calculating a sampling frequency for analog input signal

I think it is better (and may be also easier) to program one of the Timers to start the ADC12 conversion every 1/500 of a seconds. 

Yes ,I will use the timer later but as of now I am planning to use the below

void main(void)

{

ADCchannel settings and etc 

while(1)
{
for (i=0;i<500;i++)
{
ADC_Result[i] = ADC12MEM0;
}
if(i=500)
{ i=0; }
}

}

1 ADC conversion is made in around 2ms making a collection of 500 samples in 1 sec.After collecting 500 samples in 1 sec the ADCresult will start again to collect the converted ADC values.

Will this works for collecting 500 samples in 1 sec or need to switch to timer implementation.

Thanks old_cow_yellow for the information,

When conversion results are written to a selected ADC12MEMx, the corresponding flag in the ADC12IFGx
register is set.To get copied with correct converted values instead of the copies of the same conversion  can be made by checking the sate of ADC12IFGx(i.e.,  set or not)

This will result in a problem of getting lesser samples <500 in 1sec if the ADC takes more than 2ms for a single conversion as the expected was 2ms for a single ADC conversion.

Isn't it true and out to get the correct samples of around 500  so that the ADC conversion will be done in 2ms.

Is there a way?

You shall read datasheet of your chip, "ADC Timing Parameters" where you can read that slowest allowed ADC12 conversion clock is 0.45 MHz. So you simply can't get 500Hz samplerate using just slow ADC clock. As old_cow already suggested - you shall use timer instead.

You can set up the ADC12 to finish each conversion within 0.1 msec or less. Do not let it start another conversion right after it finishes the current one. Use a Timer to start a new conversion every 1/500 of a second (i.e. every 2 msec)

Using timer  to generate interrupt after every 2ms =>sampling rate of 500hz=2ms and 1 ADcsample in 1.2ms

void main(void)
{
    WDTCTL = WDTPW + WDTHOLD;                 // Stop WDT
    PXSEL |= 0x0X;              // using Px.x as an input to channel x of ADC.
	
	/*---time for 1 ADC conversion=32K/(2*13+13)=820hz=1.2ms--*/
/* Initialize ADC12_A */
    ADC12CTL0 = ADC12ON+ADC12SHT0_2+ADC12MSC; // Turn on ADC12, set sampling time 2*13 conversion cycles
                                              // set multiple sample conversion,
    ADC12CTL1 = ADC12SHP+ADC12CONSEQ_2+ADC12SSEL_1; // ACLK=32KHz ;Use sampling timer(Pulse sample mode), repeat signle channel mode
    ADC12CTL2 =ADC12RES_2 ;//12-bit (13 clock cycle conversion time)
    ADC12MCTL0 = ADC12INCH_X;//selecting the ADC channel 3
    ADC12IE = 0x01;                           // Enable ADC12IFG.0
    
	/*timer will run for 32KHz=>64*32=2ms*/
	TAR=0;
	TACCR0 = 64- 1;  //ACLK @32KHz=>64*32=2ms
	TACCTL0|=CCIE;//ccro interupt enabled
	TACTL = TASSEL_1 + MC_1 + TACLR;  //SMCLK - up_mode - clear the counter

__bis_SR_register(LPM4_bits + GIE);           //enable interrupts
__no_operation();                             // For debugger
}                   // Start conversion

// Timer A0 interrupt service routine
#pragma vector=TIMERA0_VECTOR
__interrupt void Timer_A (void)
{
  ADC12CTL0 |= ADC12ENC;                    // Enable conversions
    ADC12CTL0 |= ADC12SC;                     // Start conversion
}
#pragma vector=SD16_VECTOR
__interrupt void ADC12ISR(void)
{ 
  sample = ADC12MEM0;  //get the data and clears the interrupt flag
  ADC12CTL0 &= ~ADC12ENC;                     //disable the ADC
 }

It works as follows :

The timer starts counting from 0 to 63 making 2ms in the mean while it will jump to the ADC12ISR where the converted value in ADC12MEM0 is copied to sample and ADC is disabled and waits untill the timer completes 2ms to start the next conversion of ADC.

