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It is impossible to be precise because the wires do not have the standard colour codes that I'm familiar with, and it would be necessary to open up the panel to be certain, but I think the pairs of wires will be line and neutral (or two lines if your supply is centre-tapped to earth); and either way you need to split the pairs of wires coming through each gland and clip the c.t. on one wire of the pair only.  Of course, if you have a three-wire circuit, you need a c.t. on two of them. What you do with the orange cable that has an overall sheath, I don't know - you might fall foul of local regulations if you cut the sheath away.

You should be able to measure the output of the c.t. directly with your multimeter, you should read 50 mA per 100 A of current in the primary circuit. You need a decent multimeter to measure what might be only 30 - 50 µA.

Which capacitor do you mean? The input circuitry is described in Building Blocks » CT sensors - Introduction » CT sensors - Interfacing with an Arduino.

Thanks for the reply Robert and now I think I get it!  The capacitor in question is on both systems (voltage and current) and the (current only), but you'll see here on the link a 10uF capacitor in the needed and it is wired into the circuit.  

http://openenergymonitor.org/emon/buildingblocks/how-to-build-an-arduino...

Thanks again!

Tyler

That 10 µF provides a low impedance path to ground for the 50/60 Hz signal, otherwise you have the Thévenin equivalent of the two bias resistors in series with the input and that would affect the accuracy - as they say dig out your college notes and read, learn and inwardly digest the page I quoted ;-)  The voltage input is identical in form, instead of c.t. + burden, you have v.t. + divider network, but after that it works in pretty much the same way.

[Edit] Oops! The by-pass capacitor wasn't mentioned - it is now.

Would the accuracy of the capacitor be "inaccurate" because the current/voltage of the CT which would be AC would be mixed with the DC from the Arduino circuit?

I'm going to have to see about opening the panel on the circuit breaker box...

Is this a better placement for the CT sensors under the panel and  on one of the wires going into the actual breaker? I think I found the best place.

You need to look hard at the details of the ADC input to understand what might happen.

The ADC has a high but finite input resistance and a small capacitance - the 'sample and hold' capacitor, and because the input is switched around the 6 inputs by the multiplexer, that capacitor needs to be charged or discharged before the correct voltage can be read. If  the source that the ADC is measuring has a high resistance, there are two effects: the source resistance and the input resistance may act as a voltage divider and reduce the voltage that the ADC sees, and it takes a long time for the source to charge the capacitor to the correct voltage. Without C1 (in that diagram), the source resistance is the two bias resistors in parallel plus the series resistance of the transformer secondary (and the burden in parallel in the case of the current input), or about 235 kΩ. The data sheet recommends a maximum of 10 kΩ.

Putting C1 in parallel with that 235 kΩ equivalent resistor means that at 50 Hz, the total series resistance is now 318 Ω reactance of the capacitor, plus the transformer impedance, so less than 500 Ω in total thus well below the limit. It reaches the limit at 1.6 Hz

You posted that picture while I was typing. 

I can't read the labels on the breakers but on the face of it, if your local rules allow (and I don't see a good reason why they should not) then the best place is indeed inside the panel. If your supply is 120-0-120 V (i.e. 240 V centre-tapped to earth) and all your loads run off 240 V connected L-L, then 1 c.t. per breaker will read the correct current. If you have some 120 V loads connected L-N, then you need 1 c.t. on each line (facing opposite directions) and you need to calculate the power in each half separately and add them.

Thanks for all the input!  I'll be giving this a shot soon.  Here is one last question.  If I want to measure voltage simultaneously and do this (http://openenergymonitor.org/emon/buildingblocks/how-to-build-an-arduino...), then why do you need an ac-ac adapter?  Couldn't you just run a power cord from the wall straight to the circuit and drop the voltage across a couple resistors and get the same effect or does the adaptor change the signal somehow else?

Ahrrrrgh! If you do as you suggest, your whole circuit becomes live and, if the neutral disappears for any reason, LETHAL (apart from blowing up your laptop if it is plugged in / if you happen to plug it in).

The adapter is NECESSARY for safety. If you want something that performs a bit better than the standard ac-ac adapter, see this http://openenergymonitor.org/emon/node/1987 and take special note of the safety and isolation requirements and precautions.

Sorry, figured it was safety after I wrote it, but wanted to ask still.  It seems in theory it would be ok, but not okay in practice!  Thanks for everything Robert!

Edited:  Also thanks for keeping me from doing something stupid!  Glad to know you're looking out for me!