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michael_nuaa
michael_nuaa
Associate
April 10, 2019
Question

how can I reduce the ADC interrupt entering time?

  • April 10, 2019
  • 10 replies
  • 2314 views

The project is using MCU stm32f103C8T6 for controlling. Here we use the peripheral ADC for current sampling, and we need the time from the interrupt happening to the interrupt executing less than 2us. But the codes following spending about 6us. How can I reduce the time to 2us?

void ADC_Conf(void)

{

  ADC_InitTypeDef ADC_InitStructure;

  GPIO_InitTypeDef GPIO_InitStructure;

  NVIC_InitTypeDef NVIC_InitStructure;

  RCC_APB2PeriphClockCmd(RCC_APB2Periph_ADC1, ENABLE);

  GPIO_StructInit(&GPIO_InitStructure);

  GPIO_InitStructure.GPIO_Pin = GPIO_Pin_2;

  GPIO_InitStructure.GPIO_Mode = GPIO_Mode_AIN;

  GPIO_Init(GPIOA, &GPIO_InitStructure);

  ADC_DeInit(ADC1);

  ADC_StructInit(&ADC_InitStructure);

  ADC_InitStructure.ADC_Mode = ADC_Mode_Independent;

  ADC_InitStructure.ADC_ScanConvMode = DISABLE;

  ADC_InitStructure.ADC_ContinuousConvMode = DISABLE;

  ADC_InitStructure.ADC_ExternalTrigConv = ADC_ExternalTrigConv_T4_CC4;

  ADC_InitStructure.ADC_DataAlign = ADC_DataAlign_Left;

  ADC_InitStructure.ADC_NbrOfChannel = 1;

  ADC_Init(ADC1, &ADC_InitStructure);

  ADC_RegularChannelConfig(ADC1, ADC_Channel_2,1,ADC_SampleTime_13Cycles5);

  NVIC_InitStructure.NVIC_IRQChannel = ADC1_2_IRQn;

  NVIC_InitStructure.NVIC_IRQChannelPreemptionPriority = 2; // AD中断优先级最高

  NVIC_InitStructure.NVIC_IRQChannelSubPriority = 0;

  NVIC_InitStructure.NVIC_IRQChannelCmd = ENABLE;

  NVIC_Init(&NVIC_InitStructure);

  ADC_ClearFlag(ADC1, ADC_FLAG_EOC);

  ADC_ITConfig(ADC1, ADC_IT_EOC, ENABLE);

  ADC_ExternalTrigConvCmd(ADC1, ENABLE); 

  ADC_Cmd(ADC1, ENABLE);

}

void ADC1_2_IRQHandler(void)

{

  if (ADC1->SR & 0x02)

  {

      ADC1->SR = ~(uint32_t)0x02;

      adc.adc_samples[0] = _IQ16toIQ(ADC_GetConversionValue(ADC1));

  }

}

This topic has been closed for replies.

10 replies

S.Ma
S.Ma
Principal
April 10, 2019

Compiler options and maybe do the _IQ16tolQ outside the interrupt.

In debug mode, look at the assembly code to understand what's going on.

Interrupt priority vs others and what is the SYSCLK MHz?

michael_nuaa
michael_nuaaAuthor
Associate
April 10, 2019

11.6 Channel-by-channel programmable sample time

ADC samples the input voltage for a number of ADC_CLK cycles which can be modified using

the SMP[2:0] bits in the ADC_SMPR1 and ADC_SMPR2 registers. Each channel can be

sampled with a different sample time.

The total conversion time is calculated as follows:

Tconv = Sampling time + 12.5 cycles

Example:

With an ADCCLK = 14 MHz and a sampling time of 1.5 cycles:

Tconv = 1.5 + 12.5 = 14 cycles = 1 μs

​

according to 11.6

Tconv = 13.5 + 12.5 = 26 cycles / 12M = 2.17us​

Uwe Bonnes
Uwe Bonnes
Chief
April 10, 2019

Putting the Interrupt Routine into RAM may also spare you the Flash load time.

AvaTar
AvaTar
Senior III
April 10, 2019

Avoid the interrupt, and use DMA. Collect a number of samples, which you get with the DMA_TC interrupt.

Or, if you need the current value for cycle-by-cycle control, you can try the ADC continuous mode and read the ADC.DR register directly. Not sure if the jitter is acceptable to your application.

S.Ma
S.Ma
Principal
April 10, 2019

Interrupt vector in RAM as well as the ISR.

DMA for ADC normal channels usually reduce the ISR frequency and latency.

Tesla DeLorean
Tesla DeLorean
Guru
April 10, 2019

Reading the ADC will clear the status. If the only source you could skip reading the SR and just read the DR.

Both ISRs and vector table in RAM, as much other code as you can sustain also.

Avoid subroutine calls. Check code actually generated and optimization levels.​

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michael_nuaa
michael_nuaaAuthor
Associate
April 12, 2019

to @KIC8462852 EPIC204278916

SYSCLK frequency is 72M

I did a further test, first, I trigger the ADC by software and count the time, which is 2.45us, by setting GPIO,

second, I use ExternalTrigConv_T4_CC4 signal to trigger the ADC and count the time. which is 3.48us , from TIM4->CNT == TIM4->CCR4 to entering interrupt.

michael_nuaa
michael_nuaaAuthor
Associate
April 12, 2019

also, if I use high optimization the 3.48us will be shortened to 3.28us​

S.Ma
S.Ma
Principal
April 12, 2019

Are you using interrupt to deinterlace the ADC array samples? Or no DMA channel available? Save RAM?

Look at the ASM code on the debugger.

How is the array C declared?

As clive said, you don't need to clear the SR bit, it's done automatically by reading the DR.

michael_nuaa
michael_nuaaAuthor
Associate
April 12, 2019

to @KIC8462852 EPIC204278916

When I test the time of other interrupt entering , I get 111ns. so I think it is the ADC converting time which takes 2.45us which is OK for me.

thank you everyone for helping

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