I got the EVAL_AD9784_PMDZ and Zedboard reading data. It is differential ADC, 2 probes with 1 common ground. The ground is at the ZedBoard GND level (as expected). The voltage input is +7.5, GND, -2.5 V on connector J3 - according to specification and 3.3[V] and GND on Zedboard.

Lately, the probes are giving off voltage of 2.5 [V]. I would like to ask, whether it is faulty behaviour.

I havent measured the difference between probe and ground before, when it seemed to work properly, but I would expect the probe wouldnt force any voltage on measured signal.

EVAL_AD9784_PMDZ (please mind that jumper SL1 is in position B on my board - allegedly correct position): enter image description here

the schematics:

enter image description here enter image description here enter image description here

Could the troubles come from U8 (second picture in schematics)?


When there is analog signal 2.5 [V] (peak-to-peak), 0 [V] offset, 0.3 Hz, and -VS_EXT is disconnected the board works fine, except when signal is getting closer to 0 [V].

From my point of view, this implies that there is something wrong with the negative voltage supplied to both OP-AMPs on inputs.

  • \$\begingroup\$ What's this probe you're talking about? I don't see a probe anywhere. Does it come with some extra attachment? \$\endgroup\$
    – pipe
    Apr 10, 2016 at 19:48
  • \$\begingroup\$ probe = input (pins for analog input), Vin+, Vin- in schematics \$\endgroup\$
    – Martin G
    Apr 10, 2016 at 19:56
  • \$\begingroup\$ "the probes are giving off voltage of -2.5 [V]" under what conditions? \$\endgroup\$
    – Phil Frost
    Apr 10, 2016 at 19:58
  • \$\begingroup\$ when the board is connected (as described above, small edit). Then there is voltage -2.5[V] between any of the two analog input pins (probes) and the ground. \$\endgroup\$
    – Martin G
    Apr 10, 2016 at 20:00
  • \$\begingroup\$ If SL1 is in position B, Vin- is never used, meaning that you don't have a differential ADC anymore. It has to be in position A if you want to use it as a differential ADC. \$\endgroup\$
    – pipe
    Apr 10, 2016 at 21:08

1 Answer 1



Tracing out what you're measuring, let's start with Vin+. If nothing else is loading the input, I would expect to see 2.5 volt. Why? Because the only thing connected is VCM, AGND, and the operational amplifier U1's input.

Assuming everything works correctly, the input draws a negligible current so the current through R5 can be ignored. VCM is driven hard to 5 volt by U4, and is connected to Vin+ by a voltage divider with AGND through R1 and R3, forming a resulting 2.5 volt.


Yes, seeing 2.5 volt on each input is normal and expected.

  • \$\begingroup\$ when there is no analog signal, then: U1 pin 3 = 3.1 [V]; U2 pin 3 = 3.5 [V] \$\endgroup\$
    – Martin G
    Apr 10, 2016 at 20:14
  • \$\begingroup\$ @MartinG If pin 3 on U2 is +3.5 volt, and Vin- is -2.5 volt, R6 would dissipate 0.7 watts and burn up in smoke. How are you actually measuring this? \$\endgroup\$
    – pipe
    Apr 10, 2016 at 20:20
  • \$\begingroup\$ NO ANALOG SIGNAL : U1 pin 3 = 2.5 [V], U2 pin 3 = 2.5[V], VCM = 5[V], +VS_EXT = 7.6[V], -VS_EXT = -2.6[V] (a little bit off specification) ... board draws 45 mA ................. ANALOG SIGNAL 2.5 Vpp, 0 offset : when -VS_EXT is disconnected the board seems to work fine \$\endgroup\$
    – Martin G
    Apr 10, 2016 at 20:23
  • \$\begingroup\$ @MartinG But do you agree that according to Ohm's law, if you measure 3.5 volt on one side of R6, and -2.5 volt on the other (as you say you do), there would be a 6.0 volt difference. This would cause it to burn up. I can't help you if you don't explain exactly what you measure and how you measure it, because this does not add up. Please update your original question if you have any other information that can help us. \$\endgroup\$
    – pipe
    Apr 10, 2016 at 20:27
  • \$\begingroup\$ Sorry, my bad ... the probes are 2.5 [V] above ground ! and the U1 pin 3 and U2 pin 3 = 2.5[V] \$\endgroup\$
    – Martin G
    Apr 10, 2016 at 20:29

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