I use AD9057 with fpga in my oscilloscope project. Here is the ADC pin-out pinout

I use the following scheme to test ADC: VREF OUT connected to VREF IN, this gives internal reference voltage +2.5V

Vd +3.3 V

VDD +5 V

Encode - clock signal 3.3V (i have tried different frequencies according to datasheet)

I use constant voltage level from 2V (00000000) to 3V(11111111) according to datasheet (for Vref 2.5V), and i try to see valid bit patterns on D0-D7 pins, but it seems that only noise there.

1.) Is my test valid?

2.) I found a scheme in datasheet

ADC simple scheme from datasheet

And i don't understand why AIN is blocked with capacitor (how can i measure constant voltage in this case?) In my test i use the same scheme but without this capacitor. Maybe this is the reason of ADC's strange behavior?

  • \$\begingroup\$ "A" in AIN may stand for "audio", which is also supported by the 1 Vpp input signal. Even hi-fi audio only goes down to 20 Hz, so the DC level can be discarded. In fact, it's a good idea to get rid of frequencies that you know aren't in your signal. \$\endgroup\$ – Olin Lathrop Nov 7 '13 at 19:15
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    \$\begingroup\$ I think it stand for "analog", no? \$\endgroup\$ – Alex Hoppus Nov 7 '13 at 19:26

In the same datasheet , indeed, on the same page is the following diagram: enter image description here

The diagram you've included above is captioned "capacitively coupled"

Given you're only capacitively coupling a DC signal you should only see "noise".

  • \$\begingroup\$ Yes i saw this. But why i can't use AIN pin directly without this amplifier, and without capaciator (i mean capaciator on AIN pin in the scheme i included above)? \$\endgroup\$ – Alex Hoppus Nov 7 '13 at 19:36
  • \$\begingroup\$ The opamp there is just doing the signal inversion. That's why its called Inverted VIN. No need for that one. \$\endgroup\$ – Dzarda Nov 7 '13 at 19:51
  • \$\begingroup\$ @Dzarda Does it mean that i can transfer constant analog signal directly to the AIN without any extra elements? \$\endgroup\$ – Alex Hoppus Nov 7 '13 at 19:55
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    \$\begingroup\$ Well, you might be. The datasheet says it's got input resistance of 150kOhm, which is pretty high. Not quite as high as I would expect though. That means, if you have a signal that has high output impedance, you get noise. Take a look here. en.wikipedia.org/wiki/Operational_amplifier_applications Particularly Unity-gain Buffer. \$\endgroup\$ – Dzarda Nov 7 '13 at 20:00
  • \$\begingroup\$ @Dzarda I think big source output impedance is not my case. \$\endgroup\$ – Alex Hoppus Nov 7 '13 at 20:09

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