# Batteryless current clamps?

F

#### Fester Bestertester

Jan 1, 1970
0
I'm curious how the Fluke i200s current clamp probe can give mV output
without the use of batteries.

How is this done? If one is measuring 200A I can see how the magnetic field
could generate enough current in the probe to support some high-impedance,
low-draw circuitry.

But when measuring on the low scale, say, 2 or 3 amps, how could the probe
output a few hundred mV? (The clamp is spec'd to output 100mV / amp on the
20A low scale, 10mV on the 200A high scale.)

Can someone explain this to me? I'm fascinated to see it's possible & curious
to know how.

Thanks.

J

#### Jeroen Belleman

Jan 1, 1970
0
Bill said:
[...]
Try to find out where the caps lock is, and unlock it. At the moment
you like more like Prostheticus.

For future reference, if you don't know the answer to a question, it
is not helpful to tell people that it is in some unspecified technical
book somewhere.
[...]

You could have added a line for the OP, saying that a passive
current clamp is a transformer, or some such.

Jeroen Belleman

P

#### pimpom

Jan 1, 1970
0
Fester said:
I'm curious how the Fluke i200s current clamp probe can give mV
output
without the use of batteries.

How is this done? If one is measuring 200A I can see how the
magnetic
field could generate enough current in the probe to support
some
high-impedance, low-draw circuitry.

But when measuring on the low scale, say, 2 or 3 amps, how
could the
probe output a few hundred mV? (The clamp is spec'd to output
100mV /
amp on the 20A low scale, 10mV on the 200A high scale.)

Can someone explain this to me? I'm fascinated to see it's
possible &
curious to know how.

You seem to have a preconceived notion of what constitutes large,
small and insignificant currents levels in terms of the fields
they generate, but such categorisations are only relative. "2 or
3 amps" is quite huge in some contexts and generate an
appreciable flux in the magnetic core of the clamp. The
alternating magnetic field induces a voltage in the clamp's
pickup coil and this voltage can certainly reach "a few hundred
mV" if enough number of turns are used.

You can also think of the clamp as a current transformer. The
wire being measured for current is the primary and the pickup
coil of the DMM is the secondary.

If you're more familiar with voltage transformers, think of it
this way:
Suppose you have just 1 mV output from a microphone. Connect it
to the primary of a 1:10 transformer and you will get 10 mV at
the secondary terminals. Use a 1:100 transformer and you get 100
mV and so on, theoretically up to any voltage.

P

#### pimpom

Jan 1, 1970
0
Joel said:
5uA... nice!

Seems that someone on eBay is selling a +/-5uA movement:
http://cgi.ebay.com/Weston-Bakelite-Glass-5ua-microamp-Panel-Meter-High-Z_W0QQitemZ130344240372

Because the federales will toss your rear in jail quite
rapidly?

I don't think that's quite what John meant. Anyway, that reminds
me of a practice by some villagers in my area. They cannot
afford, or don't want to pay, the power connection charge and
monthly bills. They wire their homes for a few incandescent bulbs
and keep a pair of solid-cored wires with the ends stripped bare
and bent into a U shape, the other two ends feeding the house
wiring. When it gets dark, they use a dry bamboo pole to hook the
bare ends to the overhead power lines. Free power - until they
get caught. The power company - the government here - usually
does nothing more than reprimand the offenders, but the practice
is rare now.

M

#### Martin Riddle

Jan 1, 1970
0
Joel Koltner said:
At some level, if you wrap a transformer around a wire, you can
extract as much or as little power as you like. Consider that, say,
100mV (generated by a 1A flow in the one turn "primary" of your
current probe) fed into the 10k impedance of a multimeter is all of 1
*micro*watt, which is pretty much "nothing" in comparison to what the
primary is likely to be carrying (e.g., even 1A at 1V is a watt, a
million times higher).

