power on/off circuit for battery operated device

[circuitmangler]

Hi,

I have a device which operates off of a 9V battery, and I'm investigating removing the need for a power switch as follows: the device has an earphone jack, and so I was thinking of using the shunt lead of the jack (which is grounded when no plug is inserted and is an open circuit otherwise) to switch power on and off.

The device typically draws about 15mA. One idea that seems to be working is to use a 2n7000 with the gate pulled up by a 10 M resistor so that the stand-by current is about 1 uA. The shunt lead is also connected to the gate so that FET conducts when an earphone is plugged in. The FET's drain then becomes the GND for the rest of the circuit.

Questions: is this a reasonable way to do this? Is there a better device or method I should use? And is there a way to even further reduce the stand-by current (just use a bigger pull-up and what are the limits?)

Thanks!

 

[jimkirk]

That should work. The gate capacitance is 60 pF, so it will take a modest fraction of a millisecond to turn on. There might also be issues with noise and circuit board contamination creating unexpected leakage paths.

Also, at high temperatures MOSFET leakage increases. (Roughly doubles for every 10 Kelvin rise). At 125 C the device is specified at 1 milliamp drain source current with Vds of 48 volts and zero gate source voltage.

Gate-body leakage is -10 nA max, so you're well away from that, and I wouldn't worry about getting the resistor current any lower. An alkaline 9 volt battery (Energizer as an example) has a rated capacity of 625 mAh. At 1 uA, that drains the battery in 625 Khours, or about 71 years. The battery will self-discharge long before that.

As an alternative, you can get miniature phone jacks with single pole double throw (or even double pole) switches, which will completely eliminate the circuit leakage issue.

If you already have a jack that you're trying to use, this should work.

 

[geekEE]

If the MOSFET is connected to the jack, make sure there's some sort of ESD protection.

 

[circuitmangler]

jimkirk, geekee - thanks for the feedback!