Batteryless current clamps?
in [Design]
|
Prev: Tunable Bandpass Filter
Next: good book
From: Martin Riddle on 17 Nov 2009 22:23 "Joel Koltner" <zapwireDASHgroups(a)yahoo.com> wrote in message news:U6BMm.34563$Vr1.25172(a)en-nntp-01.dc1.easynews.com... > "Fester Bestertester" <fbt(a)fbt.net> wrote in message > news:0001HW.C7279C6F0007F514B08A39AF(a)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
From: Bill Sloman on 18 Nov 2009 03:35 On Nov 17, 3:10 pm, Jeroen Belleman <jer...(a)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
From: Proteus IIV on 18 Nov 2009 04:19 On Nov 17, 8:39 am, Bill Sloman <bill.slo...(a)ieee.org> wrote: > On Nov 17, 11:46 am, Proteus IIV <proteus...(a)gmail.com> wrote: > > > > > > > On Nov 17, 3:18 am, Fester Bestertester <f...(a)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
From: John Fields on 18 Nov 2009 08:46 On Wed, 18 Nov 2009 00:35:12 -0800 (PST), Bill Sloman <bill.sloman(a)ieee.org> wrote: >On Nov 17, 3:10�pm, Jeroen Belleman <jer...(a)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
From: John Fields on 18 Nov 2009 09:51
On Tue, 17 Nov 2009 00:18:55 -0800, Fester Bestertester <fbt(a)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 |