Determination of the ENOB in 56F8367

[Charlie]

The effective number of bits (ENOB) in the 56F8367 chip is listed at 9.6 bits.

Now, I know that this number is determined by testing and the frequency of the input signal (and sampling rate) both have an affect on the ENOB calculation.

I am working on a fundamental frequency of only 60Hz. I am relatively new to DPS but I know that this frequency is very low. So I was wondering how much, if at all, I can consider my frequency "DC" for this chip or how much such a low frequency will have an effect on the ENOB.

If the test was performed on frequency in the kHz range, I can't imagine that I would be getting a higher ENOB than 9.6.

So does anyone know what test conditions they used to calculated or measure this number?

[xiangjun_rong · Accepted solution]

2.4 How to calculate ENOB

Resolution is an important specification for an ADC, it describes the minimum ADC recognition for the analog input variation. For example, the ADC is 12bit, we say the resolution of the ADC is 12bit. But for an actual ADC, because of various reasons, it can not recognize the analog input variation as it’s resolution describe, so we should define the ENOB parameter to describe the it’s actual resolution.

    2.4.1 how to derivate ENOB parameter

 From mathematical view, quantization process is to map the continuous analog signal to a discrete definite number set, quantization error is introduced in the process, we can use the quantization error to measure the difference between the actual analog value ax(k) corresponding one ADC sample and approximated digital value {dx(k)},the difference is expressed as d{dx(k),ax(k)},the commonly used quantization error is the root-mean-squared error defined as d{dxk,axk}=( dxk,axk)2.

For a general ADC, we assume the analog signal is Uniformly Distributed, so the uniform quantization is the optimum quantization.

For a general ADC, supposing that the analog signal ranges from -V to V, the nominal resolution is L bit, the desired number of level is 2L,the quantization gap is 2*V/2L,Supposing the sample frequency is Fs, and the signal frequency is F, for meeting the Shannon  sampling theorem, the Fs should be greater that 2*F.

Supposing we quantize SINE waveform signal with the ADC, the amplitude of the Sine signal is V.

In an ideal ADC converter, the quantization error is uniformly distributed between -1/2LSB and 1/2LSB,

The sum of root-mean-squared Quantization error (or the energy of the quantization error) or the noise is

In the above formula, x denotes the quantization error between the continuous analog signal and the digital sample, p(x) is the possibility of the quantization error assumed to be uniformly distributed, so

    and   =2*V/2L

so the noise energy is

For a sine signal with amplitude V, the signal energy is ,

So 10*log(Signal energy/Noise energy)=10*log(/J)=6.02L+1.76

So the ENOB L={10*log(Signal energy/Noise energy)-1.76}/6.02.

If we know the signal energy and the noise energy,using above formula we can get the ENOB L.

2.4.2 one method to compute ENOB parameter

  From the formula L=(log(Signal/Noise)-1.76)/6.02. If we can measure the signal and noise energy, then we can get L, L is the ENOB in the data sheet.

  Suppose we input a arbitrary initial phase sine signal to the ADC, which cab be expressed as  A*sin(wt)+B*cos(wt),the amplitude of the sine signal is , assume the frequency of the signal is FREQ, the signal energy can be expressed as , if we sample the sine signal using sample Frequency FS, and let the FS/FREQ=integer for example 100, in order to increase accuracy.

  Assume we get one whole cycle sine discrete signal y(K) by sampling the sine signal, the number of point is NP which equal to FS/FREQ, you can use the formula to get Noise.

   K=NP-1

 J= SUM[ y(k)-A*sin(k/NP)-B*cos(k/NP)]**2

   k=0

we can estimate the optimum parameter A and B, then compute the noise J. let the following partial derivatives equal to 0

d(J)/d(A)=0

d(J)/d(B)=0

we can get the equation

SUM[ y(k)-A*sin(k/NP)-B*cos(k/NP)]*sin(k/NP)=0

SUM[ y(k)-A*sin(k/NP)-B*cos(k/NP)]*cos(k/NP)=0

From the two equation, we can get the A and B value.

       k=NP-1

 Noise energy = SUM[ y(k)-A*sin(k/NP)-B*cos(k/NP)]**2

       k=0

and the Signal energy =(A**2+B**2)/2

From the formula L=(10*log(Signal/Noise)-1.76)/6.02,we can calculate the L which is ENOB defined in data sheet.

We can also calculate ENOB by doing  FFT of y(k) sequence, they should have the same result.