Chip with simple program for Toy

"G-squared" <stratus46@yahoo.com> wrote in message
news:6d536181-80ae-4dc4-966a-f6a371ac2201@m73g2000hsh.googlegroups.com...
On May 4, 9:24 am, "Tantalust" <Tantal...@paradise.net> wrote:
"w_tom" <w_t...@usa.net> wrote in message
snip
We earth a 'whole house' protector AND connect all protectors
short
(ie 'less than 10 feet') to single point earth ground so that
protection inside all appliances is not overwhelmed. Simple
stuff
that so confused trader. trader *assumed* MOVs rather than read
what
was posted. trader again demonstrates insufficient technical
kowledge justifies his mockery and insult. Mythical MOV inside
appliances demonstrate that trader only reads what he wants to
see;
not what is posted.

MOVs inside appliances is another trader myth. Had trader read
what
was posted or learned technology, then trader would not invent
fictional MOVs inside appliances.

Why do you have this pompous attitude; constantly sermonizing down
to people
as if they're your little, personal kindergarten class?

You read sometimes like one of those old children's "Golden Books".

Hey, I LIKED reading Golden Books to my kids. They didn't like W-TOMs
posts at all.

GG
Click to expand...
I loved Golden Books too... I'm 55 going on 90 and I still read mine
everyday! ;-)
 
Ď "Tantalust" <Tantalust@paradise.net> Ýăńářĺ óôď ěŢíőěá
news:RPidnaZzhcrV0oXVnZ2dnUVZ_hadnZ2d@comcast.com...
"NB" <nobuyout@gmail.com> wrote in message
news:b53f2fef-00bd-40d0-9ac1-c69b3bcadf52@x41g2000hsb.googlegroups.com...
Who is W_TOM and why has he appeared in every single thread that has
contained those keywords since 2001???

He an obsessive-compulsive disorder victim, apparently driven by some kind
of bizarre fetish involving ground rods.


What kind of ground rods? I prefer steel core, copper clad ones:) I even
Click to expand...
have the special heavy hammer>


--
Tzortzakakis Dimitrios
major in electrical engineering
mechanized infantry reservist
hordad AT otenet DOT gr
 
Ď "Tantalust" <Tantalust@paradise.net> Ýăńářĺ óôď ěŢíőěá
news:RPidnaZzhcrV0oXVnZ2dnUVZ_hadnZ2d@comcast.com...
"NB" <nobuyout@gmail.com> wrote in message
news:b53f2fef-00bd-40d0-9ac1-c69b3bcadf52@x41g2000hsb.googlegroups.com...
Who is W_TOM and why has he appeared in every single thread that has
contained those keywords since 2001???

He an obsessive-compulsive disorder victim, apparently driven by some kind
of bizarre fetish involving ground rods.


What kind of ground rods? I prefer steel core, copper clad ones:) I even
Click to expand...
have the special heavy hammer>


--
Tzortzakakis Dimitrios
major in electrical engineering
mechanized infantry reservist
hordad AT otenet DOT gr
 
In article <fvn72o$rta$1@registered.motzarella.org>,
Dave <noway@nohow.not> wrote:
I wonder why, since electrical codes in North America
and Britain require a ground connection at each outlet;
computer power cords are 3 wire?
Click to expand...

(snip)

hot neutral ground
Click to expand...

or, as we call it, Live, Neutral & Earth

--
From KT24 - in "Leafy Surrey"

Using a RISC OS computer running v5.11
 
In article <fvn72o$rta$1@registered.motzarella.org>,
Dave <noway@nohow.not> wrote:
I wonder why, since electrical codes in North America
and Britain require a ground connection at each outlet;
computer power cords are 3 wire?
Click to expand...

(snip)

hot neutral ground
Click to expand...

or, as we call it, Live, Neutral & Earth

--
From KT24 - in "Leafy Surrey"

Using a RISC OS computer running v5.11
 
<spamfree@spam.heaven> wrote in message
news:4egu14daq01hro0aj8c8ugpv3emu5a3h6g@4ax.com...
On Mon, 5 May 2008 09:24:41 -0400, "Tantalust"
Tantalust@paradise.net> wrote:

"w_tom" <w_tom1@usa.net> wrote

Yes, plug-in protectors do have limited protective functions.

