So I decided to experiment with power MOSFETS (IRF510). The wife is out of town and it's easier to catch stuff on fire when she isn't here. The circuit is shown below. Eventually I would like to supply the gate with a 5V pulse from an Arduino. However, to get a feel for the chip I decided to supply the gate with a constant 5Vdc. Here is the circuit:

enter image description here

The max Vth for this FET is 4 volts. The Vd to Id curve shows about 1 amp for the setup above. I also opened up a text book and put the Id formula to use. Id = Kn(Vgs-Vth)^2. I assumed the forward transconductance would work for Kn, so I plugged it in and ended up with 1A (the datasheet shows a higher forward trans value of 2. The package reads 1).

I had a multimeter setup between vd and the drain pin. For some reason the current reported just above 2 amps before the gate was even turned on. Then it would slowly decrease to about 700mA then it would increase to about an amp and POOF! Smoke and tiny silicon ambers. I tried this with two different chips. For the second chip I tried it with a 1ms pulse followed by a 500ms delay, I didn't let this chip catch flames, but the temp was reading well over 190F.

Everything I have read points to the gate voltage not being high enough, causing the FET to not fully turn on. I guess I can kind of see how that is possible. All of the papers and books I have read discuss how the Rds decreases as the gate voltage increases. A lower voltage would in turn cause a greater resistance, generating lots of heat and wasted power. If the FET wasn't meant to operate at these voltages, why are they included in the characteristic graphs? I'm certainly no expert, but I don't think the constant gate voltage is the problem.

If this is indeed the problem would a heatsink help? Or are there any suggestions for supplying a higher voltage to the gate? I would really like to figure out how to resolve this issue before running away to another product. Eventually I am going to make two sweet ass motor controllers with these guys :) Please help me!!! MOSFETS make for expensive fire crackers.

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    \$\begingroup\$ When the MOSFET turns on it will die if driven fully on. When it does it will in some cases short from Drain to Gate wityh drain to source open circuit. In such a case it will connect +12V to the arduino drive pin. A series drive resistor and a voltage clamp on the processor side which both are designed to allow the Arduino to survive +12V to gate are required. || With no load apart from the MOSFET if you drive it fully on then you must get 10 - 20W or more dissipation in the MOSFET. | What are you trying to achieve. A hammer is often an easier way of achieving the main effect. \$\endgroup\$
    – Russell McMahon
    Aug 18, 2012 at 12:47
  • \$\begingroup\$ @RussellMcMahon Thanks for your help. In the long rung I hope to put together a stepper motor controller. I am on the kick of trying to build everything instead of buying it. I think every ee student should do this! \$\endgroup\$ Aug 19, 2012 at 5:33

2 Answers 2


Even if the graphs in the datasheet say 1 A at 5 V you should at a load to limit the current. FET characteristics can vary widely, just look at \$V_{GS(th)}\$: minimum 2V, maximum 4 V. For some FETs even a 3:1 range is given. Your 5 V may be 1 V above \$V_{GS(th)}\$, but it might as well be 3 V over it. The 1 A is typical, but \$R_{DS}\$ may already be low. Note that at 10 V this will be as low as 0.5 Ω. Now you have the full 12 V across the FET, and if \$R_{DS}\$ would go to even 2 Ω this would cause a dissipation of 72 W, which without any cooling is way too much for the poor thing.

BTW, even the 12 W of a 1 A current would be too much without cooling, though I wouldn't expect it to explode then.

So make sure you add a load so that the FET won't see the full 12 V across it.

Also, it is important to note that the value of the transconductance, G, specified in the data sheet is NOT the same as the K parameter for a mosfet. They are related by the following expression:

G = 2 * Sqrt( K * Id )

