1 Vpp video via AD8138 into AD9283: confusion about analog input voltage ranges

Hi 0xa000,

Thanks for your interest in AD9283.

0xa000 said:

Now, the AD9283 datasheet mentions in the specifications:

• Input voltage range (with respect to Ain) of +/- 512 mV p-p
• Common-mode voltage +/- 200 mV

Input voltage range (with respect to Ain) relates to the differential AC signal with typical value of +/-512mVpp or 1.024Vpp for AD9283. If you relate that one to the Diff-Amp, it means Vspp given that the Vsdc = 0V. 

0xa000 said:

Why does it say +/- 512 mV peak-to-peak (or p-p)?

The +/-512mVpp is the optimized full-scale input range. This value also the basis for the dynamic performance measurements according to footnote 5 in the datasheet.

Common-Mode Voltage means the DC voltage required by AIN pin. With reference to 0.3*Vd or 0.3*3.0V equal to 900mV which is the typical Vcm, the +/- 200mV values will give the allowable range as stated below:

• Min common-mode voltage: 900mV - 200mV = 700mV
• Max common-mode voltage: 900mV + 200mV = 1100mV

0xa000 said:

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?

Either 0.9V or 1V is allowed since it falls under 0.7V or 1.1V. You can refer to the figure below or these links below:

MT-075: Differential Drivers for High Speed ADCs Overview (analog.com)

AN-1026 (Rev. A) (analog.com)

Regards,

Xavier

Hi Xavier,

Thanks for your reply! I'll have a look at the links you mentioned.

Your DiffAmpCalc screenshot looks at bit different from mine. In my case, the input signal is always above ground. When I measure the source signal terminated into a 75 ohm resistor to ground, what I get is 1 Vpp + 0.5 Vdc. So, I used a Vspp of 2V, Vsdc of 1 V, 75 ohm source resistance, and 75 ohm termination in DiffAmpCalc to match these measurements.

Then I try to subtract the 0.5 Vdc via the inverting input, and use a gain of 0.5x and a Vocm of 1V to (I hope) match the output voltages to the range that the AD9283 expects.

But I'm not really sure what I'm doing... :-)

The voltages on the output (VoutN, VoutP) we get are the same, though. And these are in line with the AD9283 specs right?

Regards,

Joris

Hi Joris,

Sorry for the late reply. 

Here are my points from ADC's point of view (not really a DiffAmp expert :))

• The VOCM that you had set which is 1V is compliant to 0.7-1.1V range required by the ADC.
• For your signal which is +/- 490mVpp or 0.98Vpp and is still acceptable in ADC but slightly off from the optimize value of +/-512mVpp or 1.024Vpp.

Thank you and regards,

Xavier

Hi Joris,

Sorry for the late reply. 

Here are my points from ADC's point of view (not really a DiffAmp expert :))

• The VOCM that you had set which is 1V is compliant to 0.7-1.1V range required by the ADC.
• For your signal which is +/- 490mVpp or 0.98Vpp and is still acceptable in ADC but slightly off from the optimize value of +/-512mVpp or 1.024Vpp.

Thank you and regards,

Xavier

Hi Xavier,

Thanks! It's not uncommon for the type of video input signals the circuit needs to handle to go slightly over 1Vpp, so it may actually be nice to have a bit of head room I guess.

Good to know that at least in theory the circuit stands a chance of working. Now to build it and see what happens in reality... :-)

Thanks again,

Joris