Look at poor w_tom starting his back-pedalling.
Back-pedalling, back-pedalling, back-pedalling.


As I understand it, there is not "protection", or "no protection"
That is, it is not black and white, but degrees of protection, as
there are degrees of surge, or spike.

There is absolute protection of whole of house costing many thousands
of dollars, with tinfoil hats thrown in at no extra cost :)
And there is $7 protection against weeny little spikes/surges, and
then there is everything in between at varying prices.

The old saw "you get what you pay for" is generally bullshit IMHO
You get what the bastard will let you get away with IME

jack
Click to expand...
Yeah, but search google groups with the terms "plug in protectors".....
returning 33 instances clearly showing that wobbly_tom was bewildered on the
issue and he's back-pedalling now, as we expected.
 
<spamfree@spam.heaven> wrote in message
news:4egu14daq01hro0aj8c8ugpv3emu5a3h6g@4ax.com...
On Mon, 5 May 2008 09:24:41 -0400, "Tantalust"
Tantalust@paradise.net> wrote:

"w_tom" <w_tom1@usa.net> wrote

Yes, plug-in protectors do have limited protective functions.

Look at poor w_tom starting his back-pedalling.
Back-pedalling, back-pedalling, back-pedalling.


As I understand it, there is not "protection", or "no protection"
That is, it is not black and white, but degrees of protection, as
there are degrees of surge, or spike.

There is absolute protection of whole of house costing many thousands
of dollars, with tinfoil hats thrown in at no extra cost :)
And there is $7 protection against weeny little spikes/surges, and
then there is everything in between at varying prices.

The old saw "you get what you pay for" is generally bullshit IMHO
You get what the bastard will let you get away with IME

jack
Click to expand...
Yeah, but search google groups with the terms "plug in protectors".....
returning 33 instances clearly showing that wobbly_tom was bewildered on the
issue and he's back-pedalling now, as we expected.
 
"Michael A. Terrell" advised:
Use a combination of protection at the building's main disconnect,
and individual protection at each critical device.
Click to expand...

Does that mean a combination of w_tom's "whole house protection"
and individual "surge protectors" at those "critical devices"? That's
what I've always felt would be prudent - not a single method of
protection, but a combination.

*TimDaniels*
 
"Michael A. Terrell" advised:
Use a combination of protection at the building's main disconnect,
and individual protection at each critical device.
Click to expand...

Does that mean a combination of w_tom's "whole house protection"
and individual "surge protectors" at those "critical devices"? That's
what I've always felt would be prudent - not a single method of
protection, but a combination.

*TimDaniels*
 
In alt.engineering.electrical trader4@optonline.net wrote:
| On May 5, 1:44?am, phil-news-nos...@ipal.net wrote:
|> In alt.tv.tech.hdtv bud-- <remove.budn...@isp.com> wrote:| phil-news-nos...@ipal.net wrote:
|>
|> |> In alt.engineering.electrical Leonard Caillouet <nos...@noway.com> wrote:|> | <phil-news-nos...@ipal.net> wrote in message
|>
|> |> |news:fvjhvk016vr@news5.newsguy.com...
|> |> |> In alt.tv.tech.hdtv Franc Zabkar <fzab...@iinternode.on.net> wrote:
|> |> |
|> |> |>
|> |> |> The MOVs will act like conductors when they are clamping. ?The surge will
|> |> |> take both paths ... the path through the MOVs, and the path going past the
|> |> |> MOVs. ?In general, about 50% will go each way. ?That can vary at higher
|> |> |> frequencies.
|> |> |
|> |> | Why would you assume that 50% will go each way when you don't know the
|> |> | impedance of each direction? ?When conducting, or at failure, the MOV has a
|> |> | very low impedance.
|> |>
|> |> There is a distinction between "go each way" and "what comes back" due to
|> |> the impedance. ?It will be about 50% that goes each way _because_ the power
|> |> itself does not (yet) know the impedance (at a distance), until it gets
|> |> there.
|> |
|> | Another installment of Phil's Phantasy Physics using transmission line
|> | theory.
|>
|> Not understanding it is your loss.
|
|
| I have to agree that this is Phantasy Physics. We're supposed to
| believe that a surge reaching a MOV is going to split 50-50, with half
| of it going to the MOV path and half moving on down the line,
| reagrdless of the impedance of the two paths? That would render all
| surge protection about 50% effective.