The power is coming from the primary, of course: The load on the
secondary is reflected back to the primary -- multiplied by the turns
ratios of the transformer squared and all. (This load effectively
appear in series with thatever the real load on the primary is.) The
trick then, is finding sensitive enough meters that the burden on the
primary is minimized. You might be surprised at how sensitive some of
the old analog meters (galvanometers) are -- 1mA full-scale deflection
is what you find in the cheapest instruments, 100uA is found in many
mid-grade instruments, and 10uA (and even less) is found in high-end
gear.

Wrapping some turns around the power company's lines will get you
many, many watts.

---Joel

You need a loop to form an air core transformer, which this has been
done.

Cheers

F

#### Fester Bestertester

Jan 1, 1970
0
I'm curious how the Fluke i200s current clamp probe can give mV output
without the use of batteries.

How is this done? If one is measuring 200A I can see how the magnetic field
could generate enough current in the probe to support some high-impedance,
low-draw circuitry.

But when measuring on the low scale, say, 2 or 3 amps, how could the probe
output a few hundred mV? (The clamp is spec'd to output 100mV / amp on the
20A low scale, 10mV on the 200A high scale.)

Can someone explain this to me? I'm fascinated to see it's possible &
curious
to know how.

---
OK.

A passive clamp-on ammeter is essentially the secondary of a transformer
wound on a core that can be opened or closed in order to get it around a
conductor so the current in that conductor can be measured without
cutting it and using a conventional ammeter.
[...]
JF[/QUOTE]

FINALLY an answer on-topic. Thank you.

After watching the 3 Stooges act that is aee / sed...

Sheesh!

I AM FBt

P

#### Pointless Posts

Jan 1, 1970
0
Fester said:
I'm curious how the Fluke i200s current clamp probe can give
mV
output without the use of batteries.

How is this done? If one is measuring 200A I can see how the
magnetic field could generate enough current in the probe to
support some high-impedance, low-draw circuitry.

But when measuring on the low scale, say, 2 or 3 amps, how
could
the probe output a few hundred mV? (The clamp is spec'd to
output
100mV / amp on the 20A low scale, 10mV on the 200A high
scale.)

Can someone explain this to me? I'm fascinated to see it's
possible
& curious
to know how.

---
OK.

A passive clamp-on ammeter is essentially the secondary of a
transformer wound on a core that can be opened or closed in
order to
get it around a conductor so the current in that conductor can
be
measured without cutting it and using a conventional ammeter.
[...]
JF

FINALLY an answer on-topic. Thank you.

After watching the 3 Stooges act that is aee / sed...

Sheesh!

I AM FBt

Admitted that the S/N ratio on Usenet can be frustrating. But did
you stop to consider the possibility that a) you failed to grasp
other attempts to explain it to you; b) your question was so
elementary for *this* group that few people bothered; c) your
last post might be taken as a slap in the face by those who tried
to help.

F

#### Fester Bestertester

Jan 1, 1970
0
Admitted that the S/N ratio on Usenet can be frustrating. But did
you stop to consider the possibility that a) you failed to grasp
other attempts to explain it to you; b) your question was so
elementary for *this* group that few people bothered; c) your
last post might be taken as a slap in the face by those who tried
to help.

Right you are.

A big thank you to those responders who gave answers to my question. Much
appreciated.

My comment was addressed to the "noise".

FBt

J

#### John Nagle

Jan 1, 1970
0
Bill said:
I'm curious how the Fluke i200s current clamp probe can give mV output
without the use of batteries.
How is this done? If one is measuring 200A I can see how the magnetic field
could generate enough current in the probe to support some high-impedance,
low-draw circuitry.
But when measuring on the low scale, say, 2 or 3 amps, how could the probe
output a few hundred mV? (The clamp is spec'd to output 100mV / amp on the
20A low scale, 10mV on the 200A high scale.)
Can someone explain this to me? I'm fascinated to see it's possible &
curious
to know how.
---
OK.
A passive clamp-on ammeter is essentially the secondary of a transformer
wound on a core that can be opened or closed in order to get it around a
conductor so the current in that conductor can be measured without
cutting it and using a conventional ammeter.
[...]
JF
FINALLY an answer on-topic. Thank you.

