STM32F429 ADC problem

[tasabao]

I am currently using STM32CubeIDE and want to input an external sine wave. The signal has a voltage of 2.6V and a frequency of 10kHz, and I am using DMA + TIM + ADC. The following is the screenshot of my settings.

Currently, I can only capture the peak value of 2.6V, and this is only achievable when the ADC Sampling Time is set to 480 Cycles. With a lower number of cycles, I can only get lower values. Please help me resolve this issue, thank you!

DMA:

ADC:

TIM:

Clock Configuration:

Below is my code:

///////////////ADC////////////////

int test = 0;

uint32_t lastTick = 0;

uint16_t ADC_Value[4095] = {0};

float ADC_MaxValue = 0;

float ADC_MinValue = 0;

int count = 0;

double ADC_Value_Sum = 0;

int ADC_Value_AVG = 0;

float clock_AVG = 0;

uint16_t max_value = 0;

uint16_t min_value = 0;

char buffer[20];



HAL_TIM_Base_Start(&htim2);

/* USER CODE BEGIN WHILE */

while (1)

{

if (uart_received)

{

uart_received = 0;



if (uart_command == 'A')

{

check = 1;

HAL_ADC_Start_DMA(&hadc1, (uint32_t*)ADC_Value, 4095);

}

else if (uart_command == 'Z')

{

check = 0;

m = 10000;

ADC_Value_Sum = 0;

memset(ADC_Value, 0, sizeof(ADC_Value));

HAL_ADC_Stop_DMA(&hadc1);

count = 0;

high_time_Sum = 0;

count_phase = 0;

zero_remove = 0;

max_value = 0;

min_value = 0;

}

else if (uart_command == 'S')

{

check = 2;

}

}



if (adc_conversion_complete)

{

adc_conversion_complete = 0;



sprintf(buffer, "Max: %.3f V, Min: %.3f V\r\n", ADC_MaxValue, ADC_MinValue);

HAL_UART_Transmit(&huart1, (uint8_t*)buffer, strlen(buffer), HAL_MAX_DELAY);

}





if (check == 1)

{

if (HAL_GetTick() - lastTick >= 50000)

{

m += 10000;

ADC_Value_Sum = 0;

memset(ADC_Value, 0, sizeof(ADC_Value));

HAL_ADC_Start_DMA(&hadc1, (uint32_t*)ADC_Value, 4095);

lastTick = HAL_GetTick();

}

ad9850_wr(0x00, m);

}

else if (check == 0)

{

ad9850_wr(0x00, 0);

}

else if (check == 2)

{

ad9850_wr(0x00, m);

}

}





/* USER CODE END WHILE */

void HAL_TIM_IC_CaptureCallback(TIM_HandleTypeDef *htim)

{

if(TIM2 == htim->Instance){

if(HAL_TIM_ACTIVE_CHANNEL_1 == htim->Channel){

switch(capture_cnt){

case 1:

capture_buf[0] = __HAL_TIM_GET_COMPARE(htim,TIM_CHANNEL_1);

__HAL_TIM_SET_CAPTUREPOLARITY(htim,TIM_CHANNEL_1,TIM_INPUTCHANNELPOLARITY_FALLING);

capture_cnt++;

break;

case 2:

capture_buf[1] = __HAL_TIM_GET_COMPARE(htim,TIM_CHANNEL_1);

HAL_TIM_IC_Stop_IT(htim,TIM_CHANNEL_1);

capture_cnt++;

break;

default:

break;

}

}

}

}



void HAL_ADC_ConvCpltCallback(ADC_HandleTypeDef* hadc)

{

test = 1;

if (hadc->Instance == ADC1)

{

max_value = ADC_Value[0];

min_value = ADC_Value[0];

for (int i = 0; i < 4095; i++)

{



if (ADC_Value[i] > max_value)

{

max_value = ADC_Value[i];

}



if (ADC_Value[i] < min_value)

{

min_value = ADC_Value[i];

}

}

ADC_MaxValue = ((float)max_value * 3.3f) / 4095.0f;

ADC_MinValue = ((float)min_value * 3.3f) / 4095.0f;

processFFT();



if (m >= 1000000)

{

m = 10000; // Reset frequency to 10 kHz

check = 0;

}



HAL_ADC_Stop_DMA(&hadc1);

lastTick = HAL_GetTick();



adc_conversion_complete = 1;

}

}

[waclawek.jan]

> this is only achievable when the ADC Sampling Time is set to 480 Cycles.  With a lower number of cycles, I can only get lower values.

This usually means, that your input signal has too high impedance. My guess is, that you derive the sine wave using a high-resistance divider. You'll need to use a buffer (an amplifier with amplifiacation 1) before feeding that signal to ADC.

JW

[tasabao]

Hello, thank you for your response. Since the internal ADC forms an RC charging circuit, the time constant (τ) must be used to determine whether the charging is sufficient. By using 3 times the τ constant, I ensure that the capacitor is properly charged to the voltage I want to input.

So, based on the current total resistance Rtotal=RBUF+RADC=5Ω+6kΩ=6005ΩR_{total} = R_{BUF} + R_{ADC} = 5Ω + 6kΩ = 6005ΩRtotal​=RBUF​+RADC​=5Ω+6kΩ=6005Ω, and the table showing CADC=4×10−12C_{ADC} = 4 \times 10^{-12}CADC​=4×10−12, we can calculate τ=Rtotal×CADC=6005×4×10−12=24.02×10−9τ = R_{total} \times C_{ADC} = 6005 \times 4 \times 10^{-12} = 24.02 \times 10^{-9}τ=Rtotal​×CADC​=6005×4×10−12=24.02×10−9 seconds or 24.02ns.

After multiplying by 3 times τττ, the ideal sampling time becomes:

ts=3×τ=3×24.02ns=72.06nst_s = 3 \times τ = 3 \times 24.02 \text{ns} = 72.06 \text{ns}ts​=3×τ=3×24.02ns=72.06ns

Additionally, the ADC clock frequency is 36MHz, and the STM32’s internal ADC has a fixed 12 ADC cycles for internal conversion, so the minimum conversion period would be 15 cycles.

Thus, the ADC’s minimum total conversion time is:

12.4M=416ns\frac{1}{2.4M} = 416 \text{ns}2.4M1​=416ns

Therefore, there should be more than enough time for the conversion.

Is my understanding correct?

[waclawek.jan]

> By using 3 times the τ constant, I ensure that the capacitor is properly charged to the voltage I want to input.

Not really. tau means, that the capacitor is charged to 1/e i.e. cca 1/3 within the target voltage, 3x that is cca 1/10, i.e. cca 10% error (not quite, because the initial capacitor charge depends on its history).

See the input impedance values in datasheet.

JW

[SofLit]

Hello @tasabao ,

In next time, please use </> button to paste your code. I've edited your post then ..