You did not read very carefully. The reference to 50-50 split is about the
contribution of the MOVs themselves. That is an essential understanding of
the components so the whole system can be figured out. The impedance down
the paths is another separate component, which also has to be figured in
when determining the whole picture.

You have confused a component with the entire system. You need to read more
carefully. Or you need to understand the distinction of individual components
as they apply to the whole system

The whole wiring system is extrememly complex. It cannot be understood
properly without first understanding the components. And that includes
understanding that MOVs, when they conduct, do look to the propogating
energy as two paths to go down, and it will (initially) go both ways in
about an equal amount.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.engineering.electrical trader4@optonline.net wrote:
| On May 5, 1:44?am, phil-news-nos...@ipal.net wrote:
|> In alt.tv.tech.hdtv bud-- <remove.budn...@isp.com> wrote:| phil-news-nos...@ipal.net wrote:
|>
|> |> In alt.engineering.electrical Leonard Caillouet <nos...@noway.com> wrote:|> | <phil-news-nos...@ipal.net> wrote in message
|>
|> |> |news:fvjhvk016vr@news5.newsguy.com...
|> |> |> In alt.tv.tech.hdtv Franc Zabkar <fzab...@iinternode.on.net> wrote:
|> |> |
|> |> |>
|> |> |> The MOVs will act like conductors when they are clamping. ?The surge will
|> |> |> take both paths ... the path through the MOVs, and the path going past the
|> |> |> MOVs. ?In general, about 50% will go each way. ?That can vary at higher
|> |> |> frequencies.
|> |> |
|> |> | Why would you assume that 50% will go each way when you don't know the
|> |> | impedance of each direction? ?When conducting, or at failure, the MOV has a
|> |> | very low impedance.
|> |>
|> |> There is a distinction between "go each way" and "what comes back" due to
|> |> the impedance. ?It will be about 50% that goes each way _because_ the power
|> |> itself does not (yet) know the impedance (at a distance), until it gets
|> |> there.
|> |
|> | Another installment of Phil's Phantasy Physics using transmission line
|> | theory.
|>
|> Not understanding it is your loss.
|
|
| I have to agree that this is Phantasy Physics. We're supposed to
| believe that a surge reaching a MOV is going to split 50-50, with half
| of it going to the MOV path and half moving on down the line,
| reagrdless of the impedance of the two paths? That would render all
| surge protection about 50% effective.

You did not read very carefully. The reference to 50-50 split is about the
contribution of the MOVs themselves. That is an essential understanding of
the components so the whole system can be figured out. The impedance down
the paths is another separate component, which also has to be figured in
when determining the whole picture.

You have confused a component with the entire system. You need to read more
carefully. Or you need to understand the distinction of individual components
as they apply to the whole system

The whole wiring system is extrememly complex. It cannot be understood
properly without first understanding the components. And that includes
understanding that MOVs, when they conduct, do look to the propogating
energy as two paths to go down, and it will (initially) go both ways in
about an equal amount.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.engineering.electrical Jitt <tser827@yahoo.com> wrote:
| In article <74683977-6a03-4695-a5a2-
| 156ba3653409@m45g2000hsb.googlegroups.com>, w_tom1@usa.net
| says...
|> On May 3, 4:38?am, Franc Zabkar <fzab...@iinternode.on.net> wrote:
|> > Can you elaborate on this by showing us the path taken by the strike
|> > through the TV?
|>
|> Path to earth was through the network and into a third computer.
|> Through that third computer's motherboard, through modem, and to earth
|> via phone lines. Semiconductors in these paths were damaged.
|>
|> We literally traced this path by replacing ICs. Some ICs (ie
|> network interface chips) even had cracks on packages where surge
|> current entered or exiting those ICs. Absolutely no doubt as to how
|> surge currents found earth ground, destructively, via adjacent
|> computers.
|>
| I wonder why, since electrical codes in North America
| and Britain require a ground connection at each outlet;
| computer power cords are 3 wire?

