Ideas for inexpensive True-RMS Metering?
in [Design]
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From: John Larkin on 4 Jul 2010 17:21 On Sun, 4 Jul 2010 13:43:08 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> wrote: >On Jul 4, 12:52�pm, John Larkin ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> On Sun, 4 Jul 2010 10:21:33 -0700 (PDT), whit3rd <whit...(a)gmail.com> >> wrote: >> >> >On Jul 4, 9:06 am, John Larkin >> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> >> >> >Putting current transformers... >> >> But they're big, expensive, and nonlinear. > >> >Nonlinear? �That only happens with large core field, >> >and current transformers with active burdens solve that... > >> The linearity problem isn't at large fields, it's at low ones. Silicon >> steel permeability drops seriously at low fields... > >I don't understand that. The B-H curve doesn't have low >slope at low fields... is this about hysteresis? It may be hysteresis, but the effect is that effective AC permeability drops at low currents, and that adds a lot of phase shift. For an affordable 100 amp CT, things are geting pretty bad below about 2 amps, really bad at half an amp. Premium core materials fix this, at premium prices. You can compensate somewhat in software, but it's not very consistant from core to core. Not sure >about prices, but better cores (like metglas) can help, >and it can't be terribly expensive in small quantities (you >see all those strips glued in DVD packages). >Silicon iron is favored for power transformers because >of low losses due to eddy currents at high flux; if you >are making a low-flux transformer, it's not the best material. A metering-quality 100 amp 50/60 Hz CT can't be small. Just like a 1000 watt 60 Hz power transformer can't be small. Iron and copper. > >> An active burden doesn't help the copper loss problem; >> it's easy to use a passive shunt that's tiny compared to winding >> resistance. Active burdens burn power, too. > >To treat for copper loss, design with two secondaries; one for >the burden, one for sensing. That doesn't help. The burden winding has copper loss, which causes flux in the core, which produces the linearity problems. But why would you want a burden, if you're not going to use it? The sense winding would reflect the *voltage* in the burden loop, which is poorly related to the *current*. TC will be awful. The active burden DOES burn >power, no way around that. Most ADC+ processor solutions >will have low V high-ish current supplies available, though. > >The main benefit of an active burden is that the core can be >made quite small. A secondary benefit is that it fits >nicely with the amplifier stage into an ADC. A summing-point type active burden doesn't help at all, since most of the error is caused by winding resistance. And it does force you to have thousands of secondary turns, a problem in its own right. Current feedback through a third winding (the AC version of a zero-flux transformer) makes the core material not matter, but again you'll need thousands of turns, and there are serious electrical problems doing it in real life. One interesting current sensor is a high-side shunt connected to a signal-level isolation transformer. That tiny transformer can use the active burden or zero-flux tricks without a huge power penalty. John
From: whit3rd on 4 Jul 2010 19:23 On Jul 4, 2:21 pm, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Sun, 4 Jul 2010 13:43:08 -0700 (PDT), whit3rd <whit...(a)gmail.com> > wrote: > >> >> >Putting current transformers... > >> >> But they're big, expensive, and nonlinear. > >Silicon iron is favored for power transformers because > >of low losses due to eddy currents at high flux; if you > >are making a low-flux transformer, it's not the best material. > A metering-quality 100 amp 50/60 Hz CT can't be small. Just like a > 1000 watt 60 Hz power transformer can't be small. Iron and copper. Yes, of course it CAN be small. Your only size problem relates to the difference between primary and secondary inductions, and if your amplifiers balance those amp-turns, there's NO NET FLUX in the core. That's the whole POINT of the active burden. A power transformer is not at all like a current transformer. The design spec of a power transformer is to have high inductance on the primary when the secondary is open; it's getting that high inductance that makes for high winding count and a big core. > >> An active burden doesn't help the copper loss problem; > >> it's easy to use a passive shunt that's tiny compared to winding > >> resistance. Active burdens burn power, too. > > >To treat for copper loss, design with two secondaries; one for > >the burden, one for sensing. > > That doesn't help. The burden winding has copper loss, which causes > flux in the core, which produces the linearity problems. The copper loss causes heat, but you are putting a sense resistor in series with the burden winding and looking at the resistor drop due to the burden current. That resistor drop doesn't care what the copper loss is. The amplifier sense winding (a different copper winding) has only amplifier input currents in it, its copper losses are negligible too. Because you're using an amplifier to supply the burden current, there is NO effect due to the burden coil's copper losses. It just gets driven by the amplifier's output impedance so that the sense winding says the core hasn't any d(phi)/dt. The whole point of using a driven winding instead of a burden resistor is nulling of the high flux that required that big magnetic core. Same thousand-winding secondary, but on a smaller diameter and with smaller wire (because we don't care about the resistivity now).
