Power line indicator

Engineer · Dec 19, 2011

On Dec 18, 3:22 pm, Jeff Urban <jurb6...@gmail.com> wrote:

"If I'm understanding this
correctly when you "wrap" the probe wire around the conductor that
would appear to be induction, (transformer of sorts), "

That is correct if you provide a return path at the other end of the
wire. Doing that will make it sense current inductively. If you leave
it open it wil detect the voltage capacitively.

J

But the "return path" coil will then detect current, not voltage. I
think the capacitive idea is better, but you might need a bit of load
on it to avoid false detection of noise on the line.
Cheers.
Roger

Jeff Liebermann · Dec 19, 2011

On Mon, 19 Dec 2011 06:08:43 -0800 (PST), klem kedidelhopper
<captainvideo462009@gmail.com> wrote:

I think that what you may be talking about Jeff is an "automatic"
transfer operation. This sounds like a system that will automatically
start the generator on sensing a loss of power, and then activate a
huge double pole contactor.

Yes, I mentioned that in one sentence. That's the correct way to wire
an emergency generator to a house. However, I continued beyond that
sentence and attempted to reverse engineer what you're doing with the
generator, and guessed that you are back feeding the wall outlet with
a suicide cord. If this is correct, there are safety issues involved,
some of which I detailed. You might want to re-read what I scribbled.

Therefore, and
please correct me if I'm wrong but it would seem that with any manual
system, whether you use one of these manual switches or you back feed
a 240 volt circuit and manually kill the main, you would have no way
of knowing when the utility is back in service. Lenny

Again, please re-read my previous 2nd paragraph. If you are powering
a few appliances from the generator via an extension cord, and NOT
backfeeding the wall outlets, then you'll know when the power is back
on when the house lights return to normal. The reason you probably
can't tell that the power is back on is that you have switched the
main breaker to off. You'll need a neon lamp capacitive sensor.

Flipping the main breaker on, while the generator is running, is a
great way of blowing up some generators. Flipping the breaker on,
while the generator is running, while the power company is working on
the lines, is a hazard to the lineman. However, they protect
themselves by shorting or grounding the lines, which will usually blow
the output breakers on your generator.

The bottom line is that you really should spend the money on a proper
transfer switch. It does NOT need to be automatic. Manual transfer
switches are much cheaper and good enough. If you have an autostart
generator, you will need an automatic transfer switch. If not, use a
manual switch:
<http://www2.northerntool.com/generators/transfer-switches-4.htm>
<www.amazon.com/s?ie=UTF81&keywords=manual transfer switch>
<http://www.reliancecontrols.com/Documents/S1204TE.pdf>

Some transfer switches are *NOT* intended to run the entire house.
They have their own breakers and their own loads off a sub-panel.
See Fig 3 wiring diagram at:
<http://www.renovation-headquarters.com/generator-transfer-switch.htm>
When the power returns, the lights and devices that are NOT powered by
the transfer box, come back on. Meanwhile the generator is still
running the rest of the outlets. Flip the switch, and these loads are
now on utility power. Then, you can turn off the generator.

Other transfer switches switch the entire breaker panel.
<http://www.reliancecontrols.com/ProductDetail.aspx?TCA1006D>
Since these are installed between the meter and the breaker box,
you'll know that power had returned when the meter starts moving.

Note: I have 3 generators, none of which work. Yet another project.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

Dec 19, 2011

On Sat, 17 Dec 2011 10:59:52 -0800 (PST), klem kedidelhopper
<captainvideo462009@gmail.com> wrote:

Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.
I thought about the simplest way which would be to wire a small NE2
neon lamp across the 240, ahead of the main, (with the appropriate
series resistor of course). The lamp would be lit all the time there
is utility supplied power and off during a blackout. The plan is to
not have to uselessly be running my generator after power unbeknown to
me is restored. The down side of this if it is really any concern
would be that this lamp, small as it as well as its associated wiring
would be would not be protected by a breaker.
The lamp is of course the simplest way, but I was wondering if there
is some sort of inductive circuit someone may know of that that would
not require that current be flowing through a conductor, basically a
circuit that would indicate the presence of voltage.
For instance I carry a little "pen" shaped device in my tool box. I
press and hold an on button and the unit chirps when brought near a
hot AC circuit. It's a great time saver when trouble shooting a job,
but can something like this be implemented as a full time monitor
circuit? It could be powered off small rechargeable batteries and
always indicate the state of the entrance cable. Thanks, Lenny
I can see four approaches, one of which which may meet your needs.

