I am using this LDO to step down from 14V to 5V with a maximum output load current of 150mA.

enter image description here

The above is the schematic of my LDO. I want to measure the inrush current of the LDO.

But since I have an output capacitor of 1uF, I think the output capacitor will also contribute some inrush current.

I want to understand about inrush current. I have read and understood that the inrush current is caused by uncharged capacitors getting charged during start-up. Hence, the high inrush current during starting of the power supplies.

Question 1 :

My 1uF is placed at the output. So, when I need to measure the inrush current of the LDO, should I measure the current at the input pin 1 of the LDO or at the pin 8 of the LDO during start-up? (I think both scenarios are not that different, since in LDO, the input current is same as the output current as LDO just has an internal pass MOSFET or transistor- Am I correct?)

I have a doubt in this too. To measure the exact inrush current, should I take a wire (very low resistance) from the output pin 8 of the LDO and connect it to the 1uF capacitor and measure the current through this wire using a current probe and note it down as the inrush current? Is this approach correct?

Question 2 :

In DC-DC converters, for example, buck converters, say TPS54260, the input inrush current and output inrush current are different, right? In this case, how to measure the inrush currents at the input and output?

  • \$\begingroup\$ To the output of a regulator it's not really inrush current to / out of the regulator, but it can be viewed as inrush current to the load that follows. In any case, it makes sense to measure inrush current upstream, to the power source, where the said inrush current can be detrimental (tripping fuses, causing sparking in switches, etc.). It's important that you state why are you concerned with inrush current. \$\endgroup\$ – anrieff Jul 13 '20 at 11:29
  • \$\begingroup\$ Also a lot of regulators, especially newer DC-DC converters, have a soft-start feature, which specifically aims to reduce inrush current (to the downstream load, and, in effect, to the upstream power source). Google around for "soft start" \$\endgroup\$ – anrieff Jul 13 '20 at 11:30
  • \$\begingroup\$ Thank you for the answer. So, For the mentioned LDO, it is enough that I measure the inrush current along the upstream line which is connected to the pin 1 of the LDO, right? No need to measure along the output capacitor right? \$\endgroup\$ – Newbie Jul 13 '20 at 11:39
  • \$\begingroup\$ where does the inrush current in the scheme right now cause you problems? To the power source? Or to the load? \$\endgroup\$ – anrieff Jul 13 '20 at 11:44
  • \$\begingroup\$ No. I have not performed the tests yet. I am just asking theoretically for understanding purposes. And based on the answer, I am planning to perform that tests only \$\endgroup\$ – Newbie Jul 13 '20 at 11:52

As discussed in the comments, your question is somewhat theoretical and not caused by a concrete problem you are experiencing. As such, I'll need to make some assumptions to properly answer the question.

Assuming you are designing a stand-alone device, which plugs into an power source, the proper position to measure inrush current is the source power rail. In your schematic that is the 14V node, so you should put the probe on the rail, before the 220nF capacitor. This principle does not depend on whether you have an LDO, or a DC-DC regulator, or even if there's no regulator at all. You are measuring whether the power source will cope with the inrush current.

So, on your questions:

  1. Neither pin #1 nor #8 are correct, as these would ignore the charging of the 220nF capacitors, which would partly cause the inrush current. Also, your suggested probing technique may be inadequate. It really depends on what problems you have with inrush current, but assuming the given schematic is what we're concerned about, the inrush current will be caused by the capacitors on the input and output of the LDO. If the power source can deliver 0.15A (given the limit of the LDO), then the input caps (equiv. to 110nF) charged to 14V will charge in $$time(s) = \frac{voltage(V) \times capacitance(F)}{current(A)} = \frac{14 \times (110\times10^{-9})}{0.15} = 0.000010267s$$ That's around 10µs. The output caps need around 37µs (using the same formula). That's assuming the measly 0.15A capability of the power source, and if it's larger then the times decrease proportionately. 10µs is quick, so you definitely want to use an oscilloscope (in a single-shot mode) to measure the current. Then, a current probe around the wire may be inadequate, since such probes are typically bandwidth limited - if the bandwidth is, say, 100 kHz, measuring a 10µs pulse (which is right on the upper end of the bandwidth) would not be accurate as the probe limitations will interfere with the readings. A better way would be to cut the input wire and place a current shunt resistor (e.g. 1 ohm) and measure the drop across it with an oscilloscope. This way you wouldn't worry much about bandwidth, but the downside is that the addition of the shunt modifies the circuit behaviour a bit, and somewhat reduces the inrush current.
  2. The currents are different, indeed. With my assumptions I've placed already, it is the power source with does not cope with the inrush current drawn from it, so again you should measure the current on the source, same as with point 1.
  • \$\begingroup\$ Thank you very much for the detailed answer. \$\endgroup\$ – Newbie Jul 15 '20 at 3:02

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