mosfetswitches
I'm trying to create a load switch using a pair of MOSFETs
However, I'm confused about MOSFET threshold voltages. I need to switch about 0.1A at 3.3V. GPIO2 is 1.8V logic. What do I need to read in the IRLML2502 datasheet to figure out whether this 1.8V will be enough to turn Q1 on, thus pulling the gate of Q2 to ground and causing Q2 to conduct between drain and source?
By placing a zener diode (or similar) between the drain and gate, the gate voltage will be forced to rise if the drain exceeds the breakdown voltage of the diode. You are correct in thinking that this will turn the MOSFET on. This is how that protection method works. To see why you have to understand MOSFET avalanche. The body diode of the MOSFET will break down at some voltage like a zener diode (or avalanche diode to be strictly correct). A power MOSFET is made up of many cells in parallel with each other. Depending of the processing of the transistor, these cells may or may not share the avalanche energy well. Also the place within the cell where the avalanche energy is dissipated is not quite the same as where normal conduction dissipation occurs. For this reason some MOSFETs have a limited amount of energy that they can safely absorb in a single pulse of avalanche. This energy may be much less then they could take in a pulse of normal conduction. It may be zero. On the other hand, some MOSFETs have avalanche energy limited only by thermal considerations, in other words they can take any amount of avalanche energy as long as the die does not overheat.
By turning the MOSFET on before the drain-source voltage exceeds the avalanche point, the energy is dissipated by the normal conduction mechanism. This allows a MOSFET which cannot take high avalanche energy to safely dissipate a transient, for example from and unclamped inductive load. You still have to ensure that in your design the MOSFET cannot be exposed to a transient that is too large from a thermal point of view, the energy will heat the die up just the same and you must not exceed Tj max.
You have to be careful adding diodes to the gate terminal. In some circumstances diodes connected directly to the gate can result in very high frequency oscillation of the MOSFET. This results from parasitic inductance and capacitance in the circuit causing a feedback path at VHF. A diode can rectify this and pump up the DC level on the gate resulting in a stable oscillation maintaining the device in a partially on state when trying to turn it off. This is mitigated by adding a suitable gate resistor.
If your circuit topology ensures that the MOSFETs will never see excessive voltage, or if they can withstand any transients by avalanche then you may well not need to provide any protection components. This is a good solution because there can be more problems introduced by the protection devices (too big a subject to cover here). Decoupling the supply rails near to the output devices in power stages helps. Ensuring that devices in bridge configurations cannot cross-conduct is also very important.
For the type of application you mention in your question you could think of an N channel MOSFET as a relay (but without isolation between contact and coil). In other words the "relay contact" is between drain and source and the "coil" is between gate and source.
In your application, the source connects to 0 volts and the gate connects to your digital control voltage. The voltage applied between gate and source turns the "relay" on and connects drain to source. As this imaginery device is equivalent to an N channel MOSFET current should only flow between drain and source.
As you apply a gate voltage the "contact" doesn't suddenly change from open to closed - it happens gradually as gate voltage increases gradually. If gate voltage increased rapidly then the "contact" resistance changes from "open" to "closed" rapidly.
Lowest contact resistance is achieved (for an N channel MOSFET) when gate voltage is highest with respect to source AND, like a normal relay, if you put too much voltage across the coil (gate to source) you will damage your MOSFET.
If you don't apply enough voltage between gate and source you get a half-on "contact" of anything between mega ohms and a few tens of ohms. You must choose a MOSFET that can near-fully turn on with your logic control voltage of 1.8 volts. This will usually mean that the gate threshold voltage in the data sheet will need to be sub 1 volt.
So, it's like a three terminal relay with coil between gate and source and contact between drain and source.
Best Answer
You are looking for the Gate Threshold Voltage, marked Vgs(th) in the datasheet. There is a bit of a catch to watch out for here though.
For this particular MOSFET, if we look at the relevant bit of the datasheet:
We can see it's given as a minimum of 0.6V, and a maximum of 1.2V. This looks promising for your 1.8V logic output. However, the catch is that this is for a drain current of only 250uA, so it's not of much practical use, since we'll need more than that to pull the PMOS gate low with the 4.7kΩ resistor, we need to make sure the drain current is acceptable at 1.8V.
To get a better idea of what voltage we need at the gate to sink useful currents at the drain, we would usually need to take a look at the graphs for Id over Vgs and Id over Vds for different gate voltages is also useful:
Unfortunately both only goes as low as a gate voltage of 2.25V, so we don't really know how much current the drain will sink at 1.8V. Since at 2.25V Id is over 10A, it's a reasonably safe bet that at 1.8V we'll be fine sinking 3.3V / 4.7kΩ = 0.7mA, but we can't be certain, which is ideally what we need to be.
Solution options:
Personally I'd go for using an NPN, since it's cheaper - you could get a suitable part for a couple of pence, compared to the 33 pence for the MOSFET.
For sticking with this part though, unless you need fast switching, or there is a lot of noise present, I'd probably go for simply increasing R2, to around 15kΩ, so you only have to sink 3.3V / 20kΩ = 220uA. This is less than the value given at a max of 1.2Vgs, so you can be certain it will be able to pull the gate down easily.
One other possible option which I use regularly is to use an IO in open drain mode - many microcontrollers have certain pins which are tolerant of higher voltages than their supply voltage, so if your micro has a 3.3V tolerant pin which can be used in open drain mode (you don't give the part number so I can't check this, although you may have done so already) then you can do away with Q1 completely and use this instead.