Inrush current calculation

Hi Stenio,

You are correct regarding the current limit. For cases where the output capacitance is excessively large (plus load current)  the current can be limited by the internal current limit circuit.

The TPS720 seems to have a fairly well controlled startup ramp circuit.  This means that the output voltage ramp at startup will have basically the same rise time (with minor variations) no matter the load, the capacitance or the output voltage - except for cases where the LDO goes into current limit during startup. See Figure 22 & 23 of the data sheet (or Figure 3&4 of the app note). This is due to the fact that the output tracks the internal rampup of the reference voltage.  The VOUT rise time looks to be 20-30us depending on load (from the plots). So the inrush current is Irush = Cout * Vout/20us.  But note that this equation does not include the current into the load - which equals Vout(t) /Rload - which must be added to the inrush equation.

Regards

Bill

Hi Bill,

Thank you very much for your the quick answer.

Just let me add two comments:

1) I see now from the plots that this IC has a fix startup ramp time, but I don´t really understand why is that. If this rise time is fixed around 20-30us, why in the page2 of  app note it is shown a value of 1us, that results in a current drawn of 2.86A? I don´t understand what it´s trying to show.

2) In case we don´t have such a controlled startup ramp time at the IC, let say, a regulator made completely from discrete components such as transitors, opamps, resistors and so on, how can we determine the inrush current of the circuit then?

Best Regards,

Stenio

Hi Antonio,

1) I don't see any justification for that "1us" statement in the app note.  I think that the app note is using equation 1 in a too general sense inorder to illustrate the point about how the upstream supply may not be able to supply the inrush current in some circumstances.

Please note that on page 11 of the data sheet that the typical slope of the Vout ramp is shown to be 0.0454545V/us.

2) You describe the way the older parts sometimes turnon.  The answer is that the main pass fet turn full on - and sometimes huge surge currents (only limited by the Rdson of the fet) resulted for a few useconds till the output capacitor was charged.  To some extent we would use a feedforward cap across the top feedback resistor to slow down the current once the feedback rises above the reference voltage.  Sometimes we would use circuitry to precharge the feedback to limit charging current. TI has app notes on this (try slyt096) but these techniques get pretty complicated.

Regards

Bill

Hi Bill,

Thanks a lot again.

Just to summarize your last answer, you said that when the pass fet turns on, in principle, the inrush current Irush will be limited only by the fet's RDSon (or maybe + capacitor's ESR), then we have Irush = Vin/(RDSon+ESR). As the ESR gets smaller with a larger capacitor, then we can say that large capacitors create large inrush currents.

And also, the voltage rise time t_r at the capacitor Cout will be related to t_r = (RDSon + ESR) x Cout. Please, corrected me if I am wrong.

I can understand things better with equations, sorry about that!!

Best Regards,

Stenio