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It appears to be a 3-wire current loop sensor. You need to provide a 9-32V power supply and then need either a shunt resistor or the LJTick-CurrentShunt. See Figure 2 here:
I want to get these circuit onto a breadboard. I have two sensors, (i.e., one measuring inlet pressure in a condensing coil, another from an outlet pressure of a evaporating coil.) Would that mean that I would need to separate circuit for the two pressure gauges?
Also, if I go with the LJtick-currentshunt. I can just go connect the wires from the sensors to it, from Figure 4. No breadboard needed correct?
I know that the U3 can only handle a 2.4v or 3.6 Inputs. Would replacing that shunt resistor with a regular resistor with the same value work? Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
Would replacing that shunt resistor with a regular resistor with the same value work?
I don't understand what you are saying. What shunt resistor are you replacing? What do you mean by "same" value?
Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
No. If using a shunt resistor as shown in Figure 2 of the Measuring Current app note, one side of the shunt is at ground and the other side will have a max of 2.4 volts if you use a 120 ohm resistor. Similarly if using Figure 4 of the LJTick-CurrentShunt Datasheet the voltage will be very low.
I drew up a circuit of what i was talking about. I have the 140 ohms resistor there. Can I adjust the size of that resistor to get the voltage out in the range for the U3?
If you have a 2-wire sensor that is correct, but it appears to me you have a 3-wire sensor and need Figure 2.
Either way, you can use whatever shunt resistor you want to get the voltage you want. I suggest you use 120 ohms to convert 4-20mA to 0.48 to 2.40 volts.
With the 3 wire sensor, the common will go to the ground and the 120ohm resistor will be in series with the sensor and labjack. After that, labjack will read off the output readings as voltages?
I got a voltage reading from the Labjack, how would I get that voltage and get it into a pressure? Or can I just read it off on the pressure gauge display itself?
So is your sensor giving you 0.48V at 0 PSI and 2.40V at 300 PSI and is that what you expect? You can then come up with a formula that converts voltage to PSI per that relationship.
Hi...i am a new user here. I am also facing the issue. I want to know ifreplacing that shunt resistor with a regular resistor with the same value work? Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
You might want to start a separate thread or support request detailing the sensor that you are using.
In general pressure sensors can take around 9-30 V, but the output will be 0-5 V or 4-20 mA etc. So the output or the sensor is more important to match to the ADC than the excitation. U3 comes in two varieties: U3-LV and U3-HV. The HV has four analog inputs that can take ±10V. So if you are using a U3-HV then a sensor with larger output range can be used without a divider. 4-20 mA signals require a shunt resistor to convert the current signal into a voltage signal. Our CurrentShunt is set up to read 4-20 mA signals and the output is well matched to the U3's LV lines.
If you are going to create your own shunt circuit then a precision resistor should be used.
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The DPG1000DAR is a pressure sensor with display and signal-conditioning built in:
http://www.omega.com/pressure/pdf/DPG1000ADA_DPG1000DAR.pdf
It appears to be a 3-wire current loop sensor. You need to provide a 9-32V power supply and then need either a shunt resistor or the LJTick-CurrentShunt. See Figure 2 here:
https://labjack.com/support/app-notes/measuring-current
... or Figure 4 here:
https://labjack.com/support/datasheets/accessories/ljtick-currentshunt
I want to get these circuit onto a breadboard. I have two sensors, (i.e., one measuring inlet pressure in a condensing coil, another from an outlet pressure of a evaporating coil.) Would that mean that I would need to separate circuit for the two pressure gauges?
Also, if I go with the LJtick-currentshunt. I can just go connect the wires from the sensors to it, from Figure 4. No breadboard needed correct?
Yes, if you have 2 sensors you need 2 circuits.
Yes, if you use 1 LJTick-CurrentShunt you can connect 2 current loop signals directly.
I know that the U3 can only handle a 2.4v or 3.6 Inputs. Would replacing that shunt resistor with a regular resistor with the same value work? Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
Would replacing that shunt resistor with a regular resistor with the same value work?
I don't understand what you are saying. What shunt resistor are you replacing? What do you mean by "same" value?
Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
No. If using a shunt resistor as shown in Figure 2 of the Measuring Current app note, one side of the shunt is at ground and the other side will have a max of 2.4 volts if you use a 120 ohm resistor. Similarly if using Figure 4 of the LJTick-CurrentShunt Datasheet the voltage will be very low.
I drew up a circuit of what i was talking about. I have the 140 ohms resistor there. Can I adjust the size of that resistor to get the voltage out in the range for the U3?
You have drawn Figure 1 for a 2-wire current loop sensor:
https://labjack.com/support/app-notes/measuring-current
If you have a 2-wire sensor that is correct, but it appears to me you have a 3-wire sensor and need Figure 2.
Either way, you can use whatever shunt resistor you want to get the voltage you want. I suggest you use 120 ohms to convert 4-20mA to 0.48 to 2.40 volts.
With the 3 wire sensor, the common will go to the ground and the 120ohm resistor will be in series with the sensor and labjack. After that, labjack will read off the output readings as voltages?
I'm not exactly sure what you are describing, but a 3-wire sensor would be connected as shown in Figure 2:
https://labjack.com/support/app-notes/measuring-current
The reference resistor can just be 0, so that means you can just use a wire to SGND.
The 4-20 mA current will flow through the 120 ohm shunt resistor, creating a voltage from 0.48 to 2.40 volts that is measured by AINx on the U3.
I got a voltage reading from the Labjack, how would I get that voltage and get it into a pressure? Or can I just read it off on the pressure gauge display itself?
So is your sensor giving you 0.48V at 0 PSI and 2.40V at 300 PSI and is that what you expect? You can then come up with a formula that converts voltage to PSI per that relationship.
Hi...i am a new user here. I am also facing the issue. I want to know ifreplacing that shunt resistor with a regular resistor with the same value work? Also, would a voltage divider be needed because the pressure gauge takes minimum 9v to operate, and the amount of input voltage the U3 can handle?
You might want to start a separate thread or support request detailing the sensor that you are using.
In general pressure sensors can take around 9-30 V, but the output will be 0-5 V or 4-20 mA etc. So the output or the sensor is more important to match to the ADC than the excitation. U3 comes in two varieties: U3-LV and U3-HV. The HV has four analog inputs that can take ±10V. So if you are using a U3-HV then a sensor with larger output range can be used without a divider. 4-20 mA signals require a shunt resistor to convert the current signal into a voltage signal. Our CurrentShunt is set up to read 4-20 mA signals and the output is well matched to the U3's LV lines.
If you are going to create your own shunt circuit then a precision resistor should be used.