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In a digital circuit like a cell phone. If a switching ic has it’s reference point as PGND, separate from the main GND (4v Main vcc coming from battery), but they’re tied together, then

  1. What exactly makes them separate from one another?
  2. Why have two references?
  3. What would happen if the switching ic’s PGND were to be jumpered to the main vcc GND?
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    \$\begingroup\$ real wires aren't perfect and you dont want noisy ground currents from noisy switching circuits flowing through the reference of a quiet circuit. For an IC with two grounds, it might just take too much copper inside the IC so they want you to do it on the PCB instead \$\endgroup\$ – DKNguyen Nov 4 '19 at 0:19
  • \$\begingroup\$ But they’re tied together at some point? So having different return paths rather than one big ground for every circuit dissipates noise by the time they all reach each other at the same place? \$\endgroup\$ – ohmmy Nov 4 '19 at 0:23
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    \$\begingroup\$ Yes, they're tied together at some point. The trick is picking a point to connect them that minimizes the current flowing from one to the other. \$\endgroup\$ – The Photon Nov 4 '19 at 0:46
  • \$\begingroup\$ Not so much dissipates as keeps the noise contained, and away from sensitive analog components. See this question for more information, if not outright duplication. \$\endgroup\$ – TimWescott Nov 4 '19 at 1:12
  • \$\begingroup\$ @ohmmy Noise doesn't work that way. How does a noise current dissipate? It doesn't. Current that enters one end of a wire must come out the other end, no matter how long the wire is. At the same time, you don't want to force return currents to take the long way around because you have a single point they must pass through because that also makes more noise due to higher inductance which might seem contradictory. That's why you keep everything partitioned: so return currents don't stray where they aren't wanted but those that need to cross can take small, tight loops. It's nuanced. \$\endgroup\$ – DKNguyen Nov 4 '19 at 1:36

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