AD8138
Production
The AD8138 is a major advancement over op amps for
differential signal processing. The AD8138 can be used as a
single-ended-to-differential amplifier or...
Datasheet
AD8138 on Analog.com
AD9283
Production
The AD9283 is an 8-bit monolithic sampling analog-to-digital converter with an on-chip track-and-hold circuit and is optimized for low cost, low power...
Datasheet
AD9283 on Analog.com
Newbie designer here. I'm trying to design a circuit to interface a 1 Vpp (into 75 ohm) video signal DC-coupled via a AD8138 to a AD9283. I used DiffAmpCalc to design an input stage around the AD8138 (see screenshot). The AD8138 is powered by a +3.3V and a -3.3V LDO regulator; the +3.3V regulator also powers the analog side of the AD9283 (Vd).
At the output of the AD8138, DiffAmpCalc shows a 0.490 Vpp signal centered around Vocm on both the positive and the negative output. The differential voltage between the outputs, ie. VoutP - VoutN, would then be 1 Vpp centered around Vocm.
Now, the AD9283 datasheet mentions in the specifications:
I don't understand the input voltage range specification. I think it agrees with the output voltages of the AD8138 described above. Is that correct? Why does it say +/- 512 mV peak-to-peak (or p-p)?
Regarding the common mode voltage (Vocm), I've set it to 0.99 V in DiffAmpCalc, i.e. 0.3 * Vd, as mentioned here Common-mode voltage AD9283-100. However, I'm not enitrely sure how this relates to the +/- 200 mV specification for the common mode voltage. Also, what is better, 0.9V or 0.99 V, or can I simply use 1V? I guess that in practice it may be difficult to supply exactly 0.99V?
