Current drawn by on-chip ADC of Microcontroller?

[Flyback]

Hello,

We are reading a voltage rail with the on-chip ADC of the SAM3N00B microcontroller via a potential divider.
We are trying to assess what is the maximum allowable Ohmic value of the upper divider resistor.
The ADC reading will draw some current through this upper divider resistor, so we must not make this resistor too big, otherwise it will affect the potential divider voltage reading.

Do you know what is the current drawn by the ADC when an ADC reading is taken?


SAM3N00B MICROCONTROLLER DATASHEET:
https://www.electro-tech-online.com/custompdfs/2013/05/doc11011-2.pdf

 

[misterT]

The ADC is a Successive Approximation Register (SAR) converter. This means that there is a capacitor inside of the converter that needs to be charged to the input voltage. Most converters let you define the charging time.. if you charge through large resistor, you need longer charging time to get accurate sample. It is usually recommended that you use a buffer op-amp to drive SAR converter, but if your input voltage is not very dynamic (only low frequencies), then it is not that crucial.

I suggest you read couple of articles by Bonnie Baker:
https://www.electro-tech-online.com/custompdfs/2013/05/00246a.pdf
https://www.electro-tech-online.com/custompdfs/2013/05/6602451.pdf

 

[dougy83]

The resistor divider can be modeled as a Thevenin equivalent, so the output resistance is the equivalent resistance of the 'upper' resistance in parallel with the 'lower' resistance. Keep this value within the specified (ADC max source resistance or equiv.) range of the microcontroller and you should be fine.

Basically if the 'upper' resistor is that much larger than the 'lower' resistor, then the source resistance you assume will be very close to the 'lower' resistance value.

 

[Flyback]

o the output resistance is the equivalent resistance of the 'upper' resistance in parallel with the 'lower' resistance

......that sounds like harping back to BJT circuit theory......surely its only the upper divider resistor that charges the internal cap in the ADC?

 

[dougy83]

......that sounds like harping back to BJT circuit theory......surely its only the upper divider resistor that charges the internal cap in the ADC?

I don't see the relation to harps or BJTs.

Why don't you draw the model for a voltage source (including series resistor) connected to the ADC. Now draw the Thevenin equivalent circuit of the resistor divider driving your ADC. What similarities do you see?

 

[Flyback]

the thing is, the ADC input is from a voltage divider, and its basically just DC...all that thev stuff is about AC cct theory?

 

[dougy83]

the thing is, the ADC input is from a voltage divider, and its basically just DC...all that thev stuff is about AC cct theory?

Please refer to https://en.wikipedia.org/wiki/Thévenin's_theorem

 

[Flyback]

i understand thevenin, but this is just a cap inside the adc pin charging up via the upperdivider resistor.

 

[dougy83]

You obviously don't understand Thevenin. Why don't you try drawing the two circuits I described previously?

 

[misterT]

the thing is, the ADC input is from a voltage divider, and its basically just DC..

If it is basically DC, then just put a capacitor after the divider for some filtering and you'll be fine. 100nF or something..