Meaning of data from AMC1210?

And one more item of note...

Mathieu De Zutter said:

Should I only use a OSR of 256 for the AMC1210 because the AMC1203 works with a OSR of 256?

The AMC1203 works with any oversampling ratio. If you want to maximize the effective number of bits or resolution of your conversion results you should be operating at an oversampling ratio of 256. This does sacrifice your data-rate, though, since it takes more conversions to get you a result.

Additionally, this gets into the philosophy of your data-acquisition system. If you'd like to see a conversion result that represents the average value of a signal that is typically very stable a higher OSR makes better since. If you're measuring a volatile signal that you need to know the instantaneous value of with high frequency use a lower OSR.

Mathieu,

The real trick here is to determine the maximum bit width of the conversion result. The first step, then, is to determine the range of values produced by each filter architecture. These are defined by Table 10 in the AMC1210 datasheet. Once we know where the MSB is of the full conversion word we can decide where the 16 bit word LSB is and in turn know how much shift to apply to achieve meaningful 16 bit data.

Sinc¹ : Range = -x to x

Sinc²: Range = -x² to x²

Sinc³: Range = -x³ to x³

Sincfast: Range = -2x² to 2x².

Where x is the oversampling ratio.

With the range in mind we must determine the word width for each architecture and each oversampling ratio. This can be defined by:

Sinc¹ : Width = log₂(x)

Sinc²: Width = log₂(x²)

Sinc³: Width = log₂(x³)

Sincfast: Width = log₂(2x²)

Where x is the oversampling ratio.

Now that we know the maximum word width we must determine where the LSB to know the appropriate shift value...

Sinc¹ : Shift = int(log₂(x) + 0.5) - 15

Sinc²: Shift = int(log₂(x²) + 0.5)  - 15

Sinc³: Shift = int(log₂(x³) + 0.5) - 15

Sincfast: Shift = int(log₂(2x²) + 0.5)  - 15

Where x is the oversampling ratio and 'int' implies integer truncation. Only positive shift values are valid.

It is of note that as the integrator oversampling ratio increases the shift values will also need to be updated...

Hope this helps.

Hi Mathieu,

First, yes you've caught me only glancing at the AMC1203 datasheet, and I specified the full scale range incorrectly. The full scale range, however, should be +/-320mV. This is in the table on page 5 of the AMC1203 datasheet labeled "FSR - Full Scale Differential Voltage Input Range"

We've touched on everything that you need to know to convert the codes to volts, but we haven't really comprehensively tied everything together just yet. In review I wasn't particularly clear that the equation I described before assumed 16 valid bits (15 data bits + 1 sign bit). I'll try to go through the specific case of using a Sinc 2 with OSR 152 and a 110mV differential input voltage applied to an AMC1203, hopefully you can follow along and apply it in the generic case...

A sinc² filter output range is +/- x², as discussed earlier, where x is the OSR. You've chosen an OSR of 152, so we can expect to see results from -23104 to 23104.

Next, we determine the bit-width of the resulting word, log₂(x²), is 14.495855 bits. At this point you should notice that 14.495855 bits (assuming fractional bits could ever exist...) only gets you from 0-23104 and cannot obtain any negative values, so an additional sign bit must also be acquired from the AMC1210 data register (not that we mind in this case since we're forced to get all 16 bits). We know that we'll only see 14.495855 bits of valid data so we should take 16 bits to guarantee the presence of a sine bit (since we're using about half of the codes on both sides of bit 15 - it's going to move). The LSB weight should still be defined by the number of valid bits - not the number of bits we're going to use to fetch our conversion result.

In general terms there are two ways to calculate LSB size for a bipolar binary two's compliment data converter. Either:

Vref/(2^(Bits-1)-1) or 2Vref/(2^(Bits)-1)

But these equations assume the 'Bits' includes the sine bit - which we know our 14.495855 value is already missing. So our equations should look like:

320/(2^(14.495855)-1) or 640/(2^(15.495855)-1)

Using either of these values as your LSB weight on a code of 7890 yields ~109.285mV. To really verify the conversion results you'd want to look at more than a single point & average the values.

This is a challenging device to use - let me know if I need to make another attempt at explaining this...

Hi Mathieu,

Great to hear you have had success with your project. 

Mathieu De Zutter said:

I'm reading the 32-bit dataregisters with SPI. The result I get from the AMC1210 I multiply by a factor 0.320/(OSR³-1)  (sinc³-filter). But I've never wondered where the substraction by 1 come from? Can you explain this te me?

The equation you have here is a little un-conventional. For the sake of this explanation's relevance for you in things beyond the AMC1210 and oversampling lets look at the more generic form of the equation to calculate LSB weight for a bi-polar data and uni-polar data...

Bi-polar

LSB_weight = 2V_ref / ( 2^bits - 1 ) or LSB_weight = V_ref / (2^(bits-1) - 1)

Uni-polar

LSB_weight = V_ref / (2^bits - 1)

In the bi-polar case for a typical system we can resolve voltages from - V_ref to V_ref and in the uni-polar case we typically can only resolve voltages from 0 to V_ref. Bi-polar results are usually in the binary twos-complement format, so our equations have to be modified to compensate for the MSB being used for sign. If you compare the two equations I've given for the bi-polar case you'll notice they're more or less the same value for large number of bits, otherwise the second equation is the 'most correct.'

The main item of interest here is the (2^bits - 1) piece of the equation. This is quantifying how many values we can represent with some number of bits. We have to subtract one to keep the value from rolling over to an additional bit. Consider 8 bits for example... 2⁸ = 256, but what is 256 in binary? 100000000 - a 9 bit wide word. We must subtract 1 to get 255 - the highest value that can be represented with 8 bits.

Mathieu De Zutter said:

The same for the formula for calculation the number of shifts needed when a 16-bit data representation is chosen: Sinc³: Shift = int(log₂(x³) + 0.5) - 15. Where does the 0.5 come from?

We need to ensure that we have all of the bits that will be 'moving.' Adding .5 will ensure that we round appropriately to keep all of the relevant bits. Consider the example we went through earlier in this thread where we found a word-width of 14.495 bits - to get all the moving bits we actually need to ensure that we round to 15 bits plus an additional sign bit to get the full word. 

Forcing all of this into a few equations isn't really natural. It's best to look directly at the range of values that we can achieve based on the digital filter architecture and then consider how many bits we must have to represent these values in straight binary or binary twos-complement.