Question regarding extracting 12-bit two's complement out of two bytes

Projects Programming
1

Hi everyone,

I am presently in the process of experimenting with a ADC122S625 ADC from National Semiconductor. It samples two channels worth of data and sends it out in a four byte train. The data is in two's complement format, where the first byte is the MSB, the second byte is the LSB for each channel (Channel A and B). (see page 7 for a timing diagram)

Within the MSB, the first four bits are always zero. The remaining 4 bits hence presumably indicate the sign and so on. Thanks to help from others, I discovered casting the MSB results as an int in order to extract the sign. However, I wonder what the correct approach is on a number that only consists of 4 bits. Here is what I attempted, I wonder if anyone more knowledgeable could comment:

void readADC()
{
  byte a1,a2,b1,b2;
  digitalWrite(slaveSelectPin,LOW);
  a1=SPI.transfer(0x00); //MSB, Channel A
  a2=SPI.transfer(0x00); //LSB, Channel A
  b1=SPI.transfer(0x00); //MSB, Channel B
  b2=SPI.transfer(0x00); //LSB, Channel B
  digitalWrite(slaveSelectPin,HIGH);
  //now convert all four bytes into two 12-bit numbers. The first four bits of the MSBs
  //are zero, hence no need to  MSB = MSB & B00001111 to zero them out like with other ADCs.
  //The remaining 4 bits determine the first four bits of the ADC result.
  //Results are two's complement, so they can range from -2048 to +2047, hence the (int) casting

  Channel_A = (int)(a1<<4)<<4|a2;
  Channel_B = (int)(b1<<4)<<4|b2;
}

In other words, right shift the results four bits, cast into int to extract the sign, then right shift another bits, then append the LSB.

Does that seem reasonable?

2

Just assemble the parts at the top end of the integer and shift right - shift right is an arithmetic shift in C

  Channel_A = ((a1 << 12) | (a2 << 4)) >> 4 ;
  Channel_B = ((b1 << 12) | (b2 << 4)) >> 4 ;
3

I should clarify that - right shift is an arithmetic shift in C for signed variables. unsigned variables will use logical shift right, so make sure channel_A and channel_B are declared as "int".

4

Yup, they were and thanks, it works perfectly, including the signing.

Now if only I could figure out what else is off with this chip / program...