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About the ADC on F28377D MCU

Xiaowei Yu
Intellectual 510 points

Other Parts Discussed in Thread: THS4521

Hi,

The ADC on F28377D MCU is type 4 ADC. But I didn't find the information about type 4 ADC. I just find the following table, but there is no type 4. Does anybody know the range of this ADC? is that bipolar or unipolar? Thanks.

  • 0
    TI__Guru 60865 points

    Hi Xiaowei,

    I can't see your table.  Can you specify which document it came from?

    The F28377D ADC is unipolar.  In 12-bit mode it can be configured for single-ended or differential inputs.  In 16-bit mode, inputs need to be differential.

     

  • 0 in reply to Devin Cottier
    Intellectual 510 points

    Hi,

    Thanks for your help. The document number is SPRU566.

  • 0 in reply to Xiaowei Yu
    TI__Guru 60865 points

    OK, I think I know what table you are referring to.  That document is in the process of being updated to include the newest C2000 devices. 

  • 0 in reply to Devin Cottier
    Intellectual 510 points

    I have another question. I want to use the 16-bit ADC on F28377D. The 16-bit ADC only supports differential mode on .  But the analog signal is single ended. Can I just connect analog signal to ADCINyP, and connect ADCINyN to ground? Thanks.

  • 0 in reply to Xiaowei Yu
    TI__Guru 60865 points

    Xiaowei,

    No, this won't work.  The differential signal needs both inputs.  In the datasheet, we spec' that the input common mode range is +/-50mV.  This means that (ADCINp + ADCINn)/2 must be within 50mV of (VREFHI - VREFLO)/2.  

    If VREFHI = 3V, VREFLO = 0V, and the ADCINp is say 2V, then

    (2V + ADCINn)/2 = (3V + 0V)/2 +/- 50mV

    =>  2V + ADCINn = 3V +/-100mV, then

    => ADCINn = 1V +/- 100mV

    So ADCINn = 0V won't work.

    Essentially the negative signal should always be the compliment of the positive signal:

    You should be able to get whatever signal you have into differential form by using a fully differential op-amp, e.g. THS4521.  

  • 0 in reply to Devin Cottier
    Intellectual 510 points

    Hi,

    The sensor output is -5V to +5V single ended sine wave. If I want to use the this 16-bit internal ADC. for my understanding, I have to do the following steps as shown in the image.

    First, convert  -5V to +5V sine wave into 0 to 2.5V sinewave signal by decreasing signal by 4 and adding 1.25V bias.

    Then, convert 0 to 2.5V sinewave signal to differential signal using single ended to differential AMP.

    If I do these to the sensor signal, do I lost 2 bits resolution or sacrifice accuracy?

    Do you have any suggestion? Should I use an external 16-bit bipolar -5V to +5V  input range ADC?

    I am so sorry for taking a lot of time from you. Really appreciate your help.

  • 0 in reply to Xiaowei Yu
    TI__Guru 60865 points

    Xiaowei,

    (I can't see the picture that you added)

    Dividing down and then amplifying (in the analog domain) won't cause you to lose any resolution, although you may introduce additional noise, distortion, gain error, etc.

    I think the circuit that you want is: voltage divider, then single-ended to differential amplifier, then ADC (possibly with an RC filter between op-amp and ADC).  Take a look starting at page 14 of the LMH6551Q datasheet and going through the ADC interfacing section; I think this is what you want (you don't necessarily need to pick that specific op-amp).