I need to make a switch. However p-n junction drop seems to be a problem for my use. Hence I am unable to use something like this:


In this case, my load won't be getting exact 3.3V across it. Straight forward method will be to use a mechanical relay, but I want to avoid through hole and bulky components. In fact, I want to make it as cheap and small as possible - probably a few resistors, transistors or a small mosfet is what I am planning to use. Is there any good solution to such a requirement?


Load requirement:

3.2V i.e. a drop of 0.1V will work fine. Current will be approximately 200 mA.

As answered by Olin and highlighted out by JRE, a FET definitely solves my problem.

  • \$\begingroup\$ How much current do you expect the load to draw? Why do you need exactly 3.3V? \$\endgroup\$ May 18, 2016 at 18:47
  • \$\begingroup\$ Check out @OliinLathrop's suggestion. Assuming 50mOhm on resistance for the IRLML2502, that is a drop of 0.01Volts. I can't imagine any 3.3V circuit being unhappy on 3.29V. \$\endgroup\$
    – JRE
    May 18, 2016 at 18:55
  • \$\begingroup\$ I checked the datasheet for the IRLML2502, and it indeed has 50mOhm when using a gate voltage of 2.5V. So, at 3.3V it ought to be better. \$\endgroup\$
    – JRE
    May 18, 2016 at 18:57
  • 1
    \$\begingroup\$ Olin has given you the best answer, but keep in mind that you do not experience 'p-n junction drop' with a saturated transistor- for example a ZTX1049A at 2mA base drive will drop maybe 50mV. You don't say what your load is, but if it's something that is that sensitive make sure it's not really a high-side switch you need (most of the things I can think of that are fussy.. ) \$\endgroup\$ May 18, 2016 at 19:40
  • \$\begingroup\$ @Spehro - My load is a microcontroller and few other elements such as optocouplers. Do I need a high side switch? \$\endgroup\$ May 18, 2016 at 19:48

2 Answers 2


A FET should be able to do this. Of course without knowing how much voltage drop you can tolerate and what the load current is, it is impossible to say for sure. You can easily get a small FET that goes down to a few 10s of mΩ in a small SOT-23 package. Check out the IRLML2502. You can get lower on resistance by spending more money, like getting a more exotic FET or putting multiple of them in parallel.

  • \$\begingroup\$ Thanks Olin, for the answer. I mentioned switching voltage and current in the heading. I want to switch 3.3V and draw a current of approx 200mA. Preferable voltage drop is 0V however that seems impractical. Hence I am planning to settle for whatever I could get by using regular components and without spending more than half a dollar. \$\endgroup\$ May 18, 2016 at 18:40
  • \$\begingroup\$ Will this work: electronics.stackexchange.com/questions/235114/… \$\endgroup\$ May 19, 2016 at 5:56
  • \$\begingroup\$ @Whi: No, that won't work. That's a source follower. There will be significant voltage drop from the gate voltage to the source voltage. You want a plain old common source arrangement like you show in your question. Just change the NPN to the right N channel FET, and then you can get rid of R29. \$\endgroup\$ May 19, 2016 at 10:47

To add on to Olin's point of a modern FET being suitable, I'll add a couple more to the mix I recently used. Links go to Mouser (possibly Mouser NL, but Mouser's smart, if you're in Aussie or US it should easily transfer):

These are somewhat "more insane", but due to their aceptable per 10 and per 100 prices, I often stock those for power switching, just to save on losses. I stock two types each, because if a design needs it I need options for "second-sourcing".

Do you need these insane specs? No. But options allow you to make a choice.

Low Side (N-Channel) MOSFETS with a less than 50mOhm RdsOn at Vgs 2.5V:

  • IRLML6244: 20V Breakdown, absolute max 5.1A (Vgs = 2.5V), SOT23, 27mOhm or less at 2.5V gate-source]
  • DMN2041L: 20V Breakdown, absolute max 5.2A (Vgs = 2.5V), SOT23, 41mOhm or less at 2.5V gate-source.

High Side (P-Channel) MOSFETS with less than 50mOhm RdsOn at Vgs 2.5V:

  • DMP2035U: -20V Breakdown, absolute max -3.6A (Vgs = -2.5V), SOT23, 45mOhm or less at -2.5V gate-source and also specified at -1.8Vgs.
  • BSL202SN: -20V Breakdown, absolute max -5.9A (Vgs = -2.5V), TSOP-6, 36mOhm or less at -2.5V gate-source.

For good measure I'll do the same to the IRLM2502 (low side / N-Channel):

  • IRLM2502: 20V Breakdown, absolute maximum 3.6A (2.5Vgs), SOT23, 80mOhm or less at 2.5V gate-source.

The breakdown voltage means the voltage at which the MOSFET's Drain-Source may break down and start conducting, even when turned off. So this should really be quite a bit above the maximum operating voltage in most sensible designs. In your case, I'd choose at the very least 8V, but all types above are 20V, so that's fine.

The On resistance, RdsOn, is the resistance in the channel when the gate-source voltage (the voltage on the gate, compared to that on the source) is the specified number. I quoted you all the MAX figures, often modern MOSFETs will be very close to the typical value, but if you design something you should always look at the worst case numbers.

And the maximum current will greatly depend on the applied gate voltage. If the Vgs is higher, the maximum current will increase, up to the point where the pins of the device, or the internal channel cannot handle any more.

The losses inside the MOSFET will depend on the RdsOn and the current drain. If you have 50mOhm (=0.05 Ohm) resistance in the MOSFET when 200mA goes through it, then the voltage across it will be:

V = I * R = 0.2 A * 0.05 Ohm = 0.01V = 10mV

Which is absolutely negligible. I'm betting your 3.3V power supply isn't even more accurate than "somewhere between 3.2 and 3.4 volts", unless you have a stable and reliable lab-supply that it always runs off, or are using a 1% or less initially accurate regulator.

In terms of power, of course, the losses will be:

P = V * I = 0.01 V * 0.2 A = 0.002 W = 2 mW

Which again is.... waaaaay below any margin I have ever seen in something using more than 10mA.

I will conclude (As pointed out by Spehro in the comments) with the fact there are plenty of NPN and PNP bipolar transistors to be found with less than 50mV saturation voltage and even quite a bunch with that at 2mA or 3mA base currents, but they are difficult to explain properly in such a short answer, and they will always "waste" that base current in your application, unless you cascade and combine them in smart ways, but that then becomes even more complex and larger. And they are probably not going to be cheaper than the above MOSFETs anyway, so you truly are much better off using a MOSFET in this case.

  • \$\begingroup\$ Thanks Asmyldof for this great answer. I have posted a separate question. Can you please take a look and verify whether this will work or not: electronics.stackexchange.com/questions/235114/… \$\endgroup\$ May 19, 2016 at 5:55
  • \$\begingroup\$ @Whiskeyjack I find I have hardly anything to add to Spehro's answer. \$\endgroup\$
    – Asmyldof
    May 19, 2016 at 16:39

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