The 'Neon light with capacitove link' would certainly be inexpensive,
however I question if sufficient current would be generated to light a
neon bulb. It would be easy to test, and I might actually try it
sometime. If I can find my neon tester. It's probably with my tube
tester.

(Shoot, it should be possible to calculate the area of
foil required!!)

I have a Christmas Tree Light Tester. It uses capacitive pickup to
sense the voltage differential to indicate if the wires are at close
to the same voltage (good light) or at different voltages (light bulb
burned out). LEDs light to indicate the status when the button is
pressed. Less than $10 at K-Mart, as I recall. With a little
'circuit bending' it could produce the results you desire.

You didn't indicate how much current and the number of circuits the
generator is supplying. One approach would be to add an expansion
circuit breaker panel next to the present panel, and power all
'emergency' circuits through the expansion panel. The lines from the
main panel would go through a circuit breaker in the main panel and
another in the expansion panel. Switchover would be handled by
opening the breaker accepting current from the main panel and closing
the one accepting current from the generator (someone mechanically
clever could build an interlock so both could not be closed at the
same time).

Of course, there are also the commercially available switches; I find
the price of them to be excessive, even the manual ones.

PlainBill

bud-- · Dec 20, 2011

On Dec 18, 10:18 pm, Jeff Liebermann <je...@cruzio.com> wrote:

On Sat, 17 Dec 2011 10:59:52 -0800 (PST), klem kedidelhopper

captainvideo462...@gmail.com> wrote:
Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.

The neon lamp method works. You'll need one for each phase.

However, I smell a different problem. I couldn't see why you would
need to monitor the voltage before the main breaker. If you had a
proper transfer switch, it wouldn't be necessary:
http://www2.northerntool.com/generators/transfer-switches-4.htm
If you're powering some appliances with an extension cord, you can
tell if thepowerhas returned when the house lights come back on.

My guess(tm) is that you're opening the main breaker when thepower
dies, and back feeding a wall outlet with a suicide cord (apowerplug
on each end of the cable). If you're really clever, you'll backfeed
both phases with a 220VAC suicide cord.

The 220VAC version will work, but the single phase suicide cord has a
problem. Under normal conditions, the current through the neutral
wire is nearly zero. That's because the 180 degree phase difference
between each phase cancels the current in the neutral wire. However,
with only one phase powered, you're likely to see the full current
through the neutral. If you're creative and parallel both phases,
then the neutral current is even higher. Code compliant wiring can
survive this, but I've seen a few blown neutrals. That's when the
neutral wiring was rotten, and nobody noticed until the full current
was applied.

Another problem is that when you back feed a single wall outlet with a
suicide cord, you're going to be limited by the current limit of the
connected circuit breaker. In the US, that's about 15A. That should
be sufficient for most loads, but I suggest to disconnect the high
current loads to prevent blowing the breaker.

Probably the cheapest way to transfer significant load is to wire the
generator to a breaker in the service panel and use a mechanical
interlock:
http://www.interlockkit.com/CATALOG2008.pdf
These prevent both the generator breaker and service disconnect from
being on at the same time. They are NEC compliant. (The generator
breaker is back-fed and the NEC requires a device from the
manufacturer that prevents the breaker from coming out.)

Generator-to-breaker without the interlock is a serious code violation
and not safe.

