MPC5777M SIUL2 IO config

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MPC5777M SIUL2 IO config

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bwp530
Contributor IV

I have some questions for this problem:   https://community.nxp.com/thread/500624#comment-1140494 

when I set SIUL2 MSCR_IO register like this:

/* Configure pad PD14 for AF1 func: LIN2TX */
SIUL2.MSCR_IO[62].B.OERC = 3; /* Pad PD14: Maximum slew rate */
SIUL2.MSCR_IO[62].B.ODC = 2; /* Pad PD14: push-pull Output */
SIUL2.MSCR_IO[62].B.SSS = 2; /* Pad PD14: Source signal is LIN0_TXD */

but actually the result is not right as follow:

1.png

So I must set this register like this:

/* Configure pad PD14 for AF1 func: LIN2TX */
SIUL2.MSCR_IO[62].R = 0x32000002;

/* Configure pad PD15 for LIN0RX */
SIUL2.MSCR_IO[63].R = 0x0080000; /* Pad PD15: Enable pad for input - LIN2_RXD */
SIUL2.MSCR_MUX[850-512].R = 1; /* LIN2_RXD: connected to pad PD15 */

3.png

when I see MSCR_IO register table, I find a reserved bit before OERC.

2.png

My question is

1. how can I know which register can support bit setting or not? Reading reg bit can work well?

2. how does the reserved bit works when seting 0x32000002, Why OERC becomming 3?

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PetrS
NXP TechSupport
NXP TechSupport

Hi,

you can use bit access instruction, but it depends how it is compiled. 8bit and 16bit access to SIUL registers are not supported, thus if for example byte instruction is used to write/read selected bitfield nothing happen or error is generated.

If word instruction is selected then write/read operation using bit access is functional.  

Writing to reserved bits have no effect.

By setting OERC=3 a very strong drive is used for the output pin. You can set it according your needs.

BR, Petr

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