There is a broken chip on my Rpi3 motherboard labeled: 4R7. I just bought it and it broke while trying to install it inside the case. Will the board still work with that chip broken? Can I cover it with epoxy?

Question

There is a broken chip on my Rpi3 labeled: "4R7". I just bought it and it broke while while trying to install inside the case.

Will that chip: "4R7 being cracked will prevent the Rpi3 from working properly? Can I simply coat the chip with epoxy?

This is my first experience with Raspberry. I never heard of it before until I saw an advertisement for Cayenne in Facebook last month.

rew · Answer

Dear Compact,

You are thinking of a resistor. The resistors on the raspberry pi are mostly so small that they do not have a marking. There are however two inductors prominently marked "4R7". Resistors cost factions of a cent, the cheapest inductors that I can find are about 10 cents.

Most before me haven't mentioned it, but the cracked component is not a chip, not a resistor but an inductor.

The two inductors are involved in the power circuit. They make, I believe, 3.3V and 1.8V from the 5V input. Because they step the voltage down, I can predict how they are wired. If the inductor is broken, one of those power rails will simply remain at 0V. This COULD damage one of the chips on your raspberry, but it is unlikely.

The inductor consists of a wire wrapped around a core. If that core is cracked, the inductor will not work as well as it should. On the other hand, it is usually not all that critical in the power circuit, so if things change a bit due to the crack there is a possibility that things still work. The continuity of the wire is essential though. So... if you power up your pi, and it works, then that continuity is ensured. In that case encasing it with epoxy might work. On the other hand, if it fails after you epoxy it, it will be impossible to fix.

Replacing an SMD component is easy enough if you have the right tools. But having those tools and knowing how to use them is not trivial.

michaelwylie · Answer

Looks fine to me, just the package is broken. If the package is ferrite, the inductance will be slightly different. I don't think it's a big deal.

rsc · Answer

Does the PCB label read L1 or L2 next to the broken part? See lower left on image.

Those parts are part of the power supply regulator circuit, and could cause problems with all the ICs if damaged.

Take a picture of the damaged part and post it so we can see what's happening.

shabaz · Answer

Unfortunately it depends on precisely what is cracked or broken. If the pads are ripped off, or the wire inside is damaged then it won't work by securing with epoxy, it would need some soldering work and part replacement.

However, if there is just a crack in the grey colored portion then it might work if the wire inside that part isn't damaged and the pads have not ripped off the board. But, there is no guarantee that it will work (and there might be some inefficiency in the power circuit introduced too). It is worth a try to just power it up if you have no way of replacing that part (it won't be a straightforward repair unless you've got some surface-mount soldering experience and the right tools, which will cost more than the Pi itself).

If it works then yes you could apply a tiny dab of epoxy, it won't damage it any further.

COMPACT · Answer

Epoxy is not conductive so just get someone with the right equipment to replace it to ensure it'll work properly. The price of the 4k7 is very cheap - can be low as fractions of cents.

COMPACT · Answer

Michel,

Sorry, my mistake. It is actually shorthand for an Bourns SRN4018-4K7M 4.7uH (microhenry) inductor (or equivalent) and has a huge price tag of 13 US cents.

So I'd replace it if it is cracked through the centre.

No real worries if it's just a little bit of a corner.

You can test the inductor to see if its coil is intact. When measured with a multimeter it'd read 0 ohms if ok (with the Pi turned off!).

When the ferrite is in operation it is under mechanical stress from the constant charging and discharging.

Epoxy is used to reduce buzzing from inductors but the epoxy is of a type not to affect its heat dissipation.

If you play your cards right with an inductor manufacturer you may be able get one as a sample - Zero Cents!

rew · Answer

IMHO the cost of the component is irrelevant.

Before Michel is able to replace the inductor the costs are many, many times more than just the 13 cents for the inductor.

* Buy, or arrange to borrow the right desoldering equipment.

* find time to desolder the component.

* manage to solder on the new one without damage to the surrounding components.

Factor in the chances of ruining the pi for good (IMHO about 50/50), the costs are WAY more than the 13 cents for the inductor. So his idea of epoxying the component in place sounds like the better bet. Just first test if the pi still works. Then you can be confident that it is electrically OK. Then apply the epoxy. (Because Michel comes with the suggestion of epoxy, I'm assuming that's easy for him because he has all the stuff to do that.)

If the pi no longer works, it's a dead board. You can do anything with it, even: try to repair it. However, the replacing the inductor on a pi is not the easiest of "first projects" to try and learn such a task....

COMPACT · Answer

Very Compact wrote previously: Epoxy is not conductive so just get someone with the right equipment to replace it to ensure it'll work properly.

shabaz · Answer

Hi Michel,

I will not apply any power to this until I am sure there will be no further damage to the circuits. I have soldering equipment but I am not very proficient at soldering, but I have a friend how is more experienced at it than I am

If you're not confident at replacing this (it won't be easy without experience) and you cannot find someone with enough experience to replace the part (assuming you can get the part at a price you're willing to pay - most suppliers will expect a minimum order and/or postage costs) then you have not much further to lose if you apply power. Find a person to replace it and that is your best bet.

Having said that, (and of course, I take no responsibility for it not functioning and getting damaged further if you apply power to it), if you've confirmed the wire isn't broken then in this situation, typically there is little harm in applying power. It is not a heat sink, but it will affect the power supply performance, but nothing to the extent of it not functioning; most likely it will run fine.

COMPACT · Answer

Very Compact wrote: No real worries if it's just a little bit of a corner. (You can also feel free to glue on the broken bits if you wish.)

Robert Peter Oakes · Answer

I would agree with michaelwylie , based on the observed damage, as long as the windings are intact, the change in inductance could be compensated for by the power controller, it may run a little warmer but I would have fairly good confidence it will still work ok (If this is the only damage This should be fairly straight forward to replace if you are confident with surface mount soldering and if you can get the right part. Care would be needed to not damage surrounding parts or to get the board too hot. A hot air soldering gun would help here too.

rew · Answer

If the ferrite is cracked it makes sense to make the airgap as small as possible and keep it in place. Epoxy is fine for that. If the ferrite is broken that the body of the inductor is loose and it is just hanging on by the wires, then it makes sense to lock it in place. Epoxy is fine for that. But in the pictures it looks as if part of the ferrite broke off. In that case there is not much else you can do. Just power it up and see if it works. As michael says, the inductance may be a bit different, but they don't normally make anything between 2.2uH and 4.7uH, so even if they calculated the ideal value to be 3.7uH they will be using this one, and you dropped the inductance by a few percent. Nothing to worry about. Your measurement of 10 ohms is a bit high. Many multimeters cannot accurately measure low resistances. So there usually is an offset on the order of a few tenths of an ohm near zero. You can calibrate your offset by measuring the resistance from the SAME PAD of the inductor. Both leads on the same side. When you end up measuring the actual resistance you subtract what you got when you had the leads on the same side. I have one multimeter that has a 0.3 Ohm offset, but another has something like 5 ohms. Now... Even that 10 ohms reading can be considered "almost zero". 10 ohms is a tad much, but if we factor in about 5 ohms of offset on your multimeter, it doesn't sound too bad. Of course, for efficiency's sake, you would rather have 0.1 Ohms or less, but if that is unobtainable in the formfactor and/or pricepoint, then you'll have to settle for something less than ideal.