Phantom Voltage?


T

toller

I am installing a switch leg on a ceiling light that now has a chain switch.
When I measure voltage at the new switch it is H-N and H-G 120.2v with the
chain pull on; just like expected.
With the chain pull off, it measures H-N 61.8v and H-G 120.2v.

Is that 61.8v the famous phantom voltage I have read about so many times
here? It seems weird that if it doesn't exist that it reads exactly 61.8v
every time I measure it; but of course it can't be real because there is no
circuit.
What causes this?

I only measured at all because I wanted to make sure my ground was good;
which it obviously is.
 
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J

John McGaw

toller said:
I am installing a switch leg on a ceiling light that now has a chain switch.
When I measure voltage at the new switch it is H-N and H-G 120.2v with the
chain pull on; just like expected.
With the chain pull off, it measures H-N 61.8v and H-G 120.2v.

Is that 61.8v the famous phantom voltage I have read about so many times
here? It seems weird that if it doesn't exist that it reads exactly 61.8v
every time I measure it; but of course it can't be real because there is no
circuit.
What causes this?

I only measured at all because I wanted to make sure my ground was good;
which it obviously is.
This happens because you are using a digital meter with an extremely high
input impedance which allows even minute capacitive coupling, as in an open
switch, to "leak" voltage (well, technically current but lets not quibble).
If you had an old-fashioned analog multimeter there might be some measurable
voltage but it would be much lower. In either case if you put a load such as
an incandescent lamp across the lines the voltage will be pretty much
undetectable with either sort of meter.
 
T

Tekkie

toller posted for all of us....
I am installing a switch leg on a ceiling light that now has a chain switch.
When I measure voltage at the new switch it is H-N and H-G 120.2v with the
chain pull on; just like expected.
With the chain pull off, it measures H-N 61.8v and H-G 120.2v.
This worries me H-G 120v! Switch seems to be on the neutral leg. I'm not
a sparky so I could be wrong. Hope Jake or John Grabowski reads this.
 
T

toller

Tekkie said:
toller posted for all of us....


This worries me H-G 120v! Switch seems to be on the neutral leg. I'm not
a sparky so I could be wrong. Hope Jake or John Grabowski reads this.
No no. The readings are at the switch, not the light.
 
K

Kenneth

This happens because you are using a digital meter with an extremely high
input impedance which allows even minute capacitive coupling, as in an open
switch, to "leak" voltage (well, technically current but lets not quibble).
If you had an old-fashioned analog multimeter there might be some measurable
voltage but it would be much lower. In either case if you put a load such as
an incandescent lamp across the lines the voltage will be pretty much
undetectable with either sort of meter.
John -

I don't follow you. The way I read the original post, a new switch leg has
been inserted on the line side of the existing pull-chain switch, and the OP
is checking the voltage bewteen the line side of the new switch and neutral.

If he were checking voltage from the load side of an open switch to neutral
then I would understand your explanation, but I don't see that here.

Additionally, he's reading a significantly different voltage between his test
point and the EGC than he found to neutral...

I would be grateful for more explanation.

Toller -

Am I correct that the new switch leg was wired in "before" the pull-chain?
And that you were checking the voltage at the hot/line side of the new
switch? Was the new switch open or closed when you were performing these
tests? Where was your neutral test point? Your ground test point?

Thank you both for your time,

- Kenneth
 
B

Barry Mann

It's difficult to briefly explain this situation to an non scientist or
engineer. (sometimes it's even hard to convince them)

All physical measurement schemes effect the situation they are
measuring. These effects can be gross or subtile. It is up to the
experimenter to manage and interpret the results.

Let's consider a situation where there is a water leak and we would
like to measure how much water is leaking out.

One technique would be to measure the weight of a dry jar, place the
jar under the leak for a fixed period of time, measure the jar and
water, then subtract the dry jar weight. This could be a reasonable
technique in many situations, but for a small leak the water may
evaporate before it reaches the jar and we will always measure zero.

---

In your situation there is a small amount of energy leaking out of any
electrical wire into the surrounding area. Voltmeters need a small
amount of energy to make their reading. The old style pointer meters
need so much energy to move the pointer that they cannot measure this
small leakage -- they will always read zero (think about our leak
evaporation problem above). Your digital meter is much more efficient
and will react to the energy and give you a reading, but don't put too
much stock in it. As you have found, your meter always gives the same
reading when it is near an energized power conductor. Another meter may
give a different reading.

All you can say for sure is that when you see this characteristic
reading is that there is an energized circuit in the area. You can't
know how much energy is available or if you could be shocked or not by
touching the bare wire.