  • \$\begingroup\$ Thanks for your help! It's funny that you mention the 12v Vds being too high. Eventually I am going to build a stepper motor controller using these fets. The motors I am going to use are rated for 3.6 volts and 3A per phase. Do you suggest I change the Vds to 4 volts and experiment? This might be a stupid question, and I apologize for my ignorance, but how do motor controllers get away with tiny resistors? I am only familiar with using 1/4W resistors. If I cut my voltage to 4 volts that still puts about 4w on the resistor, I can see the .25W resistor bursting in flames. \$\endgroup\$ Aug 19, 2012 at 5:45
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    \$\begingroup\$ @atomsmasher - If you have the motors already and have a 3.6 V source then use the motors as load, they're made for it. I don't know why you would want 4 V instead of 3.6 V. I checked the IRF510 datasheet again, and I think you need a better one. This one has an \$R_{DS(ON)}\$ of more than 0.5 \$\Omega\$, but that's at 10 V gate voltage, at 5 V will be higher. You may lose 20 % of your 3.6 V in there, and the motor 30 % of its power. The NTD4960N has an \$R_{DS(ON)}\$ of only 12.5 \$\Omega\$ at 5 V, so it won't even get warm at 1 A, and voltage drop will be negligible. \$\endgroup\$
    – stevenvh
    Aug 19, 2012 at 9:41
  • \$\begingroup\$ @atomsmasher - You're welcome. That 12.5 \$\Omega\$ should be 12.5 m\$\Omega\$, obviously, but I guess you already knew that :-) \$\endgroup\$
    – stevenvh
    Aug 19, 2012 at 18:06

I suspect the reason you read the current as 2A before the gate was "turned on" is because the gate was floating (i.e. not pulled to ground. This is due to the gate being very high impedance (just touching it will make a difference) You need a resistor between gate and ground (e.g. 10k) to prevent this happening.

As Steven says, it would be wise to use something to limit the current. If you don't limit the current, as long as the supply is capable of much more than 2A, the full 12V will be across the FET, so 1A * 12V = 12W. At 12W, for a TO-220 package you certainly need a heatsink.
Without a heatsink, the Tja (junction to ambient) is 62°C/W, and the max operating temperature is 175°C, so assuming an ambient temperature of 25°C the FET can only handle around (175 - 25) / 62 = 2.33W without a heatsink. This is very forgiving, you would typically design for an ambient of e.g. up to 70°C and a max temp of 125°C

Note that if you want control the current accurately, you need to use some feedback rather than trying to rely on the variable Vth and transconductance parameters. A current sense resistor and opamp combined with the FET can make a simple constant current circuit (components can obviously be changed to suit, the LM741 is obsolete but people still insist on using it):

Constant current

Another thing to note is the SOA (Safe Operating Area) curves are given for different pulse widths, and often DC is omitted if the FET is designed for switching, so make sure you take all this into account when designing your circuit. Notice the SOA curves are specified for a case temperature of 25°C.

Few useful inks:
Thermal Resistance
Selecting a heatsink

  • \$\begingroup\$ Thank you for the excellent response! I wanted to give everyone a correct answer for this question, but the laws of physics won't allow it. I am defaulting to the first response, but you still get +1 and a pat on the back!! In response to the ground pin floating, is it OK to tie the Arduino gnd pin to my makeshift hacked computer power supply gnd? I keep meaning to pick up a fire extinguisher, but I always forget :$ \$\endgroup\$ Aug 19, 2012 at 5:37
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    \$\begingroup\$ No probs, glad to be of help. I was referring to the MOSFET gate floating, which just means it's not connected to anything. It only needs to be tied to local ground to keep the gate-source voltage at 0V and the MOSFET turning on (by accumulation of charge on the extremely high impedance gate - you can make a good static detector with a MOSFET - LED in series with resistor on drain and gate connected to an "antenna") \$\endgroup\$
    – Oli Glaser
    Aug 19, 2012 at 6:20
  • \$\begingroup\$ Oli, what's that "1/10" spec next to the potmeter? \$\endgroup\$
    – stevenvh
    Aug 20, 2012 at 18:29
  • \$\begingroup\$ @Steven - Yeh, I noticed that too. According to the blog I got it from, it means it's set to a ratio of 1:10 (i.e. the pot resistance doesn't really matter) I couldn't find anything better on Google images, and didn't have time to draw it in LTSpice at that point. Think I should change it? \$\endgroup\$
    – Oli Glaser
    Aug 20, 2012 at 18:47
  • \$\begingroup\$ @Oli - probably not. It's been there for 2 days and I seem to be the only one asking about it. Though I would have use concrete values if it were my schematic, like a 10k pot in series with a 1k resistor. That would have been clear. BTW, I see nor reason why you wouldn't turn the input all the way to zero. Wait, I'll fix it for you ....... There, done! :-) \$\endgroup\$
    – stevenvh
    Aug 20, 2012 at 18:54

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