What good is having the ground connection at each outlet if it is not used?
Are British power cords for computers only 2 wire?

2 of the wires are power conductors. Usually one of the is grounded somewhere
back along the path to the power system source. But it is possible for one to
have a connection with two hot wires (208V from three phase or 240V from single
phase in North America ... 400V from three phase in Europe ... I doubt any of
those 230/460 single phase systems are around anymore in Britain).

The 3rd wire is the groundING conductor. It is not supposed to carry any
current except in the case of a fault between a hot wire and the case or
frame of the computer (or whatever appliance is involved). While this is
a rare event, it is a more important protection in the case of appliances
that routinely get handled by people more than just being turned on and off.
An electric table lamp might not need the grounding conductor because of the
infrequent handling just to turn it on and off. A computer or cooker would
be handled more than a lamp. A computer would be subject to more handling
than the cooker, but the cooker would be subject to being wet. Both of them
are in far more need of the grounding protection than the lamp.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.engineering.electrical Jitt <tser827@yahoo.com> wrote:
| In article <74683977-6a03-4695-a5a2-
| 156ba3653409@m45g2000hsb.googlegroups.com>, w_tom1@usa.net
| says...
|> On May 3, 4:38?am, Franc Zabkar <fzab...@iinternode.on.net> wrote:
|> > Can you elaborate on this by showing us the path taken by the strike
|> > through the TV?
|>
|> Path to earth was through the network and into a third computer.
|> Through that third computer's motherboard, through modem, and to earth
|> via phone lines. Semiconductors in these paths were damaged.
|>
|> We literally traced this path by replacing ICs. Some ICs (ie
|> network interface chips) even had cracks on packages where surge
|> current entered or exiting those ICs. Absolutely no doubt as to how
|> surge currents found earth ground, destructively, via adjacent
|> computers.
|>
| I wonder why, since electrical codes in North America
| and Britain require a ground connection at each outlet;
| computer power cords are 3 wire?

What good is having the ground connection at each outlet if it is not used?
Are British power cords for computers only 2 wire?

2 of the wires are power conductors. Usually one of the is grounded somewhere
back along the path to the power system source. But it is possible for one to
have a connection with two hot wires (208V from three phase or 240V from single
phase in North America ... 400V from three phase in Europe ... I doubt any of
those 230/460 single phase systems are around anymore in Britain).

The 3rd wire is the groundING conductor. It is not supposed to carry any
current except in the case of a fault between a hot wire and the case or
frame of the computer (or whatever appliance is involved). While this is
a rare event, it is a more important protection in the case of appliances
that routinely get handled by people more than just being turned on and off.
An electric table lamp might not need the grounding conductor because of the
infrequent handling just to turn it on and off. A computer or cooker would
be handled more than a lamp. A computer would be subject to more handling
than the cooker, but the cooker would be subject to being wet. Both of them
are in far more need of the grounding protection than the lamp.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.tv.tech.hdtv bud-- <remove.budnews@isp.com> wrote:

| Previously you said Martzloff "flubbed the experiment".

I remember that. You were telling me about some information he had
obtained from some experiment.


| Now you agree with Martzloff that branch circuit must be 200m for
| transmission line behavior with 1.2 microsecond rise time.

That's not a result of an experiment. I'm not so sure the exact distance
is 200m for that exact rise time. But that is a subjective thing. We are
likely not that far apart. It is a matter of degree to how different the
calculations come out when figuring them with transmission line issues and
ignoring those issues.


| You say that doesn't apply because surges are faster. Martzloff uses 1.2
| us because that is a standard rise time for surges produced by lightning
| as defined in IEEE standards.

