Yesterday we presented the Smoke Tester, a board designed to safely power prototypes for the first time and test for shorts.
One feature of this design is the current measurement via the INA138 high side current monitoring IC. Below is a small tutorial on how we calculated the sense and load resistors for this IC.
We started of by deciding the highest current we’d like to measure. In the Smoke Tester this was 800mA. In INA138 datasheet it is mentioned that for most applications the maximum differential voltage across the shunt resistor should fall within 50-100mV.
With this in mind we selected the shunt resistor at 0.075 Ohm. This gives us 60mV at 800mA, which is within the suggested 50-100mA range.
Next we calculated if 800mA is less then the maximum current, given the maximum power dissipation of 0.125W (1/8W). Formula we used is I=sqrt(P/R) since P=R*I^2. The result was 1.29A which is above the maximum 800mA we envisioned. A 1206 1/8W 1% 0.075 Ohm shunt resistor was chosen.
We wanted the output voltage to mirror the output current, so 800mA of current would produce 800mV on the output of the INA138. The formula provided by the datasheet states that Vout = Is*Rs*Rl*gm. Where Is is the shunt current, Rs is the shunt resistance, Rl is the load resistance, and gm is the 200uA/V transconductance constant.
By solving the equation above for Rl we have. Rl=Vout/(Is*Rs*gm) or Rl = 800mV/(800mA*0.075Oohm*200uA/V) which is 66.666..KOhm. Since 66.666..K is not a normal resistance value we decided to build it by connecting 3 200K 1% resistors in parallel, giving us theoretically the exact 66.666..K needed. There are of course tolerances to be figured in, but we are satisfied with this result.
Hope this is helpful to anyone wanting to play around with high side current monitors.