Hi Community,
I have a hardware question.
In my application i want to sample a signal from hall effect sensor in the windows from 0.5V to 4.5V.
To improve the fidelity of the ADC i want to compress the dynamics of the ADC, and put the Vref @ 4.8V. My question is: My connection in the attached document is right? Or a voltage resistor divider is better?
Thanks to everybody in advance for our reply.
Best Regards
Hi Edward,
yes! If my calculations are correct my configuration should be good.
As i see in the table A-9 of the datasheet the admited VddA is between 2.7 -5.5V with a Delta to Vdd between -0.1 < Vdd < 0.1V.
The formula to found Delta V is: (VDD-VDDAD)^2 then in my case: (5.00-4.8)^2 = 0.04V that is the admitted range.
What you think?
Thank you!
Diego
Sorry Edward,
i don't understand. if i use the Vdd and VDDAD in mV the result of square elevation is mV^2 ...
Can you give me an example?
Thanks
I replied to your previous message via email. Community engine sometimes faills sending from browser, sometimes from email.
I meant that if delta-Vdda-to-Vdd was calculated as you specified (VDD-VDDAD)^2, then specified limits had to be specified in square millivolts units, not in millivolts. Also there could be no negative limit for this. I don't know exactly what authors of S08DZ datasheet meant, maybe they meant RMS or somethibg, but delta is still specified in mV units.
Ok, I have too many doubts. For safety I put the VDDA to 5V.
Thank you very much for your support Edward!
Best Regards.
Diego