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XTR105 Output Current Calculation

Other Parts Discussed in Thread: XTR105, TIPD161

Hello,

I'm planning to prototype with the XTR105.  I spent some time in the datasheet and the reference design guide, "Analog Linearized 3-Wire PT100 RTD to 2-Wire 4-20mA Current Loop Transmitter Reference Design".

I'm having a little trouble understanding how you calculate the output current of the transmitter.  For instance, in section A.3 of the design guide, a calibration is performed where RG is trimmed to provide an output current of 19.987mA.  How was 19.987mA calculated?  I understand there are equations provided but they depend on Vin(Vin+ - Vin-), but Vin is not quite obvious since the device manipulates the excitation current through Rlin1 and Rlin2.

I've also looked at the calculator "xtr_calc_10-23.xlsx" but loop current calculation doesn't look anything like that in the design guide or datasheet.

Any clarification on this appreciated, thanks.

Zack

  • Hi Zack,

    The equation that defines the exact output current is quite enormous which is why the look-up tables are provided in the device datasheet to simplify the calculations.  The full equations can be seen in the calculator that accompanies the analog application journal article linked at the bottom of this message. 

    In the reference design the uncalibrated output current was measured and then mathematically manipulated to remove the gain and offset errors so the shape of the final non-linearity in the design could be seen.  While the shape of the non-linearity is relatively consistent for most temperature ranges, negative temperatures and wide temperature ranges (both featured in TIPD161) cause different non-linearity shapes that are harder to calibrate. This can be seen when working with the "XTR_Calc_10-23" spreadsheet.  Once the shape of the non-linearity was known (similar to what's shown in Figure 27 below) the output current was adjusted at specific points to cause the output to behave as desired.  So as shown, the -200C intercept was found to be roughly 0.01mA below 4mA, so the output was adjusted to 3.99mA with a resistive input that correlated to a -200C temperature.  As shown, the +800C intercept is at roughly 0.013mA below the 20mA level, so the output was adjusted to 19.987ma with an input resistance that corresponded to +800C.  Then a resistor that corresponds to a +300C temperature was inserted and the gain was fine tuned to achieve exactly 12mA as desired. 

    The design was then re-tested over the full span input resistance and the results shown in Figure 27 where achieved which matched our calculated results very closely. 

    Here's the calculator: