Forums

Current transformer

Started by Sylvia Else December 8, 2015
I was looking for a cheap clip-on current transformer to use to monitor 
my power usage, and I found this:

http://efergy.com/au/products/accessories/microjackplug-extra-sensor#.VmZZQF42GVI

which I guessed (correctly) was a current transformer.

I that have no intention of using it with Efergy's products, since they 
don't seem concerned about the voltage, which can vary somewhat from 
nominal, and are certainly unconcerned about the phase.

Anyway, when it arrived, I noticed that it has a three contact plug 
which made me wonder what was actually inside the sensor. It came apart 
easily enough, and I  inferred the circuit appended to the end of this 
posting. Note that the inductor on the AC current side is really just 
the wire carrying the current to be sensed.

I supposed that the 220 ohm resistor was the required load, but when I 
tested the device, I found that the output was clearly being clipped by 
the diodes and 33 ohm resistor. In fact, a resistance of just a few ohms 
will be required to get a linear output up to the rated 70 amps.

It looks like the idea behind having the 33 ohm resistor there is that 
the rest of the electronics can determine when the sensor is overloaded 
by the resulting asymmetry in output.

I can't see what output B is for. Any thoughts?


Version 4
SHEET 1 880 680
WIRE 16 16 -96 16
WIRE 80 16 16 16
WIRE 128 16 80 16
WIRE 192 16 128 16
WIRE 256 16 192 16
WIRE 80 32 80 16
WIRE 192 80 192 16
WIRE 256 80 256 16
WIRE 448 80 368 80
WIRE -64 112 -96 112
WIRE 16 112 16 16
WIRE 80 128 80 112
WIRE 128 128 128 16
WIRE 288 160 288 80
WIRE 304 160 304 80
WIRE 320 160 320 80
WIRE 448 160 368 160
WIRE 80 224 80 192
WIRE 80 224 -96 224
WIRE 128 224 128 192
WIRE 128 224 80 224
WIRE 192 224 192 160
WIRE 192 224 128 224
WIRE 256 224 256 160
WIRE 256 224 192 224
FLAG -96 16 A
FLAG -96 112 B
FLAG -96 224 C
SYMBOL ind 240 64 R0
SYMATTR InstName L1
SYMBOL diode 112 128 R0
SYMATTR InstName D1
SYMBOL diode 96 192 R180
WINDOW 0 24 64 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D2
SYMBOL res 64 16 R0
WINDOW 0 30 49 Left 2
WINDOW 3 26 82 Left 2
SYMATTR InstName R1
SYMATTR Value 33
SYMBOL res 32 96 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 10K
SYMBOL res 176 64 R0
WINDOW 0 28 49 Left 2
WINDOW 3 21 90 Left 2
SYMATTR InstName R3
SYMATTR Value 220
SYMBOL ind 384 176 R180
WINDOW 0 36 80 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName L2
TEXT 448 120 Left 2 ;AC current
TEXT -280 128 Left 2 ;Sense output
TEXT -16 328 Left 3 ;Current Sensor


On Tue, 8 Dec 2015 15:30:53 +1100, Sylvia Else
<sylvia@not.at.this.address> wrote:

>I was looking for a cheap clip-on current transformer to use to monitor >my power usage, and I found this: > >http://efergy.com/au/products/accessories/microjackplug-extra-sensor#.VmZZQF42GVI > >which I guessed (correctly) was a current transformer. > >I that have no intention of using it with Efergy's products, since they >don't seem concerned about the voltage, which can vary somewhat from >nominal, and are certainly unconcerned about the phase. > >Anyway, when it arrived, I noticed that it has a three contact plug >which made me wonder what was actually inside the sensor. It came apart >easily enough, and I inferred the circuit appended to the end of this >posting. Note that the inductor on the AC current side is really just >the wire carrying the current to be sensed. > >I supposed that the 220 ohm resistor was the required load, but when I >tested the device, I found that the output was clearly being clipped by >the diodes and 33 ohm resistor. In fact, a resistance of just a few ohms >will be required to get a linear output up to the rated 70 amps. > >It looks like the idea behind having the 33 ohm resistor there is that >the rest of the electronics can determine when the sensor is overloaded >by the resulting asymmetry in output. > >I can't see what output B is for. Any thoughts?
Makes no sense to me. There are some nice open-loop Hall sensors in the $15 ballpark. Lem and Tamura.
"John Larkin"  wrote in message 
news:0joc6bd6nrd03csgeb07uba992hnb1ocbr@4ax.com...

