Archived Forum

Just realized this issue has already been discussed in the forums.

http://openenergymonitor.org/emon/node/1680

http://openenergymonitor.org/emon/node/1705

Your fan is probably using a diode as a half wave rectifier for the low speed perhaps?

Paul

I don't think there's a 'perhaps' at all, more like a cast iron certainty! I think your monitor is fine and telling the truth - except that the 'zero' of the current is offset because of course the transformer blocks dc - in fact you can see in pictures 1 & 3 that it starts off 'correct' and how the wave settles to the new average value.

can you elaborate on the "the transformer blocks dc" part? Please and thanks!

Edit - deleted duplicate post - BT

You didn't need to ask the question twice, and you didn't need "Hey" to get my attention.

Transformers do not pass the dc component of the voltage/current in the primary winding into the secondary winding. This property, of isolating the secondary circuit from the primary, together with the voltage/current transformation ratio are the two reasons for using a transformer.

In picture 1, before you switch the fan on, your oscilloscope shows a flat line (or it's supposed to be a flat line!) that has settled in the quiescent state at about 1.1 V. When you switch on, the first half-cycle sits correctly at that level. Thereafter, your 'scope trace settles so that the average value of the half wave ends up at that same 1.1 V level.

A pure sine wave is y = Asin(ωt).
If you do a Fourier analysis of a rectified sine wave, you find it has a d.c. component of A/π (the first term in the series):

(A is the signal amplitude). That d.c. component cannot get through the transformer, therefore the 'baseline' of the wave, as you can see in your picture, is depressed below the quiescent level of 1.1 V.

If I find the RMS value of the waveforms as they appear in pictures 1, will I be measuring the correct current RMS value?

If you use the calcIrms( ) or the calcVI( ) method in emonLib, you will indeed read the true rms current or the true real power in your load.

Awesome.

Thank you for your crystal clear explanations.

If you use the calcIrms( ) or the calcVI( ) method in emonLib, you will indeed read the true rms current or the true real power in your load.

I can believe the true real power survives the CT ordeal, but how does Irms survive it?   Won't Irms² on the secondary side of the CT be Irms² on the primary side less Idc^2, where Idc is the DC current removed by the CT (A/π).

The attached scope trace is of my hairdryer on LOW.  The green trace is a current probe, and the pink trace is the output of CT.  On the green measurements you'll see that 3.18² + 2.83² = 4.26².   If I do the calibration maths, I find the pink trace has moved south by 2.83A to be centred on 0 (note the little GND symbols in the left margin).  And the RMS values of the pink curve corresponds to the 3.18A, not the 4.26A.

You will read the true rms of the waveform that gets though the CT - within the bandwidth limit of the sampling rate of course, and not counting calibration accuracy and aliasing effects.

If you can figure out a way of restoring the d.c. offset in the software, I'd be quite interested (I think this discussion has been had here already, some while ago).

At the end of the day, half-wave rectification is bad practice because most transformers are not made to handle it (OK, a distribution transformer would never notice that much in isolation, but would be in serious trouble if a significant proportion of its load was half-wave on the same half cycle) and, although I can't recollect a specific citation, my guess is that it is illegal once the d.c. component exceeds a specified absolute current. However, it's common practice as a cheap way of getting reduced power for small appliances like that fan or hairdryer.

You will read the true rms of the waveform that gets though the CT

Yes, but that won't be the true RMS reading of the load, which I think is what the OP was asking.   At any rate, so long as he realises the difference then we're all in agreement.

my guess is that it is illegal once the d.c. component exceeds a specified absolute current

The way they regulate that in these parts at least, is to put restrictions on even harmonics.  Here's how my energy monitor sees my hairdryer in the time and frequency domains.

OK, is zero an even number and if so, where's the zeroth harmonic ?

"Yes, but that won't be the true RMS reading of the load, which I think is what the OP was asking.   At any rate, so long as he realises the difference then we're all in agreement."

I do understand the difference now, I thank both of you.