Will this work as above stated or is there anything I'm missing?

thanks

Hello Ilmars ,

I don't have the board right now with me as it will arrive next week.So,I started learning and writing the things I need in the mean while.

Thanks for suggesting a good point of verifying the sample rate using GPIO by scope.

If I want to make the above code to run for continuously acquiring 500 samples in 1 sec and resetting the values after 1 sec can I do the following

Void main()
{
ADC settings(1ms)
Timer settings(2ms)
while(1)
{
      for(i=0;i<500)
     {
     if (TACCR0==63)
        {
             Result[i]= ADC12MEM0;
              i++;
         }
     if (i==500)
       { i=0;}
      }
}
}
ISR for timer
ISR for ADC

I 'm confused in writing this as only verifying the condition of TACCR0==63(or TACCR0==0?) will satisfy the above job or wondering about the CCIFG condition.
And also I guess after the timer has done with its counting of 64 ticks making 2ms only the MEMo will be copied like it will continue copying all 500 samples and starts again.

Can you Suggest if any corrections are to be done.

Thanks.

Hi Ilmars, I just made as you suggested for counting the samples upto 500 samples and then stoppping the process 

I didn't understood this "signal main() code using global (volatile) flag - to process data."

For continuous sampling :I am planning to send the data(i.e, 500 samples) through UART after every 1 sec by inserting the uart code inside the ADC ISR where the index(i) of  result will be made'0'  after sending through UART and TACCR0==0.

Can I implement the continuous sampling like this or do I need to process with two arrays by swapping them.

stefan 12 said:

ISR for ADC

{

while(1)

{

      for(i=0;i<500)

ADC interrupt is called once per each conversion. So you can't put any while() loop waiting for data here as you did in main() routine. ISR is different.

stefan 12 said:

I didn't understood this "signal main() code using global (volatile) flag - to process data."

volatile int signal = 0;

isr() {

 if (buffer == full) {

    stop_timer();
    signal = 1;
} 

}

main() {

while (1) {

  if (signal) {

    process_data();
    signal = 0;
    restart_timer();
  }
 }
}

Hello lmars,

Can you please explain a bit clearly about the ISR(),When this will be called.

And also what is this process_data() and where should I place the result loop of

for(i=0;i<500)

Thanks in Advance.

Thanks old_cow_yellow for the neat and clean explain,

So,this is what I done after getting understood with your explanation

• uisng singelchannelsingle conversion and not uisng the MSC
• ADC cnversion of 0.05msec<0.1 uisng 1Mhz clk with div of 2 and SHT0_2
• Removed Timer ISR
• used ADCIFG0

volatile int signal = 0;
void main()
{
ADC settings with 0.05msec conversion 
Timer setting as 2ms
while (1)
	{
	  if (signal)
	  {
		  for(i=0;i<500;i++)
		       {
				  while(ADC12IFG)
				  {
					  Result[i]= ADC12MEM0;
					  ADC12IFG=0;//clear the interrupt flag after coying conversion result
				  }
		        }
	   }
	signal = 0;
	//restart timer(); 	 }
}
}
#pragma vector=ADC12_VECTOR
__interrupt void ADC12ISR(void)
{
	if (TA1CCR0=0)
	{
		ADC12CTL0 |= ADC12SC; 	}
	if (i == 500)
	{
		TA1CCR0=0;//stop_timer();
	    signal = 1;
	}
}

 But I am not clear of where to place the ADC conversion start so that it will start converting again after the timer has done with 2ms and also how to restart timer as Illmar suggested.

How to use sample-and-hold-select ADC12SHS to use one of the Timers?

Thanks.

