I want to charge my phone in a controllable manner, using a 555 and a Mosfet transistor.

The 555 is a countdown timer that gives an output (pin 3) of 5V for 1 minute, and then remains at 0V until an event occurs.

This output controls a Mosfet that enable/disable the charging (the charge is provided by a power-bank).

The schematic is this one:


The problem is:

When the the transistor turns off, the DC voltage between the + (plus) and - (minus) of the cell-phone is 0.3/0.4 Volts.

How can I make the output voltage 0V when the transistor is OFF?

I've already checked if the the output of the 555 is effectively 0V, and it turned out that it was like 0.01V HIGH, but I think this has not effect since the Vth of the 2n7000 is 0.8V.

Thanks a lot.

  • \$\begingroup\$ If you remove your circuit, is the voltage difference still 0.3 volts? \$\endgroup\$ – PkP Oct 12 '17 at 13:34
  • \$\begingroup\$ Yes, it could be just noise, the mean value is 0V (measured with an oscilloscope) but the peak to peak varies from 400mV to 700mV. \$\endgroup\$ – MellowRiki Oct 13 '17 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.