Yes. Classic AC clamp-on ammeters are simply transformers. One
"turn" through the clamp, many turns in the fixed coil for output.
The output feeds into a voltmeter.

Those are AC-only devices. There are also Hall-effect clamp-on
ammeters, and those work for both AC and DC. These have been
available for a decade or so, and pricing is now down as low as
$60. I used to have one that could read down to about 500mA DC, and it only cost$129. Very useful in robotics and controls work.

John Nagle

D

#### DaveC

Jan 1, 1970
0
[...]
There are also Hall-effect clamp-on
ammeters, and those work for both AC and DC. These have been
available for a decade or so, and pricing is now down as low as
$60. I used to have one that could read down to about 500mA DC, and it only cost$129. Very useful in robotics and controls work.

John Nagle

Which make & model would that \$129 model be? It's always useful to know
someone else's favorite tools...

Dave

T

#### Tzortzakakis Dimitrios

Jan 1, 1970
0
? "Fester Bestertester said:
I'm curious how the Fluke i200s current clamp probe can give mV output
without the use of batteries.

How is this done? If one is measuring 200A I can see how the magnetic
field
could generate enough current in the probe to support some high-impedance,
low-draw circuitry.

But when measuring on the low scale, say, 2 or 3 amps, how could the probe
output a few hundred mV? (The clamp is spec'd to output 100mV / amp on the
20A low scale, 10mV on the 200A high scale.)

Can someone explain this to me? I'm fascinated to see it's possible &
curious
to know how.
There's nothing fancy about that, the electricity meters of a medium-voltage
consumer (real and reactive energy) are powered from the two potential and
the two current transformers, without any other power supply (medium
voltage=15 kV in Crete).

F

#### Fester Bestertester

Jan 1, 1970
0
So, for a millivolt output probe, this might be as simple as 2 windings (or a
tapped single winding) with a range switch to select the winding?

D

#### daestrom

Jan 1, 1970
0
Bill said:
59:38 -0800, "Joel Koltner"
5uA... nice!
Seems that someone on eBay is selling a +/-5uA movement:
http://cgi.ebay.com/Weston-Bakelite-Glass-5ua-microamp-Panel-Meter-Hi...
Nope, it'll get you nothing.
Know why?
Because the federales will toss your rear in jail quite rapidly?
---
Nope, because the magnetic field generated by the power line will never
cut the conductor wrapped around it since the conductor will be
essentially perpendicular to the varying field.
Since the original claim was
" >Wrapping some turns around the power company's lines will get you
many, many
watts. "
This isn't the reason - lines is plural and the nett current through
the lines as a bunch balances out to zero.
---
Since that's obvious to the most casual observer, the context of his
statement must have been about wrapping some turns around [one] of the
power company's lines, which I addressed by referring to it as "the
power line".
---
Of course, you fraud, since by snipping the rest you sidestep the issue,
which is your ignorance in believing that a solenoid wound around an
alternating current carrying conductor can be used to extract power from
the varying magnetic field surrounding that conductor.

Your enthusiasm for inventing implausible straw men knows no bounds. I
never made any such claim. My scepticism about you claim was purely
based on the fact that you were ignoring what Joel Koltner had
actually said.
Nothing could be further from the truth, as demonstrated here:

liberty of emailing you the photos as soon as I post this.

Enjoy.

The pictures were perfectly clear. It was less obvious what you were
actually doing, but since I couldn't care less, this isn't any great
loss.

The joke is that even if you do extract "many watts" from the power
company's power lines, you won't be stealing from them. In order to be
able to extract power you have to be drawing power for which you will
be billed, and any extra watts you extract by transformer action is
subtracted from the power you are already paying for - your paid for
load will be seeing a lower drive voltage.

Joel Koltner made a rather good joke, which you have totally failed to
get.

If you make a 'tap' upstream of the revenue meter, even with just
transformer action, you're stealing. Revenue meters (kilowatt-hour
meters) have always had terminal voltage as one of their inputs. An
illegal tap upstream may affect the voltage at the service entrance some
small amount, but the metering will reduce the billed kWh accordingly.
So regardless of the exact voltage supplied by the utility (it often
varies slightly throughout the day), the amount of energy delivered at
the service entrance is what is billed for. Power drawn off before the
meter isn't measured and is 'stolen'.

