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@trebroNdotnet I had the same question about D190 last night on Discord, and cdhooper enlightened me:

"That is a bit confusing at first. What is happening here is that the diode, at the input to D191, is generating about 5.6V (assuming a vdrop of 0.6V due to the 1N4148 at D190). Then, D191 drops that 5.6V by 0.6V to feed 5V into RP5C01 pin 18. So it's a clever trick to avoid the voltage drop that sourcing voltage through a diode normally incurs."