AC current measurement with a current transformer

• posted

Note that this circuit will not have the usual problems with the diode drop affecting the measurements because the current transformer secondary will look like a very high-impedance device. You _will_ have to take the diode drop into account as John described, and the diode voltage will appear (attenuated by a factor of 200) at the transformer's primary leads.

You could "fix" the RMS vs. average absolute value issue by using an RMS to DC converter, but accurate ones are expensive and hard to get. You could also fix it by sampling the instantaneous current with an ADC (replace the cap with the ADC inputs) then computing the RMS -- but that's a lot of effort unless you're going into a microprocessor anyway.

I'd probably just assume a sine wave and apply the right conversion factor to read RMS, or I'd just label the voltage "average absolute". In either case I'd make sure that folks didn't think it was _true_ RMS if they looked at my schematics.

```--
Tim Wescott
Wescott Design Services```
• posted

Any current transformer has a primary volts*seconds core rating. As long as this is not exceeded each half cycle, the core should not saturate and the output current should be reasonably proportional to the input current. This rating is often expressed as a primary current rating (200 amps RMS, in this case) and a secondary maximum burden resistance (unknown, in this case), since the voltage across the burden resistor is proportional to the current passing through it. As long as your bridge rectifier, and RC network does not drop more average voltage than the rated burden resistor, the scheme should work fine and produce a voltage proportional to the average of the absolute value of the primary current (not the RMS value of the primary current).

• posted

That's right. Just as a voltage transformer maintains the voltage ratio over a fair range of currents, a current transformer maintains a current ratio over a fair range of voltages. Actually, they are exactly the same thing. Just a turns ratio on a core. The difference is that one is connected across a source of voltage and one in connected in series with a source of current.

The lower the voltage across the windings, the more ideal the transformer acts, so replacing a silicon diode bridge with a Schottky diode bridge will drop even less voltage and give slightly better linearity. But if the burden resistor (RC filter in this case) drops significantly more voltage than the bridge, the view is probably not worth the climb. But this does bring up the concept that you want to use as low a resistance as you can in the burden, (as low an output voltage as you can make use of) to get best accuracy out of the transformer.

Yes. There is about a 10% difference between average absolute and RMS for a sine wave. As long as the waveform is constant, the correction factor is constant, also. If the current has a widely varying shape under different operating conditions, then a true RMS measurement may be necessary to achieve a satisfactory accuracy.

• posted

It seems to me that you are making it a lot harder than it needs to be. I'm sure you have a DVM or two, so just connect the secondary of the CT to the amp (or milliamp) input of your DVM. Start out with something like the

1 amp range to make sure you don't overload the DVM and blow a fuse. Put (and leave) a couple of antiparallel connected silicon diodes across the secondary of the CT so that when it's disconnected from the DVM, you don't get dangerously high voltages due to an open circuit on the secondary when 100 amps is under measurement. The drop across the DVM shunt should be substantially less than the .7 volts drop of the diodes, so that when the DVM is connected, the diodes will have no effect on your measurement. You should check this with your scope.

If you make your measurement this way, all the RMS conversion and calibration issues (other than the CT ratio) will be taken care of.

• posted

I want to ask to clear up my confusion about use of c.t. to measure AC current.

I have a need to measure 60 Hz sinewave AC current up to say 100 amps.

I have the following:

- 200:0.1 current transformer

- a DC voltmeter. Actually an Omron process meter with linear conversion y=ax+b feature:

I am thinking of doing this as follows:

- put a rectifier bridge in between the contacts of the c.t., like DigiKey item DF10MDI-ND.

- Put a resistor between the + and - poles of the rectifier bridge.

- Add a capacitor in parallel with resistor to filter ripple

- Measure voltage across the cap, and use the linear conversion feature of the meter to convert this reading to AC line votls.

Is this a sensible plan?

Current xfmr 1 ----_ / \\ + - \\ / Current xfmr 2 ----~

(continued from the above diagram)

• ----------+------+----- R C=== to voltmeter

- ----------+------+-----

Would practical issues like diode voltage drop affect this setup?

i
• posted

Thanks John. I will try to get some numbers out of my CTs. Then I will know for sure, but at least I know what to look for. You were extremely helpful.

i
• posted

In layman's terms, it's about the c.t. trying very hard to produce required current and overcoming voltage drop of diodes. Right?

That's actually a factor of 2000, right? (my CTs are 200:0.1) So a 1.2 volt drop across the full bridge would appear as a 0.0006 volts drop on the actual AC line, right?

I can live with that. :)

I am not sure if I need to worry about RMS issues. The process meter that I will use has a linear conversion feature y=ax+b. In other words, I can instruct it to apply a shift and a factor conversion. The shift does not seem to be needed (b=0), but I would need to do linear conversion anyway. As long as DC voltage is proportional to AC current in the primary, I can make the meter display actual RMS current in the primary by selecting appropriate scaling factor on the meter's panel.

