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AM3352: ADC errata workaround

UNA
Mastermind 7980 points
Part Number: AM3352

Hi,

I have questions about ADC errata of AM335x.

Q1: The formula to calculate the series resistance is as follows.
　　AM335x Errata(www.tij.co.jp/.../sprz360i.pdf)


　　Rseries_max = [(ADC_ClkDiv + 1) *( SampleDelay + 2) / (116 x 10-12) * (CLK_M_OSC)] - 200 Ω - Rsource_max


　　I think 116x10^-12 shows that the input capacitance is 116pF. How was this capacity value calculated?


Q2: Similarly, how was the above 200Ω calculated?

Q3; Is it possible to substitute a diode instead of a series resistor as a workaround for this errata?
(Of course, I think it is necessary to pay attention to the forward voltage of the diode.)


Best Regards,
H.U

  • 0
    TI__Guru 62845 points

    The formula is based on an error of 1/16 LSB with a 10.5pf input capacitor.

    The 200 ohm contribution was included in the formula to account for internal resistance of the device.

    I do not understand how you expect a diode to be used to resolve this issue. The analog input can be shorted to VDDA and/or VSSA during this transient condition.  The diode would only prevent unwanted current from flowing in one direction when the analog input is shorted to the ADC supply rails, where a series resistor limits current in either direction.

    Regards,
    Paul

  • 0 in reply to peaves
    Mastermind 7980 points

    Hi Paul,

    Thank you for your reply.

    peaves said:

    I do not understand how you expect a diode to be used to resolve this issue. The analog input can be shorted to VDDA and/or VSSA during this transient condition.  The diode would only prevent unwanted current from flowing in one direction when the analog input is shorted to the ADC supply rails, where a series resistor limits current in either direction.

    I only use the single and positive input signal to the AIN.
    In this case, I think that a diode can be configured more simply than a series resistor.
    Is my idea correct?


    Best Regards,
    H.U

  • 0 in reply to UNA
    TI__Guru 62845 points

    I cannot say without knowing details of the circuit that is being used to source the ADC analog input.

    I'm concerned you do not fully understand the problem described in this advisory if you are trying to solve the problem with a diode.

    Please describe the circuit you are connecting to the ADC analog inputs or provide a snapshot of the schematic so I can review.

    Regards,
    Paul

  • 0 in reply to peaves
    Mastermind 7980 points

    Hii Paul,

    OPamp is connected to one AIN terminal, and the other terminals are unused.
    Please let me know if you need more information.

    Best Regards,
    H.U

  • 0 in reply to UNA
    TI__Guru 62845 points

    You operational amplifier is capable of sourcing current into the ADC input or sink current from the ADC input to maintain its desired output voltage on the ADC input. When the temporary condition described in the advisory occurs, the ADC input connected to your OpAmp may be shorted to VDDA_ADC, VSSA_ADC, VREFP, VREFN, or any of the other ADC inputs. When this occurs you need something in the signal path to limit current to the short and there is no way to know which short is going to occur.

    Let’s assume you insert a series diode with Anode connected to the ADC input and Cathode connected to your OpAmp output with the OpAmp sourcing 0.5V. The diode would forward bias if the ADC input shorts to the VDDA_ADC (1.8V). In this case, the only thing that would limit current during the contention period is the internal resistance of the ADC short and the OpAmp output impedance. A similar thing would occur if the Cathode was connected to the ADC input and Anode was connected to your OpAmp output when the OpAmp is sourcing 1.3V and the ADC input shorts to VSSA_ADC (0V).

     

    During normal operation, a series diode will only allow current to flow in one direction between the OpAmp output and the ADC input. The ADC input will not follow your OpAmp output when the diode is reverse biased.

     

    Hopefully this explains why a series diode is not a solution.

     

    Regards,

    Paul

  • 0 in reply to peaves
    Mastermind 7980 points

    Hi,

    In the first part, you answered that the 200Ω in the formula is the internal resistance.
    On the other hand, The data sheet shows [1 / ((65.97 × 10-12) × f)](NOM) as the input impedance.
    What is the relationship between this value and the 200Ω?


    Best Regards,
    H.U