Maths help for the terminally dim please

[aratti]

In case you will need other percentages, here how to calculate the multiplier:


Multiplier = 1 +( % / (100-%) )

where % is your value.

Example:

1 + ( 16 / (100 - 16) = 1 + (16/84) = 1 + 0.19047619 = 1.19047619 rouded 1.1905

Cheers

Al.

[retepsnikrep]

Thanks for the formula as well. Now i know how to do it

[retepsnikrep]

Another maths question.

I have an adc connected to a HV supply via a non linear isolation circuit.

It produces the below 10bit ADC readings at the given input voltage.
That works fine but i want to create a formula from these data points to allow the adc to produce the actual voltage.
I don't want to use a lookup table if possible.
Accuracy to 1V is acceptable. There should be enough adc resolution to do that with the right formula.

225v = 356
200v = 380
175v = 412
150v = 450
125v = 495
100v = 561
75v = 644
50v = 753
25v = 903
10v = 1000
5v = 1020
0v = 1023

How to convert that data into some sort of forumla to produce the voltage from a given adc reading.
It's not linear.

Thanks

[Acetronics2]

3rd degree Polynomial Fit: y=a+bx+cx^2+dx^3...
Coefficient Data:
a = 7.84284692277E+002
b = -2.42579262304E+000
c = 2.77307340820E-003
d = -1.12332698305E-006

or

Heat Capacity Model: y=a+bx+c/x^2
Coefficient Data:
a = 8.77447057194E+001
b = -1.01937213402E-001
c = 2.18878991436E+007


or use a retrieve table plus 2nd order approx.

Accuracy to 1V is acceptable

did you notice 1v is ... 0.5% ???
surely not !!!

Alain