A safe (and code compliant) way to make a temporary generator
connection to the building is to install an "inlet" fitting on the
building connected to the generator breaker.
http://www.google.com/products/catalog?q=inlet+receptacle
&oe=utf-8&rls=org.mozilla:en-US

fficial&client=firefox-a&as_qdr=y15
&um=1&ie=UTF-8&tbm=shop&cid=13690361079853906458
&sa=X&ei=QtbwTsCMG-mDsgLz6PiXAQ&ved=0CFsQ8wIwAQ
Use the appropriate rating. The prongs are never live since the
generator breaker is never on except when the service breaker is off.
A cord from the generator plugs into the inlet.

I would never wire anything ahead of the service disconnect. Nesesu
described what can happen with a "fault", and his was not directly off
the service wires. Even if fuses were used they would have to be rated
for the "available fault current". Other fuses would give the same
result nesesu got.

--
bud--

Dec 21, 2011

On Sat, 17 Dec 2011 11:50:16 -0800 (PST), nesesu
<neil_sutcliffe@telus.net> wrote:

On Dec 17, 10:59 am, klem kedidelhopper <captainvideo462...@gmail.com
wrote:
Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.
I thought about the simplest way which would be to wire a small NE2
neon lamp across the 240, ahead of the main, (with the appropriate
series resistor of course). The lamp would be lit all the time there
is utility supplied power and off during a blackout. The plan is to
not have to uselessly be running my generator after power unbeknown to
me is restored. The down side of this if it is really any concern
would be that this lamp, small as it as well as its associated wiring
would be would not be protected by a breaker.
The lamp is of course the simplest way, but I was wondering if there
is some sort of inductive circuit someone may know of that that would
not require that current be flowing through a conductor, basically a
circuit that would indicate the presence of voltage.
For instance I carry a little "pen" shaped device in my tool box. I
press and hold an on button and the unit chirps when brought near a
hot AC circuit. It's a great time saver when trouble shooting a job,
but can something like this be implemented as a full time monitor
circuit? It could be powered off small rechargeable batteries and
always indicate the state of the entrance cable. Thanks, Lenny

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

Neil S.
I did some rough calculations with this, and it doesn't look good.

The recommended resistance for an NE2 on a nominal 220V line is 220
Kohm. You are suggesting using capactive reactance instead of a
resistor. That is valid for a fixed frequency signal like a power
line.

Using the equation for capacitive reactance, you need 0.01ľF - not a
very big capacitor. I've got a bag of .01ľF 1KV ceramic caps sitting
on my desk (for another project I STILL haven't gotten to), they are
about 3/8" in diameter and 1/16" thick.

Now the calculations get a little dicey. First some big assumptions.
200A service entrance, whiich requires 000 gauge wire, and a couple of
real SWAG guesses of insulation 0.1" thick, said insulation having a
dielectric constant of 1.

000 guage wire has a diameter of .4", so a diameter of 1.25". That
means caovering a 8" long piece of wire with foil will give a
capacitance of 22pF. You would need to cover about 300' of wire with
aluminum foil.

PlainBill

Angelo Campanella · Dec 21, 2011

neil_sutcliffe@telus.net> wrote:

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

That looks like a reasonable monitoring action except that when you are
dealing with the incoming line that has no protection, you have to be sure
that whatever you connect across it does not short or catch on fire in the
long run... Aside from that, You only need to monitor one 115v side to
determine whether the power is gone. I suggest you try various physical
"capacitors" by connecting them in series with the NE-2. A 50' extension
cord would provide maybe 2500 pF (.00025 uF), which may be enough to dimly
light that lamp. For instance, with the 50' extension coil disconnected and
laying on the floor, using a plug pigtail, connect the common or ground
house wire to one lead of the NE-2. Connect the hot house wire to the hot
line of the 50' extension cord. the capacitance of the 50' coil should
supply enough brightness to be seen in a dark room.

Even simpler still and a lot safer, buy any extension cord that has a
neon light built into one end. Install it in your favorite house room. Plug
it in and hang the lit end anywhere in sight.

