My micro USB mobile charge cable is having about 98V AC voltage on the outside surface. (PFA measurement image. The black probe is connected to the Power socket Ground port. I could not take the whole picture with one hand) The DC voltage is about 0.02V w.r.t. the Ground.
Online QA websites suggested that this may be due to a faulty charging adapter. But I have inserted another micro USB cable into the same adapter and it showed only about 1.5V AC on the surface.
Then, I took another charging adapter and the same thing happened. The first cable showed about 47V AC and the second one only 1.3V.
Thus, it seems that it is not the fault of the charging adapter. My electrical knowledge is not very good. Can someone explain why is this happening?
Edit: Just to clear the confusion I created with the example of 2 wires, actually the AC voltage is always there at the adapter output. I have put my multimeter probe on the metal just inside the adapter (without touching any of the 4 wires inside). and the inside was about 98V AC. the 2nd cable's both bare end metals were not continuous. so, the voltage didn't appear on the micro USB end.
So, my question is, why is the AC voltage appearing and does this mean that the adapter has gone bad?
Best Answer
This voltage is due to standard leakage between primary (AC side) part of AC-DC converter, and secondary part. The effective impedance of this parasitic leakage is about 100k - 1000k. The leakage is allowed to be from 75-100 uA for good medical grade PSU. Conformance to UL 60601-1 dictates that the maximum allowable leakage current is 0.3mA. This kind of current is not harmful, and easily gets grounded when the plug is inserted into device. However, high-impedance probes (oscilloscope or good DMM) will show this phantom voltage when the charger is not connected to anything.
As I understand, the coupling (which causes the leakage) is intentional, because fully isolated supplies might built up substantial DC voltage if left floating, which can be really harmful.