Sensor with "current" or "voltage" output - Page 1
Forum section: electronics / general

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 10, 2010, 11:33:16 pm
content: The output of some LEM current sensors is "current" whilst that of some others is "voltage". Some engineers prefer to have a current sensor which outputs "current" because they claim for long distance transmission or in noisy environment, "current" is more robust than "voltage". Is that true? According to Ohm's law, V = I x R. "R" is constant, and if "V" changes due to external noise, then "I" should change too, right?

author: Unknown
author_rank: Unknown
timestamp: November 10, 2010, 11:47:25 pm
content: Quote from: onemilimeter on November 10, 2010, 11:33:16 pmIt's more sensitive to noise, but independent of series resistance.Quote from: onemilimeter on November 10, 2010, 11:33:16 pmWhy is R constant? Is it constant if a connector oxidizes, gets plugged in/out or a cable is replaced? Is it constant if the length of the cable changes? If the cable degrades?Following Kirchoff's current law, we can derive that as long as there's no leakage, the amount of current in any point of the system (including the transmitter and receiver) is the same, regardless of resistance. Can't say the same about voltage signals, since series resistance tends to form a voltage divider.

author: EEVblog
author_rank: Unknown
timestamp: November 10, 2010, 11:49:23 pm
content: Quote from: onemilimeter on November 10, 2010, 11:33:16 pmIt can be, yes.If a voltage output is driven into a line and then into a high impedance load at the end, pickup can be significant.On the other hand, a current source will be driving into a (usually) low impedance load located at the end of the line where it is converted into a voltage. The constant current scheme can reduce the influence the line has because the constant current is usually (relatively) high in value and is continually compensated for by the circuit. The line resistance is taken out of the equation.Voltage lines can also use low impedance loads at the end of the line, but then the line resistance matters.Dave.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 11, 2010, 12:09:47 am
content: Another reason, used on 4-20mA loop stuff is you can use a single pair of wires to supply power to a remote sensor or indicator and transmit the reading on the same wires.

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 11, 2010, 12:36:09 am
content: Quote from: EEVblog on November 10, 2010, 11:49:23 pmThanks Dave. I can see the "magic" now. I assume "the circuit" means the internal circuit of the sensor. What do you think if the "compensation circuit" do not have enough bandwidth?

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 11, 2010, 12:38:32 am
content: Quote from: alm on November 10, 2010, 11:47:25 pmHi, you got the point too... If we look at "1 second" interval, I think we can assume that the "DC transmission path" has relatively constant "R".

author: onemilimeter
author_rank: Frequent Contributor
timestamp: November 11, 2010, 12:42:23 am
content: Quote from: mikeselectricstuff on November 11, 2010, 12:09:47 amInteresting... but how can this be achieved? For example, if an output signal of the sensor at one particular moment is represented by an amount of 15mA current, what will happen to the remote system if the system itself needs 20mA to operate correctly?

author: oPossum
author_rank: Super Contributor
timestamp: November 11, 2010, 01:17:59 am
content: It has to operate on less than 4 mA.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 11, 2010, 09:38:07 am
content: Quote from: onemilimeter on November 11, 2010, 12:42:23 amThe device itself must always draw <4mA. This generally isn't a problem for sensors, and if necessary teh sensor can include a buck regulator, increasing the voltage drop across the sensor but keeping current below 4mA. 4-20 sensors will generally have a spec stating the voltage drop across them, and it is up to the installer to ensure sufficient voltage is available to supply the loop.