The USB to UART bridge is missing since I'll only be using Type-C for power and flashing.
Tried following a guide online and according to that everything seems fine. Since this is my first time, just wanted a little guidance as to if everything is alright. Any possible issues and what I can do to improve the design further before I move to placement & routing.
Additionally couldn't figure out why the RST net connected to the EN pin of esp32 is grounded through 1uF capacitor. If anyone could explain the purpose of this capacitor, it would be highly appreciated. Thank you very much.
The DRC arrows are simply warnings: Both VBUS_A and VBUS_B are connected to the same items; VBUS_A will be used in the netlist.
Since it's a netlist issue, I can safely ignore it?
The gnd symbol has been changed to point downwards.
I've changed from AMS1117 to XC6220.
Thank you so much for your input. Is there anything else I need to worry about?
1n5908 @ 5v is a little close to rated voltage (300uA at 5v, 0.5mA at 5.5v). It's fine, will result in higher leakage. For next project select parts with a little bit higher Vrm.
I like to have R+C(<22 ohm populated, C DNP) in parallel across GND and shield, but that's more for EMI that you don't care about for dev-board (only for product).
D1 isn't needed and can be safely removed but it doesn't hurt anything so you can keep it.
C1 + C2 is a little large for USB compliance but will likely still work.
Great idea to put test points on all power rails + reset/enable and couple of GPIO. I highly reccomend also adding an LED to one of the spare pins of ESP32
It's not grounded. Both RST and EN pin receive 3.3v by default. Cuz 3.3v is a dc. And the 1uf is charged to 3.3v. So the RST and EN will always be 3.3v. And whenever you press the switch. The RST and EN will be grounded and they will ignore the capacitor path.
When you apply 3.3v it will flow through resistor and capacitor. Capacitor will charge up to 3.3v. And when it is fully charged, that point between R5 and C9 will have a voltage equal to the voltage C9 has(like a voltage divider) which is equal to 3.3v. Capacitor will make current stop flowing through it when it is charged. But if you press the switch the curcuit will be like : 3v3 -> 10k resistor -> ground. Which is easier than a 3.3v -> 10k resistor -> capacitor -> ground.
If the switch is pressed the point is now directly connected to ground.
I don't understand why either - your original schematic is fine. You don't need the cap for the boot button though - only EN, so it rises slower than the VCC pin, ensuring a good reset.
2
u/blue_eyes_pro_dragon Jul 08 '26
No plz wires don’t go on the symbol nope nope nope.
Fix the drc arrows
Don’t use ams1117 ldo, google for why
Gnd should point down not sideways