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I am using a DC-DC converter which is having an inductor rating of 68uH and 2.05A.

Output Voltage of dc-dc is 23V and output current max is 300mA

Schematic :

enter image description here

I am using a current probe and measuring the current through the inductor.

When I turn ON the DC-DC converter, the Inductor draws 10.75A. Should I consider this as Inductor Inrush current or peak current? If I consider this as Inrush current, as I do this during start-up of the converter, then how to measure the peak current of the inductor?

And since the inductor is rated for only 2.05A, won't this high current damage the component even though it is for a short duration?

Current probe is set as 0.1V/A in the probe and 0.1V/A in the scope as well.

Inductor Peak current (below is the waveform captured during the start-up of the DC-DC converter):

enter image description here

Inductor RMS Current (below is the waveform captured during normal running condition of the DC-DC Converter) :

enter image description here

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  • \$\begingroup\$ Inductors resist changes of the current trough them. Inrush currents occur with empty capacitors, starting motors and also soft iron core transformers which happen to be in a magnetically saturated state since the last usage. I guess in your converter there's a big capacitor which needs to be charged to the operating point voltage and you have recorded that current pulse. A schematic could stop guessing and maybe imply some proper answers. \$\endgroup\$
    – user287001
    Dec 19 '19 at 15:42
  • \$\begingroup\$ Schematic is updated. Please check \$\endgroup\$
    – Newbie
    Dec 19 '19 at 16:24
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I assume this is a boost converter since inrush of that magnitude would be unusual in a buck or buck-boost converter. (Though you leave out a lot of information that would be helpful, like topology and schematic.)

A boost converter has a direct path from the input through the inductor and diode or synchronous rectifier to the output caps. So there is typically a large inrush current.

Technically, the peak of the inrush current is by definition the "peak" inductor current, but that's not what we usually mean in a DC-DC converter when we use the term "peak current". So you can consider that the inrush current. The peak current is the maximum current at the top of the inductor current waveform when the converter is in normal operation. It's what is meant for example by "peak current mode" control. Typical you put a short wire in series with an inductor and use a current probe to measure the inductor peak current.

When you exceed the saturation current of the inductor, the inductance drops to a very low level and you just have the parasitic resistance (L, ESR) and the diode's dynamic impedance between Vin and the output caps. So the inrush can be very high as you note.

It is short duration, and it's not likely to damage the inductor, but it could damage the output rectifier. You have to check each component including the inductor to see if it can withstand the inrush.

One trick you can use is to bypass the inductor and diode with a rectifier rated for high pulse currents directly from the input to the output. Then all the inrush current passes through the diode, which is then reverse biased when the converter starts up.

Here's an example:

enter image description here

EDIT after the schematic was added to the post:

It looks like you're using an SS34 diode, rated for 100A peak half sine current. So no problem there. I can't quite make out the inductor part number (it's also good form to add links to your component datasheets in your question) but it's unlikely that such a short duration inrush would damage an inductor. If there's any chance that the converter would start and the FET would turn on during the peak inrush you will have to check the FET SOA to see if it can handle it as well. The datasheet for the controller doesn't specify the startup characteristics very well and I can't read the FET part number so it's impossible to know for sure.

Still, it seems like you have no need for a bypass diode in your application. It will probably be fine but without doing the proper due diligence I can't say for sure.

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  • \$\begingroup\$ Updated the schematic. onsemi.com/pub/Collateral/NCP3063-D.PDF - This is the boost converter. If this helps to improve your answer, please do the needful. \$\endgroup\$
    – Newbie
    Dec 19 '19 at 16:25
  • \$\begingroup\$ If you edit your answer for adding further information, please do and also let me know, why adding the Startup bypass diode helpful when you say that, since the inrush current is only for a short duration, it won't affect the circuit components much? And my inductor current rating is 2.05A only. But 10.75A of inrush passes everytime I turn ON (I did more that like 100 times and no issues found relating to the component). So, no problem right? \$\endgroup\$
    – Newbie
    Dec 19 '19 at 16:40
  • \$\begingroup\$ @Newbie Edited with some additional information. \$\endgroup\$
    – John D
    Dec 19 '19 at 16:55
  • \$\begingroup\$ Thank you for the answer. Will surely follow \$\endgroup\$
    – Newbie
    Dec 19 '19 at 17:28
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The output capacitor C152 (=330uF) is empty and it gets charged to the input vpltage. Your inductor probably has a ferrite core which will get saturated above 2A and that allows as much current as there's available, only circuit resistances limit the inrush. The inductor does not get harmed, it operates again as inductor when there's not too much magnetization.

NOTE: There's an older answer by user John D. He saw this without a schematic.

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  • \$\begingroup\$ Thank you for the answer \$\endgroup\$
    – Newbie
    Dec 19 '19 at 17:28
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In addition to the effect of the input capacitance charging up, switching regulators present a negative input impedance.

This will cause a large input current to flow at startup.

See this App note from Vicor.

A DC-DC converter works to maintain a constant output voltage, \$Vo\$, at a certain level of power, \$Po\$. To do this, it will absorb, from the input source, a power, \$P_I\$, given by:

\$P_I = \frac{P_O}{η} =V_I ⋅ I\$

with \$η\$ being the efficiency of the converter.

Let’s assume that the input voltage decreases by a quantity \$-ΔV_I\$. In order for the converter to sustain the output power, it will increase the input current by a quantity \$ΔI_I\$ so that the input power stays constant at \$P_I\$. Assuming infinitesimal variation for \$V_I\$ and \$I_I\$, it is possible to write:

\$ \frac{dv_I}{di_I}=r{_I{_d}}\$

where \$r{_I{_d}}\$ is the incremental input resistance. For a negative voltage variation, corresponding to a positive current variation; the incremental resistance will have a negative sign. The input resistance varies according to the input conditions because it depends from the input voltage at which the converter is working, as well as the output power that it is delivering.

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  • \$\begingroup\$ Thank you for the answer \$\endgroup\$
    – Newbie
    Dec 19 '19 at 17:28

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