Martzloff did not say that was a defined standard in the statement you
quoted. He just used it as an example to come up with the 200m figure.
Do you have some other statement from Martzloff or someone else about the
1.2 microsecond rise time?


| w_' professional engineer source says 8 micoseconds with most of the
| spectrum under 100kHz.

Even with 1 nanosecond rise time, most of the energy will be present in
the spectrum below 100 kHz. That means nothing when the surge is strong
enough to have energy above some frequency that is relevant to the whole
system involved that can do damage. That frequency might be 100 Mhz for
some thing, and 1 GHz for other things. Some surges, mostly from very
close direct strikes, can have damaging energy well above 1 GHz. It will
depend on how much inductance you have between the surge and the equipment
that needs to be protected. That's why I suggest that a good protection
scheme will include added inductance on the wiring at a level compatible
with the normal use (more can be added to power than to cable TV).


| You still have *no sources that support your belief* that risetimes are
| far faster.

I have experience and observation for that. I need no more.


| Again you did not read what I wrote (what a surprise):
| "I intended 'induced' meaning produced by including the most damaging -
| strikes to utility lines."

You are saying the most damaging strikes are induced? Or are you merely
adding the most damaging direct strikes in with the induced surges as a
set of surge classes that you want to consider together?

Your wording is so bad in that one sentence that I cannot tell what you mean.
It is vague and could satisfy more than one meaning.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.tv.tech.hdtv spamfree@spam.heaven wrote:

| I'm curous to know how surge suppression can work without a ground
| (earth) of any sort. Does the "black box" detect overvoltage and
| disconnect the power like an earth leakage safety switch?

Without a ground of any sort, not all types of surges can be protected against.
But some can.

If the surge is a differential one (some use the term transverse), then what
the surge suppressor can do is cancel it out by effectively short circuiting
it to itself. A differential surge involves two wires with the voltage on
each being of opposite polarity and equal level. The MOV component inside
the suppressor will normally not be conductive. But when the voltage is high
enough, it becomes a conductor. The arrival of a high voltage differential
surge will result in the MOV between those 2 wires to become conductive.

If the surge is a common mode one, AND if the surge has a slow rise time,
then a device that is interconnected to other wires or other devices can
be protected by allowing the surge to pass to all devices at the same level.
As long as the rise is not too fast, keeping all the incoming wires, and all
the interconnected devices, at the same level results in insignificant current
flows. That surge will either reflect back from the protected equipment to
the suppressor, and from there go back through all the connected wires (which
could be more than where the surge arrived from).

Most strikes have lower energy levels at high frequencies than what would
cause damage. The exact frequency level that needs to be considered depends
on the internals of the equipment. For example, where it has inductance to
one end of a sensitive component like a CMOS chip, and no inductance to some
other end, this could result in a very brief fast rise of voltage high enough
to damage the CMOS chip. In some cases an LC circuit can actually increase
the voltage level of high frequency components (at the resonant frequency).
For example if you have energy at some voltage at 200 MHz, an LC series
circuit resonant at 200 MHz will result in a higher voltage being present
at the connection between the L and the C. So even in cases where there is
not enough energy at high frequency in a surge to cause direct damage, it
can still happen on some devices (think of them as having a lower threshold
of damage to simplify this).


| This might be fine for a TV, but surely not for a computer.

If everything the computer is connected to is protected at a common point
in the same surge suppressor, you can have this kind of protection, even
on a computer. That does mean if you have a phone line connected to a
modem, you need to protect both the phone line itself and the power to the
modem, in common with the computer.


| I don't recall any computer I've owned that did not have a three wire
| connection to the mains. That and a MOV is OK for smallish surges, but
| I believe that for a large surge, the sort that will blow a telephone
| off the wall, one needs a large, short-path earth for the surge
| detector to dump the extra power down.

Such a surge is likely to have high levels of high frequency energy. The
effective protection against these rare events is a combination of somewhere
to divert the energy (like a ground path), and something (like an inductor)
to ensure the energy does get diverted.

