I plan to use a differential AD9235 to sample a signal that will range from 0.5 to 2.5V. Currently, I am trying to interface the ADC and learn how it works. As suggested in the spec sheet, I am using a differential driver before the ADC input. I have AD8132 - a similar driver to AD8138, that is in the configuration example in Fig.35 of the ADC manual. And here's how it's connected:

enter image description here

I set the SENSE pin to ground to get 2.0V differential span (I assume that's the range of the input signal), and the MODE pin to ground to get offset binary output (I assume that the lowest input signal will correspond to 0000 and the highest input will be converted to 1111).

Thus, I expect to input a signal 0-2V and expect it to be converted 0 to 2^12. I do get all the 12bit range, however, my input range is very small: for 1V and above, the output is zero. Decreasing the input voltage from 1V to about 700mV provides 2^12 output. In other words, my input range is only 300mV, as opposed to expected 2V span.

enter image description here



REFT =~ 2V

REFB =~ 1V

Given that I set the VREF to be 1V, the calculations are consistent with the equations on page 15 of the DAC specsheet:

REFT = 0.5(3V + 1V) = 2V -> right!

REFB = 0.5(3V - 1V) = 1V -> right!

Span = 2(2V - 1V) = 2 * 1V = 2V right!

So, why is my voltage input span is only 300mV? What am I not understanding correctly? What should I do in the future to be able to input 0.5-2.5V signal?

Also, why would I even care about VREF, REFT, and REFB? Why those pins are made external?

  • \$\begingroup\$ I saw Fig. 35 on that document. There is 1Vpp source connected to input. Note, that 1Vpp (peak-to-peak) is range from -0.5V to +0.5V. Differential ADC results can be negative and positive numbers. \$\endgroup\$
    – Kamil
    May 12, 2014 at 20:18
  • \$\begingroup\$ However I coudnt find any information about data format, so I have no idea if binary output is from -2047 to 2048 or from 0 to 4096. \$\endgroup\$
    – Kamil
    May 12, 2014 at 20:24
  • \$\begingroup\$ @Kamil Well, I tried input voltages from -3V to 3V. However, the ADC seems to change the output at 700mV to 1V range only. The output TwosComplement/OffsetBianry is selected by MODE pin and described in table8. \$\endgroup\$
    – Nazar
    May 12, 2014 at 20:37

1 Answer 1


1 - I hope you're aware that using REFT and REFB to set your output common mode is not suggested by the data sheet. Those pins are brought out to allow filter capacitance to be added. I'd suggest that you derive your output common mode level from VREF - that sort of thing is why it's brought out. I suspect that you're loading down the internal levels, and that's why your full-scale input is only 300 mV.

2 - Are you aware that you've set up the AD8132 for a gain of -1? Just checking. This is why decreasing your input voltage causes an increasing digital output.

  • \$\begingroup\$ Oh, I see. REFT and REFB are understood. Thank you so much. Do you suggest to use VREF as an input to Vocm on the AD8132, or just make it AVDD/2? Now, what means "loading down internal levels"? \$\endgroup\$
    – Nazar
    May 12, 2014 at 20:49
  • \$\begingroup\$ Don't use Vref. For a span of 2.0 volts, Vref is 1.0 volts, which is too small. According to Fig 36 of the ADC data sheet, AVDD/2 ought to work fine. As for loading, think of the internal generator which produces REFT and REFB. Each of these levels will have some impedance associated with it, let's say 5kohms for each. When you attach your external 1k resistors, the outputs will be reduced by a factor of 6 (1k / 1k + 5k), producing a voltage across the internal circuits of 300 mV rather than 2V. You should be able to measure this. \$\endgroup\$ May 12, 2014 at 21:40
  • \$\begingroup\$ Ok, so now I derive Vocm for the AD8132 by halving AVDD. But the problem remains the same. \$\endgroup\$
    – Nazar
    May 12, 2014 at 21:50
  • \$\begingroup\$ At 0 volts input, what are the two ADC input voltages? What is the Vref voltage? What are REFT and REFB voltages? What are the same numbers for .2 volts in? \$\endgroup\$ May 12, 2014 at 22:01
  • \$\begingroup\$ When I ground both +in and -in of the diff. amplifier, the ADCs VIN+ = 840mV , VIN-= 2.06V , VREF = 1V , REFT = 2.12V , REFB = 1.14V. For 0.2V in VIN+ = 750mV , VIN- = 2.12V , VREF = 1V , REFT = 2.12V , REFB = 1.13V. \$\endgroup\$
    – Nazar
    May 12, 2014 at 22:22

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