STM32F4 Burned out

[thommagn]

Posted on April 18, 2013 at 11:04

Hi,

I'm using an stm32f4 on a PCB and suddenly it became extremely hot. The strange thing is that it still worked, but there was an extremely (relatively) high current running through the circuit. 

I've removed the stm32f4, concluding that my 3V domain is fine.

Now, I'm trying to sort out what has happened. Does anyone has any idea what might cause this kind of behaviour out of the processor? (High current but still ''working''). Can this happen if to high voltage is applied to any of its pins?

Should I just try to replace it and hope that it is a fluke? I have several identical PCBs and this is the only one on which I had problems with the STM32F4.

// Tom

[Andrew Neil]

Posted on April 18, 2013 at 11:45

''The strange thing is that it still worked''

It might be more accurate to say that the bits used used by your application have not been completely damaged to the extent of being entirely unusable?

Maybe they are ''barely working''?

Highly unlikely that they are fully & correctly working to specification...

A common effect of over-stress is that leakage currents (greatly) increase - which could account for your high current draw.

Note that this could also apply to other components on the board...

[John F.]

Posted on April 18, 2013 at 13:30

If your other PCBs work OK, it sounds like this PCB may have a fault - pins shorted to ground  or VDD or to each other.

Alternatively, your design may be at fault and the chip has been slowly cooking until enough damage has been done for higher currents to flow causing overheating.

If you're happy with your design, put the faulty PCB on the ''back burner'' (as you might say) and pursue your design on one of the others. If the next one fails, post your schematic ...

[thommagn]

Posted on April 18, 2013 at 14:46

Thanks for the responses.

Other PCBs has been used extensively without indication of any problems. So hopefully it's just this one. 

All current was flowing through the processor. When I removed it, power consumption went back to normal.

It may be of importance that i have a second STM32F4 parallell to the faulty one which runs on the same 3V domain. This processor seems to work just fine!

I think I'll measure all components connected to this specific processor, if they seems to be ok, I'll simply put on a new one and see what happens.

// Thom

[John F.]

Posted on April 18, 2013 at 15:35

Tom, You asked, ''Can this happen if to high voltage is applied to any of its pins?''. The answer is usually yes! If you exceed the allowed voltage, current flows into the device through parasitic or protection diodes. Depending on the silicon structure, this could cause significantly high currents to flow. If you then remove the fault voltage and power the chip down and up again all might be well ... or you may have damaged the silicon or burned out bond wires or metallisation. If you are applying voltages to any pin in excess of the supply voltage (or in excess of +5V for FT pins) then add a resistor to limit current and also maybe a diode to clamp the overvoltage to the positive supply rail. Much the same applies to negative input voltages - basically - don't.

[thommagn]

Posted on May 02, 2013 at 10:34

Thanks for the responses.

How would you dimension protective diods for the STM32F4, i.e.

what Vrwm rating should I use to protect external pins of the processor (USART etc)? 

Is this necessary though? The Discovery board doesn't seem to have any protection for its inputs, and i've never had any problems with this board.

[John F.]

Posted on May 02, 2013 at 12:55

I'd expect USART pins to be connected to an interface IC such as an RS232 or RS485 compatible device. In that case, the interface IC will have inbuilt protection. Similarly, protection devices comprising diodes or transient suppressors are available for USB inputs. Look at the NXP devices for example. For general purpose inputs, I would normally provide isolation using a bipolar transistor connected as common emitter.

See attachment for image ...

where the emitter (2) goes to 0V, the collector ( 1) goes to the GPIO input and the external logic being sensed is applied to base series resistor R5. Note that the signal at the collector is inverted with respect to the input.

Really, it all depends what

your

requirements are.

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Attachments :

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