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Category: Datasheet/Specs

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:

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

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?

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• Hi 0xa000,

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.

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
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,

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,

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

• 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

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