AD9629 Single ended

Dear all,

I am reading the Datasheet of AD9629 Rev.0.

There is one point that is not fully clear to me:

why in the single ended configuration there is a 1Vpp signal instead of a 2Vpp (Fig.39 pag. 18) ?

Does it mean that, in case of a DC value of Avdd/2 (around 1V) on both input, positive and negative,

I can only have a swing from 1.5V to 500mV on the positive input?

Thanks for your help.

Regards.

  • 0
    •  Analog Employees 
    on Dec 5, 2015 9:07 AM

    Hi Alberto,

    Thank you for your interest in the AD9629. You have a good question. The full scale input voltage range of the AD9629 is 2Vpp differential. A 2Vpp differential signal consists of two single ended 1Vpp signals that are complementary. Figure 39 in the AD9629 datasheet shows a conceptual example of with one side driven with the same 1Vpp single-ended amplitude.

    You are not limited to 1Vpp in the single ended configuration, but you cannot convert a 2Vpp single-ended signal. The reason for this is that a 2Vpp single-ended analog input signal will extend above AVDD on the high end, and below ground on the low end. At these extremes the input protection diodes could start conducting and the input signal will clip.

    Assuming AD9629 AVDD = 1.8V, theoretically your input signal would be limited to about 1.8Vpp single-ended. At 1.8Vpp single-ended you would be reaching the power supply and ground levels. Linearity might suffer a bit at these extremes, but depending on your application this could still be OK.

    Thanks again.

    Doug

  • Hello Doug,

    thanks for your answer. Just one more small question:

    I would like to use that ADC in single ended configuration.

    I have as input a pulse with a DC offset of a few tens of mV and an amplitude up to AVDD.

    It is not periodic: it is a random pulse: most of the time I have the signal fixed on the DC offset and sometimes a pulse appears.

    Do you think it is better to have the negative input connected at AVDD/2 or it is better to have it at ground?

    Thanks for all your effort.

    Regards.

  • Hello,

    I think I got the answer:

    as shown on Table 12 of the datasheet (Rev.0), with the negative input connected to GND and offset output binary code I have 1000 0000 0000 at the output, which means that I can connect the negative input to ground but, in this case, I will work only with 11 bits instead of 12.

    Regards.

  • 0
    •  Analog Employees 
    on Dec 9, 2015 11:54 PM

    Hi Alberto,

    Yes, you are correct. You will effectively be using only half of the ADC's dynamic range (11 bits instead of 12 bits) if you connect the inverting input to ground.

    The other item to be aware of is that, with only half the dynamic range you will only be able to convert a signal of up to about +1V. Above that and your outputs will digitally "clip" and you will have an overrange condition. If your application requires that you only detect the presence of a pulse and you are not concerned about the amplitude, you might be OK.

    But, if you need to measure your full pulse swing, you will need to connect the inverting input to AVDD/2.

    Thank you for using the AD9629.

    Doug