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Measuring Current (4-20mA) (App Note)

LJTick-CurrentShunt for 4-20 mA Signals

The best way to handle 4-20 mA signals is with the LJTick-CurrentShunt, which is a two channel active current to voltage converter module that plugs into the LabJack's screw-terminals.

 

Shunt Resistor for 4-20 mA Signals

The following figure shows a typical method to measure the current through a load, or to measure the 4-20 mA signal produced by a 2-wire (loop-powered) current loop sensor. The current shunt shown in the figure is simply a resistor.

Using a LabJack USB/Ethernet/WiFi Multifunction DAQ to perform Current Measurement With Arbitrary Load or 2-Wire 4-20 mA Sensor

Figure 1. Current Measurement With Arbitrary Load or 2-Wire 4-20 mA Sensor
 

When measuring a 4-20 mA signal, a typical value for the shunt would be 240 Ω. This results in a 0.96 to 4.80 volt signal corresponding to 4-20 mA. The external supply must provide enough voltage for the sensor and the shunt, so if the sensor requires 5 volts the supply must provide at least 9.8 volts.

The following figure shows typical connections for a 3-wire 4-20 mA sensor. A typical value for the shunt would be 240 Ω which results in 0.96 to 4.80 volts (use a 120 Ω to produce 0.48 to 2.40 volts).

Using a LabJack USB/Ethernet/WiFi Multifunction DAQ to perform Current Measurement With 3-Wire 4-20 mA (Sourcing) Sensorr

Figure 2. Current Measurement With 3-Wire 4-20 mA (Sourcing) Sensor
 

The sensor shown in Figure 2 is a sourcing type, where the signal sources the 4-20 mA current which is then sent through the shunt resistor and sunk into ground. Another rare type of 3-wire sensor is the sinking type, where the 4-20 mA current is sourced from the positive supply, sent through the shunt resistor, and then sunk into the signal wire. If sensor ground is connected to LabJack GND, the sinking type of sensor presents a couple of problems, as the voltage across the shunt resistor is differential (neither side is at ground) and at least one side of the resistor has a high common mode voltage (equal to the positive sensor supply). If the sensor is isolated, a possible solution is to connect the sensor signal or positive sensor supply to LabJack GND (instead of sensor ground). This requires a good understanding of grounding and isolation in the system. The LJTick-CurrentShunt is often a better solution.

Both figures show a 0-100 Ω resistor in series with SGND, which is discussed in general in the Externally Powered Signals App Note. In this case, if SGND is used (rather than GND), a direct connection (0 Ω) should be good.

 

Shunt Resistor for Other Currents

A simple shunt resistor can often be used to measure smaller currents, but for larger currents the best choice is usually the Hall effect sensor discussed below.

The shunt is chosen based on the maximum current and how much voltage drop can be tolerated across the shunt. For instance, if the maximum current is 1.0 amp, and 2.5 volts of drop is the most that can be tolerated without affecting the load, a 2.4 Ω resistor could be used. That equates to 2.4 watts, though, which would require a special high wattage resistor. A better solution would be to use a lower resistance shunt, and buy a LabJack model that can resolve the smaller signal(U6, T7). If the maximum current to measure is too high (e.g. 100 amps), it will be difficult to find a small enough resistor and a hall-effect sensor should be considered instead of a shunt.

If measuring a small current, an only minimal voltage drop can be tolerated, an amp might be needed to amplify the signal from the shunt.

A low-side shunt (bottom side of the shunt connected to ground) is preferred in most cases.  If a high-side shunt is used, the common voltage voltage must be considered and a differential measurement is required.  See the Differential app note for more information.

 

Hall Effect Sensor for Other Currents

A Hall effect is usually the best choice for current sensing, but for small currents a Hall effect might not be available and a shunt resistor will need to be considered as described above.

A hall-effect sensor is isolated so there are no concerns about grounding or common-mode voltage issues.  Low-side or high-side does not matter.  Simply connect VS/GND to the sensor to provide it power, and then connect the signal wire to an analog input.

A good place to look for Hall Effect sensors is the following link at Digikey:

https://www.digikey.com/products/en/sensors-transducers/current-transduc...

