Skip to main content
Visitor II
January 28, 2024
Solved

ADC input resistance STM32C011J4M6

  • January 28, 2024
  • 2 replies
  • 5589 views

Hi everyone,

I am using the STM32C011J4M6 in the SO8N package.

I want to use the ADC on a low frequency signal (around 1kHz). My application needs to be very low power so I decided to put a big (around 100kOhms) input resistance on the ADC pin like we can see in the picture below.

Pierre75_0-1706438661717.png

The problem is that in the datasheet the maximum value of the resistance is 50kOhms.

Pierre75_1-1706438870749.png

I want to make sure that this maximum value is given only for one ADC clock frequency (35MHz in my case) and that if we lower the ADC frequency we can increase the Rain value. Am I right ? So if I use a 8MHz ADC clock, I can put a 100kOhms input resistance.

My second question is about the voltage drop across this resistance Rain. When the Cadc capacitor is charged there will be a current flowing into Rain, so there will be a voltage drop across it.

So my question is, when we increase Rain, we lower the input current in the ADC but we increase the voltage drop across Rain so we increase the error. Is it true ?

Best regards,

 

    This topic has been closed for replies.
    Best answer by Pierre75

    Yes I think I will put a 25kOhms resistance, I can deal with µA.

    It is the voltage across a stator phase, I want to measure the speed of a generator. But the generator is really small so I don't want to consume too much current.

    2 replies

    Super User
    January 28, 2024

    The ADC input has no DC input current/resistance, so why you put a big resistor at the input ?

    +

    when we increase Rain, we lower the input current in the ADC but we increase the voltage drop across Rain so we increase the error. Is it true ?

    Yes. The ADC input is capacitive and switches some 5pF or so to the input pin, every time you start a conversion.

    Pierre75Author
    Visitor II
    January 28, 2024

    I put a big resistor to increase the charging capacitor time (T=RC), so the current to charge the 5pF capacitor will be lowered.

    Super User
    January 28, 2024

    But this will only make the input error bigger, not change the energy you need, to charge 5pF ! 

    Charging a cap to certain dc voltage always need same energy, slow or fast doesnt matter ! (cap energy = 1/2 C*U*U says physics)

    To get lower current consumption , you only can reduce the frequency of conversions and switch off the adc in between.

    Technical Moderator
    January 28, 2024

    Dear @Pierre75 ,

     


    You can refer Our Application Note here

    https://www.st.com/resource/en/application_note/an2834-how-to-optimize-the-adc-accuracy-in-the-stm32-mcus-stmicroelectronics.pdf#page48

    on how to optimize conversion in this case of high impedance Source . 

    Hope it helps you 

    Cheers,

    STOne-32

    IMG_6311.jpeg