BOR (Brown-Out Reset)

[rainman1]

Hello everyone,

I use my 8051 microcontrollre (CC2430) to operate number of units - Led driver, Relay driver, Flash memory and another chip.
Today, i had the MCU sending data wireless to the another device from the flash it connects to.
During that process, i also switched the relay, using a MCU's GP-Output pin.
When i switched the relay on (the relay was switching a 3.3KW load), I still got result correctly from MCU, but when the MCU switched off the relay, i noticed that the MCU skipped on part of the code, and kept sending me data from after that part.

Was that a case of a BOR?
Where do the MCU return to in the code, after a BOR?
Could there be a different reason for the MCU to skip part of the code?

Thanks!

 

[BeeBop]

Hello everyone,

I use my 8051 microcontrollre (CC2430) to operate number of units - Led driver, Relay driver, Flash memory and another chip.
Today, i had the MCU sending data wireless to the another device from the flash it connects to.
During that process, i also switched the relay, using a MCU's GP-Output pin.
When i switched the relay on (the relay was switching a 3.3KW load), I still got result correctly from MCU, but when the MCU switched off the relay, i noticed that the MCU skipped on part of the code, and kept sending me data from after that part.

Was that a case of a BOR?
Where do the MCU return to in the code, after a BOR?
Could there be a different reason for the MCU to skip part of the code?

Thanks!

Sounds like you don't have a protection diode across that relay. When the relay switches off it induces a huge voltage spike across the coil's teminals.

This should explain:
Inductors - Inductive Kickback

 

[electricity86]

Beebop (nice name, though roxteddy was funnier)
I read the article you put here a link to, and i wanted to ask you,
how do you explain what they say there, that for keep transferring current from A to B, the B terminal has a voltage of 1000V?

 

[BeeBop]

Beebop (nice name, though roxteddy was funnier)
I read the article you put here a link to, and i wanted to ask you,
how do you explain what they say there, that for keep transferring current from A to B, the B terminal has a voltage of 1000V?

As current is introduced to an inductor, it begins to build a magnetic field. There is little or negligible current through the inductor until this field is built. When the circuit is opened, the field collapses and produces the voltage spike. That link says up to 1000 volts, but really, the voltage will reach whatever level it has to, to jump that gap. (as they say, the switch 'blows over.')

 

[electricity86]

Thank you for answering me.