# AC current measurement with a current transformer

Discussion in 'Electronic Design' started by Ignoramus19198, Dec 14, 2005.

1. ### Ignoramus19198Guest

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:

http://tinyurl.com/882gb

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

2. ### John PopelishGuest

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).

3. ### Ignoramus19198Guest

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

i

4. ### Tim WescottGuest

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

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.

5. ### Ignoramus19198Guest

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

6. ### John PopelishGuest

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.

7. ### Ignoramus19198Guest

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:

http://igor.chudov.com/projects/17.5-Phase-Converter/

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.

run/start/balancing capacitors.

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: http://tinyurl.com/882gb, and now I
am trying slowly to learn how to make it display various stuff.

i

8. ### The PhantomGuest

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.

9. ### The PhantomGuest

Using an AC range, of course.
Diodes large enough to handle the current in the secondary. 1N4004 type
should be ok.

10. ### Rich GriseGuest

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
10. 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

11. ### Rich GriseGuest

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!

12. ### Tim WilliamsGuest

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

Tim

13. ### John PopelishGuest

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.

14. ### John PopelishGuest

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:
http://www.widebandcts.fsnet.co.uk/Index.htm

15. ### Tim WilliamsGuest

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

16. ### budgieGuest

If you'd seen the result of an OC secondary on a metering CT on a busbar, you
would paint a different picture.

The primary - at 200A rating - is undoubtedly a single pass of a conductor
through the CT core. Typical metering CT burden is 15VA max. If the secondary
is OC'ed even momentarily, this reflects an OC onto the primary conductor.
Volt drop along that conductor (busbar?) becomes "significant" and the CT
secondary voltage goes through the roof. The extent of the resultant damage
becomes evident once the smoke clears.
I strongly urge the O/P to fit a suitable low impedance resistor to the
secondary, and do all his other connections downstream of that.

17. ### John PopelishGuest

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.

18. ### The PhantomGuest

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.

19. ### Ignoramus19198Guest

That's exactly what my schematic shows.

i

20. ### Tim WilliamsGuest

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