 Ilmars here is the required thing I guess

main()
{
while (1)
	{
	  if (signal)
	  {
      Process_data();
	   }
	signal = 0;
	//restart timer();
	 }
}
#pragma vector=TIMERA0_VECTOR
__interrupt void Timer_A (void)
{
  ADC12CTL0 |= ADC12ENC;                    // Enable conversions
    ADC12CTL0 |= ADC12SC;                     // Start conversion
}
#pragma vector=ADC12_VECTOR
__interrupt void ADC12ISR(void)
{
	 while(ADC12IFG)
	{
	Result[i]= ADC12MEM0;
	ADC12IFG=0;
	 }
	i++;
	if (i == 500)
	{
		TA1CCR0=0;//stop_timer();
	    signal = 1;
	}
}

As you said in last post "process_data() - function where you do what you want to do with 1 second of data you intend to collect." doesn't this mean to send data through uart?

And how to make the timer restart the in the code //restart timer();

Thanks for reminding again and again and here I got one doubt

There are two ways to control the timer I guess as

to stop  1)TA1CTL = MC_0; and  TA1CCR0=0;

to restart 1) TA1CTL = MC_1; and 2) TA1CCR0=1;

What is the difference and which one to be used here.

Hi Jens thank for providing such a good information 

you said "IRC, 200kHz is the minimum for reliable results."

You mean the minimum  clock required for 1 ADC conversion should be >200Khz? 

If I am not wrong then CC4305137  has the maximum  clock  ADC12CLCK =1MHz I guess.If so

we need the ADC conversion to be >200Khz but the max can be calculated as 1Mhz/(13 cycles conversion time)=76Khz.

How to get >200Khz ADC cconversion clock?How to solve the charge dissipation problem.

And  what about this

to stop  1)TA1CTL = MC_0; and  TA1CCR0=0;

to restart 1) TA1CTL = MC_1; and 2) TA1CCR0=1;

Thanks.

stefan 12 said:

1)TA1CTL = MC_0; and  TA1CCR0=0;

stefan 12 said:

to restart 1) TA1CTL = MC_1; 

stefan 12 said:

to restart 2) TA1CCR0=1;

One could summarize: in up mode (MC_1) CCR0 needs to be >0 to have the timer counting and generating interrupts.
But only stop mode (MC_0) will switch the timer off and preserve power. Independent of CCR0 value.

stefan 12 said:

You mean the minimum  clock required for 1 ADC conversion should be >200Khz?
If I am not wrong then CC4305137  has the maximum  clock  ADC12CLCK =1MHz I guess.If so
we need the ADC conversion to be >200Khz but the max can be calculated as 1Mhz/(13 cycles conversion time)=76Khz.

1MHz or 1.2MHz (depends on MSP) is the maximum for the SD16 ADC.
The AD12 in the CC5137 is specified for 450kHz (not 200 on this MSP) to 5MHz operation with external reference. See Datasheet. For using the internal voltage reference with external buffering, the upper limit is 4MHz and without external buffering it is 2.7MHz. Above or below, the ADC won't be able to provide the specified performance and linearity.
ADC12OSC is ~4.8MHz typically. SO without external buffering of the internal reference, it should be used with a divider of 2.

For the timing, there is a certain minimum time required to charge the sampling capacitor (>1µs). Minimum SHTx setting is 4 clock cycles , so conversion takes at least 17 cycles. So without external buffering of the internal reference, you will be limited to 141kHz. With external reference, 294kHz conversion frequency are possible. More, if you don't care for degraded precision/linearity.

It's all in the datasheet. :)

Hello old_cow_yellow,

I just used the code you posted with my ADC12 channel being P2.2 (A2) with slight modifications like

 //P1SEL = BIT6 + BIT1 + BIT0;

//P1DIR = BIT7 + BIT6 + BIT1 + BIT0; P1DIR = BIT7;

//P2SEL = BIT0; //P2DIR = BIT0;

  P2SEL = BIT2; instead of P6SEL = BIT3;

ADC12MCTL0 = ADC12INCH_2;

It was successfully built,debugged and running continuoulsy.

Can I see the ADC conversion (ADC12MEM0) results in the debug window(as it is being constant)  or should I use the hyperterminal by uart to view the results?

