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Hello dexter,

The first thing to do is download the latest version of ZStack.

Keyswitch 6 is actually defined as bit 1 Port 0 in the hal_key.c #define HAL_KEY_SW_6_BIT    BV(1)

I believe the P1_7, P1_6 and P1_5 are being used for SPI and is set as a priority for those pins. 

For testing for HAL_KEY_SW_7, the if (HAL_KEY_SW_7_PXIFG & HAL_KEY_SW_7_BIT) in your interrupt ISR routine should work fine.

I don't think you can set the Interrupt level for both rising and falling edges, I think it has to be one or the other.

There may be a lot of noise on they key and you are getting a lot of key bounce. You would have to change the edge trigger level to the opposite value in the interrupt.

HAL_ISR_FUNCTION( halKeyPort1Isr, P1INT_VECTOR )
{

if (HAL_KEY_SW_7_PXIFG & HAL_KEY_SW_7_BIT)
{
HAL_KEY_SW_7_EDGE & = ~ (HAL_KEY_FALLING_EDGE);// 0 or 1
PICTL |= HAL_KEY_SW_7_EDGEBIT;

   halProcessKeyInterrupt();
}

Hello greenja

Thanks for your reply

So you mean cc2530 can not have a switch function, only button with pressing?

and I still confused about  EDGEBIT and IENBIT, why  EDGEBIT & IENBIT OF HAL_KEY_SW_7  is different from HAL_KEY_SW_6

is there any doc about how to decide EDGEBIT & IENBIT?

And

I do want to use the latest version of zstack

But the latest version of zstack has problem with secure

such as can not join or rejoin in the SECURE mode

if I modify the ZSecMgr.c as the forum told me,  which is about key frame count  

the coordinator will remove any successfully joined router/end device from assocDevList

etc..

So

I may choice a stable verion of zstack

Hello kidd,

I think of a switch and button as being the same thing really and so does the MCU.  The only difference is that a switch can be held in a position permanently while a button is momentary.

You can use the CC2530 for both a switch and button function.

You can find all the information on EDGEBIT and IENBIT in the CC253x/CC254x user guide.  It doesn't really explain anything in detail though and is found here:

http://www.ti.com/litv/pdf/swru191c

Take a look at the ioCC2530.h file for more clarity on the registers and how they are set up.

In general, the EDGEBIT determines if the rising edge (0 -> 1) or falling edge (1 -> 0) is used to trigger an interrupt.  The IENBIT allows an interrupt to happen based on the setting of the EDGEBIT for a particular bit.  That is the reason for the difference.

As for the ZStack, you should post that as another question for a faster answere.

Hello greenja

Thanks for your professional reply

it help me a lot

lol

as for the zstack, I think ti is working on it

so I just need wait

By the way, is there a interrupt or something in cc2530 ,to deal with long time key pressing? such as press like five seconds

or

I still need a timer by my hand to deal with the long time pressing?

And

if (HAL_KEY_SW_7_PXIFG & HAL_KEY_SW_7_BIT)
{
  HAL_KEY_SW_7_EDGE &=~(HAL_KEY_FALLING_EDGE);// 0 or 1

  PICTL |= HAL_KEY_SW_7_EDGEBIT;

  halProcessKeyInterrupt();
} // 

These codes can not be complied by IAR 8.0.3

Error[Pe137]: expression must be a modifiable lvalue D:\Texas Instruments\ZStack-CC2530-2.3.0-1.4.0\Components\hal\target\CC2530EB\hal_key.c 559

And

The default key ,sw1, HAL_KEY_SW_6 is not stable with pressing and releasing,too 

I think the problem is not just about HAL_KEY_SW_7 

Hello,

The line HAL_KEY_SW_7_EDGE &=~(HAL_KEY_FALLING_EDGE);// 0 or 1 is what is causing the error.  It is a #define statement and cannot be changed.

You can try this:

if (HAL_KEY_SW_7_PXIFG & HAL_KEY_SW_7_BIT)
{
 

  PICTL |=  ~(HAL_KEY_SW_7_EDGEBIT);

  halProcessKeyInterrupt();
} //

The problem with the stability of the HAL_KEY_SW_6 is due to contact bounce. 

You are not really suppose to use a switch that has not been conditioned as an interrupt input, it has to be properly debounced.

There is a software debounce used for the keys found in hal_board_cfg.h, you can try extending the time or create another function based on some of the other debounce routines you can find on the internet:

#define HAL_DEBOUNCE(expr)    { int i; for (i=0; i<500; i++) { if (!(expr)) i = 0; } }

There are many ways to debounce the switch and some very good articles.  I can't seem to find them on my system right now.

Hello greenja

baby ,you are so kind

thanks very much

hello greenja

I dont knon if you can see this

I tested today

the results is not good

first, I still confused about HAL_KEY_SW_1 AND HAL_KEY_SW_6

you say HAL_KEY_SW_6 is BV(1) 

so why in the key event

if ( keys & HAL_KEY_SW_1 ) and if ( keys & HAL_KEY_SW_6 )  working true both in the same time?

so what it really is ? SW_1 OR SW_6 ? OR BOTH??

Is this  wired ?

and 

when I　 press button on p1_6

It returned with two key events

 HAL_KEY_SW_1 and  HAL_KEY_SW_7

yeah' that's funny, another working true both in the same time

Cause I thought , because I define  HAL_KEY_SW_7 as BV(6) 

it might be the event of  HAL_KEY_SW_6

now I　really confused

Are these key event associaed with the names of #define?

But 

not true  for HAL_KEY_SW_6, you know, it is  HAL_KEY_SW_1 and HAL_KEY_SW_6

and I　 do not understand,why TI made this wired key 

and other funny thing

I　do the same init on HAL_KEY_SW_6 and HAL_KEY_SW_7

same  EDGE

#define HAL_KEY_SW_6_EDGE     HAL_KEY_FALLING_EDGE

#define HAL_KEY_SW_7_EDGE     HAL_KEY_FALLING_EDGE

WHEN I　press the key on P0_1 which is HAL_KEY_SW_6 

it will happend in my key event ,with if(shift) == true

not the same as HAL_KEY_SW_7

For HAL_KEY_SW_7

it will  be untrue of if(shift)

you know, there is a shift in  KEY event function.

static void GenericApp_HandleKeys( uint8 shift, uint8 keys )

The key problem 

is really killing me.