Dear TI Team,
I have go through one Application Report SLVA670A for the inrush current due to capacitance.
my question is that how measure the Inrush current(3.12A) value in below image.
if i am using the formulae which is given in the Application report for calculating the inrush current, my inrush current value is different.
i use following values for calculating inrush current(refer the image):
C= 47uF
dV= 3.3V
dt= 15.73us
formulae used:
please clear my doubt.
Hi Anshu,
You are correct, the inrush shown in the waveform does not line up with the given information. Perhaps the capacitors used were rated for a lower voltage.
When capacitors are operating in voltages close to their max specification, they function with half the capacitance as they would significantly below their maximum voltage specification.
Allow me to do further investigation. However, the correct inrush equation is:
Best Regards,
Elizabeth
Hi Anshu,
For clarity, the intended message from the app note you've cited is that the inrush will decrease for a reduction of output capacitance. Understood the waveforms do not seem to represent the appropriate loading conditions described; It appears the supply itself could be limiting the current.
In the meantime, yes let me go through one example:
The inrush current is estimated by an equation written on p.22 of the datasheet.
dt is a rise time, which is specified in the switching characteristics section as shown at p.6-8 depending on the input condition.
By combining both equations, they can estimate the inrush current value.
When Vin=3V, CL=0.1uF, dt@3V becomes ~28us.
Having 0.1uF as Cload gives I_inrush = CL x dV/dt = 0.1uF x 3V / 28us = 10.7mA.
Best Regards,
Elizabeth
