MSP430F2013: Two microcontroller damaged

Hello Johannes,

how did you measure this? And is the controller really dead?

Of course electrostatic discharge, overvoltage or incorrect polarity can always kill a semiconductor if no precautions are taken. But I could also think of, that when trying to measure your applied voltage to the pins and an existing pull up resistor for reset is existing, your controller just starts working and is consuming power, the values would give 33mA.

Jan

Hello Jan,

I measured the resistance with no supplied voltage.

The JTAG-FET can't connect the controller: Can't identify target device: No error! So, the controller ist really dead, I think.

Johannes

And now, new infos:

The MSP was soldered first and works on it's own.

A RF transceiver from nordic nRF24L01+ was connected via SPI. After flashing the MSP with new code, the resistance between Vcc and Vss is only 5 Ohm and I can't connect the board.

Any ideas?

Have a look to errata PORT16. I suspect that when you program the micro, one of those pins goes low and in your circuit there is some "high" signal on that pin that will destroy it if there is no resistor in series.

I know this is an old thread, but since it has been revived, I can’t resist (pun intended) commenting on this one:

Johannes Kreuzer98479 said:

I measured the resistance with no supplied voltage.

R=U/I. So if you want to know a resistance, then you can measure it either by injecting a current (and then you get a voltage and therefore have a voltage applied) or by applying a voltage (sic!) and measuring the current. This is how multimeters usually do it.
However, the multimeter reading is wrong. A CPU is based on semiconductors that do (best case) have a non-linear virtual resistance. In fact, when you apply a voltage, current will flow until the voltage breaks down to the breakdown voltage level of the semiconductors found between VCC and VSS. If you inject a larger current, the current will rise but not the voltage, appearing as a shrinking resistance. Until you finally fry the chip.
If your multimeter uses a measuring voltage below ~4V, you’ll simply power the MSP. If it allows a higher one (and doesn’t limit the current), then you might permanently damage the MSP.

 Nelson,
There is no PORT16 erratum for the 2013. Also, if a port pin goes low while it is source from outside with VCC, it will simply rise its ‘output’ current to maximum (saturation) current. Which won’t destroy the port pin (but can overheat the MSP if it happens for more than one pin).
But if an input voltage below VSS or above VCC is applied (e.g. the MSP is switched off while the peripherals aren’t), then current will flow through the input protection diodes. If this current is >2mA, then these clamp diodes will be destroyed, either opening (in which case the overvoltage - which has caused the current - may destroy the pin circuitry and perhaps more) or short (in which case the pin is tied to VCC or VSS, but the rest of the MSP remains intact). In the second case, trying to operate the pin with high output while tied to VSS (or v.v.) will cause excessive current. However, at power-on, all pins are high-impedance, so the MSP should respond to the FET.
It is unlikely, that both clamp diodes of a pin are shorted, creating a short between VCC and VSS. Yet it is not impossible.

Hello Jens-Michael,

I didn't know, that is such easy to get the nobel price.

But I can tell you how it works: On a pcb there was (long time ago) a power supply, which generates the 3.3V or 2.2V. When I measured the resistance, I switched this power supply off.

So please send me the nobel price and especially the money.

Best regards,

Johannes

PS: I've never read such stupid answer...

Where does it say there is no PORT16 erratum for 2013? *** EDIT ***  - I see it now. My apologies.

If a voltage is applied to a pin that is set to LOW, the current will be whatever the external circuit can supply and destroy the pin. Think as a transistor or mosfet, the current is not limited by it's saturation current but from the external circuit.

Johannes, did you make it work? I didn't realize this post was so old. Did you find the problem?

Johannes Kreuzer98479 said:

But I can tell you how it works: On a pcb there was (long time ago) a power supply, which generates the 3.3V or 2.2V. When I measured the resistance, I switched this power supply off.

Seriously? Well... you measured resistance using voltage coming from power supply of the multimeter.

Johannes Kreuzer98479 said:

So please send me the nobel price and especially the money.

I am not aware of money prize for ignoble award :) (kidding)

Hello,

are you calling a multimeter "power supply" or are you calling the LDO "power supply"?

Of course, a multimeter uses a voltage or a current (normally a current) to measure the current or voltage to calculate the resistance. But I'm very suprised, that you call this measurement voltage/current "power supply".

And regarding this discussion: I has nothing to do with the error, that I had. So I want to stopp this stupid discussion about the power from the multimeter.

Best regards,

Johannes

Johannes Kreuzer98479 said:

But I'm very suprised, that you call this measurement voltage/current "power supply".

No. I did not say this.

Johannes Kreuzer98479 said:

So I want to stopp this stupid discussion about the power from the multimeter.

Good. Please don't mislead other readers anymore.

Johannes Kreuzer98479 said:

PS: I've never read such stupid answer...

An answer isn’t stupid just because you didn’t get the point.

You said “with no supplied voltage”, not “power supply off when measuring” (which would be a matter of course for direct resistance metering)
And my point was, that you simply cannot measure resistance without applying a voltage. Not necessarily from a power supply, but from the multimeter you use for measuring resistance.

Well, maybe you could calculate the resistance by detecting the number and position of all atoms in the current path and calculating the expected electron movement. But even then, for calculating the resistance, you’d have to apply a virtual voltage in the final formula.

 The power consumption of an MSP (without “external” currents) is in the µA range – much less than what a multimeter uses for measuring. So yes, the multimeter acts as power supply. And maybe even a sufficient one.

 Besides this, my answer also contained the information that a multimeter isn’t a proper tool to measure the ‘resistance’ of a semiconductor circuit. And that the reading will be worthless.
The only parameter that makes sense is the current consumption. Which can be measured be having the supply on and measuring the current over a shunt in front of VCC.