Power On Surge I of 52A normal?

[evs]

Hi All,
I was wondering if a pwer on surge current of 52A for approx. 30mS is to high. There is a circuit I'm working on that has 5 tantalum caps that act as a charge pump. It's a Linear Tech step down regulator(LTC1735CF#PBF) 15v to 3v and 5v dual outputs. I can remove these caps and change the value and only have one capacitor leading to only a power on current of 9A. A colleague and I were discussing the merits or not of having the 52amps present and the potential for damage in the future. Let me know what you think.

TIA,
evs

 

[OutToLunch]

If the output is ramping up in a linear fashion from 0V to 3V in a 30msec span of time and you are saying 52A is being dumped into those capacitors, then you would have about 500000uF (0.5F) of capacitance on the output.

how about a schematic?

 

[Diver300]

You don't say how the charge pump is connected. As OutToLunch says, a schematic would be a big help.

If you increase the soft start capacitor on pin 3 of the LTC1735CF you can slow down how fast the output voltage rises.

 

[evs]

Yes, here's a schematic

sorry about that. Look at the caps I circled. I don't understand why it was done this way. These caps when replaced one .1uf makes the inrush current about 9A instead of the 52A measured at turn on time. It was suggested that the 52A is ok for the short turn on time. I don't like the charge pump effect for long term issues.

 

[mneary]

The surge current occurs for about 30 microseconds, not 30 milliSiemens.

That's to be expected with 108 microfarads across the input rail, if it's turned on by a switch. Will it really be a switch? Most real power sources have a finite ramp time, and some internal resistance.

 

[Diver300]

Those capacitors are simply decoupling capacitors. There doesn't seem to be anything about their arangement that could be described as a charge pump.

In the circuit that you posted, the current limit is set at 75mV / 4mhm:
(75mV comes from the LT1735 data sheet, the 4mhm: from your circuit)

That is just under 20 A which does seem high for a 2.6A circuit, even for 6 A. (I am not sure which value your circuit is rated at as both values appear on the diagram)

It also seems too high for the inductor if I read **broken link removed** correctly.

The current turns on and off at about 300 kHz, so the capacitors need to store 20/300000 = 67:mu: C of charge so 50 :mu:F near Q18 seems a good idea, as that should keep the ripple down to about a volt. There is also the ESR of the capacitors to consider. The ESR must be kept low to keep the resistive portion of the ripple low.

With all those capacitors, chosen to reduce the ripple, the inrush current will be big if the circuit is to work correctly. I wouldn't reduce the capacitors unless you redesign the whole circuit.

If you want to reduce the inrush current, limit the rate of rise of input voltage if you can, or have a resistor in series that is bypassed once the capacitor are charged.