My meter is a bit more efficient than yours and I will often measure
close to full line voltage in a situation such as yours, but there is
no shock hazard because very little energy will pass through the open
switch or insulation. Unfortunately, I can't use the simple meter
reading to determine if I have a harmless leakage situation or a
dangerous condition.

-----------------------------------------------------------
spam: (e-mail address removed)
wordgame:123(abc):<14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13> (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------
 
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C

Chris Lewis

According to toller said:
I am installing a switch leg on a ceiling light that now has a chain switch.
When I measure voltage at the new switch it is H-N and H-G 120.2v with the
chain pull on; just like expected.
With the chain pull off, it measures H-N 61.8v and H-G 120.2v.
Is that 61.8v the famous phantom voltage I have read about so many times
here?
Probably - see the explanations from John McGaw and Barry, they describe
inductive pickup pretty well, but without a better description of the circuit,
it's hard to tell for certain.

It's virtually impossible to get any voltage measurement other than "zero
or real close" or "120V or real close" on a 120V circuit. Other than thru
inductive pickup.
It seems weird that if it doesn't exist that it reads exactly 61.8v
every time I measure it;
Inductive pickup voltages can certainly repeat. Try a different meter, and
i'll be different.
but of course it can't be real because there is no
circuit.
Can't really tell from the circuit description. If you're saying you're reading
61.8V on a line that is disconnected at _both_ ends, it really has to be inductive
pickup (or damaged wire insulation).

What's the N-G voltage?
 
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J

Jeff Wisnia

toller said:
I am installing a switch leg on a ceiling light that now has a chain switch.
When I measure voltage at the new switch it is H-N and H-G 120.2v with the
chain pull on; just like expected.
With the chain pull off, it measures H-N 61.8v and H-G 120.2v.

Is that 61.8v the famous phantom voltage I have read about so many times
here? It seems weird that if it doesn't exist that it reads exactly 61.8v
every time I measure it; but of course it can't be real because there is no
circuit.
What causes this?

I only measured at all because I wanted to make sure my ground was good;
which it obviously is.

You should have been a little more specific about what you did and how
you made your measurements. But here's the most likely scenario.

You didn't say which way the "new" swich was set when you measured that
"phantom voltage", but based on what you told us, I'm pretty certain it
had to be "off" and also that when you say "H" and "N" you are
describing the "black" and "white" wires in the new switch leg cable,
AND, there really ISN'T be a "N" (neutral) in the new switch box.

I read your description as saying you started with an existing ceiling
light whose only control means was a pull chain switch, perhaps part of
one of those ubiquitous ceramic or plastic ceiling sockets, or maybe it
was a pull switch mounted directly on some part of a more sophisticated
light fixture.

The term "switch leg" implies that you've brought a piece of two
conductor (plus ground) cable from the ceiling box the light is mounted
to down to a wall box containing a conventional SPST toggle switch.

You should have "cut into" the black wire going to the existing light
and the wire colors on the ceiling end of the new cable should have been
this; black to the existing black supply wire in the ceiling box and
white to the black wire of the fixture (or the screw terminal on a
ceiling socket from which you removed the existing black wire.) The
ground wire should have been connected to the ceiling box or pigtailed
to a ground witre in that box if the box in nonmetallic.

If the SPST switch you used happened to have had its terminals marked
"line" and "load" (some are) then you should have connected the black
leg wire to the "line" terminal and the white wire to the "load"
terminal, though it would work just as well if those two connections
were reversed. The ground wire should connect to the ground screw on the
new switch.

Now, if that's what you did, there won't be any "neutral" in your new
switch box, just a continuously hot lead, the black wire, which will be
at 120 volts relative to ground and a switched lead, the white wire
which will be at 120 volts relative to ground whenever that new switch
is closed, regardless of which way the pull chain switch is set.

When both the pull chain switch AND the new wall switch are "open", the
white wire in the switch leg will be "floating", and all along it's
length it's sitting right next to that black wire which has 120 vac on
it. It is also right next to the ground wire which is at ground potential.

If the capacitance per unit length between the three wires is about
equal, then the cable will act as a "equal arm" capacitive voltage
divider and the unloaded voltage on the white wire will be about half of
the 120 vac on the black wire, so 61.8 volts is about right measured
relative to either ground OR the black wire. Your meter must have a high
enough input impedance so that it doesn't significantly load that
capacitive voltage divider.

But, with the wall switch still open, and the pull switch closed, the
resistance of the light bulb will hold the white wire down at ground
potential. (which is also what the neutral power feed lead in the
ceiling box is at.) The miniscule current through the capacitive
coupling in the cable isn't enough to develop a measurable voltage
across the resistance of the light bulb, so you'd measure a full 120
volts between the black and white wires at the switch with a high
impedance meter.

Get it?

Jeff
 

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