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.
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. :-)
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.
" >Wrapping some turns around the power company's lines will get you many, many
This isn't the reason - lines is plural and the nett current through the lines as a bunch balances out to zero.
Wrapping a clamp-on meter around one line means that there is current circulating around the clamp - the current that goes through the selected line in one direction is matched by equal and opposite current flowi g through the other lines in the other direction. The coupling coefficient is unlikely to be good, but it is finite.
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.
You asked about the Fluke i200s current clamp probe. I've never used one, and I'm not sure that I've even seen one, which discouraged me from trying to improvise an explanation.
John Field's response - now that he has finally got around to making the kind of useful post that he claims to represent the bulk of his output - does seem to be plausible.
From time to time we get responses from the people who designed the gear under discussion, but you don't seem to have been that lucky.
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.
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).
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