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AC amp meter issue

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LAN_403

IamSmooth

Wed Jan 26 2011, 01:41AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I got a 50A ac amp meter on Ebay. The seller did not realize it only measured 5A and needs a current transformer or shunt for it to read 50A. I made a simple current transformer that will put out 1A for 10A going through it.

Here is my question: If I have 50A going through the center of the toroid and I am measuring 5A from the CT, does this mean I would only be drawing 45A from my source if I did not have a shorted CT? In other words, my CT is consuming power.

My best bet would be to use a shunt, no?

Patrick

Wed Jan 26 2011, 03:36AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Others should check my reasoning, but I would say if you have a good CT transformer that your confident is accurate, then use that not a resistive shunt.

As for your question of power, you should have a resistor in pararllel with the meter movment + CT leads, so only the small power of the meter and then resistor burden current times your CT V at the output should be consumed. It should be low compared to using a direct shunt conducting all 50 amps. You could calculate it both ways to see, I dont know your numbers but its simple.

I think I have said all this correctly.

One last thing, keep the small resistor attached directly on the CT leads, the CT should never float with a no load applied or you will get high voltage and yes it will be just happy conducting several amps if you short it. Dont do that.

Dr. Slack

Wed Jan 26 2011, 08:25AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

For a 10:1 current transformer, the turns ratio is 1:10, so secondary turns NS = 10*NP.

For a transformer with a *very high* permeability core, the magnetising current is *nearly zero*, so the primary ampere turns of 50A * NP turns will be *almost exactly* balanced by the secondary ampere turns of 5A * Ns = 5A * 10*NP.

So you'll be drawing 50A from your load, and putting *nearly* 5A through your meter.

As the permeability is finite, increasing the number of turns will reduce the error between *nearly zero* and zero, and so improve the accuracy of the 10:1 current ratio.

You ought to be able to do a simple calculation. Your CT needs to generate, what, several 10s or 100s of mV to push 5A through your meter. You have Ns secondary turns, so will need a certain flux swing in the core. Given an estimate for permeability (2000 to 10000 range, from boggo power transformers to special mu-metal cores) you get the H-field needed in the core. H field is measured in ampereTurns per metre, which with the length of the core becomes the ampereTurns needed to magnetise the core.

et viola, 50*NP ampere turns in = core_magnetising current + Iout*Ns. A good current transmformer + current meter pairing will have the magnetising current <1% of the input current, for 99% accurate current transfer ratio. You can see how the current meter resistance governs how much voltage needs to be generated, which governs the magnetising current - so a poor meter, or mounting it a long way away on long leads, will degrade the accuracy of the current transfer ratio.

Your CT is consuming power, magnetising losses in the core, resistive losses in the windings, but they are *small*, there are no "amps going missing down the hole in the middle".

(for infinite permeability and zero loss, all *terms* are 0, 1 or infinity, the choice of which should be obvious from the context)

If you take the current meter off the secondary, then short the output if you want to put current through the primary.

klugesmith

Wed Jan 26 2011, 09:28PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Let me weigh in too, and try to keep it simple.

If you use a 50 A shunt in your circuit, then the meter is not isolated from the power circuit.
A standard shunt would have precisely 1 milliohm of resistance between the sense connections.
At 50A RMS (AC or DC) it would drop 50 mV and dissipate 2.5 watts.

Now consider the 10:1 current transformer. There's no problem putting it on a wire carrying 50 amps and having the output shorted, or connected to a 5A range ammeter, or connected to a burden resistor and AC voltmeter.

In the shorted CT case, the power lost will be just I2R based on 5A and the resistance of CT secondary. (plus a bit of core loss). The voltage drop in the primary circuit due to CT should be negligible.

In the 5 A meter case, let's say the ammeter has an internal shunt of 0.01 ohms, dropping 50 mV and dissipating 1/4 watt at 5 amps. Since we have a turns ratio of 1:10, the CT will add a 5 mV voltage drop to primary circuit, thus extracting 1/4 watt of power.

Suppose you had an AC voltmeter and 0.10 ohm burden resistor downstream of the CT.
5 amps would develop 500 mV to measure, and the burden R would dissipate 2.5 watts.
CT would add 50 mV of voltage drop to primary, and extract 2.5 watts of power (at 50 A).
Effective load on the primary circuit is the same as the 50A shunt we talked about.

If the burden resistor were 1 whole ohm, it would develop a voltage of up to 5 V RMS and dissipate 25 watts. Warning: 5 volts is a lot for a 60 Hz 10-turn transformer winding. If the steel core cross-sectional area isn't well over 1.5 square inches (10 cm^2), the core will be saturated before you get to 5 V.

If the CT core area is too small for primary current * burden R / secondary turns count^2 / frequency, the core will saturate and generate as much secondary voltage as it can. Which isn't that much for 10 turns on a small core -- depends on HOW much the saturation current is exceeded, so how fast the magnetic flux density switches from 2T clockwise to 2T counterclockwise. Unless the voltage is enough to give you a shock, or to break down insulation, this won't hurt anything except eventually heat up the core. (Shocks and arcs are real hazards with large-ratio CT's.) Maximum voltage drop added to the primary circuit is the voltage necessary to saturate the core -- couple hundred mV for common size 50A:5A CT's. And it's in spikes aligned with current zero-crossings, reducing the "real power" extracted from the primary circuit.
Tonight I hope to measure and report the peak open-circuit voltage from a standard 50:5 CT with various sinusoidal currents, and see when it get to the electric-shock level.

Small cores or ferrite cores may be fine if you are talking about SMPS frequencies instead of mains frequencies.

radiotech

Wed Feb 09 2011, 12:24AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

Let me add two things to the good stuff already said. In a CT
the burden range will not reduce the stated accuracy, and that
burden range is part of the specs.

The CT, (in even a 50 AMP size will develop dangerous
voltages across the secondary across the secondary if
accidently open circuited. In hugh CT's they can do major
damage, and in practice have hugh thyrites across them. (mov)

Shunts have one issue, apart from calibration. By using a shunt
to drive a remote meter, you violate a principle rule of power
systems, that is the interconnecting of different systems, IE, the
power side to the control side,

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