Help interfacing Current Transformers to ADC

In message , dated Wed,

I really wouldn't separate the burden from the CT secondary, even for that linearizing effect. If it is done, back-to-back zeners directly across the secondary are a must.

Well actually yes, but my comments were more of a rhetorical question regarding what I though I understood about the pros and cons of each of the basic methods described to me.

To be frank I am having trouble wraping my head around all of this. I get part of it, but get lost in some of the details.

So having the diodes before the "burden resistor" takes their voltage drops out of the picture. But placing the burden before the diodes is safer (in case of diode failure?) but the voltage drop of the diodes become a problem? This is just a wild guess, and at this point it should be obvious that the theory part of this is a bit over my head.

I am 100% self taught... and learning every day. I am trying to sort all of this out so that I can at least get one or two basic design on hte breadboard.

If it appears that I am somewhat dazed in a sea of informed people... well I guess that would be a fair observation. However, this is the only way I can learn (by experience and asking questions). Thanks for the reply.

A transformer is a transformer. That is, a transformer has a fixed (determined by turns ratio) ratio of input to output voltage and a fixed ratio of input to output current, neglecting losses and leakage flux between windings. If the primary is fed with a stiff voltage source (like being connected across the power lines) then the output also acts like a stiff voltage source, that varies its voltage only slightly as the load current changes.

But if the primary is fed with a nearly constant current (as when a current transformer is connected in series with a load, which is dropping almost all the line voltage, and the current transformer primary is dropping only a tiny fraction of the line voltage, so any voltage variation reflected back to it from its secondary voltage will have almost no effect on the load current) then the output of that transformer is a current, related to the load current by the turns ratio, regardless of how much output voltage the CT produces (as long as its core doesn't saturate from producing that voltage). Any inductor (including transformer windings) produce voltage in proportion to the rate of change of the flux surrounded by their windings. Larger voltages require larger flux swings. This is why CTs have a maximum burden resistor rating. Multiply that rated resistance by the rated output current (rated input current divided by the turns ratio) and you have the maximum output voltage the core can produce without saturating.

So, adding rectifier drop between a current transformer and its burden resistor forces the current transformer to produce extra voltage, to cover the diode drops, so that the total current through the burden resistor is essentially unchanged, except for the direction of alternate half cycles. You get very little nonlinearity added by the rectification process. It is very close to ideal rectification, as long as the desired output is a current.

Yes. Almost perfectly.

Yes. In that case, the voltage drop across the burden resistor becomes a signal source that is subsequently rectified, with the normal diode losses. The additional failure rate does not go up very much if the rectifiers and burden resistor are well over rated, to include the expected source current peaks any start up. Or you add peak limiting components, but those also have to handle the peak start up current, but at an even higher wattage, because of the higher voltage.

Most signals we deal with are treated as voltages, so it takes a slight adjustment to think about a signal that is inherently a current (and the voltage can be anything, over some range).

You seem to be doing just fine from where I am watching.

Ok. Usually the challenge boils down to how to get a measured value or status into the PC. Luckily you can buy USB devices for that and some come with comfortable SW support or even a nice GUI. It doesn't have to be something expensive from NI. Example, comes in several editions:

Another method to enter data into the PC at really low cost is the sound card. It's literally free because most PCs contain one, whether you want it or not. Many cards cannot be modified to accept DC levels so you may have to provide AC. One easy method it to chop the signal to be measured using, for example, a bunch of 74HC4053 chips. The signal amplitude is then proportional to the amplitude of the spectrum generated at the chopping frequency. Typically a few kHz. Now you can pick one of the freeware programs to measure spectral lines from the sound card and pipe the results to your SW. Really brazen people would probably feed in several signals in parallel at staggered (not multiples of each other) frequencies and do an FFT.

Good policy. I wouldn't ever trust a PC with critical functions if it runs Windows. Even a uC I wouldn't trust unless I knew very well that I have mastered its programming (then they can be most reliable).

An iso transformer, preferably medical grade, is a good safety measure.

I have no problem with having a rectified burden, as long as everything is conservatively rated, soldered together and connected solidly to the CT. Rectifiers usually fail shorted, anyway.

Don't forget to include any load start up in-rush current in the derating. Motor and transformer loads can have huge in-rush currents.

