MSP430F5328 won't boot if GPIO input is connected to voltage divider

Hi Steve,

It is out-of-spec to provide a voltage at greater than Vcc+0.3V on any pin (see the datasheet absolute maximum ratings www.ti.com/lit/gpn/msp430f5328 p. 40). What I think is the main problem in your case is that if you provide a voltage on the pin with no voltage at Vcc, you can sort of power-up the MSP through the pin (though not correctly of course) and have unexpected behavior - it is trying to power the MSP through that input pin. The best solution for this is to have your hardware set up to ensure the voltage on the pin is not being applied when there is no voltage on Vcc - this can be done with some sort of switch or other means depending on your application or situation.

A common area I've seen people have this problem is if they want to power down their MSP430 at some point in their application and they've provided some switch to completely cut off power to Vcc on the MSP, but they still have communication lines or other inputs that have voltages connected to the MSP. In this case it is usually recommended to put the MSP in a really low power state like LPM4 or LPMx.5 rather than completely cutting Vcc - then you won't have an issue as you still have a voltage on Vcc (not going out of spec) the part is just in a very low current state - this is using the part the way it is designed. If you really need to completely cut off Vcc for some reason, it is better to make sure that there aren't voltages being applied on the other pins to make sure you stay in the device spec.

Regards,

Katie

As as stopgap, you may try this: a) replace the top 100k with a 50M and b) eliminate the 200k bottom resistor.

0.1uA will certainly not hurt anything and hopefully will not send the chip to limbo. (But I am not sure.)

-- OCY

steve wenner said:

voltage divider running from 5V to ground (100k top resistor, 200k bottom resistor)

Change the 200k to 10k and it should work. (0.45V ) You will need to read the input with ADC. 

Alternatelly use level shifter or a NPN transistor or schottky Vcc clamp.

Peter

Vcc and RST will also be ~0.4V and the RST pins is already above threshold and will not generate a restart.

Try if adding a pull-down to the RST pin of aprox 100KOhm will work.

Try a reverse biased signal diode in series with the 5 V signal and internal pullup resistor. (Anode to input pin)

Peter

MSP digital inputs are high impedance. The maximum input leakage current is specified with 50nA. So as long as the input voltage does not exceed VCC, the maximum current is 50nA.

The input signal voltage is 3.3V. The required input voltage is Vcc (3.3V)*0.75, so a voltage drop of up to 0.8V is acceptable. At a leakage current of 50nA, you can put a 15MOhm resistor in series to the input and it should still detect the input signal.
So a series resistor of just 470k will limit any current into the unpowered MSP to below 10µA, which is low enough to not have any effect on the MSP while not influencing high/low detection.

However, the original divider already was a 100k series resistor (with additional 200k load),, so the maximum current into the I/O pin was 50µA, barely enough to raise VCC enough to have any effect on the boot process.

But a proper way to connect a 5V signal to an MSP input would be, as Peter suggested, a series diode and a pull-up to MSP’s VCC (e.g. the internal one). So the signal can’t pull the pin up to 5V, but only pull it down. In case of a Schottky diode, the low-voltage will raise only below 0.2V while the leakage current of the diode is negligible (BAT42: 50nA@5V reverse voltage).

steve wenner said:

In the first case using a scope I see about .4V

This voltage suggests an SCR latchup on the input.

I would not suspect this to be possible at such a low current. 

Peter