Ange

Robert Macy · Dec 21, 2011

On Dec 20, 5:10 pm, PlainB...@yawhoo.com wrote:

On Sat, 17 Dec 2011 11:50:16 -0800 (PST), nesesu

neil_sutcli...@telus.net> wrote:
On Dec 17, 10:59 am, klem kedidelhopper <captainvideo462...@gmail.com
wrote:
Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.
I thought about the simplest way which would be to wire a small NE2
neon lamp across the 240, ahead of the main, (with the appropriate
series resistor of course). The lamp would be lit all the time there
is utility supplied power and off during a blackout. The plan is to
not have to uselessly be running my generator after power unbeknown to
me is restored. The down side of this if it is really any concern
would be that this lamp, small as it as well as its associated wiring
would be would not be protected by a breaker.
The lamp is of course the simplest way, but I was wondering if there
is some sort of inductive circuit someone may know of that that would
not require that current be flowing through a conductor, basically a
circuit that would indicate the presence of voltage.
For instance I carry a little "pen" shaped device in my tool box.. I
press and hold an on button and the unit chirps when brought near a
hot AC circuit. It's a great time saver when trouble shooting a job,
but can something like this be implemented as a full time monitor
circuit? It could be powered off small rechargeable batteries and
always indicate the state of the entrance cable. Thanks, Lenny

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

Neil S.

I did some rough calculations with this, and it doesn't look good.

The recommended resistance for an NE2 on a nominal 220V line is 220
Kohm. You are suggesting using capactive reactance instead of a
resistor. That is valid for a fixed frequency signal like a power
line.

Using the equation for capacitive reactance, you need 0.01ľF - not a
very big capacitor. I've got a bag of .01ľF 1KV ceramic caps sitting
on my desk (for another project I STILL haven't gotten to), they are
about 3/8" in diameter and 1/16" thick.

Now the calculations get a little dicey. First some big assumptions.
200A service entrance, whiich requires 000 gauge wire, and a couple of
real SWAG guesses of insulation 0.1" thick, said insulation having a
dielectric constant of 1.

000 guage wire has a diameter of .4", so a diameter of 1.25". That
means caovering a 8" long piece of wire with foil will give a
capacitance of 22pF. You would need to cover about 300' of wire with
aluminum foil.

PlainBill

Uh, the capacitance limits the current, you don't need a resistor. You
only need the resistor if you directly connect to the 220V

Dec 21, 2011

On Tue, 20 Dec 2011 18:14:54 -0800 (PST), Robert Macy
<robert.a.macy@gmail.com> wrote:

On Dec 20, 5:10 pm, PlainB...@yawhoo.com wrote:
On Sat, 17 Dec 2011 11:50:16 -0800 (PST), nesesu

neil_sutcli...@telus.net> wrote:
On Dec 17, 10:59 am, klem kedidelhopper <captainvideo462...@gmail.com
wrote:
Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.
I thought about the simplest way which would be to wire a small NE2
neon lamp across the 240, ahead of the main, (with the appropriate
series resistor of course). The lamp would be lit all the time there
is utility supplied power and off during a blackout. The plan is to
not have to uselessly be running my generator after power unbeknown to
me is restored. The down side of this if it is really any concern
would be that this lamp, small as it as well as its associated wiring
would be would not be protected by a breaker.
The lamp is of course the simplest way, but I was wondering if there
is some sort of inductive circuit someone may know of that that would
not require that current be flowing through a conductor, basically a
circuit that would indicate the presence of voltage.
For instance I carry a little "pen" shaped device in my tool box. I
press and hold an on button and the unit chirps when brought near a
hot AC circuit. It's a great time saver when trouble shooting a job,
but can something like this be implemented as a full time monitor
circuit? It could be powered off small rechargeable batteries and
always indicate the state of the entrance cable. Thanks, Lenny

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

Neil S.

I did some rough calculations with this, and it doesn't look good.