As I said before, there are over 100 variations of MSP430 with ADC12.  There are differences between them, and I still do not know which one you are using.

Tell me which MSP430 variation you are using. I can tell you what needs to be modified.

The code I showed you is for MSP430F5419A. I was using the default clock settings. If you run it and stop it shortly afterwards, the most 100 results are saved in SRAM in the circular buffer results[0 ...99]. The oldest result is in results[oldest].

You can also insert you own code to analyze the most resent 100 results at where I marked to do so. You have to finish that task well within 2 msec. Otherwise results[oldest] will be replaced by the newest. You may have to speed up the MCLK to accomplish that in time.

Sorry old_cow_yellow for not mentioning the one I am using and  it is CC430F5137.

Can I know What are the changes to be made for this.

old_cow_yellow said:

I still do not know which one you are using.

Thanks old_cow_yellow and when I worked with the above code with replacing the

P2SEL = BIT2; and ADC12INCH_2

And used a hyperterminal output for verifying it as below

#pragma vector=ADC12_VECTOR
__interrupt void ADC12_ISR(void)
{
	ADC12IV;
	 //ADC12CTL0 |= ADC12SC;
		 r= ADC12MEM0;
    while (!(UCA0IFG&UCTXIFG));             // USCI_A0 TX buffer ready?
	UCA0TXBUF = 97;
	_delay_cycles(100);

 //results[oldest++] = ADC12MEM0; 
if (oldest >= Size)
  {
    oldest = 0;
    filled = 1;
  }
  if (filled)
  {
    __bic_SR_register_on_exit(CPUOFF);
  }
}

 Otuput:

1)Using the line "[oldest++] = ADC12MEM0;"  it is showing "error:unable to read" for ADC12 in the register list.

2)Not using (Commenting) this "[oldest++] = ADC12MEM0;" its running and is displaying a continuous alphabet a on the terminal window.

Don't know why the line  "[oldest++] = ADC12MEM0;" is making the ADC unreadable.

Can someone suggest a solution.

thanks.

Hi,

If you run the code I sent you without any change and stop it after 2 seconds or longer. Can you display the values of the variables oldest, filled, and results[0...999]? What are their values?

Which debugger are you using? After your change, what baudrate are you using for the UART? Did you get aaaaaaaaaa...?

You said the line "[oldest++] = ADC12MEM0;" Is it actually "results[oldest++] = ADC12MEM0;"?

-- OCY

Hello old_cow_yellow,

1)Running the code without any changes. Before stopping it ,the result is

2) Debugger Using is MSPez430u.Baudarte of 9600 in the uart programme and terminal baudrate also 9600.Yes ,the termianl window is showing aaaaaaa.

3)Sorry for not writing fully and yes it is results[oldest++] = ADC12MEM0;"?

I hope you can figure it out.

Thanks.

I do not use CCS (I think it stinks). But in this case, I have to give it some credits. It is telling some almost readable explanation. It says “Error; Could not read 0x023D0; Execution state prevented access” etc. I took that to mean “I cannot run and chew gum at the same time”.

I anticipated that most debuggers couldn’t do both at the same time. That is why I told you to start to run the code, stop it after a few seconds, and then display the variables after it is stopped. I did not anticipate that it tries to display the variable while running and complains that it could not do that. (I am glad that it did not swallow the gum.)

Is it possible to start to run the code -- without displaying the variable, without single-stepping, without any break-point, without trying to do anything else at the same time?

Is it possible to stop running after a few seconds?

Is it possible to display the variablea after it is already stopped?

Please give it another try. If that still does not work, I will try to write the code to send the result to a Hyperterminal. Is it possible to use 19200 b/s or higher speed?

Thanks old_cow_yellow and here is required information

1)Yes,It is possible to run the code without doing anything at the sametime.

2)When it is stopped after a few seconds it is automatically switching to the CCSedit perspective window.

3)I guess it is not possible as it swtiching to the CCs edit window may be its possible to see the variable reults on the hyperterminal.

