12VDC SMD fuse

[LancelotM]

Hello everyone,

I accidentally plugged in a 24VDC power supply to a 12VDC camera and it fried the fuse on the main board right next to the power plug. It's 3216 smd fuse and the pads and the rest of the board still looks ok so I want to try replacing that fuse. However after searching on places like digikey and mouser, there doesn't seems to be a 12VDC smd fuse anywhere. Does anyone know why that's the case? And can anyone shine some light on what kind of fuse could they had used on this board?

Thank you   

[wraper]

Why should it be with 12 V DC rating? They usually have higher voltage rating. Just random fuse exceeding 12V DC rating. https://eu.mouser.com/ProductDetail/Bourns/SF-1206HIA100M-2?qs=GBLSl2AkirsGrSf%252BgdlCmA%3D%3D. What's more important is you omitted current rating.

[LancelotM]

Why should it be with 12 V DC rating? They usually have higher voltage rating. Just random fuse exceeding 12V DC rating. https://eu.mouser.com/ProductDetail/Bourns/SF-1206HIA100M-2?qs=GBLSl2AkirsGrSf%252BgdlCmA%3D%3D. What's more important is you omitted current rating.

Because a 24VDC power supply fried that fuse, so it must be something between 12-24VDC. But the lowest rating I can find for fuse that size is 24VDC......

[T3sl4co1l]

Fuses are rated by the maximum voltage they must break.

You don't need a "12V" fuse here, you need a fuse of the same rating -- 30 to 125V is typical in that size.

Before you replace the fuse, check that the other side of the circuit doesn't have a low resistance or voltage drop.  If it does, the new fuse is just going to blow again, and you either have to fix what's there, or give up.

Tim

[exe]

Because a 24VDC power supply fried that fuse, so it must be something between 12-24VDC. But the lowest rating I can find for fuse that size is 24VDC......

Fuses trip because of the current. Voltage rating is to indicate maximum voltage they can interrupt. They may fail to interrupt voltage higher than their rating OR current higher than their breaking capacity. I.e., too much current can make fuse a conductor (I think evaporating metal can make a conductive plasma). To give some numbers, cheap glass fuses are rated for 250VAC and 400A interrupt current. More expensive fuses are sand-filled, and called HRC (high-rupture capacity).

If you think of a fuse, it's actually a small-value resistor. There is very little voltage across it because of Ohm's law (until it trips). After tripping, they have to withstand the open-circuit voltage.

It's fine to use a 1000V fuse in a 3V circuit, although its performance (=resistance) can be worse than lower voltage parts, or it can be physically larger, or much more expensive.

[ogden]

Because a 24VDC power supply fried that fuse, so it must be something between 12-24VDC.

Incorrect. Fuse is current-activated two terminal device, it does not "measure" voltage.

[LancelotM]

Because a 24VDC power supply fried that fuse, so it must be something between 12-24VDC.

Incorrect. Fuse is current-activated two terminal device, it does not "measure" voltage.

So when I plugged in the wrong power supply which has output rating of 24VDC, 1.67A max into the camera (it's original power supply has output of 12VDC, 2A max.) The current into the board far exceeded 2A and therefore blown the fuse, is that correct? Sorry I'm really a newbie to all this    

[Ice-Tea]

Because a 24VDC power supply fried that fuse, so it must be something between 12-24VDC.

Incorrect. Fuse is current-activated two terminal device, it does not "measure" voltage.

So when I plugged in the wrong power supply which has output rating of 24VDC, 1.67A max into the camera (it's original power supply has output of 12VDC, 2A max.) The current into the board far exceeded 2A and therefore blown the fuse, is that correct? Sorry I'm really a newbie to all this

That is correct. The missing piece of the puzzle here is that the current in the board exceeded 2A (this is a bit of a simplification, but it works well enough to explain things) because the circuit was not designed to handle 24V rather than 12V. The fuse has protected you and your house from, well, fire. It may have also protected the rest of the circuit but that's impossible to tell. In the moment before the fuse blew, something else may have died as well.

[LancelotM]

That is correct. The missing piece of the puzzle here is that the current in the board exceeded 2A (this is a bit of a simplification, but it works well enough to explain things) because the circuit was not designed to handle 24V rather than 12V. The fuse has protected you and your house from, well, fire. It may have also protected the rest of the circuit but that's impossible to tell. In the moment before the fuse blew, something else may have died as well.

Thanks. One last question I guess: Since a high current was drawn into the circuit from a power supply rated for 1.67A, is it possible that the 24V power supply may also be damaged?

[T3sl4co1l]

Possible, but unlikely, and easily tested (assuming you have a load for 24V ).

Tim

[ogden]

So when I plugged in the wrong power supply which has output rating of 24VDC, 1.67A max into the camera (it's original power supply has output of 12VDC, 2A max.) The current into the board far exceeded 2A and therefore blown the fuse, is that correct?

Yes. Most of low cost overvoltage protection is made out of some voltage clamp (like zener) and fuse.

Thanks. One last question I guess: Since a high current was drawn into the circuit from a power supply rated for 1.67A, is it possible that the 24V power supply may also be damaged?

Unlikely. Most likely blown fuse saved both - device and supply. Thou you shall check everything before using  again.

[fzabkar]

Show us a photo of the PCB.

[LancelotM]

Show us a photo of the PCB.

For your viewing pleasure

[sean0118]

With the board unplugged, use a multimeter to measure the resistance from each fuse side to the screw holes (I'm assuming they are GND), one side will likely be low resistance if there's a problem. Probably worth measuring the resistance over D804 and the big metal can capacitor CA801. You could also put it on diode mode and you should be able to measure the forward voltage of the diode if it's okay.

Measure the resistance across the connector pins once you replace the fuse, it should be relatively high resistance, if it's close to 0ohm you have a problem somewhere else.


As others have said, when it comes to fuses a higher voltage ratings is generally always better. But the current rating and operating speed need to be selected properly. For speed, fast blow is normally used on boards like this, slow blow are used on things like power supplies sometimes. I'm not sure how you will select a suitable current rating unless the board lists it? Might have to guess and then measure its operating current to check.

[LancelotM]

With the board unplugged, use a multimeter to measure the resistance from each fuse side to the screw holes (I'm assuming they are GND), one side will likely be low resistance if there's a problem. Probably worth measuring the resistance over D804 and the big metal can capacitor CA801. You could also put it on diode mode and you should be able to measure the forward voltage of the diode if it's okay.

Measure the resistance across the connector pins once you replace the fuse, it should be relatively high resistance, if it's close to 0ohm you have a problem somewhere else.


As others have said, when it comes to fuses a higher voltage ratings is generally always better. But the current rating and operating speed need to be selected properly. For speed, fast blow is normally used on boards like this, slow blow are used on things like power supplies sometimes. I'm not sure how you will select a suitable current rating unless the board lists it? Might have to guess and then measure its operating current to check.

Thanks. I have another identical camera and although I don't have equipment to measure operating current, I measured the power draw from the outlet which stays steady at 3W. If I assume 100% efficiency from the power supply, that'll give me 3/12=0.25A. Of course power supply is not 100% efficient (maybe 75%-90%?) so the actual current would be less than that.