help "Manny"?

If you have then perhaps you could also answer his physics questions as well
over on sci.physics


Modelling Photoelectric Effect Without Photons All 2 messages in topic -
view as tree
Manny Jul 1, 10:52 pm show options

Newsgroups: sci.physics
From: "Manny" <cktmanny@yahoo.com> - Find messages by this author
Date: 1 Jul 2005 19:52:01 -0700
Local: Fri,Jul 1 2005 10:52 pm
Subject: Modelling Photoelectric Effect Without Photons
Reply | Reply to Author | Forward | Print | Individual Message | Show
original | Report Abuse

In 1905. There is still no good model of the atoms. Why
can't one say that as the frequency of the electromagnetic
field increases, the electron orbital can no longer take the
energy and so gets ejected from the atom. Why does it have to
be a photon particle hitting the electron. What if the electron
orbital can only accept certain energies in terms of the
frequencies and when it exceed it, the electron orbital gets
destructed resulting in the electron leaving the atom?

Has anyone got an alternative atomic model whereby
photoelectric effect can occur without using the concept
of photon particles? For those who hate photons (like
Dr. Richard Perry). You must come up with how the atomic
can eject the electron by mere electromagnetic wave
energy. This also goes for the compton scattering.

Manny

 

[Jim Thompson]

On 3 Jul 2005 19:07:57 -0700, "cnctut" <cnctutwiler@wmconnect.com>
wrote:

Try google for "charge pump"--a variation of the normal doubler circuit
that can be staged as you suggest. Output of Vinsin(wt) input would be
the multiplier x Vin (DC voltage of course.) The limitation to the
idea is poor voltage regulation and low-current capability.

Good luck

Tut

See, for example...

Newsgroups: alt.binaries.schematics.electronic
Subject: Charge Pump (from S.E.D) - ChargePump-4X-Example.pdf
Message-ID: <c6oic11k134gr5u9c70re5r9ebu35iklk4@4ax.com>

...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.

 

[John Fields]

On Fri, 1 Jul 2005 17:28:42 -0500, "Old Man" <nomail@nomail.net>
wrote:

"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.

---
What equation? There's RMS current and RMS voltage, and power is just
the product of current and voltage.

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.

---
You still don't get it, huh? Who said anything about a series LC
circuit? What we're talking about is using a lowpass filter to
eliminate/attenuate harmonics in a square wave in order to extract
the fundamenal. (That is, how to turn a square wave into a sinusoid
with the same period as the square wave) That means the filter is
going to look like this: (View in a non-proportional font.)

SQIN>--+--[L]--->>--+
| |
[C] [RL]
| |
GND>---+-------->>--+

Since the reactance of the capacitor will decrease, and the reactance
of the inductor will increase as frequency increases, the high
frequency components of the input square wave will be attenuated by
the inductor and shunted to ground by the capacitor, allowing the
desired low frequency(ies) to pass through the inductor and into the
load.
---

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.

---
No. We're not talking about a resonant circuit here, we're talking
about a lowpass filter.
---

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

Exactly. Energy and power are conserved.

---
Of course they're "conserved", but that doesn't mean that the energy
in the harmonics will be magically converted into the fundamental.
What will happen is that it will be shunted to ground through the
capacitor and converted into heat. Wasted, in other words.
---

John Fields

Professional Circuit Designer

Professional ? Fields displays delusions of competence.

[[Old Man]

---
Still think so, ya goofy old fart?

--
John Fields
Professional Circuit Designer

 

[Zak]

Manny wrote:

Do you know of a design where 99% or even 100% of
the high frequency can be suppressed. I don't mind the
cost. I'm just studying inverter efficiency for
future marketing.

What you could do is generate various levels of DC and switch between them.

That would leave you with a sine approximation, which would mean you
only need to filetr out a higher freqency.

Or even better, use PWM and a small output filter - essentially
adjusting a DC power supply from 0 to max to 0 - and then reversing
polarity with a bridge.


Thomas

 

[John Fields]

On 30 Jun 2005 19:22:04 -0700, "Manny" <cktmanny@yahoo.com> wrote:

redbelly wrote:
For tungsten, the cold resistance is roughly 1/15th of the operating
hot resistance.

1.44 ohm / 15 is close to 0.1 ohm. So the 0.7 ohm measurement is
actually too high for a room temperature resistance. Weird.

