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Vcc(drop) - V?

Yasuhiko OKADA
Expert 1900 points
Other Parts Discussed in Thread: MSP430F47176

I have a question in the manual of MSP430F47176 (msp430f47176.pdf).

 

In page 35, Figure 9 and Figure 10, the y axis of the graphs are "Vcc(drop) - V".

It seems the y axis is some difference between Vcc(drop) and V.

What is the V of the "-V"?

 

Or, is this a notation saying "the unit of Vcc(drop) is in voltage"? If so, for around 1.4V

for 1 micro seconds will cause POR/Brownout Signal?

 

 

 

  • 0
    Guru 227245 points

    The -V means this is the unit of this axis. The diagrams have no units attached.

    So a voltage drop down to 0.5V will not trigger a BOR if it is shorter than 20ns. If the drop is longer than 10µs, it may not go below 1.5V or a BOR is triggered. (Fig. 9)
    This diagrams are a bit misleading because they start for VCC=3V, yet no BOR is ever triggered if you go down to 2V permanently. So in both drawings, only the part (and pulse width) is important where VCC goes below 1.6V or so. Hence the different diagram forms for different drop pulse forms.

    So yes, your second assumption is correct.

  • 0 in reply to Jens-Michael Gross
    Expert 1900 points

    Thank you very much, Jens-Michael.

     

    Jens-Michael Gross said:

    So a voltage drop down to 0.5V will not trigger a BOR if it is shorter than 20ns. If the drop is longer than 10µs, it may not go below 1.5V or a BOR is triggered. (Fig. 9)
    This diagrams are a bit misleading because they start for VCC=3V, yet no BOR is ever triggered if you go down to 2V permanently. So in both drawings, only the part (and pulse width) is important where VCC goes below 1.6V or so. Hence the different diagram forms for different drop pulse forms.

     

    What do you mean about the "misleading"?

    Do you mean "Vcc(drop)" is not the drop voltage (=Vcc - current Voltage), but the voltage after the drop (=current Voltage)?

    From your explanation, the latter applies, right?

     

  • 0 in reply to Yasuhiko OKADA
    Guru 227245 points

    OKY said:
    From your explanation, the latter applies, right?

    At least that's how these diagrams make sense to me.
    It makes no sense that a short drop of a low drop voltage will trigger a BOR, but a long one with a higher drop voltage (= lower momentary voltage) will not. The circuit would have to look into the future (will this be a short or a long drop) or have a memory (was this drop short enough to to trigger a BOR). More likely it has a capacitance that will deglitch shorter pulses. (and maybe an RC impedance pole that explains the nonlinearity in the second diagram).
    I'd say, Vdrop is rather the dropped voltage than the drop voltage.

    But as usual, I might be completely wrong and the illogical (from my view) interpretation is the truth :)

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