CODEWITHSUNDEEP
microcontrolleradctransfer-function
Jin
Jin

Reputation: 113

ADC transfer function

I took over the project from someone who had gone a long time ago.

I am now looking at ADC modules, but I don't get what the codes mean by.

MCU: LM3S9B96
ADC: AD7609 ( 18bit/8 channel)
Instrumentation Amp : INA114

Process: Reading volts(0 ~ +10v) --> Amplifier(INA114) --> AD7609.

Here is codes for that:

  1. After complete conversion of 8 channels which stored in data[9]

  2. Convert data to micro volts??

    //convert to microvolts and store the readings // unsigned long temp[], data[] temp[0] = ((data[0]<<2)& 0x3FFFC) + ((data[1]>>14)& 0x0003); temp[1] = ((data[1]<<4)& 0x3FFF0) + ((data[2]>>12)& 0x000F); temp[2] = ((data[2]<<6)& 0x3FFC0) + ((data[3]>>10)& 0x003F); temp[3] = ((data[3]<<8)& 0x3FF00) + ((data[4]>>8)& 0x00FF); temp[4] = ((data[4]<<10)& 0x3FC00) + ((data[5]>>6)& 0x03FF); temp[5] = ((data[5]<<12) & 0x3F000) + ((data[6]>>4)& 0x0FFF); temp[6] = ((data[6]<<14)& 0x3FFF0) + ((data[7]>>2)& 0x3FFF); temp[7] = ((data[7]<<16)& 0x3FFFC) + (data[8]& 0xFFFF);

I don't get what these codes are doing...? I know it shifts but how they become micro data format?

  1. transfer function

//store the final value in the raw data array adstor[] adstor[i] = (signed long)(((temp[i]*2000)/131072)*10000);

131072 = 2^(18-1) but I don't know where other values come from

AD7609 datasheet says The FSR for the AD7609 is 40 V for the ±10 V range and 20 V for the ±5 V range, so I guessed he chose 20vdescribed in the above and it somehow turned to be 2000???

Does anyone have any clues??

Thanks

-------------------Updated question from here ---------------------

I don't get how 18bit concatenated value of data[0] + 16bit concatenated value of data[1] turn to be microvolt after ADC transfer function.

data[9] +---+---+--- +---+---+---+---+---+---++---+---+---++---+---+---++
analog volts | 1.902v | 1.921v | 1.887v | 1.934v | +-----------++-----------+------------+------------+------------+
digital value| 12,464 | 12,589 | 12,366 | 12,674 | +---+---+---++---+---+---++---+---+---++---+---+---++---+---+---+

I just make an example from data[3:0] 1 resolution = 20v/2^17-1 = 152.59 uV/bit and 1.902v/152.59uv = 12,464

Now get thru concatenation:

temp[0] = ((data[0]<<2)& 0x3FFFC) + ((data[1]>>14)& 0x0003) = C2C0

temp[1] = ((data[1]<<4)& 0x3FFF0) + ((data[2]>>12)& 0x000F) = 312D3

temp[2] = ((data[1]<<6)& 0x3FFC0) + ((data[3]>>10)& 0x003F) = 138C

Then put those into transfer function and get microvolts

adstor[i] = (signed long)(((temp[i]*2000)/131072)*10000);

adstor[0]= 7,607,421 with temp[0] !=1.902*e6

adstor[1]= 30,735,321 with temp[1] != 1.921*e6

adstor[2]= 763,549 with temp[2]

As you notice, they are quite different from the analog value in table.

I don't understand why data need to bit-shifting and <<,>> and added up with two data[]??

Thanks,

Upvotes: 1

Views: 740

Answers (1)

tonypdmtr
tonypdmtr

Reputation: 3225

Please note that the maximum 18-bit value is 2^18-1 = $3FFFF = 262143

For [2] it appears that s/he splits 18-bit word concatenated values into longs for easier manipulation by step [3].

[3]: Regarding adstor[i] = (signed long)(((temp[i]*2000)/131072)*10000);

To convert from raw A/D reading to volts s/he multiplies with the expected volts and divides by the maximum possible A/D value (in this case, $3FFFF) so there seems to be an error in the code as s/he divides by 2^17-1 and not 2^18-1. Another possibility is s/he uses half the range of the A/D and compensates for that this way.

If you want 20V to become microvolts you need to multiply it by 1e6. But to avoid overflow of the long s/he splits the multiplication into two parts (*2000 and *10000). Because of the intermediate division the number gets small enough to be multiplied at the end by 10000 without overflowing at the expense of possibly losing some least significant bit(s) of the result.

P.S. (I use $ as equivalent to 0x due to many years of habit in certain assembly languages)

Upvotes: 1

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