AD8349 IQ inputs

I had a question regarding the IBBP/IBBN and QBBP/QBBN input levels as the datasheet seems to contradict itself. The pin descriptions on page six (or page 15) state that the differential inputs:

  • Must be biased to 400 mV
  • 600 mV p-p on each pin
  • 1.2 V p-p differential drive

In order to achieve all three requirements I would think IBBP and QBBP would swing from 0.4V to 1V and IBBN and QBBN would swing from -0.2V and 0.4V. This would produce a 1.2V p-p (1-(-0.2) = 1.2) drive biased to 400 mV. However, this can't be the case since the absolute maximum range for the inputs is 0-2.5V.

If both inputs were centered around 400 mV and swung from 100 mV to 700 mV I would think the differential drive would only be 600 mV. Any clarification would help. Thanks.

  • 0
    •  Analog Employees 
    on Nov 14, 2015 12:43 AM

    Hello,

    Baseband Input signal  on IBBP/IBBN/QBBP/QBBN pins swing +/-300mV centered at 400mV. So all bb signals swing in 100mV to 700mV range at different phase. Phase difference between IP(IN) and QP(QN) is 90 degree. 180 Deg between IP(QP) and IN(QN). I assumed below I lead Q, So, IP=0.3*cos(wt) + 0.4, QP=0.3*sin(wt) + 0.4, IN= - 0.3*cos(wt) + 0.4, QN= - 0.3*sin(wt) + 0.4. I Differential (IP-IN)  is 0.6*cos(wt) and Q differential(QP-QN) is 0.6*sin(wt). So I and Q differential swing 1.2Vp_p. I believe below plots help your understanding.

    Thanks

    Tony

  • 0
    •  Analog Employees 
    on Aug 2, 2018 3:15 PM
    This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

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