Output impedance of a 4mA to 20mA current loop

Question

I have the most basic of basicness circuit. The implementation is a XTR115UA IC with a 40µA - 200µA input (based on sensor voltage) which is amplified to the desired 4mA to 20mA current in the current loop.

I have been given the task to calculate / measure the output impedance of this current loop. The output impedance should be below 0.1 Ohms, but I don't see anything that gets me even close.

Hope you can help.

Answer 1

The output impedance Rx || Cx will look something like this:

^{simulate this circuit – Schematic created using CircuitLab}

Where R2 is your receiver resistance (in this case to give you 1-5V).

Vsupply must be high enough that the voltage across I1 is in the range of +7.5 +36 for the transmitter to function properly. At 12mA (mid-scale) the voltage across R2 is 3.00V and should remain at about that level for Vsupply in the proper range.

You can measure the DC output impedance by varying Vsupply from (say) 10V to 30V and measuring the change in voltage \$\Delta V\$ across R2 so

Rx = \$R_2(\frac {20}{\Delta V} -1) \approx R_2(\frac {20}{\Delta V}) \$

For example, if the voltage across R2 varies by 100uV for a 20V change in Vsupply, the output impedance would be 5M\$\Omega\$

For AC analysis, Cx will enter into the complex impedance and there may be other effects as well, but leave that for another time.

Answer 2

Look at the circuit in the data sheet: -

The output loop comprises: -

1. The chip itself and, as a current generator will have a very high impedance
2. A series voltage supply (VLLOP) that powers the chip
3. A series load resistor (RL) that is used to measure the loop current

So, from the perspective of the power supply VLOOP it will see a fairly high impedance due to Q1 being in series with RL plus a little bit of current that passes into the voltage regulator on the chip.

But from the perspective of looking at the loop across pins 7 and 4 you will see VLOOP in series with RL and, as a dynamic impedance will equal RL because VLOOP will be a voltage source with a very low output impedance that is << RL.