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Here is the circuit (Arduino Due, 1kΩ resistor, LED): enter image description here

Note how I bring the ground from the Arduino board to the breadboard, and also connect the ADC pin to the gnd rail on the breadboard.

Here is the code:

void setup() {
  Serial.begin(9600);
  #define ADC_FREQ_MAX 250000
  Serial.println(ADC_FREQ_MAX);
    }

void loop() {
  analogReadResolution(12);
  Serial.print(analogRead(0));
  Serial.print('\n');
}

Here is what I do:

  • 600cs: I unplug the LED.
  • 4000cs: I plug in the LED.
  • 6500cs: I unplug the LED.
  • 7750cs: I plug in the LED.

Here is the result:

enter image description here

I plug this same setup into an Uno, and I get no issue: enter image description here

Why is ground voltage non-zero, and why does it increase when I plug in an LED? What issue with the Due could cause this?

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  • \$\begingroup\$ What is the question? \$\endgroup\$ – Leo Jun 26 '15 at 20:11
  • \$\begingroup\$ please read the title! \$\endgroup\$ – Dave Jun 26 '15 at 20:12
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    \$\begingroup\$ LED current * wiring resistance = ??? \$\endgroup\$ – Brian Drummond Jun 26 '15 at 20:13
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    \$\begingroup\$ @Dave - Measure the resistance of each wire with a multimeter (maybe with one end on the LED lead so you include the breadboard resistance). Measure the current through the wire by putting your multimeter in series (in current measurement mode). Now you have LED current and wiring resistance. Multiply the two, and that's the expected voltage drop. \$\endgroup\$ – Justin Jun 26 '15 at 20:25
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    \$\begingroup\$ Ground is zero by definition. Hence, what you're measuring isn't ground. \$\endgroup\$ – Adam Lawrence Jun 26 '15 at 20:41
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In Breadboards with lengthy cables a 50 mV offset is acceptable and is not an issue. If it is affecting measurements, then you should consider shortening the cables and possibly use thicker cables. a 0.5 ohm resistance cable with 5mA current will give 2.5 mV drop. If there are other currents through the ground cable, assume Entire board current, then the drop will go up. For ex. 100 mA current now will cause a drop of 50 mV.

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An Ideal ground is a perfect conductor that we measure all voltage relative to. Oops there is no such thing as a perfect conductor, so the ideal ground exists only in textbooks. In the real world it is just another conductor, with its own resistance (low but present), capacitance (low, but this depends on how close it is to other conductors), inductance (not an issue at low frequencies, but it can be a problem at high frequency and is also dependent on length and proximity), and so on. I don't even understand what implication the flux would have. Most of the time we can just ignore this as it is typically not significant. But this will blow your mind, grab a small capacitor and hook one leg to the ground on your breadboard near the middle and the other leg to a unused rail. Now run your experiment again. Wild, yes?

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    \$\begingroup\$ Your answer would be more useful if you removed the sarcasm at the end and instead just explained why the voltage drop is present. \$\endgroup\$ – I. Wolfe Jun 26 '15 at 21:57
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    \$\begingroup\$ @I.Wolfe, What sarcasm? \$\endgroup\$ – hildred Jun 26 '15 at 22:03
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    \$\begingroup\$ The last 3 sentences come across as sarcastic. Might not have meant it so, but it helps be clearer if you just state what you mean. \$\endgroup\$ – I. Wolfe Jun 27 '15 at 4:11
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    \$\begingroup\$ It will be much clearer still if you do as he says and pay attention to what happens. The general idea for this site is that we answer questions and make suggestions, not that we spoon-feed your mind. \$\endgroup\$ – WhatRoughBeast Jun 27 '15 at 13:44

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