Ideas for inexpensive True-RMS Metering?
in [Design]
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From: JosephKK on 5 Jul 2010 15:39 On Sat, 03 Jul 2010 17:24:05 GMT, nico(a)puntnl.niks (Nico Coesel) wrote: >Winfield Hill <Winfield_member(a)newsguy.com> wrote: > >>EnigmaPaul wrote... >>> >>> I'm looking for the best ideas for implementing an AC power >>> instrumentation front end for a microcontroller or FPGA. >>> >>> What I need to accomplish is to measure 50-60HZ AC in the range of >>> 120V upwards of 600VAC with about 1% accuracy, TRUE RMS. I need to >>> extract: >>> >>> 1. Voltage >>> 2. Current (via current transformers) >>> 3. Frequency >>> 4. Power Factor >>> 5. Phase difference >>> >>> This needs to work with reasonably distorted waveforms, like what >>> might be found in small generators driven by engines. I also need >>> something that will work for single phase or for 3phase and be >>> scalable to instrument up to two 3-Phase sources. >> >> Don't waste your time trying second- and third-best solutions. >> Go to Analog Devices' web site and acquaint yourself with their >> wide selection. We're featuring the ADE7753 in H&H AoE III, >> you could start there. 20-pin ssop, $4.61 at DigiKey. It's >> single phase, but they have more complex 3-phase versions with >> the same architecture, so what you learn on one is extendable. > >Still very expensive. The same amount of $ buys you a complete micro >with ADCs. Be that as it may, sometimes you need to tune your answer to the querant.
From: JosephKK on 5 Jul 2010 15:58 On Sun, 04 Jul 2010 16:52:00 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >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 Fault tolerance is the problem with all approaches.
From: John Larkin on 5 Jul 2010 16:34 On Mon, 05 Jul 2010 12:58:03 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Sun, 04 Jul 2010 16:52:00 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>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 > >Fault tolerance is the problem with all approaches. A CT driving a shunt isn't bad. You can use series resistance after the shunt to protect an opamp. At extreme currents, the CT will saturate, limiting the power dumped into the shunt. John
From: Grant on 5 Jul 2010 17:22 On Mon, 5 Jul 2010 05:43:46 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >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. 'Scuse me for butting in with a query. I have some small CTs to measure AC current up to 10 Amps, would putting a bidirectional 16V TVS (1.5KE16CA) across the secondary be enough to protect the CT from overload or disconnection? Would they affect normal measurements much? Long time ago I worked on a power system that used a standard 5A burden CT to measure current feeding 200A into a transformer across 415V --> high heat salt bath for steel hardening, the secondary was like the business end of a fork lift truck :) My small CTs are intended to help measure mains power going into (generally switchmode) power supplies converting between around 100W to 1kW. Grant.
From: John Larkin on 5 Jul 2010 18:22
On Mon, 5 Jul 2010 05:43:46 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"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 Big amp stuff is measured with 5-amp-secondary CTs. http://en.wikipedia.org/wiki/Current_transformer http://www.gedigitalenergy.com/products/specs/44224.pdf John |