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some question about AD9248


I have a 100mVp-p signal and the frequency range is between 100KHz to 10MHz.

the AD9248 has some Reference configuration to select Vref (it could be 0.5V or 1V).

*if I want to sample the signal with AD9248, which voltage reference is better to get best resolution?

*I have attached a pic in the below that shows the interface circuit. How can I calculate the input resistance and capacitance of the interface? (please help me in details,in calculating and what's the role of VINCT_A? )

Best Regards

  • Hello,

    My comments are as follows:

    1) Set ADC full-scale range to 2V

    2) Remove R1 (49.9 ohm resistor) since this should be considered the termination resistor and placed differentially across secondary side of T2 (pin 6,4).

    2)  Not sure if you require two cascaded 1:1  transformers if "overkill" if desired signal is up to 10 MHz.  One should look at the amplitude balance and phase balance of the transformer to see if it is good.  Also, you may want to consider a 1:4 transformer since this will give you voltage gain of 2 when you terminate with 200 ohms across its secondry.

    3)  For a  100mVp-p signal with a  frequency range  between 100KHz to 10MHz, I would have recommended using a single-ended to differential ADC driver like the AD4930 to provide gain of 5 to 10.  (Refer to figure 15 for HD3/HD3 vs gain)

  • Dear

    I have purchased Terasic ADA card bcoz I couldn't find a good ADA card for my application.

    the ADC input interface is the above pic.

    actually I want to measure the voltage across R8. the differential voltage signal is a 100mVp-p with frequency range  between 10KHz to 10MHz.

    I used two unity gain OPA657 amplifier to buffer the signals. then I can connect Vo1 and Vo2 to a difference amplifier such as AD8130 (actually this topology is an instrument amplifier). finally I want to connect the result signal to the above interface!

    I would be grateful if you could help me to modify the circuit and interface Please.

    for example we could remove two OPA657 and AD8130 and use ADA4930 with gain of 5 and connect the result to the interface! or we can remove T1 and so on.

    this is my email:

    Best Regards

  • plus I have replaced T1 and T2 with ADT1-6T: 

    I think the insertion Loss in the desired bandwidth is good.

  • Can I drive the AD9248 with ADA4930-1 such a way?

    note that AD9248 doesn't have Vcm pin. This is why I've used two resistor voltage divider.

    (AVDD_AD9248 is the positive supply for AD9248)

  • Hello,

    It would appear that you are trying to modify an existing board.  Anyways, the attached circuit should allow you to use the differential op amp configuration your are using with out need for transformer.  The differential op amp is configured for gain of 10 to amplify 100 mVpp. differential to 1.0 Vpp output with the AD9248 configured for a 1 Vp.p. input with its VCM set to 1.5 V using voltage divider network.   The op amps are biased around AGND via the 10 kohm resistor such that each output will swing +/-1 V p.p. out-of-phase with each other thus resulting in 2 Vp.p. differentially.


  • Dear

    I have a simple question.

    when I set the Vref to 0.5v, the resulting differential span be set to 1Vpp.

    on the other side Vcm is 1.65V  (Bcoz 3.3v/2=1.65v)

    it means if I have a 1Vpp sine signal, the signal at the input of ADC pins is 1Vpp+1.65vDC.

    now how much the lower and higher amount of digital output is?

    it returns 0 for (1.65-0.5)Volt and 16384 for (1.5+0.5)Volt?

    I'm puzzled

    principplly it should return 0 for 0volt and 16384 for 1volt