analog input error calcuation

[david90]

Im trying to calculate the measurement error when using a 4bit resolution input module with a 0-5V range to measure a 5V signal.

What is the measurement error if the voltage at the input channel is 5V?

here is my calculation

5V / 16 (from 4 bits) = 0.3125 voltage resolution

with 0.3125 voltage resolution, the maximum voltage that the input module can represent is .3125 * 15 = 4.6875V.

Therefore the error % is
[(5-4.6875)/5] * 100 = 6.25 %

Is my calculation correct? Also, what do you call the 4 bit digital value that represent the analog input signal voltage?

 

[fou]

Resolution over the full range =

0.3125/5= 6.25 %

The input will see his maximum at 5v not 4.6875V

 

[david90]

can u elaborate?

 

[iant]

the input range of an analogue card in this case is 4 bit
- this means the ratio 0-5v is equil to 0-15 (2^4)
so 15 = 5 v

 

[TConnolly]

Resolution is Range/(2ⁿ-1) where n is the number of bits of resolution. Binary 1111 = 15.

So your resolution is 5/15 or .3333 volts per count.

At full range .3333volts/count * 15counts is 5 volts.

 

[david90]

Resolution is Range/(2ⁿ-1) where n is the number of bits of resolution. Binary 1111 = 15.

So your resolution is 5/15 or .3333 volts per count.

At full range .3333volts/count * 15counts is 5 volts.

why did you divide 5 by 15 instead of 16? Shouldn't u divide 5 by 16 since a 4 bit number has 16 different values (including 0000)?

let say the resolution is 1 bit. If I follow your way, then I would divide 5/1 or 5 volt per count. But the thing is, you are not taking into consideration that a 1 bit number has two possible numbers: 0 and 1.

 

[Mickey]

There are 15 steps or 15 incremental steps with four bit resolution. Zero is not one of them.

Zero to one is one of them, one to two is another and so own...

 

[iant]

I am curious - what are you using for this analogue device

 

[danw]

Range/(2n-1) can be clarified when the intervals are printed out like in the spreadsheet image, but I could only do that for 4 bits, not for 16 bits.

Accuracy:
Assuming the A/D converter converts at exactly the midway point between counts, where the resolution between counts is 0.33Vdc, and that the conversion is not off by 1 count (a typical A/D spec) because the input range is a "high level" signal; then the accuracy would be 0.333/2 = ±0.166V, or ±3.3%

 

[TConnolly]

I divided by 15 because 2⁴-1 = 15.

If you have to then write it all out. As you write it out notice the units and how they algebraically cancel, leaving you with volts as the sole unit remaining.

0 counts * .3333 volts/count = 0 volts.
1 counts * .3333 volts/count = .3333 volts
2 counts * .3333 volts/count = .6666 volts
.
.
.
15 counts * .3333 volts/count = 5 volts.


Now do it with your original 0.3125 volts/count.
0 counts * .3125 volts/count = 0 volts.
1 counts * .3125 volts/count = .3125 volts
2 counts * .3125 volts/count = .625 volts
.
.
.
15 counts * .3333 volts/count = 4.6875 volts.
4.6875 is wrong. If you have a 5 volt range then your scale function should return 5 when the A/D value is 1111.


If that still doesn't help then answer this question:
If you have a stick and you want to divide it into 16 pieces then how many cuts will you make?

 

[iant]

sorry but 15 * 0.3333 = 4.9995


the calculations are at the maximum bit
you are saying that @ 5 volts the number is 1111 (4 Bit) = 5 @ 15 = 5/15 = 0.3333r
number 16 is binary 10000 (5 bit) - that is why you do not use 16 in your calculations

 

[danw]

TConnolly had a typo in his 7:06PM post:

>15 counts * .3333 volts/count = 4.6875 volts.

I'm sure he meant
15 counts * .3125 volts/count = 4.6875 volts.

 

[TConnolly]

Thats what I get for cutting and pasting. Thanks for the catch Dan.

15 * 0.3333 = 4.9995

Ian is a funny guy. And 5/15 isn't .3333 but if I typed out the exact answer this thread would never end.

 

[david90]

so I guess the text that I have reading is wrong or did I miss something?
Here is a copy

Number of Bits of Resolution
The resolution of an A/D operation determines the number of digital values that the
converter is capable of discerning over its range. As an example, consider an analog input
with 4 bits of resolution and a 0-10 volt range. With 4 bits, we will have 16 voltage steps,
including zero. Therefore, zero volts will convert to binary 0000 and the converter will
divide the 10 volt range into 16 increments. It is important to understand that with four
binary bits, the largest number that can be provided is 11112 or 1510. Therefore, the largest
voltage that can be represented by a 10 volt 4-bit converter is 10 / 16 * 15 = 9.375 volts.
In other words, our 10 volt converter is incapable of measuring 10 volts. All converters are
capable of measuring a maximum voltage that is equal to the rated voltage (sometimes
called VREF) times (2n-1)/(2n), where n is the number of bits. Since our converter divides the
10 volt range into 16 equal parts, each step will be 10/16=0.625 volt. This means that a
binary value of 0001 (the smallest increment) will correspond to 0.625 volt. This is called
the voltage resolution of the converter. Sometimes we refer to resolution as the number
of bits the converter outputs, which is called the bit resolution. Our example converter
has a bit resolution of 4 bits. It is important to remember that the bit resolution (and voltage
resolution) of an A/D converter determines the smallest voltage increment that the
converter can determine. Therefore, it is important to be able to properly specify the
Chapter 7 - Analog I/O
7-3
converter. If we use a converter with too few bits of resolution, we will not be able to
correctly measure the input value to the degree of precision needed. Conversely, if we
specify too many bits of resolution, we will be spending extra money for unnecessary
resolution.

 

[david90]

oh and also check out this link
http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Interfaces/ConvAD.html

The text in this page also says to divide the max voltage range by 2^n to get the voltage resolution.

for me personally, it makes sense to divide by 2^n instead of (2^n)-1