STM32F042 Reading ADC with Analog MUX 74HC4067

Thread Starter

doompeac3

Joined Apr 25, 2019
7
Hello everyone,

I have designed a circuit that will be read 15 analog inputs. Because I use the STM32F042K6T6 as a MCU, I needed to use an analog MUX to read each inputs. I use the ADC peripheral with continuous conversion mode, 239.5 Cycles. And I have written a code to each inputs with 74HC4067 like following:

void Read_ADC_From_74HC4067(void)
{
for ( uint8_t z = 0; z <= 15; z += 1 )
{
//Z'nin degerine göre Kanal secimi yapilacak.
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT0_Pin, ( z >> 0) & 0x01 ? GPIO_PIN_SET : GPIO_PIN_RESET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT1_Pin, ( z >> 1) & 0x01 ? GPIO_PIN_SET : GPIO_PIN_RESET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT2_Pin, ( z >> 2) & 0x01 ? GPIO_PIN_SET : GPIO_PIN_RESET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT3_Pin, ( z >> 3) & 0x01 ? GPIO_PIN_SET : GPIO_PIN_RESET);

//Enable'i Low ve High yapip yukarida secilen kanali ortak çikisa aktariyor.
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT_EN_Pin, GPIO_PIN_RESET);
Temporary_ADC_Value[z] = HAL_ADC_GetValue(&hadc);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT_EN_Pin, GPIO_PIN_SET);
}

}

But the problem is that when I use this reading method, I face several problems such as wrong results. However, when I read a specific output by setting selecting pins like following, I have no problem.:

HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT0_Pin, GPIO_PIN_RESET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT1_Pin, GPIO_PIN_SET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT2_Pin, GPIO_PIN_SET);
HAL_GPIO_WritePin(GPIOA, ADC_CH_SELECT3_Pin, GPIO_PIN_RESET);

//Enable'i Low ve High yapip yukarida secilen kanali ortak çikisa aktariyor.
Temporary_ADC_Value[z] = HAL_ADC_GetValue(&hadc);

It may be a problem about timing or something else? Could you please help me about this issue?

Thanks in advance.
 

Thread Starter

doompeac3

Joined Apr 25, 2019
7


Here is the schematic. ADC_COMMON_OUTPUT is connected to MCU's analog pin (PA0). The others are connected to digital output. I have been trying to measure all the pins that you can see at the attachment.

The link : https://imgur.com/a/sS9AtSC
MOD: Added that image.E
AA1 30-Apr-19 11.19.gif
 
Last edited by a moderator:

Thread Starter

doompeac3

Joined Apr 25, 2019
7
I have detected that when I decrease "for repetition count" to 3, namely for( uint8_t z = 0; z <= 2; z += 1 ), the problem fades away. I guess , by using this repetition technique, I can not read appropriate values. Which method should I use to handle this problem? For example, If I use a timer interrupt, would I deal with that problem? Thanks in advance.
 

MrChips

Joined Oct 2, 2009
20,156
Have you tried inserting a short delay just before reading the ADC?

Temporary_ADC_Value[z] = HAL_ADC_GetValue(&hadc);

By the way,
z += 1
is used to increment z. That is ok.

You can also use ++z or z++.
++z is pre-increment, i.e. z is incremented first before using z.
z++ is post-increment, i.e. z is used in the expression and then incremented.
 

Thread Starter

doompeac3

Joined Apr 25, 2019
7
Yes, I have tried to insert a short delay, but the situation was not different.

By the way, thank you for your briefing. Actually, I know it is plausible to use z++ or ++z , I just want to use z += 1 because I just have learnt that. I want to stick on my mind.


Have you tried inserting a short delay just before reading the ADC?

Temporary_ADC_Value[z] = HAL_ADC_GetValue(&hadc);

By the way,
z += 1
is used to increment z. That is ok.

You can also use ++z or z++.
++z is pre-increment, i.e. z is incremented first before using z.
z++ is post-increment, i.e. z is used in the expression and then incremented.
 

MrChips

Joined Oct 2, 2009
20,156
Be aware that each compiler may execute the two differently.
z +=1;
may be compiled as

fetch z
add 1
save z

which would take three instructions and more since z has to be taken from the function stack.

An optimizing compiler is likely to execute z++ in one intrinsic instruction using a CPU register.
(The end result is the same except the latter takes fewer CPU cycles.)
 

Thread Starter

doompeac3

Joined Apr 25, 2019
7
OK. I have applied your suggestions to my code. But, the situation didn't change.


Be aware that each compiler may execute the two differently.
z +=1;
may be compiled as

fetch z
add 1
save z

which would take three instructions and more since z has to be taken from the function stack.

An optimizing compiler is likely to execute z++ in one intrinsic instruction using a CPU register.
(The end result is the same except the latter takes fewer CPU cycles.)
 

MrChips

Joined Oct 2, 2009
20,156
As I said, this has nothing to do with your problem and will not change your situation.

As a test, I tried all three versions:

z += 1
++z
z++

on my compiler (IAR EW) with optimization turned off and got the exact same assembly code (using a CPU register).
 
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