Calculate it for yourself - the wire will act as a resistor to lower the fed in voltage at its end; different to the impulse transmission where the microcontroller port just needs to "see" a detectable voltage drop here we deal with the actual voltage itself; with the wire's resistance with length the voltage detectable at its end will decrease; the lesser the resistance the lesser the voltage drop; the voltage drop then has to be set into relation with the fed in voltage that is proportional to the amperage measured at the mains; and as we deal with a Vref of 1V1 at the FLM any millivolt counts...
Happy physics revision... http://en.wikipedia.org/wiki/Electrical_resistance_and_conductance
gebhardm |
To answer the question: If you use a supra-conducting wire, the length may be infinite ;-)
ygeffens |
Thanks for the answer, but I don't understand anything about it.
gebhardm |
Ouch, that is hard ("Strom kommt aus der Steckdose") - to explain it simpler:
A current clamp measures the (alternating) current running through a wire inductively; it "translates" the measured current into a proportional voltage - in case of the FLM clamp, for example 0 to 50A are "translated" to 0 to 5V, thus, if 1A is floating the clamp delivers 0.1V
Within the FLM a voltage divider furthermore "translates" the delivered 0 to 5V from the current clamp to 0 to 1.1V actually measured by the FLM's analog-digital-converter (ADC); thus 1A at the clamp is actually "equal to" 0.022V at the ADC input.
Why Ampere, not Watt? Because 1W for resistive loads "equals to" 1VA, calculated from the Amperes measured multiplied with the "assumed constant" voltage of the mains
The clamp's voltage is transported through a wire; a wire consists of a material that has an inherent resistance; copper, for example, has a specific resistance of approximately 0.017 Ohm * mm^2 / m; thus, 1 meter of copper wire
with 1mm diameter has a resistance of 0.02 Ohms (which is rather insignificant) - BUT with length (and decreasing diameter) the resistance increases
The resistance of the wire with the connector at the FLM forms a voltage divider; thus with increasing resistance there is a voltage drop at the FLM input, thus an "error" occurs on the measurement
Homework now it calculating the wire length of significant impact using the given boundary conditions ("Selber denken macht schlau").
Fluc |
Thats a very good explanation Gebhardm, i hope Ygeffens will understand.
I think it is a good idea to check the link to the Wikipedia site above, who is very instructive for novices in that matter.
ygeffens |
Ok, thanks for that. It's a lot more clear now.
So the thing is to add a 10 meter wire to the clamp (what I have is an SVV 0.8mm), use eg. a lamp of 100watt, and see what is measured. So it's possible that I'll read 90watt in the FLM.
Correct?
petur |
That's all nice and clever (just don't rub it in), but without knowing the current that the FLM draws from the clamp (ie mostly the current going in the divider), you still don't know the influence.
Bart will have made a compromise there between cable voltage drop due to higher current, and picking up more noise along the way (if current is too small).
I assume the values of the divider are somewhere on this site...
Fluc |
Keep the cable away of other cables from high energy sources or you will get noise.
A shielded 2 core wire is a good solution in these cases.
petur |
And the shield of that shielded cable should be grounded on one side (or two if you are sure there is no chance of a ground loop) ;)
I just had a quick peek at the schematic and it is near the limit of what I still remember from school, I only see 1 1K91 resistor so the rest of that divider must come from the cable or clamp, in which case the cable might actually be more important? It also makes me wonder if that ADC really uses 1V1 as stated above.
Vbg = 1.1V corresponding to the spec of the AVR ATmega168; help yourself. The clamp has an impedance of 6k8 (so you get to 1.1V at full clamp voltage) and the current drawn by the AVR is insignificant. (http://www.atmel.com/Images/doc2545.pdf page 250ff.) - it's all in the sources.
Calculate it for yourself - the wire will act as a resistor to lower the fed in voltage at its end; different to the impulse transmission where the microcontroller port just needs to "see" a detectable voltage drop here we deal with the actual voltage itself; with the wire's resistance with length the voltage detectable at its end will decrease; the lesser the resistance the lesser the voltage drop; the voltage drop then has to be set into relation with the fed in voltage that is proportional to the amperage measured at the mains; and as we deal with a Vref of 1V1 at the FLM any millivolt counts...
Happy physics revision... http://en.wikipedia.org/wiki/Electrical_resistance_and_conductance
To answer the question: If you use a supra-conducting wire, the length may be infinite ;-)
Thanks for the answer, but I don't understand anything about it.
Ouch, that is hard ("Strom kommt aus der Steckdose") - to explain it simpler:
with 1mm diameter has a resistance of 0.02 Ohms (which is rather insignificant) - BUT with length (and decreasing diameter) the resistance increases
Homework now it calculating the wire length of significant impact using the given boundary conditions ("Selber denken macht schlau").
Thats a very good explanation Gebhardm, i hope Ygeffens will understand.
I think it is a good idea to check the link to the Wikipedia site above, who is very instructive for novices in that matter.
Ok, thanks for that. It's a lot more clear now.
So the thing is to add a 10 meter wire to the clamp (what I have is an SVV 0.8mm), use eg. a lamp of 100watt, and see what is measured. So it's possible that I'll read 90watt in the FLM.
Correct?
That's all nice and clever (just don't rub it in), but without knowing the current that the FLM draws from the clamp (ie mostly the current going in the divider), you still don't know the influence.
Bart will have made a compromise there between cable voltage drop due to higher current, and picking up more noise along the way (if current is too small).
I assume the values of the divider are somewhere on this site...
Keep the cable away of other cables from high energy sources or you will get noise.
A shielded 2 core wire is a good solution in these cases.
And the shield of that shielded cable should be grounded on one side (or two if you are sure there is no chance of a ground loop) ;)
I just had a quick peek at the schematic and it is near the limit of what I still remember from school, I only see 1 1K91 resistor so the rest of that divider must come from the cable or clamp, in which case the cable might actually be more important? It also makes me wonder if that ADC really uses 1V1 as stated above.
Maybe Bart can clear it up?
https://github.com/flukso/flm02/blob/develop/avr/b/main.c --> line 445:
Vbg = 1.1V corresponding to the spec of the AVR ATmega168; help yourself. The clamp has an impedance of 6k8 (so you get to 1.1V at full clamp voltage) and the current drawn by the AVR is insignificant. (http://www.atmel.com/Images/doc2545.pdf page 250ff.) - it's all in the sources.