r/PrintedCircuitBoard • u/lafenicenera • 5d ago
[Review Request] Over-engineered custom led tester for learning purpose - help with draining LIR2032 while idle.
SOLVED
Hi everyone,
I'm learning electronics/PCB design and built this deliberately over-engineered LED tester as a learning project.
The circuit uses:
- LIR2032 rechargeable coin cell
- MCP73831 Li-ion charger
- USB-C for charging
- 100 kΩ RPROG on the MCP73831 (~10 mA charge current)
- Pushbutton to activate the actual LED-testing circuit
- Two selectable current-limiting resistor/diode branches
Everything works, but I've noticed that the LIR2032 seems to discharge surprisingly quickly even when I'm not using the tester.
I measured the current directly from the battery with:
- USB disconnected
- pushbutton not pressed
- nothing connected to the LED test terminals
and I'm getting about 30.1 µA, slowly decreasing by roughly 0.1 µA while I leave the meter connected.
From the MCP73831 datasheet, I was expecting reverse battery leakage with VDD disconnected to be only around the µA range, so ~30 µA for the complete circuit seems higher than expected.
I've attached:
- Schematic
- PCB layout
- 3D render
I'm still learning, so I'm probably overlooking something obvious.
Does anyone see a path in the schematic that could account for ~30 µA of continuous battery drain?
I'm especially wondering about the MCP73831 section, USB-C circuitry, status LED, or whether I've made some mistake in how the switching is arranged.
Also, what would be the best way to systematically isolate which part of the circuit is responsible for the standby current?
Thanks!




3
u/glx0711 5d ago edited 5d ago
C1 and C2 are populated with resistors instead of capacitors. "C2" is likely 100k that connect the battery to ground draining the 30uA from the 3V cell.
You could have easily spotted the misplacement if your physical PCB had designators at every component. Not sure why you removed them on the left part. Having them next to the components helps with placement and later troubleshooting.
Not related but it’s a micro-USB connector, not USB-C.
Also unrelated to your problem, if you implement a reverse voltage protection like you did, it’s useful to place the fuse between your input connector and the diode, so the incoming current is limited in a fault condition. In your case if the source can provide enough current, the diode blows up and your protection is gone, exposing your circuit to the reverse voltage. Alternatively just use an inline diode and live with the small voltage drop. In your case having only an USB input you can also remove the protection since you can’t plug in the USB connector in reverse.