The recommended resistance for an NE2 on a nominal 220V line is 220
Kohm. You are suggesting using capactive reactance instead of a
resistor. That is valid for a fixed frequency signal like a power
line.

Using the equation for capacitive reactance, you need 0.01ľF - not a
very big capacitor. I've got a bag of .01ľF 1KV ceramic caps sitting
on my desk (for another project I STILL haven't gotten to), they are
about 3/8" in diameter and 1/16" thick.

Now the calculations get a little dicey. First some big assumptions.
200A service entrance, whiich requires 000 gauge wire, and a couple of
real SWAG guesses of insulation 0.1" thick, said insulation having a
dielectric constant of 1.

000 guage wire has a diameter of .4", so a diameter of 1.25". That
means caovering a 8" long piece of wire with foil will give a
capacitance of 22pF. You would need to cover about 300' of wire with
aluminum foil.

PlainBill

Uh, the capacitance limits the current, you don't need a resistor. You
only need the resistor if you directly connect to the 220V
Do the math. You can't get enough capacitance by using any reasonable

size piece of foil.

PlainBill

Dec 21, 2011

On Tue, 20 Dec 2011 22:10:21 -0500, "Angelo Campanella"
<a.campanella@att.net> wrote:

neil_sutcliffe@telus.net> wrote:

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

That looks like a reasonable monitoring action except that when you are
dealing with the incoming line that has no protection, you have to be sure
that whatever you connect across it does not short or catch on fire in the
long run... Aside from that, You only need to monitor one 115v side to
determine whether the power is gone. I suggest you try various physical
"capacitors" by connecting them in series with the NE-2. A 50' extension
cord would provide maybe 2500 pF (.00025 uF), which may be enough to dimly
light that lamp. For instance, with the 50' extension coil disconnected and
laying on the floor, using a plug pigtail, connect the common or ground
house wire to one lead of the NE-2. Connect the hot house wire to the hot
line of the 50' extension cord. the capacitance of the 50' coil should
supply enough brightness to be seen in a dark room.

Even simpler still and a lot safer, buy any extension cord that has a
neon light built into one end. Install it in your favorite house room. Plug
it in and hang the lit end anywhere in sight.

Ange
Do the math. You need 10,000pF at 220 volts, roughly double that at

120 volts to drive a NE2 lamp to full brightness. Treat it like a
parallel plate capacitor; the formula isC= K*0.2248*A/d. K is the
dialectric constant (most materials have a dielectric constant between
1 and 10), A is the area of the plates, d is the distance between the
plates.

If you wrap foil around a 000 Ga insulated wire you will be getting
roughly 1 square inch of capacitive area per linear inch of cable.
Heavy gauge extention cord has 14 Ga wire, being generous you will be
getting .06 square inches of capacitive area per linear inch of the
cord.

PlainBill

klem kedidelhopper · Dec 23, 2011

On Dec 20, 7:10 pm, PlainB...@yawhoo.com wrote:

On Sat, 17 Dec 2011 11:50:16 -0800 (PST), nesesu

neil_sutcli...@telus.net> wrote:
On Dec 17, 10:59 am, klem kedidelhopper <captainvideo462...@gmail.com
wrote:
Our area suffers from frequent blackouts, and many people including
myself use generators, (manually) during these periods. It would be
really helpful to know when the utility side of the circuit is once
again live so that I can put the generator away. Basically I need to
monitor the entrance cable side ahead of the main.
I thought about the simplest way which would be to wire a small NE2
neon lamp across the 240, ahead of the main, (with the appropriate
series resistor of course). The lamp would be lit all the time there
is utility supplied power and off during a blackout. The plan is to
not have to uselessly be running my generator after power unbeknown to
me is restored. The down side of this if it is really any concern
would be that this lamp, small as it as well as its associated wiring
would be would not be protected by a breaker.
The lamp is of course the simplest way, but I was wondering if there
is some sort of inductive circuit someone may know of that that would
not require that current be flowing through a conductor, basically a
circuit that would indicate the presence of voltage.
For instance I carry a little "pen" shaped device in my tool box.. I
press and hold an on button and the unit chirps when brought near a
hot AC circuit. It's a great time saver when trouble shooting a job,
but can something like this be implemented as a full time monitor
circuit? It could be powered off small rechargeable batteries and
always indicate the state of the entrance cable. Thanks, Lenny

