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ADC12DJ3200: Absolute Maximum Ratings

Massimiliano Foggia
Prodigy 120 points
Part Number: ADC12DJ3200

Hello

looking at the datasheet of the ADC in object, Section 6.1 Absolute Maximum Ratings,  I was struggling a bit trying to understand the relationship between  "Peak RF input power (INA+, INA–, INB+, INB–)" and "Terminal Voltage Range, INA+, INA–, INB+, INB–"

If I try to convert the Peak RF input power in voltage it looks to me that the values are not matching in the datasheet.

Please could you clarifiy how the values are linked within the datasheet?

thanks in advance

Best regards

Max

  • 0
    TI__Mastermind 38110 points

    Hi Massimiliano,

    The Peak RF input power and Terminal voltage power range are correct in terms of the numbers specified in the datasheet.

    Peak power is reference differentially with 100ohm or single-ended with 50ohms. In calculation, this means +/-1V on each pin. The Terminal voltage power is references single-ended only, so the voltage number seems correct, however the reference to ground doesn't seem correct. I will look into this and get back to you.

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 120 points

    Hi Rob

    thanks for the feedback.

    Unfortunately I'm still missing the relationship. Normally if I convert the peak RF input power into voltage I proceed as follows:

    16.4dBm = 10*log10 ( ((Vrms)^2)/R) + 30dBm

    Assuming a differrential resistance of 100Ohm this leads to a differental peak voltage of 2.95V therefore I still don't get the +/-1 V reported on the datasheet.

    It would be great to get a feedback on this.

    Thanks

    regards

    Max 

     

  • 0 in reply to Massimiliano Foggia
    TI__Mastermind 38110 points

    Hi Max,

    I spoke to product engineer on this to get you a more clear answer on how they specify this. The confusion is likely due to the ESD diodes (e.g. protection diodes). There are three things that factor in. The peak voltage, the peak current and the peak RF power. The RF power is calculated from the peak voltage and current. The ESD diodes kick in around +/-1 V (as the DS suggests) with +/-50 mA current draw. If we assume a very high amplitude sine wave is applied to the part, such that the ESD diodes kick on, and the source is capable of driving 50 mA, then the ESD diodes will effectively turn it into a +/-1 V square wave with constant +/-50 mA current draw. The RF power calculation then is:

     

    PRF-PEAK = 10*log10(1V * 0.05A / 0.001) [dBm] = 17 dBm

     

    Not quite sure what accounts for the 0.5 dB difference (17 vs 16.4..)… the data I have includes the current through the termination resistance, with a linear slope between +/-0.9V.

     

    Hope that helps.

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 120 points

    Hi Rob

    Thanks a lot for the detailed the explanation. Now it's all clear to me.

    Just to be sure I would like to get a feedback on a similar topic, that is the conversion of VIN_FSR (specified in section 6.5 "Electrical Characteristics - DC Specifications" on page 12) to input RF power.

    I'm currently assuming valid the equation:

    P(dBm) = 10*log10 ( ((Vrms)^2)/R) + 30dBm

    therefore in case of VIN_FSR = 800mVpp the RF power would be -0.9691dBm, isn't it?

    thanks in advance for your support

    best regards

    Max

  • 0 in reply to Massimiliano Foggia
    TI__Mastermind 38110 points

    Hi Max,

    Yes, this is correct. See attached.

    Keep in mind that is with 100ohms and a differential signal. So therefore, each leg/AIN+ or - will be half of that.

    Regards,

    Rob

  • 0 in reply to Rob Reeder
    Prodigy 120 points

    Hi Rob

    thanks for the support.

    Issue is therefore resolved.

    best regards

    Max