DRV10963: The max continuous output current and peak current

Hey Ricardo,

Thanks for helping us out, let me answer your questions below:

1. The output current of DRV10963 is 500mA according to the description of datasheet. But the Table 7 (Pg. 14) shows that ILIMIT could go up to 875mA. Could the continuous output current of DRV10963  go up to 875mA?
   1.  We state 500mA as a typical output RMS current value (otherwise known as continous current). If you transform output current (RMS) to output current (peak) you get about 707mA (peak). 500*sqrt(2) = 707mA.
   2. This is important because ILIMIT refers to peak current. Since we don't necessarily characterize the output current as an electrical characteristic, 875mA is chosen to leave some margin during operation (to incorporate min and max values of other electrical characteristics of our device)
2. The value of short circuit current protection is 1.8A when ILIMIT [2:0] = 4 (limit current is 500mA). My question is what's the value of short circuit current protection when ILIMIT [2:0] = 7 (limit current is 875mA)?
   1. Short circuit current is a constant of 1.8A independent of ILIMIT. ILIMIT specifically refers to current that will be outputed to the phases of the motor. Our device prevents from drawing 1.8A from the supply (whether that be an external or internal issue)
3. In our product introduction page, the DRV10963 was described as "5W, 5V 3-Phase Sinusodal sensorless motor driver". Could you please explain the calculation process of this result/value 5W ? 

1. 1. That you for point us out to this, this is a mistake. The DRV10964 is a sibiling device of the DRV10963 which correctly displays 4W. We get this value from taking 500mA(RMS) and converting to peak for 707mA then multiplying by 5.5V (our maximum voltage range) for about 3.9W (or 4W).
   2. Thank you for pointing out the mistake, will correct this in the future.

Hope this helps,

-Cole