4) Yes, it is possible to use >19200b/s using the SMCLK= 1Mhz as mentioned in the user(slau259e.pdf)  upto a max of 115200 with UCBRx=8.

And also the diffrence between the below two

main()
{
 while(1)
    {
      __bis_SR_register(CPUOFF + GIE);
     }
}
ADCISR()
{
   ADC12IV;
   ADC12CTL0 |= ADC12SC;         // Start conversion
   //result=ADC12MEM0;

   result=597; 

    uart_send(result);
}

It is giving the ADC result value of 597 on the terminal window continuoulsy.

main()
{
 while(1)
    {
      __bis_SR_register(CPUOFF + GIE);

      result=597; 

      uart_send(result);
     }
}
ADCISR()
{
   ADC12IV;
   ADC12CTL0 |= ADC12SC;         // Start conversion
   //result=ADC12MEM0;
    //uart_send(result);
}

This is running in the debugger window but is not displaying the ADC result value(597) on the hyperterminal even waiting for a long time also.

I guess this will give you some idea.

Thanks.

Hi  old_cow_yellow ,

It is not going into the while loop I guess.

Can I know what is the reason.

thanks.

In both cases, the inclusion of ADC12CTL0 |= ADC12S: is wrong.

Running the code and showing the CCS  "terminal window" at the same time can potentially mess up the code. When the code is running, you should not do anything else. And that include showing the CCS terminal window.

Your ISR never uses _ _bic_SR_register_on_exit(...); so the code after _ _bis_SR_register(...); can never be executed.

I do not use CCS. I cannot help you.

Thanks old_cow_yellow for your valuable help.

Can someone help me out in figuring out the problem as why the ISR never uses _ _bic_SR_register_on_exit(...) and why the code after _ _bis_SR_register(...); is not executing.

Shouldn't I use the ADC12CTL0 |= ADC12SC.

what is the prblem with the code.

stefan 12 said:

hy the ISR never uses _ _bic_SR_register_on_exit(...)

stefan 12 said:

why the code after _ _bis_SR_register(...); is not executing.

Entering LPM is not a 'stop CPU until there is an interrupt' but rather a 'disable CPU for anything except ISR execution' action.

Hi Jens,

Now I used the __bic_SR_register_on_exit(CPUOFF) as shown

main()

{

ADC and UART settings

while (1)
  {
  	result=333;
    conv(result);

  __bis_SR_register(CPUOFF + GIE);   }
}
#pragma vector=ADC12_VECTOR
__interrupt void ADC12_ISR(void)
{
	ADC12IV;
	//ADC12CTL0 |= ADC12SC;    // Start conversion
	//result=ADC12MEM0;
	//result=567;
    //conv(result);
    __bic_SR_register_on_exit(CPUOFF);
}

Now it is running and showing the value 333 only once on the hypertermianl and not displying anything on the termianl after first diaply of 333 .

And when I use both like to display 333 in while and 567 in ADC ISR it is only displaying 567  at a very faster rate but not the 333(the baudaret used for UART is 9600) but as per the code posted by old_cow_yellow says

 while (1)
  {
    __bis_SR_register(CPUOFF + GIE);
// Do data processing here
// The oldest result is in results[oldest]    
    P1OUT &= ~BIT7; //new result processed
  }
}

If displaying a value of 333 on terminal is my "Do data processing here" won't it show 333 or is this a problem with the UART settings of 9600.

Will increasing the baudrate will solve this problem or is the timer isn't doing the job of 2ms.

Thanks in advance.

stefan 12 said:

And when I use both like to display 333 in while and 567 in ADC ISR it is only displaying 567  at a very faster rate but not the 333

your conv() function takes so much time that after sending the '567', another ADC result is already there, so the ISR is entered right again. And main() has no time to do anything, let alone sending another time-consuming '333'. Not to mention that you cannot call conv() inside and outside ISR as this means that the ISR could interrupt execution of conv() and call conv() again while it is already being called by main. I doubt that conv() is reentrant.