Hi All,

I totally new to electronics so please be forgiving for me for my lack of knowladge.

I am working around someone's electronic circuit to setup the LabVIEW program which previous version was lost.

I have been wondering if anyone could help me out to understand the circuit and tell me how can I read current between J6G$5 and J6G$6. (It's two wire pressure sensor 4-20mA).

Please see attached file.

J6 connection are inputs to National Instrument Card (analog inputs read voltage -10V - +10V).

Also if anyone could explain to me why pressure transducer inputs are connected into RC circuit (to filter the noises?).


Thanks.
Greg

 

Also if anyone could explain to me why pressure transducer inputs are connected into RC circuit (to filter the noises?).
Greg

It's a low pass filter. Take into account that RC network does not act as a filter on the input signal unless fed by a current source.

If you would like to measure the loop current (4-20 mA), place a meter in series with J6G$5 (for example).

 

It's a low pass filter. Take into account that RC network does not act as a filter on the input signal unless fed by a current source.

If you would like to measure the loop current (4-20 mA), place a meter in series with J6G$5 (for example).

Thanks for you reply.

Do you know why there is inductance in the line J15 on the diagram. That is a power supply for the sensors.

J6G$5 and 6 are the plus and minus terminal for input signal on NI device (is limited up to 10V). If those are connected straight from the power supply that would damaged the hardware. I can't see any divider on the diagram unless I read it wrong.

Someone did managed read current on this configuration by calculation of voltage reads across this two connectors. Any ideas?

 

Decoding someone's else circuit is not an easy task. And even harder without much experience in electronics. Do you have extra information about the pressure sensors he is using? It would be helpful.

Do you know why there is inductance in the line J15 on the diagram. That is a power supply for the sensors.

They are there to filter unwanted AC components.

J6G$5 and 6 are the plus and minus terminal for input signal on NI device (is limited up to 10V). If those are connected straight from the power supply that would damaged the hardware. I can't see any divider on the diagram unless I read it wrong.

Why would you connect the analog inputs straight from the power supply? The power supply has its own connector apart from the analog inputs connectors. As you said, J15 is the power supply for the sensors, but not source for J6G$5 and 6.

According to your diagram, it corresponds to NI USB-621x DAQ M Series. For the NI-6218, the pinout is as follow:

J6G$5 and J6G$6 correspond to pins 19 and 20, which are AI 2 and AI 10 respectively. They are Analog Input Channels.

Someone did managed read current on this configuration by calculation of voltage reads across this two connectors. Any ideas?

It's kind of ambiguous. We need more details about the purpose of the circuit. Why would you want to measure current in analog inputs?

I don't know the origin of this circuit. I'm trying to analyse with you what it seems reasonable.

 

Decoding someone's else circuit is not an easy task. And even harder without much experience in electronics. Do you have extra information about the pressure sensors he is using? It would be helpful.



They are there to filter unwanted AC components.



Why would you connect the analog inputs straight from the power supply? The power supply has its own connector apart from the analog inputs connectors. As you said, J15 is the power supply for the sensors, but not source for J6G$5 and 6.

According to your diagram, it corresponds to NI USB-621x DAQ M Series. For the NI-6218, the pinout is as follow:

J6G$5 and J6G$6 correspond to pins 19 and 20, which are AI 2 and AI 10 respectively. They are Analog Input Channels.

It's kind of ambiguous. We need more details about the purpose of the circuit. Why would you want to measure current in analog inputs?

I don't know the origin of this circuit. I'm trying to analyse with you what it seems reasonable.

Thank you for your time Xavier.

Sensors are Druck precision pressure transmitters 450bar 9-32V in 4-20mA.

What I was trying to do to read current which is going from Transducer by measuring voltage across analog inputs with shunt resistor connected across them.

I am just wondering if there is any other method according to attached circuit as I can’t see how they calculated current by reading voltage.

If I would use shunt resistor it would temperature raise much with such a small current level going through it?

Thanks.

 

Hi, Gregorian, Your circuit seems to be a 4-20mA current loop transmitter. From looking at the circuit it seems that you need to connect 24V from J15 to the positive of the transmitter then connect the negative side(of transmitter) to J6G$5 then connect the J6G$6 to the negative supply on J15. Current will flow through the transmitter , to J6G$5,then through the resistor(R9 470ohm) and back through J6G$6 to negative. This will form a series loop(standard two wire current signal system).
The current through the resistor(precision value) will develop a voltage drop(usually 1 to 5 volts for 4-20mA). This in your input signal.
If you want to measure transmitter current just connect your meter in series(set to measure current) in any part of the circuit.
PS don't forget to insert a 100mA fuse in the transmitter supply.
PPS Inductor in supply may be to reduce switch on current surge. I had a previous issue with transmitters blowing fuse in intrinsic safety barriers on switch-on.

 

Now I'm more clear about what you want. As @probuxton said, I think this is what you're doing:

Just that your case is a differential one.

V = I x Rshunt , Let's say a standard precision 250 ohms shunt resistor
For 20 mA,
V = 20mA * 250 = 5V

For 4 mA,
V= 1V

There is a directly proportional relationship.

 

If I would use shunt resistor it would temperature raise much with such a small current level going through it?

Power Rating (Watts) = Max Current (mA)2 x Shunt Resistance (Ohms)
For Imax = 20mA and R=250 ohms
P = 100mW

A 1/2W resistor shouldn't be a problem.

 

Please note that there is already a resistor in the system(R9) which is generating the voltage signal for the AI input. If you want to you can measure the voltage across JG$5 and JG$6 which will give you the measurement required. I am not sure why they use (R9)470 ohms as it doesn't give
2 -10volts with 4-20mA. If 2-10 volts is required then R9 would be 500ohms. There is actually no need to use another external resistor if the R9 gives you the signal required.

 

Thank you guys.

I am really appriciated your help.