I don't understand that. The B-H curve doesn't have low slope at low fields... is this about hysteresis? 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.
To treat for copper loss, design with two secondaries; one for the burden, one for sensing. 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.
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
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.
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
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.
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.
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.
True. It's "little opamp plus two buffer transistors" turf. This amounts to And if there's a load fault, the power kicked back into the drive amp
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.
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 =3D more flux =3D 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.
formatting link
Tim
--=20 Deep Friar: a very philosophical monk. Website:
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.
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
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.
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.
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!
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.
Big amp stuff is measured with 5-amp-secondary CTs.
It's troublesome to put high fault currents, possibly with fast risetimes, through some fine wire in a secondary; you might want to put your bidirectional protector onto a purpose-built protective (tertiary?) winding. Maybe just a few windings of stiff wire into an antiparallel diode pair?
Small wires and fast current risetimes is the 'exploding-wire' scenario.
One can imagine a washer-like chunk of MOV semiconductor, with the core going through its midhole, that only conducts when a big EMF (beyond the calibration range of the circuit) comes along. That's be a very cheap and reliable kind of protection.
"John Larkin" wrote in = message news: snipped-for-privacy@4ax.com...
No, limited flux, unlimited power/energy. The amps delivered during the = pulse are proportional to the fault current. Only the voltseconds are = limited.
Blown fuse ~1kA? Maybe not a big deal. Short circuit ~10kA? A = challenge. Lightning strike ~100kA+? Ball up into the fetal position..
Tim
--=20 Deep Friar: a very philosophical monk. Website:
e pulse are proportional to the fault current. =A0Only the voltseconds are = limited.
The saturation of the core makes the coupling of primary and secondary go away; only the primary side gets the fault current, the secondary current is no longer proportional (unless you choose a really odd winding scheme that stays highly flux-coupled when the core is removed).
I've seen too many exploded MOVs, exploding wiring, closest I've seen is three phase cables strapped together in a 1500A converter to stop them jumping about while waiting for the fault interrupters to open.
Except it goes boom and nothing there for next time?
The energy transfer is limited by the saturation of the core, still; taking 95% of the high-slew energy available before the core saturates is unlikely to cause a 'boom' event.
MOVs have wire joints that seem to fail before the material does (don't have full data on dead MOV properties, though, so that could be wishful thinking).
I'm wondering if some of the nonlinear interference-suppression ferrites have some current percolation internally... grains of conducting ceramics (B4C) and low-breakdown semiconductors (InSb) could be quite easy to add to a ferrite bead mix.
Yes, which is why the energy delivered is proportional to flux as well. = It only delivers power until the transformer stops transformering.
Say you get 1A fault current from a CT (i.e., secondary referred), which = throws your circuit into overload, so the voltage on the windings jumps = to 5V (clamped by a perfect 5V supply, assuming ideal clamp diodes). = The delivered power is evidently 1A * 5V =3D 5W, going into your supply. = If the winding has a saturation flux of 1mWb, this fault current will = flow for 1mWb / 5V =3D 0.2ms. The energy is 5W * 0.2ms =3D 1mJ, or 1A * =
1mWb.
Increase the fault to 10A. The winding is clamped at 5V, so 50W is = delivered, and the fault again lasts for 0.2ms, because the flux is =
1mWb. The energy is 10mJ.
Increase the fault to 1kA. Now you get 5kW and 1J, beyond the capacity = of a 1.5KE6.
Strike it with a bolt of lightning. 100kA gives 500kW peak and 100J, = assuming the transformer doesn't fail first; if it has an internal = resistance of 0.01 ohm, it will drop 1kV, probably breaking down the = meager insulation in a CT.
Tim
--=20 Deep Friar: a very philosophical monk. Website:
only delivers power until the transformer stops transformering.
your circuit into overload, so the voltage on the windings jumps to 5V (clamped by a perfect 5V supply, assuming ideal clamp diodes). The delivered power is evidently 1A * 5V = 5W, going into your supply. If the winding has a saturation flux of 1mWb, this fault current will flow for 1mWb / 5V = 0.2ms. The energy is
5W * 0.2ms = 1mJ, or 1A * 1mWb.
and the fault again lasts for 0.2ms, because the flux is 1mWb. The energy is
10mJ.
1.5KE6.
the transformer doesn't fail first; if it has an internal resistance of 0.01 ohm, it will drop 1kV, probably breaking down the meager insulation in a CT.
The usual practise in electronic metering is to have the CT secondary drive a low-resistance wirewound or manganin strip shunt. The shunt resistance is considerably less than the winding resistance. Outrageous CT overloads don't damage the shunt... most of the power dissipation is in the winding. The signal conditioning opamps or whatever are protected by high value resistances between them and the shunt.
It's also common to buy "current sensors" which are just CTs that come with internal burden resistors. 0.25 and 0.333 volts RMS out at rated current are the two common standards.
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