topic Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem in Sensors

MM9Z1_638 Microcontroller GPIOB Input Set Problem

mfatihkoseoglu — Tue, 03 Feb 2026 21:32:35 GMT

Hello All,

I'am programming MM9Z1_638 microcontroller. I set output and input PORTA, but i don't set input GPIOB.

I wrote this function:

void GPIOB_Init(void)
{
B_GPIO_CTL = (B_GPIO_CTL_PE1M_MASK |0);
B_GPIO_CTL = (B_GPIO_CTL_PE2M_MASK |0);
B_GPIO_CTL = (B_GPIO_CTL_PE3M_MASK |0);
B_GPIO_CTL = (B_GPIO_CTL_PE4M_MASK |0);

B_GPIO_CTL = (B_GPIO_CTL_PE1M_MASK |0);

B_GPIO_CTL = (B_GPIO_CTL_DIR2M_MASK|0);
B_GPIO_CTL = (B_GPIO_CTL_PE2M_MASK |B_GPIO_CTL_PE2_MASK);
B_GPIO_CTL = (B_GPIO_CTL_DIR3M_MASK|0);
B_GPIO_CTL = (B_GPIO_CTL_PE3M_MASK |B_GPIO_CTL_PE3_MASK);
}

and i wath B_GPIO register and B_GPIO_CTL register is 0x0000.

Why cant i set port and how can i use this port?

Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem

StanoA — Mon, 20 Sep 2021 14:05:24 GMT

Hello Muhammed,

No problem, just note the B_GPIO_CTL register has different bit’s meaning from the commonly used standards.

The *_MASK bits block (“0”) or allow (“1”) to write value into data bit. So if you want to have all port’s bits accessible you need use:

void GPIOB_Init(void)
{
B_GPIO_CTLH = (B_GPIO_CTLH_PE1M_MASK | // enable write to PE1;

B_GPIO_CTLH_PE2M_MASK | // enable write to PE2;

B_GPIO_CTLH_PE3M_MASK | // enable write to PE3;

B_GPIO_CTLH_PE4M_MASK | // enable write to PE4;

B_GPIO_CTLH_DIR1M_MASK | // enable write to DIR1;

B_GPIO_CTLH_DIR2M_MASK | // enable write to DIR2;

B_GPIO_CTLH_DIR3M_MASK); // enable write to DIR3;

// and then write values to low byte of B_GPIO_CTLL register;

B_GPIO_CTLL = (B_GPIO_CTLL_PE1_MASK | // PE1 enabled;

B_GPIO_CTLL_PE2_MASK | // PE2 enabled;

B_GPIO_CTLL_PE3_MASK | // PE3 enabled;

B_GPIO_CTLH_DIR1_MASK | // PE1 as output;

B_GPIO_CTLH_DIR2_MASK | // PE2 as output;

B_GPIO_CTLH_DIR3_MASK ); // PE3 as output;

}

Sorry I don’t have the board with MCU for test, but I hope my suggestion could help you.

Best Regards,

Stano.

Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem

mfatihkoseoglu — Mon, 20 Sep 2021 14:32:39 GMT

Hello StanoA,

I tried the way you said but it is not work.

My init function:

Not: I want to use input GPIOB pins.

void GPIOB_Init(void)
{
B_GPIO_CTLH = (B_GPIO_CTLH_PE1M_MASK | B_GPIO_CTLH_PE2M_MASK | B_GPIO_CTLH_PE3M_MASK | B_GPIO_CTLH_PE4M_MASK
| B_GPIO_CTLH_DIR1M_MASK | B_GPIO_CTLH_DIR2M_MASK | B_GPIO_CTLH_DIR2M_MASK);
B_GPIO_CTLL = (B_GPIO_CTLL_PE1_MASK | B_GPIO_CTLL_PE2_MASK | B_GPIO_CTLL_PE3_MASK | B_GPIO_CTLL_PE4_MASK);
}

After the work, B_GPIO_C register value is 0x0000.

My init function:

Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem

StanoA — Tue, 21 Sep 2021 15:48:35 GMT

Hello Muhammed,

Please reset the MCU in debug mode, set the breakpoint to line “for(;;) { …” and run the code. When it will stop on breakpoint read the variables:

value and value2.

This means select them in window:

And check the values.

Best Regards,

Stano.

Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem

mfatihkoseoglu — Wed, 22 Sep 2021 07:26:58 GMT

Hello StanoA,

I reset mcu in debug mode and set break point in for loop, i see the all values is 0.

Re: MM9Z1_638 Microcontroller GPIOB Input Set Problem

StanoA — Wed, 22 Sep 2021 12:18:46 GMT

Hello Muhammed,

I asked our specialist for help. The result is:

The port bit must be written together with their mask bit = 1.

If the associated mask bit = 0, the write is not allowed.

The masked bits (e.g. GPIO...PE1M and GPIO...PE1) have to be written in a single access. In this case as 16bit write. Please see the attached gpio.h file also for this example:

//-------------------------------------------------

// enable PTB1 as output

B_GPIO_CTL = GPIO_CTL_PTB1_ENABLE | GPIO_CTL_PTB1_OUTPUT;
B_GPIO_OUT1 = OUT_PTBx;

// set PTB1
B_GPIO_OUT1 = OUT_PTBx;
// clear PTB1
B_GPIO_OUT1 = 0;

//----------------------------------------------------

Please see also this AN for this product:

https://www.nxp.com/docs/en/application-note/AN12301.pdf

I hope it could help you with your task.

Best Regards,

Stano.