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I'm trying to wire up an ADS1115 Analog to Digital Converter so I can use it with my FPGA Nandland Go Board.

Can anyone confirm if this is the correct wiring?

I think I've set it up so that I can read in the voltage of the potentiometer on A0 (through the LED, to see it working)

I'm concerned that I've connected the unused A1, A2, A3 to ground, and left Alert hanging...is that ok?

The pull up resistors (4.7k) are both connected to SCL & SDA lines

enter image description here

Thank you

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  • \$\begingroup\$ It looks like A0 is shorted to the blue rail on your breadboard. Check it with a multimeter. \$\endgroup\$
    – ErikR
    May 26, 2021 at 21:08
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    \$\begingroup\$ Where's the schematic? \$\endgroup\$
    – user16324
    May 26, 2021 at 21:11
  • \$\begingroup\$ Schematic. A schematic shows what you mean to do. The breadboard photo shows what you actually did. \$\endgroup\$
    – DKNguyen
    May 26, 2021 at 22:56

2 Answers 2

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I'm concerned that I've connected the unused A1, A2, A3 to ground

  • I'm concerned why you shorted the Pot to the ground ( along with A0).
  • When the wiper goes to V+, you will be shorting the LED between V+ and 0V without a current limiting R in series. Confirm cathode (-) on LED is on correct polarity.
  • you will get slightly better signal quality for > ~10cm wires using a twisted gnd. around signals. This gives some protection from stray pulse noise. 1k pullups improve even more for I2C.

+1 for neat layout/ -1 for no schematic {even on a paper napkin;}

let's see if YOU can follow my layout edits without a schematic enter image description here

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There are a few mistakes.

Your intention is to measure the potentiometer wiper voltage. You've connected the wiper to A0, which is correct. But then it seems that you were tempted to close that circuit by connecting A0 to ground. Doing so is misguided: that will just short that entire branch of the circuit to 0V. You don't need to close that circuit because you want to peek at the wiper voltage, not draw any current from it.

As a matter of fact a very tiny amount of current will be drawn from A0, through the ADC, and back to ground. This is an undesired side-effect of the circuit the ADC uses to measure the voltage. ADC designers do their best to keep this very small, in the order of micro or nano amps, because that current could modify the voltage you're trying to measure.

For the same reason you can't measure the voltage "through" the LED to see it working, because almost no current should flow through that branch, so it would not turn on. It would also modify the voltage you measure.

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