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I am designing a system that incorporates the ADC12DL040 ADC. Page 27 of the datasheet gives the following example schematic for converting a single-ended signal into a 1V +/- 0.5V differential signal as required by the inputs to the ADC:

With the following table for the values of the unspecified resistors:

My signal input ranges from 0-0.5V, so I used the resistor values in the second row. I am trying to simulate the amp in LTspice with the following schematic:

However, the simulation produces the following output, rather than two 1V +/- 0.5V signals:

Any idea where the circuit is going wrong? Thanks!

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Something seems really out of whack with that table/ spec sheet... or else I am not getting how that thing interprets differential.

Note it says

VCM, should be in the range of 0.5V to 2.0V and be a value such that the peak excursions of the analog signal does not go more negative than ground or more positive than 2.6V.

And the signal difference should not exceed Vref = 1V

That makes the table plain wrong....

Try these...

R1 1500
R2 1500
R3 1500
R4 1500
R5 2500
R6 2500

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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  • \$\begingroup\$ I knew something was up with the datasheet. Thanks! \$\endgroup\$ – Billy Kalfus Oct 18 '17 at 21:32
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If you look carefully you will realize that the non inverting input of U6 is connected to ground rather than a voltage divider as shown in the reference schematic, that may well be the cause of the unpredicted behavior

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  • \$\begingroup\$ The table of resistor values says that one of the resistors in that divider should be zero ohms and the other should be open. Hence a zero ohm connection to ground and an open connection to 2V. \$\endgroup\$ – Billy Kalfus Oct 18 '17 at 18:46

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