Measuring current issue

Hi guys,
Im using the YHDC sct-0013-000 current type CT and the sample code and hardware provided on this site.
I have the CT around the live wire of a small test circuit I created consisting of a 40watt filament bulb.
For some reason the serial monitor shows values like 160 and .7 for the apparent and current draw, when it should be more like ~40watt and ~.17A. Im on a 50Hz 230V system using the correct circuit on breadboard with the same components i.e 2 x 10k resistors for the voltage divider, a 10uF cap polarised and a 33ohm burden resistor.
It make no difference whether the load is on or off the values dont make any sense.
Im using an Arduino UNO R3
Thanks guys
Appreciate the help

Exactly your problem is explained in the Frequently Asked Questions.

Hi,
Yeah thanks for the reply.
I think the load is too small, for the moment I will wrap the live around the CT a few times and see what happens
Thanks Robert
Regards

I’d suggest 20 for a reasonably accurate reading, bearing in mind your max power reading will be 23 kW, so you’re trying to measure below 0.2% of the maximum your system is capable of.

[BTW - put a sensible country in your profile. It helps if we know where you are when you ask about the public electricity supply.]

Hi Robert,
I should give 20A+ a go. Ive tried with a 700Watt ~3A vacuum. The emon calibration of 111.1 gave me a reading of around 1400W. I tried a plug in energy meter and it averaged around 690w. I changed the calibration to something like 51.8 and the readings are more accurate.

Also, my query about an AC/AC adapter which I’m using to measure voltage, might be a stupid question but, the 9V AC adapter should still output at 50Hz or mains frequency on the output, its a Mascot 9580. I am also measuring the frequency of the mains.
Thanks

We used to use the Mascot 9580. There is still a lab report in the Learn section. The output was 11.2 V @ 240 V in, so a slightly different voltage calibration constant is needed.
The adapter is only a transformer, so you should get and exact replica of the mains waveform, only the amplitude gets smaller. See the pictures of the output voltage in the report.

The current calibration expects a single turn primary winding. If you have multiple primary turns, you change the transformer ratio (from 100 A : 50 mA to 50 A : 50 ma, 33.33 A : 50 mA, 25 A : 50 mA and so on, and the calibration constant changes likewise from 111.1 (18 Ω burden) to 55.56, 37, 27.78 etc.

Ah ok,
Im just using a single turn with a 700watt load, I am using the 33 ohm burden resistor.
thanks for your help
regards

The full explanation of the calibration constants, and how to calculate them, is in “Learn”. But you will still need to make small adjustments due to component tolerances.

Hi Robert,
Ive been reading the calibration procedure listed here: Learn | OpenEnergyMonitor
Yes it seems that using a 33 ohm burden requires a 60.6 calibration factor,
CT Ratio / Burden resistance = (100A / 0.05A) / 33 Ohms = 60.6 (for the emonTx Shield)
Well that is from the theoretical calibration of the CT.
Thanks
regards

Hmm, ive been trying to calibrate the load of a vacuum cleaner I have. It seems to hover around 650watts, i tried the calibration of 60.6 but it doesnt seem right. So i tried a value of 49.9, i calculated the error and adjusted the value accordingly. At the moment it seems to be a few Watts off what the meter reads, I also tried a 2000W electric kettle and it seems to have the same accuracy. I only ask as I was wondering whether the need to vary the calibration so much from the 60.6 for a 33R burden was normal?
thanks

You most probably won’t be able to calibrate with a motor as a load, because it will have a power factor of something less than unity. This is why the calibration instructions insist that you use a resistive load.

I would believe the results using the electric kettle as the load for calibration. The current calibration constant depends on many variables. For the calculated value to be correct, all those must be exactly right, or if there are differences, the errors must all cancel out. There is an article in Learn where all the factors are listed.

However, one thing is not clear to me: you called this thread “Measuring current issue”, but you are talking about reading values of power. The voltage calibration - or the voltage default value - will also affect the power calculation, so it is confusing and wrong to talk about power when you are trying to calibrate the current input. If you are measuring voltage and current, you should calibrate the two separately. If you are measuring current only, then you must set the assumed voltage accurately or the current calibration will have the voltage error added to it.

Hi Robert,
Sorry yes I am using an assumed RMS value of 230V, I have not yet implemented the Mascot AC AC adapter.
I am also using 5% tolerant resistors, I will switch it to 1% tolerant resistors and try again. Yes, I should have stuck with the electric kettle. What kind of resistive load should i use to say achieve a current draw of 20Amps, I realise that the 100A CT isn’t accurate under 1A, I’m thinking like a 5kW heater?

I also wanted to post a question about power factor but I’ve been reading that you cant tell if its leading or lagging, so phase angle cant be calculated using the Emonlib? I was thinking of using it to represent a graphical single phase phasor diagram.
The power factor is a measurable quantity as well as real power, apparent power using emonlib which is more than sufficient. What about reactive power does the emonlib calculate that, sorry if that a stupid question I’m just thinking of it now.
thank you for all your help, Robert!! I really appreciate it.

But is that the true voltage that you are seeing when you are trying to calibrate the current using a power meter? If not, the difference is part of your error.

You can use your kettle, and tell lies to your c.t.!
The c.t. actually responds to ampere-turns, so 3 turns of wire through the c.t. should give you about 26 A.

EmonLib uses the definition of power factor - the ratio of real power to apparent power - to calculate it by calculating separately real power, rms voltage and rms current. It does not attempt to extract the quadrature component of current (which will give you leading or lagging, plus greater accuracy at very low power factors). So it cannot get a reliable value for reactive (imaginary) power. The only way that I can think to do that would be to delay the voltage wave samples by 90° and multiply the current samples by those, and average the result.

Phase angle is only meaningful when you have a pure sine wave for voltage and current. Once you have harmonics present, only the phase angles of the corresponding harmonic components, or the phase of a harmonic relative to its fundamental, has a meaning.

Oh yeah, i forgot about the windings, I already tried that with the 40watt bulb.
I will try again with the kettle.
The power factor in the emonlib, is it accurate to within a certain tolerance i.e I havent looked through the library but can I extract the PF value as a value of between 0 -1 etc?
if so, cant the phasor angle not be calculated by using Tan-1(PF)? i.e Cos 45 degree = 0.707, tan-1 .707=45.
Or will it still not matter as I wont know if its leading or lagging.
Im really only interested in real power, apparent power and power factor. maybe a calculated reactive power
thank you for all your help and understanding with me.
Kind regards

Power factor will be as accurate as the numbers that are used to calculate it. It is a unitless number.

As I said, phase angle only has a meaning when both current and voltage are pure sine waves. Only then can phase angle be calculated.

You cannot know from the definition, hence you cannot know from emonLib, whether the power factor is leading or lagging. Usually, it is lagging. But if you have many small loads that all use capacitors to drop the mains voltage to a low value to run the electronics, then you might well have a leading power factor. You might also have a leading power factor if you over-correct with power factor correction apparatus - but that is bad because you will have a higher mains voltage than normal.

You can apply Pythagoras to get the value of reactive power (but not the sign):
reactive power = √ ([Apparent Power]2 - [Real Power]2)

Hi,
Thanks for all your help, i was thinking I wouldn’t know the sign. I saw a zero crossing detector could be used to measure the time between v and i. I’m already using one in a simple way to measure the mains frequency. It’s a 4n25 that counts high and low times over a sample time and it’s not rectified so just the half wave.
Thank you for all your help