I would calibrate it by using my old trusty Weston AC current meter. (military) I would make sure that when the Weston reads, say,

42 amps, the process meter displays "42".

Is that right?

i
• posted

Using an AC range, of course.

Diodes large enough to handle the current in the secondary. 1N4004 type should be ok.

• posted

That's a great explanation. I was wondering about the same thing.

I see. Makes perfect sense.

That's pretty much the case. It's going to be either regular AC from the wall, or third leg of a phase converter.

You see, I just took apart my old 10 HP phase converter, which did not provide enough power for the welder, and made a 17.5 HP phase converter, with two idlers:

It has a nice starting arrangement of starting one idler first (10 HP), using start caps between legs 1-3, and after the first one spins up, the second idler is started. When the second idler is turned on, another pair of capacitors between legs 2-3 is also turned on, therefore the caps turn into balancing capacitors.

See

It has a few loose odds and ends (like needing some butt splices and cord grips and a pushbutton for the second motor), but it is already working fine.

Because it is enclosed in such a nice enclosure (see pictures), I want to add a nice display of various things, like line voltage, third leg voltage, line current, third leg current, etc. So I bought this process meter for \$19.99 on ebay:

and now I am trying slowly to learn how to make it display various stuff.

i
• posted

Related question: does a current transformer work with a variable frequency source?

Tim

-- Deep Fryer: a very philosophical monk. Website:

• posted

Because it acts as an inductor, right?

I once put some turns on a small toroid and dropped it over the ground return line on my open-loop induction heating circuit (which I'm guesstimating, skewed by all sorts of parasitic inductances and wirewound resistance, has no more than 30A peak), and read the voltage across a resistance (2.2 and 10 ohms, giving readings in the 0.5Vp-p range). The thing is, voltage response remained relatively constant across the frequency band. Not only didn't it taper off at high frequencies, but it didn't register the series resonance peak, IIRC.

I don't see what volt-seconds has to do with it, so long as it stays below saturation. This is maybe 30A in one half-assed turn (half turn?). I've measured toroid cores saturating above 100AT, I haven't measured the one I used but it was used for a buck converter, it's probably better than most toroids then, in regards to Bsat.

-- 5 to 100kHz, plus whatever harmonics a 1us tr/tf squarewave will generate. Probably a good place to use ferrite. :)

Ooo, my pants are tingling. I bet their stuff costs a fortune, too...

Tim

-- Deep Fryer: a very philosophical monk. Website:

• posted

My holy yikes, I'm about to contradict John Popelish. From my meagre experiences with CTs, I wouldn't put a bridge right at the secondary. From the 200:0.1 spec, it sounds like 100 mA per 200A in the primary. I'd pick a value - if the CT is, say, the size of a microwave oven magnet, I wouldn't be afraid to use a 100 ohm, 2 watt resistor - that translates to 10V at 100 mA, doesn't it? E = I * R, so 100 * .1 is

1. That's a watt.

Then, since it's isolated, I'd go ahead and just go to an opamp precision rectifier, (high impedance, you see, it doesn't load the circuit much) and do whatever with it I wanted - peak detector, RMS circuit, or if I wanted to be really off-the-wall, use a uP to sample it at a few KHz and calculate the RMS in software. That way, it wouldn't even need to be rectified, although I'd take a look at input scaling. If your ADC can only handle 5VPP, I'd use a 50 ohm burden resistor, and so on. ;-) (or whatever - I'm answering a USENET post, not doing arithmesthenics!) ;-)

[maybe more like 14.1 ohms - what's the P-P of 200 ARMS? -- Thanks!]

And, with another ADC on the volts, you can calculate Real Power and Power Factor! :-)

Good Luck! Rich

• posted

And, just for the record, it's surprisingly easy to get a really good approximation for RMS with, say, 32 samples per cycle, and LUTs for the square and square root of 8-bit numbers. ;-)

When I did that in my 68HC11-powered SCR phase-controlled 24V 40A battery charger design, it turned out that it didn't make very much difference in the regulation whether I went through all that RMS rigamarole or just used the average. I was proud of the short circuit shutdown, however. If the input of the HC11 current-sensor ADC hit the rail (0FFh), I just had it not fire the SCRs next cycle. Or again, until the user disconnected the cables from the short.

Turned out to be too expensive for the market, though. )-;

Cheers! Rich

Cheers!

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What's the big deal? I didn't design the circuit, I just described how it would work. I am always glad to see alternate approaches and opinions. Even if they are not all perfect, they generate educational discussion.

Plus and minus 283 amps. If you wanted to keep that inside 0 to 5 volts, with a bit of head room for starting surges (I would clamp the signal with a pair of 1 watt 4.2 volt zeners) You might use a burden of 6.8 ohms (+-1.9 volts at 200 amps RMS through the primary).