> On Tue, 8 Dec 2015 15:30:53 +1100, Sylvia Else > <sylvia@not.at.this.address> wrote:
>>I was looking for a cheap clip-on current transformer to use to monitor my >>power usage, and I found this: > >> http://efergy.com/au/products/accessories/microjackplug-extra-sensor#.VmZZQF42GVI > >>which I guessed (correctly) was a current transformer. > >>I that have no intention of using it with Efergy's products, since they >>don't seem concerned about the voltage, which can vary somewhat from >>nominal, and are certainly unconcerned about the phase. > >> Anyway, when it arrived, I noticed that it has a three contact plug which >> made me wonder what was actually inside the sensor. It came apart easily >> enough, and I inferred the circuit appended to the end of this posting. >> Note that the inductor on the AC current side is really just the wire >> carrying the current to be sensed. > >>I supposed that the 220 ohm resistor was the required load, but when I >>tested the device, I found that the output was clearly being clipped by >>the diodes and 33 ohm resistor. In fact, a resistance of just a few ohms >>will be required to get a linear output up to the rated 70 amps. > >>It looks like the idea behind having the 33 ohm resistor there is that the >>rest of the electronics can determine when the sensor is overloaded by the >>resulting asymmetry in output. > >>I can't see what output B is for. Any thoughts?
> Makes no sense to me.
> There are some nice open-loop Hall sensors in the $15 ballpark. Lem and > Tamura.
It is standard practice to use diodes across a CT to protect against open circuit and overload conditions, to limit the output to under 1 volt. The series resistor in one diode leg would create an unbalanced square wave with a net DC component, which could be detected. Maybe that's what the 10k is for, or maybe the measuring device uses it to determine presence of the probe using a DC signal. If so, maybe the 33 ohms in one leg balances this additional DC offset. Paul
> > > >>I can't see what output B is for. Any thoughts? > >
The B output may be simply to ID the tyoe of sensor. This type of sensor has a 10K. Another type may have a 1K etc. Mark
On Tue, 8 Dec 2015 01:47:00 -0500, "P E Schoen" <paul@pstech-inc.com>
wrote:

>"John Larkin" wrote in message >news:0joc6bd6nrd03csgeb07uba992hnb1ocbr@4ax.com... > >> On Tue, 8 Dec 2015 15:30:53 +1100, Sylvia Else >> <sylvia@not.at.this.address> wrote: > >>>I was looking for a cheap clip-on current transformer to use to monitor my >>>power usage, and I found this: >> >>> http://efergy.com/au/products/accessories/microjackplug-extra-sensor#.VmZZQF42GVI >> >>>which I guessed (correctly) was a current transformer. >> >>>I that have no intention of using it with Efergy's products, since they >>>don't seem concerned about the voltage, which can vary somewhat from >>>nominal, and are certainly unconcerned about the phase. >> >>> Anyway, when it arrived, I noticed that it has a three contact plug which >>> made me wonder what was actually inside the sensor. It came apart easily >>> enough, and I inferred the circuit appended to the end of this posting. >>> Note that the inductor on the AC current side is really just the wire >>> carrying the current to be sensed. >> >>>I supposed that the 220 ohm resistor was the required load, but when I >>>tested the device, I found that the output was clearly being clipped by >>>the diodes and 33 ohm resistor. In fact, a resistance of just a few ohms >>>will be required to get a linear output up to the rated 70 amps. >> >>>It looks like the idea behind having the 33 ohm resistor there is that the >>>rest of the electronics can determine when the sensor is overloaded by the >>>resulting asymmetry in output. >> >>>I can't see what output B is for. Any thoughts? > >> Makes no sense to me. > >> There are some nice open-loop Hall sensors in the $15 ballpark. Lem and >> Tamura. > >It is standard practice to use diodes across a CT to protect against open >circuit and overload conditions, to limit the output to under 1 volt. The >series resistor in one diode leg would create an unbalanced square wave with >a net DC component, which could be detected. Maybe that's what the 10k is >for, or maybe the measuring device uses it to determine presence of the >probe using a DC signal. If so, maybe the 33 ohms in one leg balances this >additional DC offset. > >Paul
The asymmetric diode thing liiks like a great way to magnetize the core. Once you do that, the low-end phase shift and gain go to hell.
On 9/12/2015 2:39 AM, John Larkin wrote:

> The asymmetric diode thing liiks like a great way to magnetize the > core. Once you do that, the low-end phase shift and gain go to hell. > >
Are you thinking in terms of permanent magnetisation? Sylvia.
On Wed, 9 Dec 2015 12:08:37 +1100, Sylvia Else
<sylvia@not.at.this.address> wrote:

