Help identifying and replacing a blown component - Page 1
Forum section: electronics / beginners

author: qoole
author_rank: Contributor
timestamp: January 21, 2013, 06:53:53 pm
content: Hi all,I'm new to the forum but have been watching the EEVBlog videos for some time! Excellent work! Very enjoyable and informative!Recently someone plugged the wrong power supply (correct voltage/connector but reversed polarity) into my rather expensive (however, out of warranty) AA/AAA battery charger/conditioner.Luckily the person realised their mistake when they heard the sizzling and saw the smoke and unplugged the device pretty quick.Having inspected the board I can see one failed component. It appears to be an SMD Diode it has the marking "PJ8D9" on the first line and "SX34" on the second line.I have googled these and have not managed to find a datasheet for it.Attached is a rather poor camera-phone image of the component.Does anyone have any idea what the component is and what I can replace it with?If anyone's interested this is a Maha Powerex MH-C9000 charger/conditioner.Thanks,Alex

author: PA0PBZ
author_rank: Super Contributor
timestamp: January 21, 2013, 07:12:53 pm
content: Looking at the board the diode is just there for reverse power protection, it sits on the power input and it will conduct when you reverse the power.While I think there should have been a fuse so the diode blows the fuse at least it protected your charger from further damage (we hope!).So just get a 1N400X and replace it with that, no need for an smd really. You can solder it on the power input connector.

author: qoole
author_rank: Contributor
timestamp: January 21, 2013, 07:37:15 pm
content: Thank you muchly!Something I find interesting is that the two solder points directly above the diode in question is an electrolytic cap on the other side of the board.If the supply had been reversed any longer that cap would probably have blown too...It seems like a rather half-hearted attempt at reversed polarity protection. Although, I suppose two blown components would be better than taking all the ICs out.Thanks again!

author: SeanB
author_rank: Super Contributor
timestamp: January 21, 2013, 07:42:51 pm
content: Yes, a reverse protection diode. Before you power it up again check the 5V regulator next to it is still working, or that there are no additional shorts on the input side.

author: qoole
author_rank: Contributor
timestamp: January 28, 2013, 10:06:51 pm
content: This setup confuses me.In correct operation the diode is not used but when the polarity is reversed it creates a short between the +V rail and ground. This would obviously blow the diode and if it failed 'open' then would not protect the circuit at all. They seem to be relying on the fact that it would fail 'short' and divert the current away from the chips.Would it not be more sensible to have the diode directing current in normal operation and when polarity is reversed act as a 'no pass' for the current. Thus a) not burning out b) not putting other components at risk.Would it not be the _most_ sensible to use a bridge rectifier on the in put so that whatever the polarity the input it would always be correct by the time it made it to the electronics.I am quite inexperienced at the practical side of electronics so there may well be reasons against doing either of these options.I would love to hear both sides, why their method is better and why my methods are bad!Thanks,AlexP.S. I replaced the Diode with a 1N4001 (possibly the ugliest bit of soldering ever (an SMD Diode a 1N400x isn't) but it works).

author: Paul Moir
author_rank: Frequent Contributor
timestamp: January 29, 2013, 04:40:52 am
content: There are advantages to both ways. Usually the reverse protection diode is paired with a fuse, causing the fuse to blow if polarity is reversed. The disadvantage of using a series diode is that it wastes power from the diode's voltage drop when the circuit is operating normally, whereas the parallel one doesn't do anything when the circuit is operating normally. Note that the diode voltage drop increases with increased current. On something like a 1N4001 it'll be up over a volt at 1 amp. Using a bridge rectifier will double the loss due to the two diodes. But there's no "right" way to do it: it's application dependant.

author: digsys
author_rank: Unknown
timestamp: January 29, 2013, 05:53:24 am
content: In really well made equipment, like some Motorola radios I service, they use these -www.vishay.com/docs/88383/sm6a27.pdfor the 33V version. They can take up to 600A !! They give you BOTH reverse protection and spike protection and are FAST.Usually when you use these type of devices, you make sure that there's a LOT of copper right up to the Power socket ie 80z Cu.I've had cases where they've held down 9-16A P/Supplies and wouldn't let go .. and you are supposed to have a fuse.