One problem is that at the point of use, an alternate ground path is not
practical. The grounding wire of the power circuit coule be as much a source
of the surge as the neutral wire would be. The place to put the diversion
system is at the entrance to the building. Most surges that come in by other
paths besides the entrance to the building are induced surges that will not
have so much energy and even less at high frequencies.

Still, I have seen three incidents in which an induced surge damaged a device
that was not connected to anything at all (in two cases they were battery
powered devices, and in the third, it was disconnected before the storm but
suffered damage anyway).


| I've got a few plug in protectors here and there to sop up a small
| spike, but when a storm is within a few km, I pull the phone wire out
| of the ADSL router, and the plug out of the mains. If I'm working at
| the time, I might just keep a watch on the weather radar and count
| lightning fashes to thunder times. It's rare that I get interrupted. I
| have underground power and phone lines so that gives a little extra
| protection, I believe. I've been told that Australian phone lines are
| the most vulnerable, and the most urgent to protect or disconnect.
| I hope to be going wireless soon which obviates this problem.

Disconnecting provides even better (but still not 100%) protection. Yes, the
underground wiring helps. I don't know the issues with Australian phone lines.
I do the wireless thing myself and feel much more comfortable with it. Most
of the past damaging surges I've seen come in do that on phone and cable wires,
and much less often on power wires. That may be due to the more sensitive
aspect of equipment where it connects to these wires.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.tv.tech.hdtv bud-- <remove.budnews@isp.com> wrote:

| The last standards for simulating typical surge waveforms I have seen
| (IEEE) were
| 1.2 us rise time, 50 us duration
| 8 us rise time, 20 us duration
| a ring wave with a frequency about 100kHz.

So now you are saying these figures represent a typical surge waveform,
as opposed to the worst case waveform you said a long time ago.

The term typical is generally accepted as a median. That means half of
the surges would have a slower rise time, and half would have a faster
rise time.

My concerns are not the typical surges. I suggest that half the surges
don't even need protection at all; they won't cause damage even if there
is no protection. But that also means half can be damaging and need the
protection. And a fraction of those surges need _substantial_ protection.


| All are long relative to 0.2 microsecond, so wave propagation should not
| be relevant for household circuits.

Maybe for the typical surge. How about for the most energetic 1% that are
the ones I'm most concerned with because they are hard to protect against.


| A favorite article from w_ also uses a "8x20 us impulse as a very rough
| representative pulse" with most harmonic content from 20kHz to 100kHz.
|
| Martzloff, using the shorter rise time, has written: "For a 1.2/50 us
| impulse, this means that the line must be at least 200 m long before one
| can think in terms of classical transmission line behavior."

And this statement is only using 1.2/50 us as an example. If you think
such a timing is the standard, why not offer a quote that actually says
that?

What does the "/" mean in that case, anyway? I never got to ask you that
before. Does it mean "divide 1.2 by 50"?


| What reason is there to believe wave propagation is relevant to house
| circuits?

The most damaging surges (not the typical ones) have substantial fast rise
high frequency energy (such as due to a very close direct contact strike).
In these cases, even if you can remove all of the low frequency energy, there
is still damaging energy in the higher frequencies that do follow transmission
line behaviour not only in wiring lengths of typical homes, but even in wiring
lengths inside a small appliance like a computer modem.


|> As to the advantage of "whole house" vs local surge protection, "whole house
|> protection depends on distances to all "protected" items being small.
|
| Longer distances make the system more subject to effects like direct
| induction from lightning into the wiring. I don't see why, in general,
| the distance has to be small.

I believe he was referring to the distance between the whole house protection
and the ground/earth electrode.

