eblc1388 said:
It would be an handy alternative if one want to reads the AC voltage from say a transformer secondary.
Can you also provide a way/formula to calculate those resistance values so that others will benefit from using this method?
I don't know if I can provide a simple formula, but the way I calculated it is like this.
The first thing to know is that a potential divider produces a voltage, (obviously) and that the impedance of that voltage is the parallel combination of the two resistors (less obvious). So the pull up and the pull down (the 4166.714611
hm: and 7142.857143
hm: produce a voltage of 3.1579 V that has an impedance of 2631.598
hm: . It is just the same as a 3.1579 V source with a 2631.598
hm: in series.
I calculated the 3.1579 V and the 2631.598
hm:, and from that worked back to the 4166.714611
hm: and the 7142.857143
hm:
I chose 10k, so the 10k to 2631.598
hm: divider needs to reduce 24V swing (±12V) needs to 5V, which is how the 2631.598
hm: is decided.
10000
hm: /(24-5)*5 equals 2631.598
hm:
When the input is 0V the output needs to be 2.5V, and this is how the 3.1579 V is decided.
2.5/10000*(10000+2631.598) equals 3.1579 V
Then the 4166.714611
hm: and 7142.857143
hm: can be found because
7142.857143 / ( 7142.857143 + 4166.714611 ) * 5 = 3.1579
and
1/7142.857143 + 1/4166.714611 = 1/2631.598
That actually involves a simultaneous equation, and I'm not going to tell everyone how to do that. However, it is OK to guess a value for the resistors and see what voltage and impedance you get, then modify the guesses. It only takes a few tries to get more accurate guesses than the resistors can be.