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?

enter image description here

  • \$\begingroup\$ The truly brave use a MOSFET for Q2 which is adequately off at Vgs = 1.7V and adequately of at Vgs = 3.3V - then drive it directly from GPIO2. This is doable but due care needed. Worst case values for the 3V3 supply and GPIO2 output swing and the FET on and off Vgs values. Overall probably not worth trying as saving in cost is minimal - but useful to note possibility. \$\endgroup\$
    – Russell McMahon
    Feb 2, 2013 at 12:52
  • \$\begingroup\$ Note that in this case use of almost any jellybean NPN bipolar transistor for Q1 makes design easier than with a MOSFET for Q1, as the low input voltage swing means MOSFET has to be suitably low Vth.|| FWIW information only look at CES2310 and family of MOSFETS. CETSEMI Taiwan - availability varies. Some excellent low Vth parts. \$\endgroup\$
    – Russell McMahon
    Feb 2, 2013 at 12:54

2 Answers 2


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:

Id over Vgs Id over Vds

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:

  • Use a different N-ch MOSFET with a guaranteed drain current of over 0.7mA at 1.8V Vgs given in the datasheet.
  • Use any small NPN transistor which has a Vbe of ~0.7V and will be turned on easily with 1.8V (with current limiting base resistor of course. For a gain of e.g. 100, assuming you want to pull as low as possible, calculate using (1.8V - 0.7V) / (0.7 / 100) = ~150kΩ, then halve this to compensate for gain reduction at saturation, e.g. 75kΩ or less. Another conservative rule of thumb is to assume a gain of 20-30)
  • Do your own tests with a few IRLML2502s and confirm there is more than enough drain current at a Vgs of 1.8V.
  • Increase R2, so less current is needed to pull the gate down.

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.

  • \$\begingroup\$ Thanks - I'd say your answer is just as good, mine just waffles on a bit more :-) Certainly worth +1 too. \$\endgroup\$
    – Oli Glaser
    Feb 2, 2013 at 4:01
  • \$\begingroup\$ When using an NPN with hFE = 100, assuming the collector current is 0.7mA and GPIO2 HIGH is 1.8V, does an R_b of 150Kohm sound about right? \$\endgroup\$
    – Vanush Vee
    Feb 2, 2013 at 4:21
  • \$\begingroup\$ @VanushVee - I'd go for a bit less than this as gain drops off at saturation. Assume a gain of ~20 or so for safety. I'd go for 30k or less. It's very conservative, as you don't even need to pull all the way to ground to turn on your P-ch for 0.1A, so you could go higher if you want to keep current consumption as low as possible. Just make sure design for worst case and you test thoroughly. \$\endgroup\$
    – Oli Glaser
    Feb 2, 2013 at 4:28
  • \$\begingroup\$ @OliGlaser The irlml6402 mentioned as Q2 is only specified in the graphs starting beyond 2 Volts. The Vgs data states 250 microamps at 1.2 Volts. Will it even work as a power switching MOSFET as per this circuit? \$\endgroup\$ Feb 2, 2013 at 4:41
  • \$\begingroup\$ @AnindoGhosh - Yes, as long as the gate is pulled below e.g. 3.3 - 2.25V = ~1V or so, which will give an Id of a few amps according to the graphs. The OP only wants 100mA, so all should be okay. \$\endgroup\$
    – Oli Glaser
    Feb 2, 2013 at 4:44

In order to pull gate of Q2 to ground, the resistance Rds needs to be sufficiently low to overcome the pull-up effect of the 4.7 kOhm resistor R2.

Therefore, in the IRLML2502 datasheet, my first step would be to look for a graph of On Resistance versus Gate Voltage (Figure 11 on Page 6), and locate the point on the graph where Rds drops sufficiently.

This MOSFET is not even graphed for Vgs lower than 2 Volts, so an alternative approach is called for:

  • R2 would pass 702 µA if Q1 were replaced with a short circuit.
  • Hence, if the MOSFET were guaranteed to pass a fair bit more than that current at desired Vgs, the gate of Q2 would be guaranteed to be pulled low.
  • From the datasheet, Vgs ranges from 0.60 to 1.2 Volts, specifying an ID of merely 250 µA at that Vgs.

Thus, there is no data in the datasheet which trivially indicates that the IRLML2502 is suitable for the purpose. It is likely to still work, since the ID will increase rapidly with Vgs going beyond the 1.2 Volts specified above, but this would not be my part of choice for the design.

I would look for a MOSFET which switches on hard (the Rds goes past the knee of the graph) at the expected Gate voltage, OR increase R1, OR use one of the other excellent ideas in the answer by Oli Glaser.

Within the constraints of not changing the design, NXP's BSH105 meets the requirements, at a single unit price of $0.40 on Digikey.

See this graph:

MOSFET transfer characteristics

The knee of the graph is at around 1.2 Volts, and the part is nicely on at 1.8 Volts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.