For things like the service drop distance and the branch circuit distance, it
can be a tradeoff between different kinds of surges. The longer wiring will,
through its self-inductance, reduce the high frequency energy and slew the
rise time of the wavefront ... especially for common mode surges. However,
that same longer distance increases the potential level of induced surges
where the wire is effectively an antenna.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
In alt.engineering.electrical bud-- <remove.budnews@isp.com> wrote:
| phil-news-nospam@ipal.net wrote:
|> In alt.tv.tech.hdtv bud-- <remove.budnews@isp.com> wrote:
|>
|> | The last standards for simulating typical surge waveforms I have seen
|> | (IEEE) were
|> | 1.2 us rise time, 50 us duration
|> | 8 us rise time, 20 us duration
|> | a ring wave with a frequency about 100kHz.
|>
|> So now you are saying these figures represent a typical surge waveform,
|> as opposed to the worst case waveform you said a long time ago.
|
| Still missing - your source that indicates nanosecond rise times and
| 100MHz spectrum.

Observation.


|> What does the "/" mean in that case, anyway? I never got to ask you that
|> before. Does it mean "divide 1.2 by 50"?
|
| It is standard notation in the surge field. 1.2 us risetime and 50 us
| duration

And what does the duration time have to do with high frequency energy?
Hint: nothing

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
<jfisher864@comcast.net> wrote in message
news:3otu149vd4te4f523tsj3m97njt0tbf5vg@4ax.com...
On Mon, 05 May 2008 13:10:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 05 May 2008 14:18:39 -0500, entropy429@yahoo.com wrote:

I'm confused about the use of terminators. I'm sending pulse signals
from some 93 ohm output equipment through 93 ohm coax to my scope
which is 50 ohm. I have a t at the scope. Should I be puting a 93 ohm
terminator on the other end of the t or a 50ohm? Is there a rule of
thumb where to place terminators? thanks jk


If the generator is really a 93 ohm source, you don't need a
terminator at the far end, no matter what impedance the scope is. Your
system is "source terminated" and any reflections from the scope will
be re-absorbed when they bounce back into the generator.


But if you want to soak up any reflections at the scope (say, the
generator isn't truly a 93 ohm source, and you don't want stuff
sloshing around) you need a series resistor of 43 ohms between the end
of the coax and the 50 ohm scope input... not a tee.


John



Does source termination only refer to periodic signal or does it
also apply to one shot signals also?? What is source termination? I
haven't heard that before,( shows you my level) :) jf
Click to expand...
"Source termination" can be used when you have a driver, a transmission
line, and a single high-impedance load at the very end of that transmission
line. It works with any type of signal. The need for termination depends
upon your situation. Giving this subject proper treatment, here, would take
too much time and effort.

The idea for source termination is that if your driver will put out V into
an open, then when you connect the transmission line (with same
characterstic impedance as the driver's net source impedance) then V/2 will
be "launched" into the transmission line (toward the load). When the singal
hits the open end of the transmission line then V will be present at the end
of the line (into your high-impedance load) and V/2 will be reflected back
to the source. When the reflection arrives back at its source, it will be
completely absorbed by the driver's source impedance.

So, the advantages of source termination are that it is simple to implement,
and you get the entire open-circuit drive voltage at the load. The drawbacks
are that if you have any discontinuities along the transmission line (other
than the intentional one at the end) you will get multiple reflections and
(relatively) poor signal integrity at the load, and with source termination
you cannot daisy chain loads because the ones not at the end will "see"
multiple copies of the signal.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==
 
<jfisher864@comcast.net> wrote in message
news:3otu149vd4te4f523tsj3m97njt0tbf5vg@4ax.com...
On Mon, 05 May 2008 13:10:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 05 May 2008 14:18:39 -0500, entropy429@yahoo.com wrote:

I'm confused about the use of terminators. I'm sending pulse signals
from some 93 ohm output equipment through 93 ohm coax to my scope
which is 50 ohm. I have a t at the scope. Should I be puting a 93 ohm
terminator on the other end of the t or a 50ohm? Is there a rule of
thumb where to place terminators? thanks jk


If the generator is really a 93 ohm source, you don't need a
terminator at the far end, no matter what impedance the scope is. Your
system is "source terminated" and any reflections from the scope will
be re-absorbed when they bounce back into the generator.