>On 9/12/2015 2:39 AM, John Larkin wrote: > >> The asymmetric diode thing liiks like a great way to magnetize the >> core. Once you do that, the low-end phase shift and gain go to hell. >> >> > >Are you thinking in terms of permanent magnetisation? > >Sylvia.
Yes. If it's an iron core CT, and it gets magnetized, it gets ugly on the low end. The fix is to unburden it and run a bunch of AC amps in from a variac, turn it up into saturation, and gradually crank it down to zero. It can get magnetized by a big asymmetric current surge, or more likely by running it without a burden resistor. That lopsided 33 ohm thing looks bad. What's the turns ratio on that thing? What's the winding resistance? Is it a true CT? Some of the cheap power meters use a core and winding that's meant to be run unloaded. So the sensor isn't a CT, it's a transformer, with a 90 degree phase shift and vague calibration. -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
On 9/12/2015 12:17 PM, John Larkin wrote:
> On Wed, 9 Dec 2015 12:08:37 +1100, Sylvia Else > <sylvia@not.at.this.address> wrote: > >> On 9/12/2015 2:39 AM, John Larkin wrote: >> >>> The asymmetric diode thing liiks like a great way to magnetize the >>> core. Once you do that, the low-end phase shift and gain go to hell. >>> >>> >> >> Are you thinking in terms of permanent magnetisation? >> >> Sylvia. > > Yes. If it's an iron core CT, and it gets magnetized, it gets ugly on > the low end. The fix is to unburden it and run a bunch of AC amps in > from a variac, turn it up into saturation, and gradually crank it down > to zero. > > It can get magnetized by a big asymmetric current surge, or more > likely by running it without a burden resistor.
Oops! Given that I did. I'll have to see whether it has any noticeable magnetisation.
> That lopsided 33 ohm > thing looks bad. > > What's the turns ratio on that thing? What's the winding resistance?
Don't know about the turns ratio, although it's a fairly substantial winding. The resistance is about 33 ohms.
> > Is it a true CT? Some of the cheap power meters use a core and winding > that's meant to be run unloaded.
That can't be the case here, given that doing so with even a moderate current (5 amps, versus it's rated 70 amps) created an output voltage high enough to turn both diodes on. Sylvia.
On Wed, 9 Dec 2015 12:26:01 +1100, Sylvia Else
<sylvia@not.at.this.address> wrote:

>On 9/12/2015 12:17 PM, John Larkin wrote: >> On Wed, 9 Dec 2015 12:08:37 +1100, Sylvia Else >> <sylvia@not.at.this.address> wrote: >> >>> On 9/12/2015 2:39 AM, John Larkin wrote: >>> >>>> The asymmetric diode thing liiks like a great way to magnetize the >>>> core. Once you do that, the low-end phase shift and gain go to hell. >>>> >>>> >>> >>> Are you thinking in terms of permanent magnetisation? >>> >>> Sylvia. >> >> Yes. If it's an iron core CT, and it gets magnetized, it gets ugly on >> the low end. The fix is to unburden it and run a bunch of AC amps in >> from a variac, turn it up into saturation, and gradually crank it down >> to zero. >> >> It can get magnetized by a big asymmetric current surge, or more >> likely by running it without a burden resistor. > >Oops! Given that I did. I'll have to see whether it has any noticeable >magnetisation. > >> That lopsided 33 ohm >> thing looks bad. >> >> What's the turns ratio on that thing? What's the winding resistance? > >Don't know about the turns ratio, although it's a fairly substantial >winding. The resistance is about 33 ohms. >
33 ohms is huge for a classic CT, but then it may just have a very high ratio, namely lots of turns. It's not hard to measure the turns ratio.
>> >> Is it a true CT? Some of the cheap power meters use a core and winding >> that's meant to be run unloaded. > >That can't be the case here, given that doing so with even a moderate >current (5 amps, versus it's rated 70 amps) created an output voltage >high enough to turn both diodes on.
Back-to-back diodes are a reasonable protection against runing it unburdened. Except for the 33 ohm mystery. -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
In article <ue1f6b9gqr2agp8n27a3sdrqs0qal8s9bl@4ax.com>, 
jjlarkin@highlandtechnology.com says...
> >> > >> Is it a true CT? Some of the cheap power meters use a core and winding > >> that's meant to be run unloaded. > > > >That can't be the case here, given that doing so with even a moderate > >current (5 amps, versus it's rated 70 amps) created an output voltage > >high enough to turn both diodes on. > > Back-to-back diodes are a reasonable protection against runing it > unburdened. Except for the 33 ohm mystery. >
That's one action you really don't want to do, that is running a CT unloaded. In this case here, the 33 ohms is most likely a calibrated value to generate a specific V to the monitoring device. Diodes are protection from over current for the monitoring device and internal shunt R. Jamie