Disabling parts of circuit to save power mosfet vs transistor

I am trying to make a very low power sensor board that should run for years from a coin cell. Several of my sensors require 3.3v but the coin cell only provides 3v so I am using a boost chip. I want to be able to turn the boost chip on/off as most of the time it is not required. I also have a RF device I wish to turn off when not in use.

I am unsure if it is best to use a P-channel mosfet or a pnp transistor. I need the devices to be high-side switched as they require a good ground plane.

I want the power consumption to be as small as possible. I have attached an initial schematic.

3V logic level p-channel small power MOSFET (this will be SMT), driven via
a CMOS inverter which is permanently powered - that way when the arduino
is sleeping the invert input is LOW, the inverter output is then 3V and keeps the
p-channel switched off.

Alternatively find a boost converter with an enable input and very low off-current
(1uA or less ideally).

Be aware coin cells can only provide a few mA before the voltage droops.

Would a pull up resistor do the same job as the powered CMOS inverter? Or would the CMOS inverter use less power?
I tried to find a very low off-current boost converter but they are a lot more expensive and using a MOSFET to turn it off must use less power.

I have been looking for mosfets on farnell but they all seem to be for negative voltages…
Such as:
http://uk.farnell.com/nxp/nx3008pbkw/mosfet-p-ch-30v-200ma-sot323/dp/2069551

Will this work or do I need one for positive voltages? I am confused as most are for negative which seems strange.

Do you think these will do the job as the inverter?
Farnell: 1201258
http://uk.farnell.com/nxp/74ahc1gu04gw-t1/74ahc-single-gate-smd-74ahc1gu04/dp/1201258

On Resistance Rds(on): 2.8ohm - that is too high.
Rds(on) Test Voltage Vgs: -4.5V

Find a Logic level, Low Rds, P-channel MOSFET.
You can find them with very low gate levels. Since you are planning to run on 3V, find a 1.8V gate part.
Negative voltage just means the gate must be lower than the supply voltage to turn it on.

Be aware that control signals to devices with their power removed must be taken low - otherwise they can be powered via control signal voltage “leaking” thru the input protection diodes.

Ah thanks for the info. I am a bit confused between threshold voltage and test voltage.

Is this one any better? http://uk.farnell.com/nxp/nx2301p/mosfet-p-ch-20v-2a-sot23/dp/1894738 Transistor Polarity: P Channel Continuous Drain Current Id: -2A Drain Source Voltage Vds: -20V On Resistance Rds(on): 0.1ohm Rds(on) Test Voltage Vgs: -4.5V Threshold Voltage Vgs: -750mV

I would ideally like a lower drain leakage current but 1uA seems to be standard.

Thanks!

With 3V battery, you're going to need lower gate drive. http://www.digikey.com/product-search/en/discrete-semiconductor-products/fets-single/1376381?k=p-channel%20mosfet&stock=1 See some of the 1.2, 1.5, 1.7, 1.8V parts here.

That MCP1640 has an EN input and a shutdown current of < 1 μA. No need for a FET there unless 1uA is still too much.


Rob

For MOSFETs the threshold voltage is nothing to do with switching it on, its the point it switches off. Unless you are planning to use a non-zero voltage to switch it off its irrelevant.

You look for something like "Rds(on) = 0.04 ohms for Vgs=4.5V", meaning it needs 4.5V of drive to switch on to that resistance. The lowest Vgs quoted is the minimum gate drive voltage.

A CMOS inverter takes no current when the inputs are static (well, a few nA only), any pull-up resistor will take orders of magnitude more current, unless in the 10M ohm range, when noise pickup will be a show-stopper.