But if you want to soak up any reflections at the scope (say, the
generator isn't truly a 93 ohm source, and you don't want stuff
sloshing around) you need a series resistor of 43 ohms between the end
of the coax and the 50 ohm scope input... not a tee.


John



Does source termination only refer to periodic signal or does it
also apply to one shot signals also?? What is source termination? I
haven't heard that before,( shows you my level) :) jf
Click to expand...
"Source termination" can be used when you have a driver, a transmission
line, and a single high-impedance load at the very end of that transmission
line. It works with any type of signal. The need for termination depends
upon your situation. Giving this subject proper treatment, here, would take
too much time and effort.

The idea for source termination is that if your driver will put out V into
an open, then when you connect the transmission line (with same
characterstic impedance as the driver's net source impedance) then V/2 will
be "launched" into the transmission line (toward the load). When the singal
hits the open end of the transmission line then V will be present at the end
of the line (into your high-impedance load) and V/2 will be reflected back
to the source. When the reflection arrives back at its source, it will be
completely absorbed by the driver's source impedance.

So, the advantages of source termination are that it is simple to implement,
and you get the entire open-circuit drive voltage at the load. The drawbacks
are that if you have any discontinuities along the transmission line (other
than the intentional one at the end) you will get multiple reflections and
(relatively) poor signal integrity at the load, and with source termination
you cannot daisy chain loads because the ones not at the end will "see"
multiple copies of the signal.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==
 
In alt.engineering.electrical bud-- <remove.budnews@isp.com> wrote:
| phil-news-nospam@ipal.net wrote:
|> In alt.tv.tech.hdtv bud-- <remove.budnews@isp.com> wrote:
|>
|> | Previously you said Martzloff "flubbed the experiment".
|>
|> I remember that. You were telling me about some information he had
|> obtained from some experiment.
|>
|> | Now you agree with Martzloff that branch circuit must be 200m for
|> | transmission line behavior with 1.2 microsecond rise time.
|>
|> That's not a result of an experiment.
|
| "*From this first test*, we can draw the conclusion (predictable, but
| too often not recognized in qualitative discussions of reflections in
| wiring systems) that it is not appropriate to apply classical
| transmission line concepts to wiring systems if ..."
|
| As usual, you don?t know what was written.

What what kind of surge did Martzloff use to carry out that test?


|> I'm not so sure the exact distance
|> is 200m for that exact rise time. But that is a subjective thing.
|
| Quit equivocating. Where is your cite. Like for nanosecond risetimes.

Observation.


|> | You say that doesn't apply because surges are faster. Martzloff uses 1.2
|> | us because that is a standard rise time for surges produced by lightning
|> | as defined in IEEE standards.
|>
|> Martzloff did not say that was a defined standard in the statement you
|> quoted. He just used it as an example to come up with the 200m figure.
|
| He used it because 1.2/50 (voltage) is an IEEE standard. The 8us from
| w_?s engineer is another standard (8/20 current).

The standard for what? The typical surge?


|> | w_' professional engineer source says 8 micoseconds with most of the
|> | spectrum under 100kHz.
|>
|> Even with 1 nanosecond rise time, most of the energy will be present in
|> the spectrum below 100 kHz. That means nothing when the surge is strong
|> enough to have energy above some frequency that is relevant to the whole
|> system involved that can do damage. That frequency might be 100 Mhz for
|> some thing, and 1 GHz for other things.
|
| Still missing ? your source. Nanosecond risetime. 100MHz spectrum.

Observation. Of course this is a concept you cannot understand.


|> | You still have *no sources that support your belief* that risetimes are
|> | far faster.
|>
|> I have experience and observation for that. I need no more.
|
| Lots of people have experience and observation with flying saucers.
|
| The rest of us want a source.

The only flying saucers I have seen are the ones I've tossed.

--
|WARNING: Due to extreme spam, I no longer see any articles originating from |
| Google Groups. If you want your postings to be seen by more readers |
| you will need to find a different place to post on Usenet. |
| Phil Howard KA9WGN (email for humans: first name in lower case at ipal.net) |
 
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