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The driver circuit of AD9633 or AD9681

Hi Guys,

We are using many AD9218 chips, and the driver using AD8138, which is ideally suited for broadband dc-coupled applications. The AD9218 datasheet also gives the reference circuit by using AD813x devcies.

I read the datasheet of AD9633 and AD9681, and found these commnet:" becuase the noise performance of most amplifiers is not adequate to achieve the true performance of the AD9633." Is this means that i can't use amplifiers like AD8138 to drive these ADCs?

Also, in AD9633 datasheet, which gives two refierence circuits. One uses double balun input configuration, another uses differential transformer-coupled configuration. Both are showed as figure 57 and figure 58 in datasheed respectively.

Now, we are going to modify our design to using more channels integrated, however, our design is of dc-coupled application. How to apply AD9633 or AD9681 in our system?

  • Hi Coyoo,

    The statement about the noise performance of amplifiers means that any amplifier will add noise and distortion to the analog signal so that overall performance will be reduced when compared to the performance of the ADC alone.

    You can still use an amplifier to drive the AD9633 or AD9681. If you are DC coupling you need to make sure that the output common mode voltage of the amplifier meets the input common mode voltage requirements of the ADC. In the case of the AD8138, you might need a negative supply to achieve this.

    Have you seen the ADI Diff-Amp Calculator? This is a handy tool with which you can experiment with different ADC drivers and their configurations ( http://www.analog.com/en/design-center/interactive-design-tools/adi-diffampcalc.html ).

    You will also need to take care that the ADC inputs are protected from voltages exceeding the maximum allowable limit.

    Thank you.

    Doug

  • Hi Doug, thanks for you reply.

    In our system, we need mulitiple AD9633 in single board. So synchronization among devices is a big problem for us, someone said in here, the input clock should be 2x or 4x of the sampling rate. For the internal divider will help to synchronize among devices. Is it right? However, we can only provide 100MHz clock to ADC, and 100MHz is our sampling rate.

  • Hi Coyoo,

    In the default state (clock-divide-by-1, 1x Frame Mode), multiple AD9633s will automatically be synchronized if they receive the same sample clock. Having only a 100MHz clock is OK.

    The conditions that require using the SYNC function for synchronization of multiple AD9633s are:

    • If using the clock divider (divide by 2 or greater), the SYNC function is needed to synchronize multiple AD9633s.
    • If using 2x Frame Mode,  the SYNC function is needed to synchronize multiple AD9633s.

    Thank you.

    Doug

  • Hello Dong,

    Thanks alot!

    Actually, our clocks for ADC are from clock distributor, where the clocks come from and branch to each ADC device. Is it still ok without using SYNC to apply synchronization function for all ADC devices?

    If AD9633 can automatically be synchronized in default state(clock-divide-by-1, 1x Frame Mode), this seems depnend on the input clock of AD9633, instead of AD9633 device own, am i right? Otherwise, how doesn't AD9633 automatically synchronize by itself?

  • Hi Coyoo,

    In the case where your clocks come from a clock distributor, the AD9633 SYNC function will not provide additional synchronization.

    Yes, in the default state (clock-divide-by-1, 1x Frame Mode), the synchronization depends on the applied sample clock. If all the ADC receive the same sample clock, they will be synchronized.

    Thank you.

    Doug