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XTR117: 2 wire current loop design

zabbza
Prodigy 140 points
Part Number: XTR117
Other Parts Discussed in Thread: REF200, , TIPD126, PGA309

Tool/software:

Dear TI Team,

Is there any problem with the following circuit?

When I change potantiemeter Vin will change So Io will change.

How Vin may calculate on this circuit?

When the circuit designed like this Vin calculation is easy, Is there any problem with the above circuit?

Best Regards.

  • 0
    Guru 316450 points

    In a two-wire loop, the transmitter is powered from the loop, and the IRET pin is not at the same voltage as the lowest loop voltage, so you must not connect IRET to ground.

    Please note that IRET is the same voltage as the 5 V regulator's ground, i.e., VREG is 5 V above IRET. (But not with a load; see figure 5-4.)

    It might be helpful to replace the upper resistor with a constant-current source, e.g., REF200.

  • 0 in reply to Clemens Ladisch
    TI__Mastermind 27360 points

    Hi Zabbza,

    Clemens is correct. The second schematic showing the potentiometer and IRET being tied to ground is not recommended in 2 wire 4-20mA current loop designs. Think of IRET as a local ground. 

    Best Regards,

    Robert Clifton 

  • 0 in reply to Clemens Ladisch
    Prodigy 140 points

    Dear Ladisch,

    Thank you for your advice.

    Constant-current source is a good opportunity. But ;

    İf Vreg = 5V and pin 3 is 5V regulators ground. Is below equations right?

    Vin = 5 / (R1 + Rpot) * Rpot

    OR

    Vin = (Vreg-Vret) / (R1 + Rpot) * Rpot

    Vreg-Vret = 5

    Best Regards.

  • 0 in reply to zabbza
    TI__Mastermind 27360 points

    Hi Zabbza,

    Ideally, the voltage drop across the Rpot and Rin will be the same since the op amp will want to ensure that the inputs have the same voltage. So you can consider Rin and Rpot to be in parallel to each other. Now you can calculate the Vin voltage to be:

    Vin = 5 *(Rpot || Rin)/(R1 + Rpot || Rin)

    Best Regards,

    Robert Clifton 

  • 0 in reply to Robert Clifton56
    Prodigy 140 points

    Hi Robert,

    You are right, but As you say Ideally, the voltage drop across the Rpot and Rin will be the same.

    How much effect opamp V- and V+ pins voltage difference to output current?

    Is there any problem with this circuit?

    Best Regards.

  • +1 in reply to zabbza
    TI__Mastermind 27360 points

    Hi Zabbza,

    zabbza said:
    How much effect opamp V- and V+ pins voltage difference to output current?

    We actually provide this information in the datasheet! Because the XTR's need to have as closely matched inputs, the offset voltage is optimized to be very low, in this case the worst case at room temperature is 500uV (see below): 

    zabbza said:
    Is there any problem with this circuit?

    Besides what's been mentioned before from Clemens and I, I don't see any other potential problems. 

    Best Regards,

    Robert Clifton 

  • 0 in reply to Robert Clifton56
    Prodigy 140 points

    Hi Robert,

    Thank you for help,

    "Clemens is correct. The second schematic showing the potentiometer and IRET being tied to ground is not recommended in 2 wire 4-20mA current loop designs."

    If Iret pin is not connect to ground there is no problem.

    I can not find Vreg pin maximum output current value. How much is maximum current of Vreg pin?

    Best Regards.

  • 0 in reply to zabbza
    Guru 316450 points

    Figure 5-4 shows that the regulation is quite bad even for small currents:

    This is why I have recommended a constant-current source.

  • 0 in reply to Clemens Ladisch
    Prodigy 140 points

    Hi Clemens,

    The Circuit will be design as 2mA minumum current,

    But how much is maximum current value.

    I can not find Vreg pin maximum output current value.

    Is maxiumum current value is 4mA.

    Best Regards.

  • 0 in reply to zabbza
    TI__Mastermind 27360 points

    Hi Zabbza,

    The maximum current that the VREG can supply is shown in it's short-circuit current, which shows a typical value of 12mA. 

    However, we only characterized the VReg voltage in relation to Ireg current up to 4mA. 

    Since this is a 2 wire loop you want to avoid having the XTR117 and devices being powered by Vreg to consume over about 3.8mA of current. This is due to the current loop for the signal also being used to power the XTR117 and the other devices. If they require more than 4mA of current to operate, then your current loop can't go down to the minimum current. 

    Is there a reason in particular you are wanting to know the maximum current for Vreg? 

    Best Regards,

    Robert Clifton

  • 0 in reply to Robert Clifton56
    Prodigy 140 points

    Hi Robert,

    Thank you for your explanation, its useful,

    Max drawn current of Vreg is important to output current as you explain.

    Vreg maximum drawn current is 3.8mA, OK for me.

    I see PGA309 (TIPD126) referance design. PGA309 supply pin connect to directly Vreg. And PGA309 drawn max. 10mA.

    Best Regards.

  • 0 in reply to zabbza
    TI__Mastermind 27360 points

    Hi Zabbza,

    The quiescent current of the PGA309 is typically 1.2mA but maximum of 1.6mA. 

    The output current from the PGA will only be 40 µA to 200 µA to get the proper 4mA to 20mA output of the XTR117.

    Is it possible that the confusion came from the  input current absolute maximum rating in the datasheet? 

    Best Regards,

    Robert Clifton