0
\$\begingroup\$

The markings are:

1112
 1Z

The package is 2010 (5025 metric). Any clues will be much appreciated.

photo od the SMD diode in question

diode in the context of the PCB

Apparently it is located right at the 12V input. On DMM / diode test - it measure same in both diections, so I gather it is most likely dead. LM2596S gives 0V on the output (5V variety). This is all I can tell, I am afraid. Sorry for poor quality images.

\$\endgroup\$
6
  • 4
    \$\begingroup\$ A picture of the part (on the board it came from) might help to identify it. \$\endgroup\$
    – JRE
    Commented May 19, 2019 at 15:59
  • 2
    \$\begingroup\$ user41224 - FYI these are the current "Component Identification Question Guidelines". The more information you can supply from that list (including the photos kindly requested by JRE), the better the chances of identification. Thanks. \$\endgroup\$
    – SamGibson
    Commented May 19, 2019 at 17:03
  • \$\begingroup\$ Question edited. Pictures/info added. \$\endgroup\$
    – user41224
    Commented May 20, 2019 at 17:55
  • \$\begingroup\$ You can't test parts with them still on the board. \$\endgroup\$
    – JRE
    Commented May 20, 2019 at 19:04
  • \$\begingroup\$ "The same in both directions." Low (less than 1 Volt,) or open ("OL"?) \$\endgroup\$
    – JRE
    Commented May 20, 2019 at 19:07

2 Answers 2

1
\$\begingroup\$

From the pictures and your description, I think it is likely that your diode is fine.

If the regulator isn't working, the problem is going to be somewhere else.

Diode D1 (the one you asked about) is there to protect the regulator in case you accidentally connect the 12V to the barrel jack with the polarity reversed.

The cathode of D1 goes to the 12V line, and the anode to ground. In normal conditions, it will not conduct at all.

If your power supply has the connections to the barrel jack backwards, though, D1 will conduct.

This causes a short circuit. The LM2596 only gets -0.7V on its input - it can survive that.

The short circuit also causes the green polyswitch F7 (marked p 150) to open, and disconnect the module completely from the powersupply.

If D1 is bad, you should also check F7.

D1 isn't a critical part. I couldn't find out what its markings mean, but given its job, any diode with a reverse voltage well over 12V (say, like 50V) and a forward current of a couple of amperes should do fine - if you need to replace it, and I really don't think you need to.


You measured "less than 1V in both directions" for D1.

Since you measured that with the part still on the board, it doesn't mean much. You have the whole regulator in parallel with the diode.

You need to raise one end of the diode and test it again.

I expect you will find it is OK.

\$\endgroup\$
3
  • \$\begingroup\$ Thank you for great answer. I have saved it as a learning reference. I will replace LM2596 and see if it would get me any further. \$\endgroup\$
    – user41224
    Commented May 21, 2019 at 19:03
  • \$\begingroup\$ Regarding D1. I think yur guess about the characteristics of the diode is right on the money. The only type of the diode I found so far, that would fit at least the physical description is schottky diode series from Vishai (SS12, SS13, SS14, SS15, SS16) 20V - 60V 1A link \$\endgroup\$
    – user41224
    Commented May 21, 2019 at 19:13
  • \$\begingroup\$ I don't know if the the LM2596 is toast, or if something else is shot. I'm just pretty sure that D1 is OK. \$\endgroup\$
    – JRE
    Commented May 21, 2019 at 19:20
0
\$\begingroup\$

The significant marking is "IZ".

The package is SMA (standard size + J lead; we dont completely see the lead shape in the solder surface so the lead goes probably under the plastic). With that marking code, on can find that it is a SMAJ24A diode. This is a generic part number, a lot of manufacturers make these parts.

After a search in google I eventually found that the manufacturer of your model is LGE. (No marking in the datasheet unfortunately)

But in the following picture from LCSC you can see that the marking is similar to yours:

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.