very low voltage -- dangerous?

Not necessarily "very near". In fact in a lot of places it's just about impossible to escape induction of power line current into almost any other conductor, simply because power wiring, and hence the fields from it, are ubiquitous.

Where is this very near conductor? It has to be *very* near! (Okay, it could also be very large... ;-)

Reply to
Floyd L. Davidson
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To avoid possible confusion, it's a good idea to use as load something that doesn't have a lot of electronics in it. Incandescent bulb, motors, etc. Some devices may present very high impedance (and fail to short out the stray/phantom voltage) except if there's enough voltage to activate the electronics. Eg: electronic ballasts may not do the trick.

Reply to
Chris Lewis

Certainly good for hum pickup in audio circuits. The OP originally reported millivolts, which could be induction. Later post was about 1V which sounds high for induction. Most posts are more like 20V or higher which I don't think is possible with induction. Magnetic fields in one wire are almost entirely canceled by the opposite field in the adjacent return wire. If misswired with the 'supply' and 'return' taking different paths, at high current the 'loop' can reportedly produce a magnetic field that can screw up the picture on a CRT near the loop.

A common source is probably a run to a switch which is only hot feed and switched return. In any case, wires in the same cable or pipe.

Reply to
bud--

Sure it is. I've seen over 50 volts from induction.

That would be true if the cable was twisted pair and the circuit is common mode, but it isn't. The longitudinal balance is no where near good enough to cancel "almost entirely", and the current in the two lines does not have a 180 degree phase at any particular location anyway, because it is a single ended circuit not a common mode balanced circuit.

If wired normally, it can do that. It is *not* a balanced circuit and the two lines do not cancel each other.

The distance required to get much capacitive leakage is huge. The distance required to get induction is relatively small.

Reply to
Floyd L. Davidson

Induced voltage will generally be less than that of voltage across the inductive reactance of whatever is inducing it, and that is L times di/dt. di/dt will normally be current times 2 times pi times 60, and L of house wiring is generally in the hundreds of microhenries. I don't see much of a transformer being likely to be accidentally made of household wiring. Also, induced voltages in wires will generally have fairly low impedance and not be loaded down by common analog meters.

As for capacitive - the capacitance between adjacent conductors in a cable is in the 10's of pF per foot ballpark. Multiply that by 2 times pi times frequency times voltage and that is how much current can flow capacitively from one conductor to another - a few microamps per foot! Enough current flows through the plug of a switched-off floor lamp to make a neon test lamp visibly glow. An unconnected wire adjacent to a hot one in a cable easily picks up enough capacitively to generate a fairly high voltage reading on high impedance meters, and even an impressive reading on many analog meters.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

That's nonsense. (What you wrote is; but that may not be what you were thinking. Please restate that point.)

The impedance of the circuit is significant, *and* it affects the voltage seen (either with a relatively low impedance analog meter or with relatively high impedance digital meter). How much it affects it depends on the ratio of circuit impedance to meter impedance, but even with an analog meter that is a relatively high ratio for many circuits (i.e., for any circuit that is not an "open circuit").

True for communications cables, but much lower for power cable. Regardless, that is nothing like the capacitance between adjacent _cables_, which is what we would be interested in.

But that does not deal with the cited case, which would be capacitive coupling between different cables, not between the conductors in a single cable.

Reply to
Floyd L. Davidson

Measured capacitance for 14/2 Romex - 15 pF/ft hot-to-neutral, ground open. I would call that 10's of pF per foot *ballpark*.

May be what you are interested in, but I am interested in "wires in the same cable or pipe". For example a 2 wire run to a switch with "only hot feed and switched return" or 3 wire run with hot, neutral, swiitched.

It deals with my cited case exactly. You eliminate capacitive coupling by eliminating the places where it occurs.

Reply to
bud--

I wouldn't. That's 1.5 * 10, not even 2x.

You do realize how much reactance 15 pF/ft is at 60 Hz, right? How many feet of cable does it take to even become slight, much less significant? A 100 foot run, at 15 pF/ft simply isn't much.

Even so, it is less than typical induction.

I'm just pointing out that capacitive coupling is very slight, compared to inductive coupling, which is virtually always relatively significant.

Reply to
Floyd L. Davidson

Refreshing to see a knowledgeable person post information in a clear and concise way that fits the situation so well and negates the misinformation so prevalent on this group lately.

KUDOS for a factual, yet simple response.

Twayne

Floyd L. Davids>> Floyd L. Davids>> Certainly good for hum pickup in audio circuits. The OP originally

Reply to
Twayne

An analog meter cannot have a high input Z? They can easily be high enough to read phantom voltages, as they're usually called.

Twayne

Reply to
Twayne

Kind words. Thank you.

Reply to
Floyd L. Davidson

Gee - the reactance could actually be calculated.