From: John Larkin on 4 Jul 2010 19:52 On Sun, 4 Jul 2010 16:23:36 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> wrote: >On Jul 4, 2:21�pm, John Larkin ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >> On Sun, 4 Jul 2010 13:43:08 -0700 (PDT), whit3rd <whit...(a)gmail.com> >> wrote: > >> >> >> >Putting current transformers... >> >> >> But they're big, expensive, and nonlinear. > >> >Silicon iron is favored for power transformers because >> >of low losses due to eddy currents at high flux; if you >> >are making a low-flux transformer, it's not the best material. > >> A metering-quality 100 amp 50/60 Hz CT can't be small. Just like a >> 1000 watt 60 Hz power transformer can't be small. Iron and copper. > >Yes, of course it CAN be small. Your only size problem relates >to the difference between primary and secondary inductions, >and if your amplifiers balance those amp-turns, there's NO >NET FLUX in the core. That's the whole POINT of >the active burden. > >A power transformer is not at all like a current transformer. The >design spec of a power transformer is to have high inductance >on the primary when the secondary is open; it's getting that >high inductance that makes for high winding count and a big core. > > >> >> An active burden doesn't help the copper loss problem; >> >> it's easy to use a passive shunt that's tiny compared to winding >> >> resistance. Active burdens burn power, too. >> >> >To treat for copper loss, design with two secondaries; one for >> >the burden, one for sensing. >> >> That doesn't help. The burden winding has copper loss, which causes >> flux in the core, which produces the linearity problems. > >The copper loss causes heat, but you are putting a sense resistor >in series with the burden winding and looking at the resistor drop due >to the >burden current. That resistor drop doesn't care what the copper >loss is. The amplifier sense winding (a different copper winding) >has only amplifier input currents in it, its copper losses are >negligible too. > >Because you're using an amplifier to supply the burden current, >there is NO effect due to the burden coil's copper losses. It just >gets driven by the amplifier's output impedance so that the sense >winding says the core hasn't any d(phi)/dt. > >The whole point of using a driven winding instead of a burden resistor >is nulling of the high flux that required that big magnetic core. >Same thousand-winding secondary, but on a smaller diameter and >with smaller wire (because we don't care about the resistivity now). Of course you care about the resistance of the feedback winding; the more resistance, the more voltage you'll need to drive it. Consider a 100 amp CT. 1000 turns on the feedback winding will need at least 141 mA peak drive, 200 or more with reasonable headroom. If a small toroid with a 1000 turn winding has 20 ohms of resistance, it will need over 4 volts at 200 mA drive. This is *not* little opamp turf. And if there's a load fault, the power kicked back into the drive amp will be ghastly. It will need serious protections. John
From: whit3rd on 5 Jul 2010 05:53 On Jul 4, 4:52 pm, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Sun, 4 Jul 2010 16:23:36 -0700 (PDT), whit3rd <whit...(a)gmail.com> > wrote: > >The whole point of using a driven winding instead of a burden resistor > >is nulling of the high flux that required that big magnetic core. > >Same thousand-winding secondary, but on a smaller diameter and > >with smaller wire (because we don't care about the resistivity now). > Of course you care about the resistance of the feedback winding; the > more resistance, the more voltage you'll need to drive it. > > Consider a 100 amp CT. 1000 turns on the feedback winding will need at > least 141 mA peak drive, 200 or more with reasonable headroom. If a > small toroid with a 1000 turn winding has 20 ohms of resistance, it > will need over 4 volts at 200 mA drive. This is *not* little opamp > turf. True. It's "little opamp plus two buffer transistors" turf. This amounts to <1W dissipated in the measurement circuit when the customer consumption hits 20 kW or so... 50 parts per million of waste energy. Alternatives like the shunt resistor into a transformer are going to have similar consumption. My home meter wastes about 3W, and the usage is circa 1kW. > And if there's a load fault, the power kicked back into the drive amp > will be ghastly. It will need serious protections. I think that's not the case any more. Remember, we started with a large core so it could put 50 mV onto the burden resistor; then we went to a smaller core which only needs drive an op amp to saturation with maybe 0.5 mV... The core is now two orders of magnitude smaller than the one you are familiar with, and the energy transfer in case of fault currents is likewise smaller.
From: Tim Williams on 5 Jul 2010 06:43
"whit3rd" <whit3rd(a)gmail.com> wrote in message news:2fbf8ba2-33cb-483c-a560-7ff7e3fc66a0(a)g19g2000yqc.googlegroups.com... > The core is now two orders of magnitude smaller than the one you > are familiar with, and the energy transfer in case of fault currents > is likewise smaller. Once the op-amp saturates (along with whatever protection diodes you've put in the circuit), the transformer is free to transform, so it quickly fills up with flux, saturates, and then you simply get zero additional EMF. Zero EMF means zero induced current, so your protection diodes and output stage don't even have to be very big, they need only contain the energy until saturation. Now, the energy in that gulp is proportional to current, so if you get a 100kA lightning strike or something, and the flux is maybe 1000uWb, that's an easy 100J showing up in your project. That's real energy which gets dumped into your supply rails (if it doesn't fuse the protection diodes right away, or arc through the transformer windings for that matter), which would do a fine job exploding most circuits. Fortunately, many things aren't expected to survive a lightning strike, so a consumer device (Kill-A-Watt?) doesn't need this level of brick-shithousery. Of course, a power distribution system *does* need this. Incidentially, note that bigger cores = more flux = proportionally more energy before saturation. So if you think 100J from a lightning strike is bad, try telling that to the conventional CT that may not saturate at all. Bye bye resistors, meters, ADCs, everything! Since the current induced in the CT windings themselves will cause so much voltage drop across the fine wire used (we've been talking, what, like 1000T of 34AWG wire inside a ~1" toroid?), it will surely break down before deliving full energy to the circuit, limiting itself fatally. I don't see a big CT surviving a lightning strike anyway (100kA --> many kV even across a small burden resistor), so this failure mode isn't any different. It's interesting that this limit is defined by the size of the core (how much flux causes saturation, and how much winding area is available to stuff full of copper) and the properties of the materials used (like copper's resistivity, and the insulation breakdown voltage). As a result, for the same geometry, it will scale equally. It's no surprise that big amp stuff is measured with Rogowski coils. http://en.wikipedia.org/wiki/Rogowski_coil Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |