Hi

I know that i can use a voltage shunt regulator for offset voltage (bias, subtract), e.g. 5V.

Can i use also the ADR5045BKSZ-REEL7 ?

I haven an STM32 with 0-3.3 Input. My calculations says that the total current is betwwen 75uA (6V input, 0.3V on ADC) and 825uA (16V, 3.5V on ADC).

Ist this schema correct?

Thank you very much for every input

Regards

Andy

Hello,

without analog subtraction and a properly dimensioned divider, you would map the range 0..16V to your ADC. I assume you ADC has 10bit, so the voltage resolution refered to Vinput would be 16V/2^10=15,6mV.

With subtraction and a proper divider you map a range 11V to the input of your ADC (from Vinput=5V to Vinput=16V). The voltage resolution referred to Vinput is 10,7mV. The improvement in resolution beween both scenarios is a factor 1,5, i.e. less than a factor of 2 (so less than one bit of resolution). The disadvantage is the additional error introduced by the subtraction (though this may be almost negligible for the ADR5045).

The subtraction would help more in a case, where the subtracted voltage is large and the remaining voltage range is small. You have such situations often e.g. in high-side current measurements. But then I would normaly use a diffamp with adequate input voltage range to execute the subtraction, not a voltage reference in series.

"i have calculated with an impedance of 50k for the stm32" Typically the DC impedance for an ADC input will be much higher. I do not have the STM datasheet in front of me, but maybe you calculated the 50k based on the biggest possible leakage current at the pin? Well, the typical leakage current should be much smaller than the biggest possible value, which occurs only at extreme temperature, voltage and process corner. Typically the DC input current is much less important than the pulsed input current which occurs, when the sample and hold stage is switched to the input. The source impedance msut be low enough to charge the S&H-capacitor in time. An external capacitor of some nF will provide the charge for this pulsed current and eliminate this problem.