I and v
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Liam Moran
Lifetime Supporting Member
A simple matter of economics, really, and Kirchoffs laws and just simplicity.
If you have a current loop say 5 kilometers long,
and you inject a current of say 10 milliamperes into that loop from a source,
then if you break into that loop at any point and measure the current
it must be 10 milliamperes.
With voltage measurement, over the same distance,
the cable resistance comes into play, however tiny it may be,
which would depend on the device reading the voltage signal.
It would be required to use a second pair of wires to measure the voltage at the source,
an example of this practice being a loadcell.
These sensing wires would also have a resistance, therefore a tiny voltage drop.
A current source on the other hand will automatically compensate for cable resistance.
If you have a current loop say 5 kilometers long,
and you inject a current of say 10 milliamperes into that loop from a source,
then if you break into that loop at any point and measure the current
it must be 10 milliamperes.
With voltage measurement, over the same distance,
the cable resistance comes into play, however tiny it may be,
which would depend on the device reading the voltage signal.
It would be required to use a second pair of wires to measure the voltage at the source,
an example of this practice being a loadcell.
These sensing wires would also have a resistance, therefore a tiny voltage drop.
A current source on the other hand will automatically compensate for cable resistance.
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In addition to Liam Moran's explanation: using current rather than volts, also means that wire breaks are detectable.
A wire break does not effect a voltage signal going out (except for the devices beyond the broken wire), so an signal monitor located close to the signal source, will not detect the fault, but in a current loop, the break stops the current in its entirety, and thus an alarm can be raised regardless of the location of the signal monitor device.
A wire break does not effect a voltage signal going out (except for the devices beyond the broken wire), so an signal monitor located close to the signal source, will not detect the fault, but in a current loop, the break stops the current in its entirety, and thus an alarm can be raised regardless of the location of the signal monitor device.
Sherlock4me
Member
http://www.iqinstruments.com/iqshop/technical/signals.html
Of course voltage signal applications exist in process instrumentation as well, (0-10v, 1-5v).
Forgive me if you are already familiar with that.
Congratulations on your quest for learning!
Regards : )
Of course voltage signal applications exist in process instrumentation as well, (0-10v, 1-5v).
Forgive me if you are already familiar with that.
Congratulations on your quest for learning!
Regards : )
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geniusintraining
Lifetime Supporting Member + Moderator
danatomega
Lifetime Supporting Member
A ton of information on electricity and electronics on this site:
http://www.allaboutcircuits.com/
http://www.allaboutcircuits.com/
ndzied1
Lifetime Supporting Member
Actually, without a special meter this is NOT true. Using a regular multi-meter you MUST break the circuit and put the meter in series into the circuit to measure the current signal.
For a VOLTAGE signal you do not have to break into the circuit to take measurements.
Well yes, that is why I said CAN
.I love this piece of equipment:
http://en-us.fluke.com/products/ma-loop-calibrators/fluke-773-process-calibration-tool.html
ndzied1
Lifetime Supporting Member
Well yes, that is why I said CAN
I love this piece of equipment:
http://en-us.fluke.com/products/ma-loop-calibrators/fluke-773-process-calibration-tool.html
Who wouldn't like it but at > $1000.00 it is out of the reach of a lot of people.
http://www.newark.com/fluke/fluke-7...09R2351?ost=fluke+773&categoryId=800000038002