Original question:
Hello dears,
We are trying to develop a receiver that allows reading 4 – 20mA current from industrial transmitters and converting it to voltage to be processed by the ADC of a Microcontroller. I am attaching the schematic diagram, I need to know if it is correct or if you could recommend any parts for our application.
I appreciate your help.
Hi Ivan,
looks a bit like overengineering with the INA190. Also, the wiring of 2-wire transmitter looks a bit weird.
I would do it this way:
C2 and the TVS form a EMI filter and protection circuit against ESD, Surge and Burst. Only a low leakage TVS must be just here, for instance from ST. Eventually a SMBJ8.5A may also be used. D1 is a protection against negative input voltages and reverse currents. R1 transforms the 4..20mA input current into a voltage of 1..5V. And R2 and C1 form a charge kickback filter for the ADC and serve as input low pass filter (anti-aliasing filter). R2 in combination with the internal protection diodes of ADC to the supply rails also protects the ADC against overvoltages by limiting the input current. If the ADC has no internal input protection diodes, add external ones from C1 to the supply rails.
Three more points:
1. R2 = 10k is a bit high and can make issues, if the ADC has high input leakage currents. Then R2 should be decreased.
2. The supply voltage must be able to absorb the reverse current during an overvoltage event. Eventually a suited TVS from the supply voltage to signal ground has to be added to absorb the reverse current and to limit the supply voltage.
3. If the ADC has a different input voltage range, R1 must be modified accordingly.
Kai
Hello Peter,
Thanks for the reply, I have changed the INA190 to the INA185 as you suggested.
I have the following queries:
1) Are the 10R resistors necessary at the input?
2) Do you recommend placing a filter on the INA185 OUT pin?
3) According to the attached image, is the connection of the 4-20mA transmitter correct?
Hello Kai,
Thank you very much for the answer, then I answer your questions.
1) The microcontroller used is AVR32DA28, I attach configuration images:
uC Configuration:
2) A single load resistor will be used, what load resistor do you recommend to use to get an output voltage (0.5V >= Vout <= 4V) ?
Thanks for your help.
Ivan
Hello Ivan,
In the application note I sent, it goes through the equations you can use to approximate a first-pass Rfilt and Cfilt for ADC input pin. Based upon the ADC setting you have sent I tried to calculate what the acquisition time will be and plug this into the calculator this application note is based around. The calculator is called the Analog Engineer's Calculator.
Here are the first pass results.
Unfortunately, the processor datasheet does not detail what the typical input capacitance is for the ADC (Csh), so maybe contact maufacturer to see if they can give you a value. Also, know that the MUX is going to add some capacitance to the Cfilt value, so the Cfilt you use may actually need to be the Cfilt calculated minus the Cmux. If you can get these capacitances, you can go through with a SPICE simulation procedure (outlined in the aforementioned application note) to determine the optimal Rfilt and Cfilt such that the ADC voltage falls to under 0.5LSB.
The other way to fine tune the Rfilt and Cfilt is through actual system testing. You could measure the ADC input voltage with accurate volt-meter and compare to ADC readings. You could also make the board with a operational amplifier buffer (e.g., OPA320) in between the current-sense amplifier (CSA) (i.e., INA185) and test if there is any significant degradation of accuracy when ADC has the buffer and when it does not.
Even if you do use a buffer, you will still want the have pads for this Rfilt, Cfilt as you may end up needing them anyway as it is usual best practice.
As for setting the Rsh. You need to understand what is the maximum shunt voltage drop allowed and/or what is the maximum shunt power dissipation allowed in your system, this will set the upper bounds of Rsh. Then to get the best error, set the Rsh to highest possible value based upon max current and CSA gain and CSA max Vout. I usually choose max Vout to be 4.85V (note consult Figure 19 to see if Vout could go lower if Iout can exceed 1.5mA. Also note that Iout can peak for high value of Cfilt, another reason to use simulation to determine this transient Iout).
So if Vout_max is 4.85V, then Vin_max for INA185A2 (50V/V) is 4.85/50 = 97mV. Thus Rsh_typical = 97mV/20mA = 4.85-Ω. Round this down to nearest available resistor.
Sincerely,
Peter
Hello Peter,
Thanks for the detailed explanation.
So I have modified the circuit, I will also replace the OPA320 with an OPA325 since my supplier does not have stock.
Ivan.