First go to the "Voltage - Supply" column, ctrl-select all ranges that include 5 volts, and then click "Apply Filters".  Then go to the "Current - Sensing" column and ctrl-select ranges that might work.  Also consider checking the "In stock" box.  That usually results in about 1 page of results.

When selecting ranges in the "Current - Sensing" column, it is not always straightforward.  For example, if the max current you want to measure is 10 amps, that does not necessarily mean you should select a sensor rated for exactly 10 amps.  You have to consider:

    - The signal output voltage of the sensor at different currents.

    - The signal input range of the LabJack.

For example, consider the SCD10PUN.  Following are its signal output at different currents:

0A        0.3V

10A      2.3V

20A      4.3V

The U3-HV, U6, UE9, and T7, all have analog inputs that can measure up to 5 or 10 volts, so the SCD10PUN is good for measuring up to 20 amps with those devices.  The analog inputs on the U3-LV have an input range of 0-2.4 or 0-3.6 volts, so it is good for 0-10.5 or 0-16.5 amps with the SCD10PUN.

Note that these hall-effect sensors do not do any math on the signal. If you are measuring DC current, then you get a simple DC output signal ... very straightforward.  If you are measuring AC current, however, you will get an AC output signal waveform.  If you want RMS, you need to acquire this waveform using stream mode and do the appropriate math, or use the T7 which can do all that in hardware so you can just read the true-RMS value.

For AC current measurement, current transformers are also an excellent option to consider.  Current transformers are low-cost and isolated, and provide a waveform output just like a hall-effect sensor does for AC.

There are some AC current sensors available that do math for you and provide a simple DC output signal that is proportional to RMS of the AC current.  Most of these do simple RMS not true-RMS, so are only accurate if the current is a pure sine.

 

6 comments

Hello!

Today is my first day using a labjack and I'm overall very new to automation in every sense.

I am using a labjack U6 and have wired up thermocouple's on the AIN0/1 analog inputs and have scaled them correctly.

Now I want to record pressure from these sensors: http://www.automationdirect.com/adc/Overview/Catalog/Sensors_-z-_Encoders/Pressure_Sensors/Pressure_Transmitters/Stainless_Steel_Sensing_Element_-_DIN_Connector

They output 4-20ma and require a 9-36Vdc power supply. 

Looking around this site, I haven't found a clear way to make this work, and would love to know!

Any help would be greatly appreciated.

That sensor also has a 0-10V output option that would be more straightforward.

Looking at the 4-20mV output option, it appears to be a 2-wire sensor, so Figure 1 above is what you want.  An alternative is to use the LJTick-CurrentShunt, in which case look at Figure 6 of the LJTCS datasheet.

Thanks for the super quick reply!

Looks like all the transducers we have are the 4-20ma version so looks like we'll have to get a resistor and go off of figure 1.

Any tips on scaling to PSI?

Yes, figure 1 is what you should follow. Connect our device to the output of the sensor as figure 1 suggests using a resistor less than or equal to the maximum resistance value that they specify in their data sheet as the series resistor. 

Use a LabJack to measure the voltage drop across the resistor as figure 1 also suggests. 

You can use ohm's law V = IR to calculate the current being outputted from the sensor.  With this information you can follow their data sheets to correlate current values to PSI readings.

darteaga's picture

Do you expect good resolution and accuracy after the current to voltage signal adaptation?

LabJack Support's picture

The LJTCS is purely analog so will not affect resolution.  Resolution will be limited by the resolution of the DAC in your sensor (if it uses a DAC) and the resolution of the ADC in the LabJack DAQ device.

Since the LJTCS is analog, one concern would be whether it will add noise that reduces the effective resolution of your measurement.  I doubt it.  I would not expect the LJTCS to add noticeable noise.

As for accuracy, the absolute gain accuracy of the LJTCS is 0.5%.  Any LabJack DAQ device is more accurate than that, so not the limiting factor.  Your sensor could be less accurate and could be the limiting factor.  In any case, if you want higher accuracy you should do some sort of in-system calibration where you look acquire 2 (or more) pairs of parameter (pressure or whatever you are measuring) versus voltage reported by the LabJack, and then you can come up with a calibration equation you apply to voltage readings.

https://labjack.com/support/app-notes/resolution-and-accuracy