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TMS320F28035: Cannot generate real "low level" when connect two GPIO pin which configured as output pin

Brian Chang90901
Intellectual 2540 points
Part Number: TMS320F28035

Hi Champs,

My customer had an application. Their product will be used  in parallel.

They use GPIO pin to connected an signal bus directly as bellowing block diagram.

The GPIO pins were configured as output pin and the pull up module was disable.

They need to have one GPIO was set high initilally and the another will pull down the bus.

We use GPIO module to test firstly at single unit and we fund the voltage was not zero.

The dropping voltage will be kept at 1.49V as the bellowing waveform.

The original was 3.3V and the voltage will be dropping to 1.49V when we clear another GPIO DAT value.

At this condition ,  one is GPIO12 and another GPIO31 

  

However , Only GPIO18 will pull down the voltage bellowing 1V.

Customer want to know why the GPIO can not really pull down the voltage to zero and why there was difference between pins.

But also what  was the block diagram of the GPIO module at C2000 ?

Did you have any suggestion for the configuration?

BR

Brian

  • 0
    Guru 30330 points
    Brian,

    When you connect and drive the two GPIOs as shown, there will be contention. As one of the GPIOs is driving the voltage high, the other is trying to pull it low, since both GPIOs have the same drive strength, the resulting voltage will be about ~1.65 V depending on other connections.

    Unfortunately, the F28035 does not have an open-drain configuration of the GPIOs, so directly tying the two pins together will not work. The customers options are either some software logic to detect the state and drive a single GPIO appropriately, or adding external combination logic to produce the necessary output based on the state of the two driven IOs.

    Which diagram are you specifically asking about? Have you looked at the System Control and Interrupts User Guide? (www.ti.com/.../sprugl8)

    Thanks,
    Mark
  • 0 in reply to Mark Labbato
    TI__Intellectual 2540 points
    Hi Mark,
    If that was the same driving strength , it is meant that the equaivalent resistance was the same when output high or output low. What value it was ?
    The user guide was very brief diagram to show the GPIO module. They would like to know the circuit of output stage and the real resistance. That was impact their design with the fianl end customer. They would need to address how much load their end customer can put to connected with GPIO pin directly.

    Based on the datasheet , the minimum VOH was 2.4V when IOH as 4mA.
    Can we say the quaivalent resisitance was around 225 ohm when output high ?
    The maximun VOL was 0.4V when IOL was 4mA.
    Can we say the quaivalent resistance was around 100 ohm when output low?

    please give us your suggestion.
    Thanks

    BR
    Brian Chang
  • 0 in reply to Brian Chang90901
    TI__Expert 5240 points

    Brian,

    Are both F28035 devices in GPIO output mode directly connected?  This is not a good configuration.  Use this instead.

    • Place an external Pull-up resistor on this signal, typically 1k ohm can be used.
    • Keep GPIO's normally not in output mode
    • GPIO can turn to Output drive Low when the net needs to pull down

    Hopefully that addresses the need.

    To answer your other question, the VOH and IOH are minimum values and the output impedance can be stronger (lower) than just a simple calculation.  We don't specify output impedance.  The IBIS model can be used to see the I-V characteristics of the IO, though I'm not sure that is needed in this case.

    Best regards,

    Jason

  • 0 in reply to Jason Whiles
    Guru 30330 points
    Brian,

    Have you been able to resolve this? Jason's suggestions appears to be the best bet in accomplishing this.

    Thanks,
    Mark
  • 0 in reply to Mark Labbato
    TI__Intellectual 2540 points
    Hi Mark and Jason,

    Thanks for your suggestion.
    However, in customer's application , they can not add the pull high resistor at pin because there will be leakage current from their product at connector.
    The final decision was to use different pin and use external MOSFETs to achieve that.

    BR
    Brian