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ADF4113 convert V to dbm

I am conused withe the descriotion in the ADF4113 datasheet. Could anyoneone clarify and give me any advice?

In the ADF4110/ADF4111/ADF4112/ADF4113 Rev,F datasheet on page 3, the Reference Input Sensitivity is specified as 0.4/AVDD V p-p min/max for AVDD = 3.3 V and 3.3/AVDD V p-p min/max for AVDD = 5V.
Isn't it 0.4 / 3.3(=AVDD) V p-p min/max for AVDD = 3.3V and 3 / 5(=AVDD) V p-p min/max for AVDD = 5V?

I did convert V to dbm.

It case,
AVDD=3.3V : 0.4 / 3.3(=AVDD) V p-p min/max → -3.979dBm / 14.349dBm
AVDD=5V : 3 / 5(=AVDD) V p-p min/max → 13.521dBm / 17.958dBm

However, It is different operating rage.
AVDD=3.3V : 0.4 / 3.3(=AVDD) V p-p min/max → -8dBm / 20dBm
AVDD=5V : 3 / 5(=AVDD) V p-p min/max → -4dBm / 20dBm

I know wrong, but I don't know reasons.

Please. Can someone tell me?

Best Regards

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  • Hi.

    The maximum allowable voltage range for each condition is the AVDD. So convert 3.3 V and 5.0 V to dBm using a 50 Ohm load. Assuming a sinusoidal waveform:

    3.3V p-p = +14.35 dBm

    5.0V p-p = +17.95 dBm.

    For AVdd = 3.3V operation the minimum voltage 0.4V p-p translates to -3.98 dBm. So a sine wave with -3.98 dBm power, or 0.2V amplitude, or 0.4 Volt pk-pk will meet minimum requirements.

    For AVdd = 5.0V operation, the minimum voltage 3.0V translates to 13.52 dBm. 5.0V operation needs a lot more power compared to 3.0V operation, so it is +13.52 dBm or 1.5V amplitude of 5V pk-pk.

    I don't understand how you derive the second part of the question.

  • Hi, icollins
    Thank you for You answer

    I know wrong the second part of the question.
    however I don't know reasons.

    My customer did Test of ADF4113.
    Test result of ADF4113
    AVDD=3.3V : 0.4 / 3.3(=AVDD) V p-p min/max → -8dBm / 20dBm
    AVDD=5V : 3 / 5(=AVDD) V p-p min/max → -4dBm / 20dBm

    It is different operating rage of Datasheet.

    Please. Can tell me?

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