Current transformer

I was looking for a cheap clip-on current transformer to use to monitor my power usage, and I found this:

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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

Reply to
Sylvia Else
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Makes no sense to me.

There are some nice open-loop Hall sensors in the $15 ballpark. Lem and Tamura.

Reply to
John Larkin

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

Reply to
P E Schoen

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.

Reply to
John Larkin

Are you thinking in terms of permanent magnetisation?

Sylvia.

Reply to
Sylvia Else

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  
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Reply to
John Larkin

Oops! Given that I did. I'll have to see whether it has any noticeable magnetisation.

Don't know about the turns ratio, although it's a fairly substantial winding. The resistance is about 33 ohms.

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.

Reply to
Sylvia Else

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.

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  
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Reply to
John Larkin

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

Reply to
M Philbrook

The 33 ohms in series with one of the diodes, so I can't see it having a calibration function.

There is 220 ohms across the coil, but it's so large compared with the actual required burden resistor (a few ohms) that I have to wonder what it's for.

Sylvia.

Reply to
Sylvia Else

Is it possible that the main burden resistor is in the instrument end?

Reply to
John S

It would appear so. But then why is the 220 ohm resistor there? The designer presumably had a reason, given that redundant components waste money.

Sylvia.

Reply to
Sylvia Else

But adding these components makes the engineer more valuable and irreplaceable in the eyes of management, because "only s/he" knows what they are for. Making something more complex than need be is common in software, so maybe the engineer was a M$ programmer. If all the bloat were removed, the entire Windows OS would fit on a single CDROM. MSDOS fit on a single floppy!

Paul

Reply to
P E Schoen

I've never noticed management showing the slightest sign that they recognise the value of an engineer.

Sylvia.

Reply to
Sylvia Else

Perhaps to provide some sort of burden at levels below the turn-on level of the diodes?

Reply to
John S

I can't see the need.

Sylvia.

Reply to
Sylvia Else

Yes, I saw my error just after hitting send. Sorry.

Reply to
John S

Don't you hate it when that happens? ;)

Sylvia.

Reply to
Sylvia Else

Of course I do. It is hard enough to climb out of the stupid pit I'm in.

Reply to
John S

That's why engineers shouldn't have managers.

Reply to
John Larkin

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