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First off, I'm fairly new to electronics, I don't have much knowledge so bear with me. I saw a schematic of an ADC connection to an MCU, and I noticed that the grounds are split into two - Analog and Digital ground. I read up on why they are split. It's to prevent the high frequency noise of the digital signals to interfere with the analog components.

In theory I can understand but practically, if I were to make any product where I use an ADC, then if I were getting power from the mains or a battery, ultimately won't the grounds meet at one point? So then what is the point of separating the ground lines. How does this whole thing work?

Thank you

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  • \$\begingroup\$ Remember that all PCB tracks present some impedance to signals and there will be a slight voltage drop along the tracks due to this impedance. Analog circuitry is generally less tolerant of shift in voltages on the ground line - it will appear as an offset or noise and may affect the output, digital noise on an audio circuit, for example. By only connecting them at one point this can be avoided. \$\endgroup\$
    – Transistor
    May 23, 2016 at 7:42

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The simple answer is that it keeps digital and analog ground currents separated. Digital ground current is typically "noisy". If the analog and digital ground are intermixed, noisy digital ground currents can induce noise into the analog parts of the circuit.

But if the circuit is configured properly, the digital and analog parts of the circuit are kept independent except at one controlled point configured so that digital ground currents don't enter any parts of the analog circuit. This is often back at the bulk smoothing/filter capacitor at the output of the power supply.

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  • \$\begingroup\$ So ultimately they are connected at one point then. So the grounds are merged? \$\endgroup\$
    – user108320
    May 23, 2016 at 6:41
  • \$\begingroup\$ Yes. The system would not work properly unless the grounds all connect together at SOME point. But one can select WHERE that point is to minimize digital noise getting into sensitive analog parts of the circuit. \$\endgroup\$ May 23, 2016 at 6:55

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