Lenny, try taking a plain NE-2 with no resistor and connecting about a
foot of insulated #22 solid insulated wire to one lead and wrapping
that around a piece of insulated wire the same size as your service
entrance wire and connecting the other lamp lead to neutral. Energise
the heavy wire with 120V and see if the NE-2 glows just from the
capacitance of the large wire to small wire. That is about as simple
as it gets and is adequately insulated from the service cable to not
need any current protection.

Neil S.

I did some rough calculations with this, and it doesn't look good.

The recommended resistance for an NE2 on a nominal 220V line is 220
Kohm. You are suggesting using capactive reactance instead of a
resistor. That is valid for a fixed frequency signal like a power
line.

Using the equation for capacitive reactance, you need 0.01ľF - not a
very big capacitor. I've got a bag of .01ľF 1KV ceramic caps sitting
on my desk (for another project I STILL haven't gotten to), they are
about 3/8" in diameter and 1/16" thick.

Now the calculations get a little dicey. First some big assumptions.
200A service entrance, whiich requires 000 gauge wire, and a couple of
real SWAG guesses of insulation 0.1" thick, said insulation having a
dielectric constant of 1.

000 guage wire has a diameter of .4", so a diameter of 1.25". That
means caovering a 8" long piece of wire with foil will give a
capacitance of 22pF. You would need to cover about 300' of wire with
aluminum foil.

PlainBill

So then if you are capacitively coupling to the main it will require
some type of amplification to activate a signal of some kind. My
little "pen" I mentioned earlier blinks and beeps when held near a
live line. I have never opened it but it runs off a small button cell
of some kind. I haven't tried the inductive, (wrapping the probe wire
around one leg of the entrance cable) method yet.

Years ago I designed a little circuit to monitor compressor run time
on a refrigerator. A small coil was wound around one of the wires to
the compressor. That pickup coil, through a step up transformer
introduced sufficient signal into the input of a UA709. The output
turned on a relay driver which ran an elapsed time indicator. I
thought of trying to modify that circuit to include rechargeable
battery back up, modify the pickup, different turns ratio on the
transformer etc. However depending on what was on in the house when
power failed, the total load would always be different. I never try to
run my electric stove, well pump or electric hot water heater from the
generator. However once utility power is restored they would need to
be turned back on. Only one thing would be certain though. After an
extended blackout the chip would be blasted with an overload. So
trying to implement this old circuit doesn't seem feasible. I suppose
I probably ought to look more at the capacitive "pen"approach too.
Does anyone know exactly what is inside that little "pen"? Lenny

Jeff Liebermann · Dec 23, 2011

On Fri, 23 Dec 2011 07:34:25 -0800 (PST), klem kedidelhopper
<captainvideo462009@gmail.com> wrote:

I suppose
I probably ought to look more at the capacitive "pen"approach too.
Does anyone know exactly what is inside that little "pen"? Lenny

Look on your pen and see if there's a patent number. Then search for
the patent with google:
<http://www.google.com/patents?hl=en>
There's usually a schematic or at least a block diagram.

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

William Sommerwerck · Dec 23, 2011

I probably ought to look more at the capacitive "pen"approach, too.
Does anyone know exactly what is inside that little "pen"?

I believe they use a FET to drive a lamp or other indicator. As FETs (can)
have a very high input impedance, it's possible to pick up 60Hz AC through
capacitive coupling.

