cancel
Showing results for 
Search instead for 
Did you mean: 

ADC input resistance STM32C011J4M6

Pierre75
Associate II

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,

 

1 ACCEPTED SOLUTION

Accepted Solutions

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.

View solution in original post

7 REPLIES 7
AScha.3
Chief

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.

If you feel a post has answered your question, please click "Accept as Solution".

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

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.

If you feel a post has answered your question, please click "Accept as Solution".

I agree with you, the energy will be the same but the current spike lower if we increase the charging time. And the only way to increase the charging time here, is to increase the input resistance.

To be more precise about my application, I don't need to lower the energy that I consume but the current spikes.

So I need to find a compromise between input current spike and error.

 

STOne-32
ST Employee

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

You do lot of complications to yourself by placing this 100k resistor in series. What a kind of voltage source it is if it doesn't tolerate switching 5pF load ? If you realy need low loading consider using opamp as buffer (some STMs have integrated one).

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.