I have a differential signal in a range of 2V( between 0.5V and 2.5V) . The average voltage around which the differential pair swings is set to 1.65V.
May I use the ad9220 ADC?
Yes, the input circuit is flexible and should accomodate this range. However, I have a question about what you describe above. You say that the differential input range is 2V, but also 0.5 to 2.5V. A single ended signal swinging bewteen 0.5 to 2.5V is 2Vp-p, a differential signal with this swing on each input would be 4Vp-p. Also, if the swing is 0.5 to 2.5V, then the expected common mode would be 1.5V. Could you calrify this ?
As I said above, the input is quite flexible, and is the configuration of the input pins and reference for a variety of conditions are described on pages 10-19 of the AD9220 data sheet. Once you give me some clarification, I can probably point you to a more specific portion of this section to help you with your application.
I moved this question about the AD9220 to the High Speed ADCs Community. Someone here should be able to assist.
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my input is differential, each signal range is from 0.65 and 2.65, therefore the input span is 5Vpp. I'm not sure if I understood well the meaning of Vref. May I set it to 2V or 5V 8I'm using the configuration with two external resistor (Vref= 1V* (R1/R2))
thanks a lot
VREF sets the full scale input range of the ADC . which will be 2 X VREF (see page 12 of the data sheet, under REFERENCE OPERATION). It can be used to bias the analog inputs as well, but the primary function is to set the input span. It looks like to me that each phase of the differential signal is 2Vp-p, making the differential input signal of interest 4Vp-p, not 5Vp-p. If you want to set the full scale input range to 4Vp-p, then you will need R1=R2, and VREF will be 2V. It looks like your input signal has a common mode of 1.65V, this should be fine as is and not need any additional biasing.
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