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After searching for a BJT solution, I decided to replace my Darlington pair with a single MOSFET transistor. I have a microcontroller with max I/O pin current of 4 mA at 3.3 V. I have a FQP30N06L MOSFET with 2.5 V VGS threshold, and if i look at the graphs:

  1. Figure 1 shows me that on VGS 3 V and ID of 0.5 A, I have really low VDS, does it matter and is it fine?
  2. Figure 2 shows on the minimum ID of 1 A, the VGS is 2, which is good for me, right?
  3. Figure 3 shows RDS of 30-40 mΩ, what does that mean? I read it is important info, but could not understand how I can use that.
  4. I could not find how much minimum current I need to pull on the gate, and with that info I want to calculate the resistor value on the base.

I know it is a lot of questions, but I want to make sure I am not going to burn my microcontroller or my door lock.

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    \$\begingroup\$ These are all good questions, and I remember that all these figures were very confusing to me when I first started to use MOSFETs. Some people may not like that you ask so many questions at once, but I really think that they somehow belong, and should be answered as one unit. \$\endgroup\$ – pipe May 14 '16 at 14:03
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    \$\begingroup\$ I agree. And shall attempt so \$\endgroup\$ – Asmyldof May 14 '16 at 14:04
  • \$\begingroup\$ Not so sure you can drive a MOSFET with 3.3V, usually the VGS is rated 10V. You would better use a MOSFET switch, instead. \$\endgroup\$ – Marko Buršič May 14 '16 at 14:28
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    \$\begingroup\$ This specific MOSFET if low gate as 2.5V.. \$\endgroup\$ – gabi May 14 '16 at 14:30
  • \$\begingroup\$ That MOSFET might not be a good choice. Vgs(th) is 2.5V, but that is specified at 250 uA. Rds is not specifiied at any Vgs below 5V. You should look for a MOSFET that specifies Rds(on) at Vgs of 3V or lower. You want the Rds(on) to be, let's say 100 mOhm or less. \$\endgroup\$ – mkeith May 14 '16 at 14:43
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Let me start by saying, that yes, your question is broad and/or complex, but you're sticking to the rules of doing some research first and asking a (set of) specific question(s). This as a disclaimer to those who might see your question as a reason to ask "So, what MOSFET should I use?".

I will answer your questions in a somewhat different order than you ask them, as this helps me try to build up to what a MOSFET really is.

First, the current needed for the gate (q. 4):

The gate in a MOSFET is different from a standard bipolar (junction) transistor. In a "normal" BJT there is a diode-like connection between the base and the emitter, which needs some current to make more current flow, it works as a current amplifier, of sorts.

In a MOSFET the gate is "floating". The name stands for: "Metal-Oxide Silicon Field Effect Transistor". The Metal Oxide is what insulates the gate from the Silicon channel. The field effect is what causes a voltage to open a current-capable channel in the transistor.

So, the gate has a capacitance inside it, which will require an amount of energy, but once it is at a given voltage, it will have a microscopic leakage current. So a gate resistor is a good idea to keep the peak current low enough to not destroy anything, but once the gate is at a given voltage, there will be virtually no voltage falling across even a 100k resistor.

Of course, if you choose a 10M resistor, your capacitor in the gate will need time to charge, and that will stretch out the switch time of the MOSFET, which will cause unnecessary heat.

For most low-frequency applications, you can just use 2k and rely on that being quick enough. Your controller can probably easily support 500 Ohm as well, if you want to, and even a direct connection might work well long term, with no resistance, but good design would be putting at least a few hundred Ohm in there.

The Rds-on: What does it mean? (q:2)

Continuing on the talk of channels and field effects, the MOSFET really acts like a resistor to the flow of current.

There is a channel inside the MOSFET (hence N-Channel or P-Channel). A more common Enhancement MOSFET (like yours) has a normally closed channel, because the internal make-up of the semiconductor "pushes it closed", disallowing current to flow (significantly). By applying a voltage to the gate, the extra charge on the gate electrode forces the channel slowly open.

Through the control of that voltage the MOSFET will be either a little bit open, in a region where that voltage nearly linearly controls the flow of current at a fixed voltage across it. Or it will be completely turned on, where the resistance is incredibly low and with a small voltage across it lots of current will flow. (I am ignoring a few even more complex situations in this, but don't worry about it yet).

For switching you want to be in the second part, as on as it can be. In that area there is an Rds-on value, which dictates the apparent resistance of the channel at a certain gate voltage. 40mOhm is pretty low. You can easily get much lower these days, but value for money, the region of 20 to 100 mOhm seems to be where you find most N-types.

It basically just means, when it's on, with this gate voltage, normally you need to think of there being a 40mOhm resistor in series with what you're switching, like this:

schematic

simulate this circuit – Schematic created using CircuitLab

In the case of 0.5A that means the MOSFET has 20mV across it and dissipates a tiny 10mW once it's on. That's perfectly fine and you won't notice such a small voltage drop either.