Yes. If you already use PIC's or some other processor that comes with an A/D, that would be a useful project.

• posted

You don't think saying "dangerously high voltages" is sufficiently explicit, I guess.

He should be plenty scared now.

A pair of antiparallel connected diodes *is* a suitable (non-linear) resistor, and I told him to *leave* them connected to the secondary, thus making measurements "downstream" of them.

• posted

Just as with any transformer, there is a practical lowest and highest usable frequency. As the frequency goes down, you also have to lower the burden resistor to keep the primary volt*seconds per half cycle from growing. The upper end is usually limited by the thickness of the laminations in the core. But there is also a leakage inductance limit that rolls off the coupling between primary and secondary as frequency rises. That said, good quality current transformers with generous core masses (for the low frequency end) and very thin, very high permeability tape wound cores can be reasonably flat well beyond the audio spectrum. For instance:

• posted

Yes. An inductor produces voltage proportional to the rate of change of flux passing through the coil. If that lasts a long time, this represents a lot of flux accumulating before the next half cycle undoes that flux and builds the same amount the opposite way round.

The current to voltage conversion is essentially independent of frequency, since the transformer is converting primary current to secondary current, just by the turns ratio. The secondary current develops voltage across a resistor, that has a flat frequency response. Only the flux swings in the core vary inversely to frequency.

Volt*seconds is what drives saturation. You might think that the current passing through the primary would instantaneously saturate the core at some low value, like it would with DC, but this doesn't happen as long as the flux can swing enough to drive an equal and opposite ampere turns through the secondary, so that the net amperes passing through the core approaches zero as long as the core permeability stays high.

This is maybe 30A in one half-assed turn (half turn?).

If it passes through the only hole in the core, it counts as a full turn.

Are These DC amperes? 100 AT is pretty high for a closed magnetic path of a few inches through a high permeability material. For a gapped core or a powdered iron one, the gap determines the saturation AT.

No doubt.

• posted

Yup. Same reason you can draw a hundred amps from a transformer rated

10A... well, not for long :) Primary current rises in proportion so that total magnetic burden for the core is dependent on inductance.

Yeabbut, the relation between volts, amps and seconds depends on the inductance, and I don't know the inductance of that strip of wire as it goes through the core (it can be calculated, of course). Isn't the basement figure the amp-turns (magnetic field) presented to the core?

Volts and seconds would depend on the inductance, which depends on amps, turns and (integral?) permeability.

Ok, so if I make one loop around the core, that counts as two? I'd better revise my turns ratio markings...

- Well, peak, since it's hard to measure saturation other than dynamically. (Okay, so you can make a permeability measuring setup, but hell, I don't feel like winding a saturable reactor and all that.)

Yeah, not very permeable in other words ;)

Ya. Most pot cores I measure start in the 100-200AT range ungapped (measured as where the dI/dt slope increases for constant applied V).

Tim

-- Deep Fryer: a very philosophical monk. Website:

• posted

C'mon, guys, this is a home project, and he's already got all he needs except the diode bridge!

Yeah, you let the output voltage get too high, and you get inaccuracy.

Yeah, you let the output go open, and you get lots of pretty sparks, smoke, and maybe flame. And a dead CT.

Yeah, you can put in lots of circuitry and maybe (but I doubt it) get a tiny improvement in performance.

Yeah, you can use it at outrageously inappropriate frequencies and lose a little accuracy. Or you can try it at rf or dc and get real strangeness.

But he's got a 200:0.1 CT and a programmable process meter. The process meter (I looked at his URL :-) has a 0 -- 199.99mV input range.

Now, if you want some extra protection, follow Phantom's advice and put

8 diodes, 4 forward in series and 4 backward in series, across the secondary. I'd put them at the CT itself, so you won't be tempted to remove them while you're experimenting. This gives about 2.8VAC maximum on the secondary when you do something stupid. It also avoids diverting significant current away from your bridge circuit.

John Perry

• posted

I

secondary

He was already appropriately warned. Igor has been working on his welder for some time and is careful of high powers and associated effects.

None of the commercial suppliers of CT's go to the extremes you do to describe the effects of open circuiting the secondary of a CT. For example:

says: "Care must be taken to ensure the secondary leads are connected at all times when current is passing through the primary conductor."

says: "...A turns ratio step-up would result in even higher secondary voltage. Any circuitry beyond the secondary load resistor could be subjected to high voltage, possibly resulting in circuit damage. Because of this potential high voltage, the load resistor should never be removed from the secondary when the current transformer is being powered."

Those of us who work in power have seen molten copper a time or two, and a word to the wise is sufficient.

resistor,

Because I'm suggesting he use the current input of his DVM as the burden for the CT, just like the clamp-on CT's that are designed to plug into the current terminals of a DVM. See: