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The shielding layer of the inductor in the DC-DC circuit is broken, causing the hot-swap circuit to have no output. [Copy link]

As shown in the hot-swap circuit below,
when the inductor shielding layer is broken (verified by a multimeter, the inductor is not disconnected), the hot-swap circuit has no output, that is, the VCC_12V network in the figure outputs 2.8V (normal output is 12V). U11's 4th pin outputs a low level (about 1, 2V), and U11's 4th pin outputs a high level when it is normal.
If you replace a good inductor, the VCC_12V network in the figure can output a normal 12V. How do you explain this phenomenon? Thank you all!

Note: 1. The broken inductor is L10 in the figure.

2. There is another phenomenon: after power-on, there is a 12V output at VCC_12V, and after 1~2 seconds, the VCC_12V network outputs 2.5V.


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Read the manual carefully, the inductance is reduced.   Details Published on 2022-4-15 13:45

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"The shielding layer of the inductor in the DC-DC circuit is broken, causing the hot-swap circuit to have no output"

From the photo, it looks like the inductor's core is broken.

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Senior, the core crack only causes the inductance to decrease, why does it cause the hot-swap chip to not output? Thank you!  Details Published on 2022-4-15 10:51
 
 

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After the inductor is damaged, you can test the inductance

If there is a change, it may be that the inductor is saturated when the DCDC circuit is working.

This causes the input current to increase and U11 enters protection mode.

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maychang posted on 2022-4-14 18:26 "The shielding layer of the inductor in the DC-DC circuit is broken, causing the hot-swap circuit to have no output." From the photo, it seems that the inductor's core is broken.

Senior, the core crack only causes the inductance to decrease, why does it cause the hot-swap chip to not output? Thank you!

This post is from Analog electronics

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The reduction of inductance may cause the input current of this stage to be too large, and the previous stage will enter the protection state.  Details Published on 2022-4-15 10:57
 
 
 
 

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xiaxingxing posted on 2022-4-15 10:51 Senior, the core crack only causes the inductance to decrease, why does it cause the hot-swap chip to not output? Thank you!

The reduction of inductance may cause the input current of this stage to be too large, and the previous stage will enter the protection state.

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Read the manual carefully, the inductance is reduced.

This post is from Analog electronics
 
 
 
 

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