Be careful, that this MOSFET's figure you point to doesn't include 3V, and that it starts at 2A, so real assumptions are hard to make about 3V and 0.5A based on that graph, but the graph does suggest that it's likely that for 3V on the gate you might be looking at 40 to 60 mOhm at low currents.

Figure 2: Vgs vs Id (q.2):

That figure does imply that even at the worst possible temperature that MOSFET can still easily turn on your 0.5A load. It's not the best graph to get that from, but, yes, at a conservative 2.5V Vgs at unuasably cold temperatures it still should conduct 1A. But at what voltage drop? 25V... might be a bit high for you to invest in just your switch.

Which brings me last, to your first question:

Figure 1 shows low Vds at my operation, that's good? (q.1):

Yes, that's good.

What you want, ideally, is infinite current at 0 Vds. Of course, ideally doesn't exist in the real world, so as low as possible with a high enough current is what we make of it.

Be aware though, that they add the caveat of a 250μs pulse to that graph, so it may mean it will not be able to keep performing like that for longer pulses if you go right to the edge. The edge being, for example, 8A drain current at 3V gate and expecting 1V or less across Drain and Source.


All in all, this MOSFET is easily capable of driving your load, and probably still suited well enough (given a little bit of heat-sinking) to still drive ten of them and as long as you don't switch them very fast and often, can be driven by your 4mA capable 3.3V controller with a small resistor on the gate.

EDIT: Though, comments to your question are right in suggesting there are better ones for the task of using 3.3V to drive it. Regardless, this MOSFET will work for your situation.

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  • \$\begingroup\$ @mkeith while Fairchild didn't put "at 3V it'll be 80mOhm" at the top in the table, using the graphs supplied in this datasheet it's easy enough for me to promise you, on whatever penalty you wish, that this MOSFET will have less than your 100mOhm Rds-on at 3V gate voltage, with any current 1A or less. \$\endgroup\$ – Asmyldof May 14 '16 at 14:51
  • \$\begingroup\$ Sorry. I deleted my first comment because I realized we were conversing in two separate threads. All the graphs are typical, not guaranteed. I agree the part will work. But I would select a different part for this application, unless the OP already has these on hand. Then might as well use them. \$\endgroup\$ – mkeith May 14 '16 at 14:55
  • \$\begingroup\$ Excellent! So, if i have this part already, i can use it, right? Although it's not the best part for the job. I will use 2k resistor on the gate. About the Rds, 10-20mW is not notifiable, right? What if i got 500mW? When does it start to be a problem? About Vds, if i need for example exactly 12V on the lock (just example, not this case) , and i know the Vds is 1v for example, i need to get 13v power supply for the job? \$\endgroup\$ – gabi May 14 '16 at 15:43
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    \$\begingroup\$ @gabi The MOSFET will work in the parameters you gave. If you want 12V on your lock, then you should have a 12V supply and a switching element that uses less than a volt for itself. Having 10% of your supply across the switch is bad design. This transistor, standing in free air can probably be trusted up to about 1W or 1.5W, but to be sure a lot of specifics come into play. \$\endgroup\$ – Asmyldof May 15 '16 at 15:25
  • \$\begingroup\$ @mkeith I understand your point, and I would select a different part as well. But I found, quite litterally, in the question the "I already have this, so...", which I answered. I find many people here answer questions that aren't asked for the purpose of making the perfect answer (and get loads of points in the process?), whereas I want to answer the actual question in the best way possible so that the asker will actually secretly learn (hopefully) something in the process. In this case "Nah, you need something different, use X or Y with a 2.5Vgs curve showing" would, I feel, be uninformative \$\endgroup\$ – Asmyldof May 15 '16 at 15:28
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While this FET probably would work, it's not a good choice. The main problem is that it is not specified at 3.3 V gate drive. You can take a few guesses from the graphs and figure it will be OK, but guessing is not a good way to get specs, especially when there are alternative parts that provide them outright.

Take a look at the IRLML2502. That has a guaranteed maximum on resistance 80 mΩ at 2.5 V gate drive. Since your load draws 500 mA, you know the dissipation with 3.3 V gate drive will be less than (500 mA)2(80 mΩ) = 20 mW. That's no problem at all for that transistor. You won't even be able to notice it getting warm if you put your finger on it.

To be clear, here is the circuit I'm talking about:

Note the diode. It is not optional. Without the diode, the FET will get abused every time the solenoid current it turned off.

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schematic

simulate this circuit – Schematic created using CircuitLab

You can calculate the RG to limit the current Rg=U/I=3.3V/4mA=825ohm. WIth such low current the rise time of the turn on will be tr= Qg/Ig = 20nC/4mA= 5us (if 3.3 Vgs is enough). Then you need an additional resitor to discharge the gate between G and S (leak resistor). I think you should stay with darlington power transistor.

EDIT: The schematics, if the MCU hasn't totem pole output, you will never be able to turn the MOSFET off. So I added the leaking resistor 10k, once tho MCU output goes of the gate is dischargeed (very slowly!)

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