Q: REG710 - Reducing inrush current by 1817208

I am using the REG710 to boost from 3.3V to 5.0V to drive a blue LED at 20mA. However, I am experiencing inrush current in excess of 300mA when I enable the REG710.

How can I reduce the inrush current? I am considering using a ferrite bead and more capacitance on the input. Does this sound like a solution?


A: Re: REG710 - Reducing inrush current by 1822974

The inrush current is strongly dependent on the output capacitance: I = Cout*dv/dt. So, reducing the output capacitance will decrease the inrush current, and increasing the input capacitance will help to supply the extra energy needed at startup. I'd recommend trying both of these first. 

I'd caution against the ferrite filter; it could lead to L *di/dt voltage spikes on the 3.3V rail during startup.

Another option might be to leave the REG710 enabled and use a FET between the LED and ground to turn on the LED. The inrush current due to the load change from 0 to 20mA should be much less than if you were to use the Enable pin.