MCP3301 A/D converter SPI reading

[flotulopex]

Hi there,

I'm trying to read a 13 bits MCP3301 A/D converter who's output value should go from 0 to 8191.

Unfortunately, even if I can read something, it looks to be only the "upper half" of the expected full value.






Could I have a hardware (wiring) issue or is there something in my code that would be wrong?

' ====== DESCRIPTION ===============================================================================
' MCP3301: SPI 13Bits ADC

' ====== FUSES =====================================================================================
' PIC 16F690 Fuses - Internal oscillator (activate OSCCON register)

'@ __config _FCMEN_OFF &_IESO_OFF &_CPD_OFF &_WDT_OFF &_INTRC_OSC_NOCLKOUT &_BOR_OFF &_CP_OFF &_PWRTE_OFF &_MCLRE_OFF

#CONFIG
    __config _FCMEN_OFF &_IESO_OFF &_CPD_OFF &_WDT_OFF &_INTRC_OSC_NOCLKOUT &_BOR_OFF &_CP_OFF &_PWRTE_OFF &_MCLRE_OFF
#ENDCONFIG

@ ERRORLEVEL -306
@ ERRORLEVEL -202

' ====== REGISTERS =================================================================================
'             76543210
OPTION_REG = %10000000 'PORT A&B Pull-Ups disabled (see WPUA & WPUB)
OSCCON     = %01100000 'Internal RC set to 4MHZ
ANSEL      = %00010100 'Select analog inputs Channels 0 to 7 (AN2 + AN4 selected)
ANSELH     = %00000000 'Select analog inputs Channels 8 to 11
ADCON0     = %10000000 'A/D Module (Bit5:2 select channels 0:11)
ADCON1     = %00000000 'A/D control register
CM1CON0    = %00000000 'Comparator1 Module is OFF
CM2CON0    = %00000000 'Comparator2 Module is OFF
INTCON     = %00000000 'INTerrupts CONtrol (TMR0 OFF)
TRISA      = %00000100 'Set Input/Output (0 to 5)
PORTA      = %00000000 'Ports High/Low (0 to 5)
TRISB      = %00000000 'Set Input/Output (4 to 7)
PORTB      = %00000000 'Ports High/Low (4 to 7)
TRISC      = %00000001 'Set Input/Output (0 to 7)
PORTC      = %00000000 'Ports High/Low (0 to 7)

' ====== DEFINES ===================================================================================
DEFINE OSC 4
DEFINE NO_CLRWDT 1   'Don't waste cycles clearing WDT  
DEFINE HSER_CLOERR 1 'Automatic clear overrun error 
DEFINE SHIFT_PAUSEUS 100

' ====== VARIABLES =================================================================================
SD          var PORTB.4 'SPI Data pin
CS          var PORTB.5 'SPI Chip Select
SCK         var PORTB.6 'SPI Clock pin
Serial_Out  VAR PORTB.7
Mode        Var WORD
MCP3301     Var WORD

' ====== INITIALIZE VARIABLES ======================================================================
CS          = 1
Mode        = 84 'serial com
MCP3301     = 0

' ====== PROGRAM ===================================================================================
MAIN:

    CS = 0                          'initiate 3301 comm
    SHIFTIN SD,SCK,1,[MCP3301\13]   'shift 13 bits of data in, lowest bit first
    CS = 1                          'stop 3301 comm
    
    Serout2 Serial_Out,Mode,["MCP3301: ", BIN13 MCP3301," - ", DEC4 MCP3301,13,10]
    
    PAUSE 1000

    GOTO MAIN
END

[richard]

according to the data sheet the data read is 12 data bits + sign bit

[flotulopex]

Really?



I don't get it

[richard]

i'm looking at the data sheet you posted, that's what i see

[tumbleweed]

Keep reading. It's a 13-bit differential converter.

From the datasheet section 5.1:

and the device uses the collected charge to produce a serial 13-bit binary two’s complement output code.

So that's 12 bits of data + sign. See Section 6.1.

[flotulopex]

Okay then, didn't see that one 😵

So I'll get a 12 bits "value" and the last bit will indicate either "+" or "-", right?

[richard]

the last bit will indicate either "+" or "-", right?

if by last bit you mean the msb then yes
not forgetting its a 13-bit binary two’s complement number

[flotulopex]

not forgetting its a 13-bit binary two’s complement number

...and that means what exactly please?

[Ioannis]

Your ADC is differential one. So, that means it can convert positive and also negative (in relation to the two inputs + and -) voltages.

If the + input is more positive than the - input of the converter then the output will be a plain 12 bit binary number.

But if the - input is more positive than the + input, then as you can understand the result of the ADC should be negative. This is shown in binary with the most significant bit (MSB) of the value being 1. That is the sign bit.

But the rest of the 12 bits follow a specific rule then. All bits are reversed and also -1.

For example, 1 1111 1111 1111 means negative number and this number is the reverse of 1111 1111 1111 plus one, or just 1. So in total -1.

Or, 1 0000 1111 0000 is also negative and the value is -3856

Hope this clear a bit of the mud!

Ioannis

More here https://www.cs.cornell.edu/~tomf/not.../twoscomp.html

[flotulopex]

Thanks Ioannis and sorry for the late reply,

So, if I would connect GND to the "-" input and measure i.e. a + signal from 0 to 5 Volts, I should have as 12 bits result, right?

Or, 1 0000 1111 0000 is also negative and the value is -3856

This is confusing to me. I get the positive and negative thing but I can't figure out how you come to this 3856 number?

[richard]

to expand a 13 bit negative two's complement number to a 16 bit number bits 13,14 an 15 need to be set accordingly
in general the "sign bit" needs to be extended to the new width
so 1 0000 1111 0000 becomes 1111 0000 1111 0000 qed -3856

[Ioannis]

Yes, Richard is correct. I forgot to mention you have to adjust the value to the byte, word or whatever width you have.

So the sign is the left most bit but the rest should also be 1. Remember we reverse the bits to get the absolute value disregarding the left most, sign bit.

Seems confusing at first, but has its logic.

Ioannis

[flotulopex]

Got it now!

Thanks a lot

[tumbleweed]

Remember we reverse the bits to get the absolute value disregarding the left most, sign bit

To convert a 2's compliment negative value to a positive number, invert all the bits (NOT) and add 1.

For example 2's compliment -1 = FFFF
invert = 0000, add 1 = 0001

[Ioannis]

I was reffering to the left most bits but yes. Absolutely correct, to convert a positive to negative after reversing all bits you add one.

Ioannis