Thanks.Because the NANO’s internal resistance changes the value of the divider.
As long as the input impedance of the NANO remains constant.Thanks.
Is that means recalibration after connect will be OK?
Thanks,Is the Nano running on 5V or 3.3V? Your difference seems to be exactly the same ratio as 5V/3.3V. (reading 4v vs 2.7V)
Check the reference voltage of the Nano....
ThanksHas the I/O pin on the Nano been set as an input and configured for analogue mode ?
Les.
Thank you DANA.The analog input looks like this -
View attachment 176544
Analog input DC resistance typical in datasheet 100 Meg ohms. Your divider not
affected by its loading. DC......
Regards, Dana.
ThanksIs pin 15 connected to anything?
Rather than asking you how each pin s connected or what your code is doing, I strongly suggest that you post a schematic and your code (in code tags).
You don’t need Vcc/2. I think what Dana has shown is actually internal to the Nano.Thank you DANA.
Found a 22ph, how to get half Vcc then? use two R take the mid?
Do you think the large ohms better? my result shown different?
ThanksYou don’t need Vcc/2. I think what Dana has shown is actually internal to the Nano.
First, the grounds of the Nano and the device from which you’re reading the voltage must be connected.
Secondly, I have to ask because you haven’t shared your code, how are you calculating the voltage? The analogRead() returns a value between 0 and 1023. To get the actual voltage, you have to multiply the number returned by analigRead() by 0.00488. Have you done this?
by Aaron Carman
by Jake Hertz
by Jake Hertz
by Jake Hertz