# Latch power on-off circuit

See the circuit below:

I am trying to make a shutdown circuit for a low power application. Basically, I want the system to be able to be turned on by a button push, and then turned off by a microcontroller. But I also want to be able to read the button state while the circuit is on.

In the circuit shown, when the circuit is off, the "E" input is high. This means that Q=D, which defaults to low when the button is not pressed. When the button is pressed, D (and consequently Q) go high, turning on the FET and enabling the circuit. This also brings "E" low, preserving this state.

Then, when it's time to turn off the circuit, the microcontroller pulls "E" low and turns everything off. It's important to note that the pin that "E" is connected to is normally a high impedance input.

My question is to what the state of "E" will be in the off state. This circuit is based on the assumption that "E" is high when the FET is off. However, I'm not 100% sure this is the case. My next step is to build this circuit, but I wanted to see if anyone could find something wrong before I buy the parts and try it out.

Thanks!

Part numbers:

D-latch: SN74LVC1G373DCKR

MOSFET: NTMFS4927NT1G

Battery: 18650 Li-ion battery

• Where MCU Button wire goes? if it somewhere inside LOAD, then when circuit goes off, it possibly will be driven high by MCU's protection diodes and circuit will turn on again. – Vovanium Apr 22 '14 at 16:10
• When "E" is high, the circuit is off (unless the button is pressed) which is the desired behavior. The resistor to the drain of the FET actually acts as a pull-up in this case, because transistor is off. – Joshua Franz Apr 22 '14 at 16:57
• Small comment - I've seen issues with button inputs to latches and F/Fs. Specifically, the digital hash produced by a bouncing button can violate setup and hold times and cause all sorts of issues. Your mileage may vary of course but it's worth knowing :-) – carveone Apr 23 '14 at 0:17