So I was playing around with my amica nodemcu. So here’s the story. I first powered up my nodemcu with a power bank which outputs max of 5v 1A so then after that I was playing around with the pins of the nodemcu. I saw a pin named ‘VIN’. I thought it outputs 5v of power and then I took a 9v battery and connected the live terminal to the vin and neutral to Ground and immediately the nodemcu built in led was so brightly lit up. I figured out what was happening and checked whether my nodemcu was working. It worked fine. Should I worry about damage to my board. I do not see any visible damage.
2 Answers
You stressed every component (the 9V battery, power bank and node MCU) in your setup (most likely) beyond their specs.
You introduced a short circuit between 9V and 5V potential (stressing the battery and the power bank) and applied some voltage between 9V and 5V to the node MCU:
simulate this circuit – Schematic created using CircuitLab
The resistors reflect the wires you used. Normally, we assume that wires have no resistance, but when a "short circuit" (really just a low-ohm connection where none should be) happens, we have to take that into account to estimate what’s happening.
Since we don’t know the exact resistances, one can only guess. But you did most likely not apply 9V to the VIN terminal, but it was most certainly above 5V.
If R7=0, R3=0, R5=0, R8=0, R4 == R6 and the internal resistances of the power bank and battery are neglectible (e.g. by being equal), you applied (9V+5V)/2 = 7V.
In the worst case (all Rs = 0, except R4>0), you applied 9V.
If R3, R5, R8, or R7 are non-zero (they are non-zero), the voltage is further reduced (this is easy to overlook).
I don’t know a lot about battery chemistries and circuitry, but I suspect that the 5V source might be unhappy with having current sinked into its output (which is what happened, too).
If the MCU still works, good for you. Not all damage is visible though, and not all damage might be apparent immediately. Some peripherial you don’t use right now but may use in the future (for example, an ADC) might be more sensitive to excess voltage than other components. It could also be that the part is still functional, but does not perform as well (for example, a capacitor or resistor might now fail closed, which could for example increase analog noise or some current consumption).
You provided 9 v in parallel to 5 volts, meaning it only saw 9v not 11. Parallel voltages do not add up.
You may have damaged it, you might have not. You can't tell unless it's not working the way you think it should be. If anything, just buy a new one, they're cheap.