Jeff Liebermann · Dec 24, 2011

On Fri, 23 Dec 2011 08:47:43 -0800, Jeff Liebermann <jeffl@cruzio.com>
wrote:

On Fri, 23 Dec 2011 07:34:25 -0800 (PST), klem kedidelhopper
captainvideo462009@gmail.com> wrote:

I suppose
I probably ought to look more at the capacitive "pen"approach too.
Does anyone know exactly what is inside that little "pen"? Lenny

Look on your pen and see if there's a patent number. Then search for
the patent with google:
http://www.google.com/patents?hl=en
There's usually a schematic or at least a block diagram.

Foundit.
<http://www.google.com/patents/US5877618>
Schematic in sheet #4. No FET on the input stage. It uses a 4069
CMOS hex inverter through a 10Meg resistor. Cheap.
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# 831-336-2558
# http://802.11junk.com jeffl@cruzio.com
# http://www.LearnByDestroying.com AE6KS

William Sommerwerck · Dec 24, 2011

"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:k8paf7h1n278p77pcntr9r3hapos7t0kmm@4ax.com...

http://www.google.com/patents/US5877618
Schematic in sheet #4. No FET on the input stage. It uses
a 4069 CMOS hex inverter through a 10M resistor.

Uh... CMOS uses MOSFETs.

Jeff Liebermann · Dec 24, 2011

On Sat, 24 Dec 2011 04:11:49 -0800, "William Sommerwerck"
<grizzledgeezer@comcast.net> wrote:

"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:k8paf7h1n278p77pcntr9r3hapos7t0kmm@4ax.com...

http://www.google.com/patents/US5877618
Schematic in sheet #4. No FET on the input stage. It uses
a 4069 CMOS hex inverter through a 10M resistor.

Uh... CMOS uses MOSFETs.

Yes, but you said "a" FET which I read as a single JFET or MOSFET used
as an analog 60Hz amplifier. The CMOS gate has far more than one FET
in the input stage.

Pedantic-R-Us

--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

William Sommerwerck · Dec 24, 2011

"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:400cf7pa4r872kdh4k1hhb8f6grlofn62v@4ax.com...

On Sat, 24 Dec 2011 04:11:49 -0800, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:
"Jeff Liebermann" <jeffl@cruzio.com> wrote in message
news:k8paf7h1n278p77pcntr9r3hapos7t0kmm@4ax.com...

http://www.google.com/patents/US5877618
Schematic in sheet #4. No FET on the input stage. It uses
a 4069 CMOS hex inverter through a 10M resistor.

Uh... CMOS uses MOSFETs.

Yes, but you said "a" FET which I read as a single JFET or MOSFET
used as an analog 60Hz amplifier. The CMOS gate has far more than
one FET in the input stage.

Pedantic-R-Us

<mumble-mumble grumble>

Robert Macy · Dec 26, 2011

On Dec 24, 9:59 am, "William Sommerwerck" <grizzledgee...@comcast.net>
wrote:

"Jeff Liebermann" <je...@cruzio.com> wrote in message

news:400cf7pa4r872kdh4k1hhb8f6grlofn62v@4ax.com...

On Sat, 24 Dec 2011 04:11:49 -0800, "William Sommerwerck"
grizzledgee...@comcast.net> wrote:
"Jeff Liebermann" <je...@cruzio.com> wrote in message
news:k8paf7h1n278p77pcntr9r3hapos7t0kmm@4ax.com...
http://www.google.com/patents/US5877618
Schematic in sheet #4. No FET on the input stage. It uses
a 4069 CMOS hex inverter through a 10M resistor.
Uh... CMOS uses MOSFETs.
Yes, but you said "a" FET which I read as a single JFET or MOSFET
used as an analog 60Hz amplifier. The CMOS gate has far more than
one FET in the input stage.
Pedantic-R-Us

mumble-mumble grumble

LOL !!!

David Lesher · Feb 16, 2012

a) Some transfer switches are whole-house, being the first
step downstream from the meter.

b) Others are fed by the main panel, and feed a subpanel of
priority loads.

In the case of A, there is a specific metering tap to go to the
line monitoring, even if that's solely a "LINE POWER" lamp.

In either case, stupidity like suicide cords should be
tossed.

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& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

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