2.6.3.7 - Measuring Current (Including 4-20 mA) with a Resistive Shunt [U3 Datasheet] | LabJack

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# 2.6.3.7 - Measuring Current (Including 4-20 mA) with a Resistive Shunt [U3 Datasheet]

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 UE9 screw-terminals.

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.

Figure 2.6.3.7-1. 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 120 Ω. This results in a 0.48 to 2.40 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 7.4 volts.

For applications besides 4-20 mA, 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 1.0 volts of drop is the most that can be tolerated without affecting the load, a 1.0 Ω resistor could be used. That equates to 1.0 watts, though, which would require a special high wattage resistor. A better solution would be to use a 0.1 Ω shunt, and then use an amplifier to increase the small voltage produced by that shunt. 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.

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

Figure 2.6.3.7-2. Current Measurement With 3-Wire 4-20 mA (Sourcing) Sensor

The sensor shown in the above figure 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 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 U3 ground, the sinking type of sensor presents a problem, as 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 U3 ground (instead of sensor ground). This requires a good understanding of grounding and isolation in the system. The LJTick-CurrentShunt is often a simple solution.

Both figures show a 0-100 Ω resistor in series with SGND, which is discussed in general in Section 2.6.3.4. In this case, if SGND is used (rather than GND), a direct connection (0 Ω) should be good.

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 U3 screw-terminals.

### Are there any similar pages

Are there any similar pages for measuring current at higher amperages? I am particularly interested in 0 - 5 amps with limited precision.

Thanks!

### For larger currents I suggest

For larger currents I suggest you look at hall effect sensors.  They provide a nice output, and are isolated so you don't have to worry about grounding and common-mode voltage.  We usually search here:

http://www.digikey.com/product-search/en/sensors-transducers/current-transducers/1966573

Start by going to the "Voltage - Supply" column and only select supply ranges that include 5.0 volts.  Then go to "Current - Sensing" and select ranges that might work for you.

Consider the SCD10PUN.  It outputs 0.3V at 0 amps and 2.3V at 10 amps, so is a good fit to connect to a low-voltage analog input such as FIO4:

http://www.digikey.com/product-detail/en/SCD10PUN/102-1052-ND/388821

### Thank you for the very

Thank you for the very helpful response and sorry for the simple question. Mechanical engineer here, some of us need you electrical guys to explain things many times and slowly.