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TMS320C5505 GPAIN leakage

AM44866
Intellectual 275 points
Other Parts Discussed in Thread: TMS320C5505

Page 63 of the TMS320C5505 datasheet (rev B) indicates the input leakage current of any pin can be as high as +/- 5uA. Additionally, page 128 specifies an analog input impedance of 1 M-ohm. When configured as an analog input, do both of these parameters need to be taken into account when calculating the accuracy of the input signal measurement incident to a given GPAIN pin, or are they dependent on the configuration of the pin?

Thanks - AM

  • 0
    TI__Expert 8065 points

    AM,

    Could you explain more about "accuracy of the input signal measurement"?

    I believe that INL and DNL will tell you the accuracy is the ADC. Since its input impedance is 1M, it only takes very small current to carry voltage. 

    Best Regards,

    Peter Chung

     

  • 0 in reply to Peter Chung
    Intellectual 275 points

    Sorry for the lack of detail on the actual application; we're planning to use the analog inputs to measure the various voltage rails inside of the device. Each of the analog inputs are connected to a voltage divider which divides each rail voltage down to approx 0.9V. The resistors used are fairly large to minimize the amount of power the divider consumes.

    Using the 5V rail as an example, we're using a resistor divider consisting of a 200K resistor in series with a 45.3K to result in an ideal divider voltage of 923mV when the rail is 5V. When the ADC's 1M ohm resistor is factored in, the voltage could drop as low as 890mV, which is still tolerable for rail monitoring purposes. However, if you also factor in the leakage spec outlined in the data sheet of +/- 5uA, the value seen at the input of the ADC could be as low as 712mV, which would be too far from the ideal voltage to make the measurement useful.

    Therefore, I was wondering if I was interpreting the parameters from the data sheet correctly. When using the analog inputs, does one need to factor in both the 1Mohm resistor and the +/- 5uA leakage current for doing such calculations?

    Thanks - AM

  • 0 in reply to AM44866
    TI__Genius 14965 points

    Hi,

     

    Sorry for the late response on this. Have you looked SAR user guide http://focus.ti.com/lit/ug/sprufp1/sprufp1.pdf ?

    There is an explanation about battery measurement case at section 2.4.

    Regards,

    Hyun

  • 0 in reply to Hyun Kim
    TI__Genius 14965 points

    hi,

    It's 890mV correct. Leakage current causing voltage is already calculated when you add 1M ohm.

    Regards,

    Hyun