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Power line indicator

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.

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:

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:

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.

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
Reply to
Jeff Liebermann
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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

Reply to
PlainBill

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:

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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.

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5 &um=3D1&ie=3DUTF-8&tbm=3Dshop&cid=3D13690361079853906458 &sa=3DX&ei=3DQtbwTsCMG-mDsgLz6PiXAQ&ved=3D0CFsQ8wIwAQ 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--

Reply to
bud--

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 0.01µF - not a very big capacitor. I've got a bag of .01µF 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

Reply to
PlainBill

. I

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

Reply to
Robert Macy

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

Reply to
Angelo Campanella

Do the math. You can't get enough capacitance by using any reasonable size piece of foil.

PlainBill

Reply to
PlainBill

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

Reply to
PlainBill

. I

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

Reply to
klem kedidelhopper

Look on your pen and see if there's a patent number. Then search for the patent with google:

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
Reply to
Jeff Liebermann

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.

Reply to
William Sommerwerck

Foundit.

Schematic in sheet #4. No FET on the input stage. It uses a 4069 CMOS hex inverter through a 10Meg resistor. Cheap.

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

Uh... CMOS uses MOSFETs.

Reply to
William Sommerwerck

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
Reply to
Jeff Liebermann

Reply to
William Sommerwerck

LOL !!!

Reply to
Robert Macy

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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Reply to
David Lesher

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