And a 100 ft cable at 15 pF/ft gives a reactance of 1.77meg ohms. Assume you have one cable wire connected to hot and you are measuring from the other wire to neutral ? classical measurement of ?phantom voltage? on a switch run. A good digital meter would have an input impedance of at least 10M. The voltage measured would be over 100V.

With 20 ft - common to have that much cable - the reactance would be

8.8M and the voltage would be over 55V [over 90V if you add the resistance and reactance correctly].

Using a good analog meter - a Simpson 260 has an input impedance of 5000 ohms/volt on AC scales. On the 250V scale the input impedance is 1.25M. The reading for 20 ft of cable would be almost 15V - an ?impressive reading?. (On the 50V scale the reading would be 3.3V - falling voltage on lower scale indicates high impedance source.)

A random piece of zip cord had a measured capacitance of 17 pF/ft. The capacitive leakage current calculates as about 0.75 microA per foot. (Compare to Don?s value of ?a few microamps per foot?.)

Reply to
bud--

A reality check is in order. Since those numbers are *not* what people actually read, something is wrong with your calculation.

Reply to
Floyd L. Davidson

Some people _do_ see voltages as high as that.

The calculations are "ideal", with known conditions and ignoring other factors. In the real world, instead of 20ft of consistent per-foot capacitance of conductors in the same cable and ignoring everything else, you're seeing capacitance between conductors in different cables with varying capacitance per foot, capacitance to other things, resistive leakage, inductance effects, and others.

Simple calculations are good to help you get an idea of the magnitudes you might see, but real world instances of this nature are almost impossible to analytically predict _exactly_ what the voltage is going to be. Easier to just measure ;-)

Reply to
Chris Lewis

According to Twayne :

They can, but as analog meters are usually 50K ohms (or less) per volt, and digitals are often in the 10s of megaohms, the effect is usually a lot more drastic with the latter.

Reply to
Chris Lewis

And that's why they report seeing millivolts?

Reply to
Floyd L. Davidson

How do you know what people actually read. A brief look at voltages people on this newsgroup thought were ?phantom? - 33, 75, 27, 104, 60. The voltages in this thread are extremely low. Almost everyone would ignore voltages that low.

What is wrong with the calculations. Maybe you could find an error.

If I did a test setup matching my calculations you would say I got inductive coupling.

A couple other possible sources - a circuit with an open neutral a 14/3 from switch to lite with capacitive divider from switched wire to hot, neutral, ground.

Don gave an engineering reason why inductive coupling will not be very high which you ignored.

You are long on opinion and short on science.

And from NEMA (the National Electrical Manufacturers Association, a very active organization): ?Phantom Voltages ..... Conductors that are installed in close proximity to one another, and are capacitively coupled to each other, can cause this a.c. voltage reading. Such a reading could be 2 or 3 volts, or it may be as high as the voltage on the adjacent conductors.?

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from Jeff Wisnia]

Reply to
bud--

120V 60 Hz across 1500 pF will result in 69 microamps of current. Full scale reading of many common analog meters is 50 microamps.

Guessing velocity factor of 2/3 at low frequencies, that 15 pF/foot for

14/2 romex works out a characteristic impedance of 102 ohms and an inductance of .155 microhenry per foot (current going out along the hot and back along the neutral). Put 15 amps through a 100 foot run of this and the EMF around that loop is about .09 volt. Any secondary 1-turn loop not perfectly coinciding with the primary loop in this example will have less than .09 volt induced in it.

You can get more inductance ang greater induced voltage with more current or if the current going one way and its return take paths farther apart than the hot and neutral in a 14/2 romex cable are.

Try connecting the black wire of a 20 foot run of 15 pF/foot romex to

120V hot, have ground open, and have an analog meter on an AC voltage range go from neutral (neutral is otherwise open) to something grounded. Let's say the meter has a sensitivity rating of 10K-ohms per volt on AC voltage ranges and you are using a 250V range, for a meter resistance of 2.5 megohms. That 300 pF will allow enough current through to get a 33 volt reading.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

When people get voltage readings from capacitive coupling, it is usually from something with less coupling than hot to neutral of same cable with the grounding conductor open or absent and length 10's of feet.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

I have heard people report volts and even occaisionally quite a few volts and 10's of volts before - for example, voltage from a switched hot to ground with the switch off, when the switch is at the end of cable having a switched hot and an unswitched hot.

In rooms that are especially bad for "electrostatic hum", I have seen a long wire, over a foot from any wall/floor/ceiling and not connected to anything but an oscilloscope input pick up about a volt. More common is several millivolts (usually 10's) to a tenth or two of a volt. The waveform was a distorted 60 Hz sinewave.

Keep in mind that most AC analog meters will not give a reading at all on just a few millivolts. It takes at least one or two tenths of a volt to get even a fraction of a microamp through most silicon diodes.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

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