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Topic: how to connect CURRENT TRANSDUCER LEM HTFS 200-P (Read 3785 times) previous topic - next topic


Nov 23, 2010, 01:03 am Last Edit: Nov 23, 2010, 01:10 am by fca Reason: 1
Hi i want to measure some DC current non invasive, so i'm thinking on something like this...

reading the specs .
I don't understand the need of Vref..
i was thinking on using +5v from arduino, ground and output connected to a analog pin of arduino...
Where do i connect this vref pin ?

also didn't get the output formula:
VOutput voltage (Analog)@Ip    ===  Vref+/-±(1.25·IP/IPN)

i'm not very experienced in reading this datasheets so please any advice will be welcome.


ps: i need to measure up to 60amp DC current max 30V.


Nov 23, 2010, 01:47 am Last Edit: Nov 23, 2010, 01:48 am by RuggedCircuits Reason: 1
Let's say the current is 0. Then the output will be Vout=Vref.

Now let's say that 60A of current is flowing and you have the HTFS 200-P so that Ipn is 200. Then the output will be Vout = Vref + 1.25*60/200 or about Vref+0.375V.

If -60A of current is flowing (other direction) then Vout = Vref - 1.25*60/200 or about Vref-0.375V.

Note that even when current is 0 then Vout=Vref+/-0.025V so there is some uncertainty in the measurement.

The Vref pin is useful to be able to measure both positive and negative currents when you have a single-supply (0V-5V) system rather than a split-supply (-5V - +5V) system. If you set Vref to 2.5V then you can consider voltages between 2.5V and 5V to be "positive current" and voltages between 2.5V and 0V as "negative current". If you know current is only ever going to be flowing one way I suppose you could set Vref=0.

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thanks, but how do i set the ref   voltage ? do i need to connect that pin to arduino ? where ?


It looks like the reference voltage is already set to the supply voltage divided by 2, so 2.5V. You don't have to do anything.

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hi & thanks for your help the best price i have found for this is ~16?+shipping.

Do you know any other similar products cheaper ?



Hi another question regarding this if i use this with a cable that comes from a wall plug, the person who connect to the wall the plug may reverse the livewire, is this important ?
i mean those these hall sensors have to be on the live wire or can measure also on the neutral (ac corrent) ?



You can measure on either live or neutral line.  Bear in mind that only one of them should pass through the "core".  If you feed both through, the currents (which are flowing in opposite directions) will cancel each other out.  

Also be aware that if you have bought a 200amp unit and you want to measure only 60 amps you can increase the sensitivity of the system by feeding the wire three times through the "core".   You do this by feeding the wire into the core then once it leaves the sensor's core, loop it back around the sensor and feed it back in etc.  This will triple the output slope since the currents effectively add and the core interprets this as 180 amps.  I used this technique to calibrate a 200 amp unit by winding 100 turns of fine wire through the core and passed 2 amps through it.  100 feeds @ 2amps = 200amps.



For now i will no measure more than 50amps in one DC ..
And not more than 10amp ac 250V on other...
Do you mean it will not read well this low amps ?
So you are saying if if i loop the cable 2 times between the hole it will measure 2times the load ?
example passing 2times the cable with 10amp will read 20amp ?

Why should i do this (confused).



The accuracy is defined by the manufacturer's data sheet as +/- 1% of IPn.  For a 200amp unit this means an accuracy specification (or typical error) of +/- 2 amps.  Contrary to what might be intuitive, measured error is not prorata over the device range.  With 0 amps of current flowing the output could read + or - 2 amps and still meet the manufacturer's specified accuracy !!

So, if you are measuring a maximum of 10 amps the possible error could be +/- 2 amps.  It could read anywhere between 8 amps to 12 amps.  This is a measurement error of +/- 20%.  For obvious reasons the lower your measured current the larger the percentage of error.

By increasing the number of core passes, say 5 times, your 10 amps is measured as 50 amps and the error tolerance is still +/- 2 amps which represents a +/- 4% error.  Not good but very much better than +/- 20%

Therefore the larger the number of turns you can pass through the core, within the limits of the device range, the better the measurement accuracy and the larger the signal change for your measured current.



thanks a million for the tip, never tought about that.
wil the acs756 or the acs758  have this error ? and if yes how to solve it ? do you know.


I would suggest you are worse off with the ACS sensor for several reasons :
a) They require direct connection to the current flow and hence present a danger if working with high voltage circuits. In effect they defeat your first requirement "to be non-invasive"
b) The specified error is 2%
c) You cannot multi-feed the current to reduce error when measuring low currents

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