Mark

I didn't subtract the resistances in my multi-meter leads. Is
there a resistance there. But when I short it, there reading
says 0 ohms. I just directly measure the positive and negative
lead of the 12 volts halogen lamp and it gives out 0.7 ohms.
Where did I get it wrong.

About tungsten. There is something I wanna know. Thick tungsten
has low resistance while thin tungsten has high resistances.
This is regarding the thickness per section. If I make
the thick tungsten longer in length, can it increase the
resistance?? Or is thickness the main factor in resistances
considerations?

Also suppose I want a certain resistance. How do I know
what optimum thickness of the tungsten to use with good
allowance for heat dissipation. If I use too thin, there
may be so much heat and it may burn. If I use too thick,
I may have to make the length longer to increase
the resistance. How do you know what thickness and length
is the optimum (is length a factor at all or is it just
the thickness?). This is just for the sake of theoretical
understanding and I'm not building bulbs ok.

Thanks.

Manny

--
John Fields
Professional Circuit Designer

 

[Jan Panteltje]

On a sunny day (Mon, 04 Jul 2005 11:53:08 -0500) it happened John Fields
<jfields@austininstruments.com> wrote in
<n5nic19sfc6iuskapkjgldm5o8p1snovju@4ax.com>:

What we're talking about is using a lowpass filter to
eliminate/attenuate harmonics in a square wave in order to extract
the fundamenal. (That is, how to turn a square wave into a sinusoid
with the same period as the square wave) That means the filter is
going to look like this: (View in a non-proportional font.)

SQIN>--+--[L]--->>--+
| |
[C] [RL]
| |
GND>---+-------->>--+
John, this is a bit dangerous if that thing really has low output impedance.

It will cause the switcher to current limit or blow up...

Enter it in spice, with square wave input, look at current in C.
It is set by the Ri of the source (and Rs cap).

in --L------- out
|
C RL
|
----------
is a bit safer I think, but without load can give very high voltages in
resonance.

 

[Jim Thompson]

On 4 Jul 2005 10:58:57 -0700, rgregoryclark@yahoo.com wrote:

Jim Thompson wrote:
On 3 Jul 2005 19:07:57 -0700, "cnctut" <cnctutwiler@wmconnect.com
wrote:

Try google for "charge pump"--a variation of the normal doubler circuit
that can be staged as you suggest. Output of Vinsin(wt) input would be
the multiplier x Vin (DC voltage of course.) The limitation to the
idea is poor voltage regulation and low-current capability.

Good luck

Tut

See, for example...

Newsgroups: alt.binaries.schematics.electronic
Subject: Charge Pump (from S.E.D) - ChargePump-4X-Example.pdf
Message-ID: <c6oic11k134gr5u9c70re5r9ebu35iklk4@4ax.com

...Jim Thompson

Google didn't pull that one up, since Google doesn't archive binaries.

Use a good news server. Doesn't your ISP have a news server?

Do you have another source for that?

I designed it ;-) It's now on the S.E.D/Schematics page of my
website.

Bob Clark

...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.

 

[John Fields]

On Mon, 04 Jul 2005 17:32:23 GMT, Jan Panteltje
<pNaonStpealmtje@yahoo.com> wrote:

On a sunny day (Mon, 04 Jul 2005 11:53:08 -0500) it happened John Fields
jfields@austininstruments.com> wrote in
n5nic19sfc6iuskapkjgldm5o8p1snovju@4ax.com>:

What we're talking about is using a lowpass filter to
eliminate/attenuate harmonics in a square wave in order to extract
the fundamenal. (That is, how to turn a square wave into a sinusoid
with the same period as the square wave) That means the filter is
going to look like this: (View in a non-proportional font.)

SQIN>--+--[L]--->>--+
| |
[C] [RL]
| |
GND>---+-------->>--+
John, this is a bit dangerous if that thing really has low output impedance.
It will cause the switcher to current limit or blow up...

---
I was just thinking of a way to get a lowpass circuit down for Old
man, but you're right. I didn't even think of that. Thanks!
---

Enter it in spice, with square wave input, look at current in C.
It is set by the Ri of the source (and Rs cap).

in --L------- out
|
C RL
|
----------
is a bit safer I think, but without load can give very high voltages in
resonance.

---
Yes. Passive voltage gain!

--
John Fields
Professional Circuit Designer