I see in E table the VDO=500mV(max) when Iout=1A;(VOUT + 0.5 V ≤ VIN ≤ 6.5 V, VIN ≥ 2.5 V, IOUT = 1 A, VFB = GND or VSNS = GND)
I have question:
if my application VIN has a accuracy of 3%. and VOUT has a accuracy of 3.92%.
when we set the VIN=5.5V; VOUT=4.984V
then we can calculate the actual difference between VIN and VOUT. Vin(Min)-Vout(Max)=5.5*(1-3%)-4.984*(1+3.92%)=155mV<500mV;
does that mean we can't use this VIN and VOUT because of the impact of the system accuracy? and we need to leave enough margin for the system accuracy?