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AD9649 Maximum Differential Analog Input Voltage Range

Hello I am Jitendra Chaurasiya.

I am using AD9649 14 bit ADC in my circuit. The internal voltage reference of ADC is 1V and the differential analog input on VIN+ and VIN- 2Vp-p. 

Also the absolute voltage on differential analog input VIN+, VIN− to AGND is −0.3V to AVDD + 0.2V given in datasheet.

So my question is: Whether the ADC input protected against its absolute rating? Because my input range is from -1.7V to 1.2V and after 1V Internal reference my input become from -0.7V to 2.2V which is out of it's range (as compared to -0.3V to (AVDD = 1.8) +0.2) which is -0.3 to 2V.

Kindly suggest me if any external protection required to protect my ADC analog input pins. 

In the image below I used ADA4932 differential ADC driver with unity gain configuration.

Thanks and Regards,

Jitendra Chaurasiya

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  • Hi Jitendra,

    Thank you for your interest in AD9649. The device inputs are not protected beyond the absolute maximum voltage ratings in the datasheet.

    So you must protect the analog inputs from overvoltage. One possibility is to add protection diodes on the board. This will add to the BOM cost and it will affect performance, but it will work.

    Another possibility to consider is to use an amplifier that is designed specifically for driving low voltage ADCs. For example I believe the ADA4930 powered with a single 3.3V supply will keep tis output voltage within an acceptable range for the AD9649. Are you able to use the ADA4930?



  • Thanks for your suggestion as well as solutions.

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