Batteryless current clamps?

[Fester Bestertester]

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.

 

[Proteus IIV]

On Nov 17, 3:18 am, Fester Bestertester <f...@fbt.net> wrote:

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.

CURIOUSITY KILLED THE CAT
GO TO SCHOOL AND HEAR IT FROM THE HORSES NOUTH

OR GO TO YOUR NEAREST TECHINAL BOOK STORE AND PURCHASE TEST METERS
FOR DUMMIES

I AM PROTEUS

 

[Jeroen Belleman]

Bill Sloman wrote:

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

 

[Bill Sloman]

On Nov 17, 11:46 am, Proteus IIV <proteus...@gmail.com> wrote:

On Nov 17, 3:18 am, Fester Bestertester <f...@fbt.net> wrote:

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.

CURIOUSITY KILLED THE CAT
GO TO SCHOOL AND HEAR IT FROM  THE HORSES NOUTH

OR GO TO YOUR NEAREST TECHINAL BOOK STORE AND  PURCHASE TEST METERS
FOR DUMMIES

I AM PROTEUS

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.

If you can identify a specific book that has a specific reference to
the problem - with the ISBN for the book and the page or chapter
reference for the helpful bit - you can earn brownie points without
providing a direct answer.

Unhelpful abuse counts as a waste of bandwidth.

Raise you game or expect to be plonked. But don't worry if Jim
Thompson plonks you - he plonks everybody who disagrees with him,
which is probably one of the reasons he believes so many things that
don't happn to be true.

--
Bill Sloman, Nijmegen

 

[John Fields]

On Tue, 17 Nov 2009 05:39:40 -0800 (PST), Bill Sloman
<bill.sloman@ieee.org> wrote:

Unhelpful abuse counts as a waste of bandwidth.

Raise you game or expect to be plonked. But don't worry if Jim
Thompson plonks you - he plonks everybody who disagrees with him,
which is probably one of the reasons he believes so many things that
don't happn to be true.

---
Typical Slomanesque two-faced rhetoric; you damn unhelpful abuse as a
waste of bandwidth and then, in the same breath, engage in it yourself.

JF

 

[pimpom]

Fester Bestertester wrote:

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.

 

[Tim Williams]

On Nov 17, 2:18 am, Fester Bestertester <f...@fbt.net> wrote:

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

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.

They also read DC, and have an offset knob to account for the
ferrite's hysteresis.

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

Tim

 

[John Larkin]

On Tue, 17 Nov 2009 02:46:44 -0800 (PST), Proteus IIV
<proteusiiv@gmail.com> wrote:

On Nov 17, 3:18 am, Fester Bestertester <f...@fbt.net> wrote:
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.

CURIOUSITY KILLED THE CAT
GO TO SCHOOL AND HEAR IT FROM THE HORSES NOUTH

OR GO TO YOUR NEAREST TECHINAL BOOK STORE AND PURCHASE TEST METERS
FOR DUMMIES

I AM PROTEUS

Hey, it's been too long. The only time we hear from you is when your
kid sister kicks you off the Xbox.

John

 

[Joel Koltner]

"Fester Bestertester" <fbt@fbt.net> wrote in message
news:0001HW.C7279C6F0007F514B08A39AF@news.eternal-september.org...

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.)

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.

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

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

---Joel

 

[John Fields]

On Tue, 17 Nov 2009 09:32:34 -0800, "Joel Koltner"
<zapwireDASHgroups@yahoo.com> wrote:

"Fester Bestertester" <fbt@fbt.net> wrote in message
news:0001HW.C7279C6F0007F514B08A39AF@news.eternal-september.org...
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.)

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.

---
news:7ar5g59hdrcdpu3icb3rlmdn31iqiqfa67@4ax.com
---

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

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

---
Nope, it'll get you nothing.

Know why?

JF

 

[Joel Koltner]

"John Fields" <jfields@austininstruments.com> wrote in message
news:sir5g5h9h69vfurapjd9e2kn8efeod8qat@4ax.com...

news:7ar5g59hdrcdpu3icb3rlmdn31iqiqfa67@4ax.com

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

Nope, it'll get you nothing.
Know why?

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

---Joel

 

[pimpom]

Joel Koltner wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:sir5g5h9h69vfurapjd9e2kn8efeod8qat@4ax.com...
news:7ar5g59hdrcdpu3icb3rlmdn31iqiqfa67@4ax.com

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

Nope, it'll get you nothing.
Know why?

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.

 

[John Fields]

On Tue, 17 Nov 2009 10:59:38 -0800, "Joel Koltner"
<zapwireDASHgroups@yahoo.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:sir5g5h9h69vfurapjd9e2kn8efeod8qat@4ax.com...
news:7ar5g59hdrcdpu3icb3rlmdn31iqiqfa67@4ax.com

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

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.

JF

 

[Bill Sloman]

On Nov 17, 4:29 pm, John Fields <jfie...@austininstruments.com> wrote:

On Tue, 17 Nov 2009 05:39:40 -0800 (PST),Bill Sloman

bill.slo...@ieee.org> wrote:
Unhelpful abuse counts as a waste of bandwidth.

Raise you game or expect to be plonked. But don't worry if Jim
Thompson plonks you - he plonks everybody who disagrees with him,
which is probably one of the reasons he believes so many things that
don't happn to be true.

---
Typical Slomanesque two-faced rhetoric; you damn unhelpful abuse as a
waste of bandwidth and then, in the same breath, engage in it yourself.

Unhelpful abuse? I told him that if he wants to claim that the answer
to a question is availlable in a textbook, he's got to tell us which
text-book and whereabouts in that text-book.

People who use text-books know about this stuff. Try and remember back
to when you did.

