Hello, all. Please take a look at this example schematic:


The chip in question is the PIC18F4550 and as you can see power is provided on both sides (with a 100nF cap to smooth noise I guess). Is this strictly required or could you put power into one side and just ignore the other? I know that I have just put power in one side and everything still seems to work fine but I feel uneasy with doing that. I feel that they would not waste pins that could be doing something useful by having an extra pair of power in/out pins.

So my question, at its heart is: for what purpose does this PIC MCU (and others like it) have multiple power pins? Thanks in advance.[/img]
olaf123 wrote:
So my question, at its heart is: for what purpose does this PIC MCU (and others like it) have multiple power pins? Thanks in advance.
It's usually to ensure the current requirements can be met with respect to current capacity of per-pin connections.

As an extreme example, consider a high-power Intel processor (section 7.9), which can draw more than 100A. In the documented 2011-pin socket, there are hundreds of pins allocated to power (section 8) just to satisfy the machine's needs.

You'll often see systems with separate analog and digital grounds, too. The point there is to isolate the analog components from digital noise (which requires careful board-level design).
I've had issues when getting a USB HID up and running with the PIC18F4550 without both VDD/VSS and as Tari suggested you will likely require it if you want to connect anything else to the MCU.

This is the basic circuit that I used (I think):



I did connect an ICSP header as well as per your circuit.

What are you building?
  
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