Current Activated Diode Switch

This is variation on my answer to this question. Like Mike, I am also using a P-channel MOSFET. The OP stated: "The Voltage regulator requires Vin +/- 0.3V on the Enable Pin", and "MicroController Output Pin High state is 2.7V Max". Therefore when the battery voltage is above 2.7v, the µC will be unable to supply the required voltage to meet the voltage spec for the Enable lead. So I am using the control lead of the µC to control the MOSFET, which in turn switches the battery voltage (or Vin) to the Enable lead.

Initially, the control lead from the µC is configured as an input. When the button is pressed, battery power is applied through a Schottky diode (keep Vf below 0.3v) to enable the regulator. The µC then configures the control lead as an output and grounds it, keeping the enable lead high through the MOSFET (which keeps the voltage at the battery level). Meanwhile, the button can be used as an input.

The circuit should draw very little power (less than 1 µA) when the µC is powered down, assuming the regulator shuts down completely when the Enable lead is low.

This eliminates the cost of the LTC2954 chip which runs over $2 in 500 unit quantities. This circuit should be under 10 cents in quantity (except for the switch and regulator).

To return to your transistor solution (as others have opted for a different mechanical switch), it might look like this. I included Wouter's hint for using a diode instead of the second transistor. Also I changed the bipolar transistor to a MOSFET. Reason for a MOSFET is because the base current of a bipolar transistor gets rather high with regular power transistors. Reason for P-channel high side switching is that this way the load can be attached to ground. If this is not a requirement, the circuit can be slightly changed to support an N-channel MOSFET.

• D1 should be rated > 2A
• T1 should be rated Id > 2A (P-MOSFET)

^{simulate this circuit – Schematic created using CircuitLab}