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.
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.
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.
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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
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.
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.
--
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.
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.
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