# Differential ADC - voltage on probes

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):

the schematics:

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.

• What's this probe you're talking about? I don't see a probe anywhere. Does it come with some extra attachment? – pipe Apr 10 '16 at 19:48
• probe = input (pins for analog input), Vin+, Vin- in schematics – Martin G Apr 10 '16 at 19:56
• "the probes are giving off voltage of -2.5 [V]" under what conditions? – Phil Frost Apr 10 '16 at 19:58
• 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. – Martin G Apr 10 '16 at 20:00
• 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. – pipe Apr 10 '16 at 21:08

## Analysing

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.

## Conclusion

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

• when there is no analog signal, then: U1 pin 3 = 3.1 [V]; U2 pin 3 = 3.5 [V] – Martin G Apr 10 '16 at 20:14
• @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? – pipe Apr 10 '16 at 20:20
• 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 – Martin G Apr 10 '16 at 20:23
• @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. – pipe Apr 10 '16 at 20:27
• Sorry, my bad ... the probes are 2.5 [V] above ground ! and the U1 pin 3 and U2 pin 3 = 2.5[V] – Martin G Apr 10 '16 at 20:29