Hello Yuriy,

Further above in the thread the OP had stated his desire to learn the technology. It's not about money here. Except that he probably cannot exceed a cartain budget.

You can do it without the active filter. That's cheaper :-)))

Another one dawned on me just now: If your sound card supports 60Hz (it should) your could process the output of the current transformers directly. Must be scaled and isolated properly of course. And once more, don't forget the burden resistor or you sound card is going to be roasted.

Now use a freeware routine to measure the amplitude of the 60Hz signal.

As a multiplexer you can either use 74HC mux switches and scoot the DC level to their center or use little relays.

VERY GOOD POINT! And simplifies things a lot!

See...

R1 is there to ensure OpAmp stability.

D1 is there in case there is enough signal to rail the OpAmp and lift the summing node.

...Jim Thompson

So in the time I have wracked my brain over this, you have breadboarded it in a spice program and plotted the results for differing input levels? I guess I am looking at a single channel of my desired (24)? The diode is there for over drive protection, and the rectifier/filter is acting as the burden without a seperate shunt? Looking back over others recomendations, would the back to back zeners still be needed for safety before the bridge? The op-amp and the RC network are used to scale and buffer the voltage to that of my A/D correct?

Will any old op-amp do? Will any random single sample from the A/D input be a reasonable approximation of the load current that the device on the primary is drawing, or do i need to average several samples -AND- if I do, is the timing critical?

I think I am close to understanding all of this and moving towards the math and trying to figure out component values. If sample timing is not critical, I will simply toss these onto the dallas DS2450 1-wire A/Ds and grab random samples when needed via my logging software. If timing is an issue, then I will have to educate myself on A/D with a PIC and learn how to bit bang the results via serial ( i imagine there a lot of examples out there that will help).

It may not appear that I have absorbed much, but I do feel smarter (then again it just could be the thickening of my head and due to confusion!)

Most power-metering CTs are low ratio transformers. 5 amps secondary is the metering standard, and "electronic" rated CTs often output 100 mA or more; nobody wants to put 20,000 turns on a toroid. So maybe add a low value resistor from the bridge output to ground, and use a voltage amplifier.

Better yet, amplify and then rectify, to keep the transformer burden down.

Better yet, amplify and digitize and true-RMS rectify in software.

John

Yes.

Yes.

No. Unless you fear that the bridge could fail open. Not likely.

Yes.

Yes. At 60Hz even an LM324 will look ideal ;-)

How often are you sampling? The best approach would probably be just more filtering.

You're doing fine. Unlike the typical poster here, you ARE exhibiting an ability to think a problem through.

Good luck!

...Jim Thompson

Hello John,

In a pinch you can use pretty mundane stuff here. Don't remember where it was, oil rig or something. Some place where there was absolutely no way to get to a Radio Shack. Offered someone a few cool $20 bills if he'd sell us his old transistor radio, followed by an enthusiastic "Here, take it!". Cracked it open, ripped out the audio transformer, scraped the secondary off of it with a knife (without dismantling the core or anything), ran one turn of the power line through there and bingo, we had a CT. It probably wasn't a very good one but it worked just fine.

Sometimes you can find bags of these little old audio xfmrs in surplus stores. Five bucks per pound or so.

RMS? That's where a MSP430F2013 would really hit the spot because it's got the 16bit converter on chip.

Low-pass filer does not have to be active. I did not think that I should state such trivial things...

But an active rectifier is also unneeded.

John Perry

(snip)

I would recommend a 1000:1 current transformer similar to this series of different primary current ratings by Amveco, from Digikey:

Nice, but it's still a lot of current to slam into the summing point of an opamp. 20 amps RMS divided by 1000 is still 28 mA peak, and metering systems should have a heap of headroom too.

I wonder what phase shift and linearity are like on these parts; they're fairly small.

John

$6.50 @ qty.20 vs. CSE187L @ $2.34 @ 20 pcs.

They are pretty mediocre as far as linearity goes, unless you keep the burden resistor about as low as a few times the winding resistance.

I would try to design a burden and averaging filter that did not need any opamps, especially if the peak load current is much higher than 5 amps.

Fair enough. What is the turns ratio and burden limit for these?