TMS320F2800137: Why is 9.5-bit effective DACs on CMPSS_LITE instances?

Wayne Huang

Genius 16585 points

Part Number: TMS320F2800137

Tool/software:

Hi champs,

I am asking this for our customer.

1. The user wonders why the effective bit of DAC is 9.5 bit rather than a integer like 9 bit or 10 bit?

2. Even though the effective bit of DAC is 9.5 bit, the user should still need to input 12-bit to the register in codes like that for CMPSS DAC, but only the lower 10 bits of the input matters.

Is it right?

Wayne

8 months ago

0 Wayne Huang 7 months ago

TI__Genius 16585 points

Hi champs,

Any update?

0 Nima Eskandari 7 months ago in reply to Wayne Huang

TI__Guru 52040 points

Wayne, while I find a person to help you with this, I can give you some answers.

Effective Number of Bits: https://en.wikipedia.org/wiki/Effective_number_of_bits

It is calculated based on SINAD. That's why you get 9.5.

Yes you still have a 12bit number to input register. based on the analog circuitry and the noise, the number of effective bits are different.

Same concept exists for ADC.

+1 Fleenor 7 months ago

TI__Mastermind 23753 points

Hey Wayne,

Sorry we missed getting you an answer earlier.

1) As Nima previously mentioned, the ENOB calculation includes fractional parts that can result in a non-integer value.

There is a great TI video regarding ENOB for ADC, the same concept applies to the DAC. The Basics of the Effective Number of Bits - YouTube

2) The user still needs to input a 12-bit value, but only the 10 MSB (Upper 10 bits) of the input matter, the LSB's become invalid because the noise will have a greater impact on the output signal then changes to the 2 LBS (Lower 2 bits).

Best Regards,

Zackary Fleenor

C2000™︎ microcontrollers

C2000 microcontrollers forum

TMS320F2800137: Why is 9.5-bit effective DACs on CMPSS_LITE instances?