怎样缩短中断进入时间

michael_nuaa 提问时间：2019-4-10 16:42 /

使用芯片stm32f103c8t6

ADC采样配置如下：
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);
}

AD中断服务函数如下：
void ADC1_2_IRQHandler(void)
{
if (ADC1->SR & 0x02)
{
ADC1->SR = ~(uint32_t)0x02;
adc.adc_samples[0] = _IQ16toIQ(ADC_GetConversionValue(ADC1));
}
}

问题：
从发生中断到进入中断服务函数时间过长（不计算中断执行时间），大概在6us左右。
请问各位大神，有没有办法把这个进入时间缩短到2us左右。

赞收藏评论19 发布时间：2019-4-10 16:42

19个回答

michael_nuaa 回答时间：2019-4-10 17:19:22

谢谢 @tanic 提醒
我这边程序里面AD采样时钟为12M
      RCC_HCLKConfig(RCC_SYSCLK_Div1); // AHB 72M
      RCC_PCLK2Config(RCC_HCLK_Div1);    // APB2 72M
      RCC_PCLK1Config(RCC_HCLK_Div2);    // APB1 36M
      RCC_ADCCLKConfig(RCC_PCLK2_Div6); // ADC 12M

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

按照11.6计算的话：
Tconv = 13.5 + 12.5 = 26 cycles = 2.17us

这里还有其他什么原因吗？

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michael_nuaa 回答时间：2019-4-12 12:06:42

问题算是找到了
if (ADC_GetITStatus(ADC1, ADC_IT_EOC) == SET) 占用了较长时间，后面估计要改成汇编了

谢谢各位

if (ADC_GetITStatus(ADC1, ADC_IT_EOC) == SET) //if (ADC1->SR & 0x02)
{
      GPIO_SetBits(GPIOB, GPIO_Pin_5);
   ADC_ClearITPendingBit(ADC1, ADC_IT_EOC);
}

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