Sorry to hear that. What voltage were you supplying it?
Sadly, there is no fuse on either board.
Are you sure the Edison is fried, or could it just be the voltage regulator on the breakout board? It looks like the voltage regulator on the mini breakout board supports up to 3A, so I don't think you should have been stressing it.
1 of 1 people found this helpful
I was supplying it 12V. What I should've done is limited the current... Sigh, costly mistake.
Luckily, I participated in an IoT hackathon recently and have another board albeit the Arduino version. So I swapped the fried Edison with that one and verified it was not powering up. However, putting the working edison on the mini break out didn't work either. It looks like I fried both the mini break out and edison.
Well, lesson learned... But I still can't understand why a USB camera can cause this... BTW, I was using a Mobius Actioncam in usb webcam mode. Stay away from this camera on this board...
after i had read this article from "Tage" i became rather skeptical about the max. usb current:
It sounds not like you can get up to 500 mA current on USB permanently. Maybe the reason for your bad luck?
The question for me is:
Having a good power supply connected at the 5-15V input, what is the maximum current limit on one or both usb (mini breakout board)?
1 of 1 people found this helpful
I just briefly looked at the power conversion system on the mini breakout board, but what I can see from that "paper investigation" is that the power on USB J16 is limited to 500mA by the MIC2039. However, this part is able to put out up to 6A (typical max current 3.2A) for up to 0.2s. The reason for having this overload capability would be to avoid that the voltage on the USB connector drops when you plug in a gadget, and to allow brief overloads on USB.
The other USB connector J3 has no power as far as I can see, it is instead receiving power from the outside to power the FT232R.
If the Edison board failed while the mini breakout board was powered from 12V, my guess is that the TPS62133 first failed in a way that its output voltage which should have been 5V became 12V. This could kill the BQ24074 which has an input range of 4.35 to 10.5V. This is only a wild guess. I may be wrong. Perhaps it was the BQ24074 that failed. But is seems logical that in order for the Edison module to fail it must have been subject to overvoltage on V_SYS (maximum voltage is 4.4V). I think it is likely that if the BQ24074 failed with a short circuit from IN (5V) to OUT (V_SYS) then it is unlikely that V_SYS would have been so high that the Edison module failed. at least, it is my guess that applying 5V is not high enough to cause a failure. that is only 0.6V above the rating.
I should take my min breakout board to the lab bench and make actual measurements to confirm my suspicion that the inductor L1 used in the TPS62133 circuit is too small, and study what happens during an overload. Otherwise I am just guessing. (That is where I was hoping that Intel had already some verification data on the power system that they would like to share). Inductors of this size usually have a saturation current of just above 1A but usually much less. which means the circuit is ok as long as the total load on 5V_SYS is not more than 1A. if the load is high enough to saturate the inductor, there will be high stress on the TPS62133. this circuit can drive (typically) 4.2A short term, but if the inductor saturates this current level can rise a lot higher. when the inductor saturates it basically disappear and is replaced by a short circuit, electrically. this is because the magnetic material saturates and cannot store any more energy, so the current builds up very quickly if you have 12V on the input and 5V on the output. to fix that problem one needs to use a physically larger inductor.
so it is logical to think that if there is a load on the USB connector and the load is such that it occasionally draws more than 500mA (it is allowed to by the MIC2039), and the Edison module draws close to 500mA, we have an overload on the TPS62133 circuit. The BQ24074 has an input current limit set at 0.96A, so if we also have a battery connected and it is being charged, there is definitely a possibility that we overload the buck regulator on the 7V-17V input.
Thanks very much for the detailed explanation! Let me try to say it shortly:
To be on the (hopefully) sure side in an average environment:
Let's say the Edison module itself can use 500 mA (short spikes up to 700 mA with wlan) then additionally you can add a total load of max. 500 mA (summing up usb, battery loads, etc.).
So the mini breakout board should not permanently exceed a total load of 1 A. Am i right?
if we look at the maximum continuous load on the 5V output of the buck regulator that is powered by the 7-17V input, 500mA can go to the USB and another 1A can go to the regulator that feeds the Edison and charges the battery. that is a total of 1.5A continuous load. I do not really know what the maximum current draw of the Edsion module is. I am just looking at the current limit settings of the different power controllers.
but as I mentioned, the circuit allows a large transient load on USB, and that was my main concern when I looked at the inductor L1.
I will try to find time to make actual measurements to see what happens with the regulator on the 7V-17V input and report back. it would be better if Intel could do this type of testing, because it is likely to destroy my board, and after all it is an Intel design..