--
Bill Sloman, Nijmegen

 

[Martin Riddle]

"Joel Koltner" <zapwireDASHgroups@yahoo.com> wrote in message
news:U6BMm.34563$Vr1.25172@en-nntp-01.dc1.easynews.com...

"Fester Bestertester" <fbt@fbt.net> wrote in message
news:0001HW.C7279C6F0007F514B08A39AF@news.eternal-september.org...
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.)

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.

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

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

 

[Bill Sloman]

On Nov 17, 3:10 pm, Jeroen Belleman <jer...@nospam.please> wrote:

Bill Slomanwrote:
[...]
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.

Unfortunately, I don't know that. There are several ways in which a
current clamp can work, and not all of them depend on on the clamp
acting as a transformer.

--
Bill Sloman, Nijmegen

 

[Proteus IIV]

On Nov 17, 8:39 am, Bill Sloman <bill.slo...@ieee.org> wrote:

On Nov 17, 11:46 am, Proteus IIV <proteus...@gmail.com> wrote:





On Nov 17, 3:18 am, Fester Bestertester <f...@fbt.net> wrote:

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.

CURIOUSITY KILLED THE CAT
GO TO SCHOOL AND HEAR IT FROM  THE HORSES NOUTH

OR GO TO YOUR NEAREST TECHINAL BOOK STORE AND  PURCHASE TEST METERS
FOR DUMMIES

I AM PROTEUS

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.

If you can identify a specific book that has a specific reference to
the problem - with the ISBN for the book and the page or chapter
reference for the helpful bit - you can earn brownie points without
providing a direct answer.

Unhelpful abuse counts as a waste of bandwidth.

Raise you game or expect to be plonked. But don't worry if Jim
Thompson plonks you - he plonks everybody who disagrees with him,
which is probably one of the reasons he believes so many things that
don't happn to be true.

--
Bill Sloman, Nijmegen- Hide quoted text -

- Show quoted text -

THIS IS NO GAME
YOUR HEADS UP MAY HELP THE GROUP THOUGH


I HAVE ALREADY GIVEN MY INPUT TO THIS TOPIC

I AM PROTEUS

 

[John Fields]

On Wed, 18 Nov 2009 00:35:12 -0800 (PST), Bill Sloman
<bill.sloman@ieee.org> wrote:

On Nov 17, 3:10 pm, Jeroen Belleman <jer...@nospam.please> wrote:
Bill Slomanwrote:
[...]
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.

Unfortunately, I don't know that.

---
Finally owning up to your ignorance, huh?
---

There are several ways in which a
current clamp can work, and not all of them depend on on the clamp
acting as a transformer.

---
Really?

How about some examples, then, Mr. Bullshit Artist, and don't forget
that the keyword here is "passive".

JF

 

[John Fields]

On Tue, 17 Nov 2009 15:01:11 -0800 (PST), Bill Sloman
<bill.sloman@ieee.org> wrote:

On Nov 17, 4:29 pm, John Fields <jfie...@austininstruments.com> wrote:
On Tue, 17 Nov 2009 05:39:40 -0800 (PST),Bill Sloman

bill.slo...@ieee.org> wrote:
Unhelpful abuse counts as a waste of bandwidth.

Raise you game or expect to be plonked. But don't worry if Jim
Thompson plonks you - he plonks everybody who disagrees with him,
which is probably one of the reasons he believes so many things that
don't happn to be true.

---
Typical Slomanesque two-faced rhetoric; you damn unhelpful abuse as a
waste of bandwidth and then, in the same breath, engage in it yourself.

Unhelpful abuse? I told him that if he wants to claim that the answer
to a question is availlable in a textbook, he's got to tell us which
text-book and whereabouts in that text-book.

---
As usual, more of your evasion since, clearly, the reference was to your
waste of bandwidth caused by the unhelpful abuse you heap on JT while
condemning PROTEUS for the waste of bandwidth caused by his unhelpful
abuse of Fester Bestertester.

True hypocrisy.
---

People who use text-books know about this stuff. Try and remember back
to when you did.

---
An insult in an attempt to change the subject and put me on the
defensive, huh?

It won't work, you contemptible, fatuous ass.

JF

 

[John Fields]

On Tue, 17 Nov 2009 00:18:55 -0800, Fester Bestertester <fbt@fbt.net>
wrote:

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.

A transformer is used to transfer _power_ from a source into a load,
thus the power, P2, required by the load, will be that power, P1,
supplied by the source.

In an ideal transformer there will be no losses, and then P1 and P2 will
be equal.

Next, the voltages on the primary and the secondary will be directly
proportional to the ratio of the number of turns on the primary to the
number of turns on the secondary, and the currents in them will
inversely proportional to the turns ratio.

With that in mind, let's say we have a transformer with a one turn
primary and a 1000 turn secondary, across which is connected a 1000 ohm
resistor which is dropping one volt.

The current in the load will then be:

E 1.0V
I = --- = ------ = 1e-3 ampere = 1 milliampere
R 1e3R

and the power dissipated by the load:


P = IE = 1e-3A * 1V = 1e-3 watt = 1 milliwatt


Now, since the turns ratio is 1000:1, the current in the primary is
inversely proportional to the current in the secondary, and since the
current in the secondary is 1 milliamp, the current in the primary must
be:


Is * nS 1e-3A * 1000t
IP = --------- = -------------- = 1 ampere
nP 1t

If we now double the current in the primary, the current in the
secondary will be doubled as well, causing the 1000 ohm resistor to drop
2 volts.

If we triple the primary current, the secondary current will be tripled
as well, the resistor will drop 3 volts, and so on...

So, what we have is a device which will have an output voltage which is
directly proportional to the input current and which we can use to
determine the input current by measuring the output voltage.

JF