ADC driver for AD9255

I'm planning to use the AD9255 single 14 bit ADC with DC coupled input signals in the dc - 20 MHz range.

Because of the DC coupled requirement I am looking at ADC driver amplifiers instead of a transformer.

The AD9255 web page recommends the ADA4937-2 and ADA 4938-2 parts.  The AD9255 data sheet shows the ADA4938-2 in figure 66.  However, these ADC drivers are dual parts and I don't understand why these are recommended, the AD9255 is a single ADC.  Is it a typo or am I missing something?

Can I just substitute the the ADA4937 in place of the ADA4938 as shown in figure 66?  If so that may be the driver and circuit that I need. 

Thanks.

  • Hi,

    Thank you for considering ADI products for your design. The dual parts are probably recommended for marketing reasons as they are cheaper compared to two individual singles. The other unused channel could be used for other diff amp requirements of the system for other applications. But technically, for the specific purpose of driving the AD9255 which is a single ADC, a single driver chip is perfectly fine to use.

    I am not sure what tradeoffs you are considering, but yes, you can use ADA4937 instead of the ADA4938. ADA4937 offers better AC specification but ADA4938 has a wider supply range and lower input bias currents among others. You can use the free ADI DiffAmpCalc software to better evaluate ADI diff amps.

    I hope this helps. Please let me know if you have more questions.

    Best regards,

    Neil

  • Thanks very much for the reply and for the link to the Diff Amp Calculator. 

    This application will be single ended input, dc coupled,  ground referenced input with frequency range DC to 20 MHz, the common mode output voltage of the differential amplifier will be set to 0.9 V by the AD9255, the diff amp gain = 1. 

    After using the Diff Amp Calc tool I think the ADA4938 will work fine for my application.  There is an Output LPF box which shows the importance of the LPF in between the diff amp and the ADC to limit the noise.

    So how exactly do I get a single pole 50 MHz LPF in between the ADA4938 and the AD9255?   Do I use the circuit shown in fig 66 of the AD9255, how do I calculate the component values? 

    Thanks.

  • Hi,

    You are correct that the LPF greatly improves the noise performance of the system. Yes, you can use the filter topology of Figure 66 to build your LPF. The approximate bandwidth is calculated using this formula:

    Where

    R1 is the series resistor at the output of the amplifier

    C2 is the capacitor accross the differential output

    C1 is the capacitor from output to ground

    You can set C1=15pF and R1=33ohms. These values are typical for this kind of application. Calculating for C2 given BW=50MHz, C2 should be around 40pF. The series resistors at the inputs of AD9255 help lower the effect of "kick back" from the ADC when it switches. You can set it to 15ohms each.

    One more thing, to get the best possible performance of your ADC driver, you need to design your PC board in such a way you minimize stray or parasitic elements. Page 12 of AN-1026 discusses PC Board Layout for High Speed ADC drivers. You can also get further information on high speed ADC drivers on that application note.

    I hope this helps. Good luck!

    Best regards,

    Neil

  • Excellent answer!  I looked at several ADI notes and documents but did not see this differential bandwidth equation anywhere.  

    Thanks again.