Of course if you just 'wrap some turns around the power line' without
orienting the coil properly in relation to the line, you're not going to
get any power because transformer action won't work when your turns of
wire are parallel to the power line's magnetic field (i.e. 'wrapped
around' the power line). And I think that was John Field's point.

daestrom

T

#### Tzortzakakis Dimitrios

Jan 1, 1970
0
? "John Larkin said:
Current transformers are usually dumped into a load resistor aka
burden resistor, to convert their output current into voltage. I'm
sure the Fluke clamp-on has an internal burden resistor, and they may
switch that to change ranges.

Without a burden resistor, the output voltage will be proportional to
frequency and very dependent on core reluctance, which would be fatal
for a clamp-on meter with a hinge and a non-repeatable air gap.

Coreless Rogowsky coils are used unloaded, but need a downstream
integrator to accurately measure current.

http://en.wikipedia.org/wiki/Rogowski_coil

The coolest current transformer is a second-harmonic DCCT, accurate to
parts-per-million from DC to many kilohertz.

http://www.gmw.com/electric_current/Danfysik/866_867/867.html
Anyway, current transformers must always be operated with the secondary
shorted. In the generating facilities in Kozani, West Macedonia, where 400
kV current transformers were involved, the operators of the plant had a
special indicator whether the secondary was shorted.

D

#### daestrom

Jan 1, 1970
0
Tzortzakakis said:
Anyway, current transformers must always be operated with the secondary
shorted. In the generating facilities in Kozani, West Macedonia, where 400
kV current transformers were involved, the operators of the plant had a
special indicator whether the secondary was shorted.

Some old switchboard CT's I worked on in the Navy had very thin
insulator between two spring clips. Whenever we wanted to remove a
meter for cal, we slip the insulator out so the two clips would short
together, shorting the CT. Then we could open circuit the meter and
remove it from the panel. I don't remember exactly what the blade was
made of, but it's surface wasn't perfectly smooth like polished
material, more porous like unglazed ceramic (of course it wasn't any metal).

The reason they built the insulator so thin was that if one accidentally
open-circuited the CT without removing the wafer first, the high voltage
developed by the CT would just 'punch thru' the wafer and safely short
the CT. Then all you had to do to repair things was make sure you
closed the circuit and replace the wafer-thin insulator blade.

Was kind of surprised when I moved to commercial power systems that they
didn't use something similar. Just has to have a breakdown voltage that
is low enough to avoid damaging the CT.

daestrom

J

#### JosephKK

Jan 1, 1970
0
Is that the one with the 10/100 switch and a green LED?

I'm pretty sure most of the weight is not ferrite, it's a battery
somewhere.

1/3 either of those. No battery whatsoever and AC only.
They also read DC, and have an offset knob to account for the
ferrite's hysteresis.

Only active probes do that.
The passive probes only read AC, and as I recall, are 1 or 10 mV/A.

Tim

I can make any output ratio i want, i know how they work.

J

#### JosephKK

Jan 1, 1970
0
Lower-power CT, like residential-metering size, 100 amps or so, will
generally tolerate being unloaded. They will saturate and make two
not-too-huge voltage spikes per cycle and not get very warm. The nasty
part is that, once the burden is reconnected, they are very likely to
wind up magnetized, which will mess up low-current accuracy.

John

There is no CT in residential metering, the energy meter is connected
directly. Even commercial / light industrial you do not see CTs in
the meter circuit until 600 A, and before that you are typically at
480 V 3-phase (in the US).

J

#### James Sweet

Jan 1, 1970
0
There is no CT in residential metering, the energy meter is connected
directly. Even commercial / light industrial you do not see CTs in
the meter circuit until 600 A, and before that you are typically at
480 V 3-phase (in the US).

My uncle's house (in the US) has a 400A service with current
transformers. They're not common but they do exist. IIRC 200A is the
largest residential meter.

G
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