IRF3205 MOSFET gate threshold voltage higher than expected

Question

I have been trying to drive a 12 V LED array with an IRF3205 MOSFET. I wanted to do this:

^{simulate this circuit – Schematic created using CircuitLab}

However, when I tested it, the LED was only being given 6 V and the MOSFET was dropping 10 V across the drain and source. Going through the things that could be wrong, I found that giving the gate 5 V causes the problem to be solved, and the LED gets the full voltage of the 12 V power supply. The LEDs draw about 100 mA when they are being driven correctly in this way.

I looked up the datasheet, and saw that the gate threshold voltage is 2-4 V, meaning that this should work fine with the 3.3 V Raspberry Pi GPIO.

What is going on, and is there a way I could drive this chip with a 3.3 V Raspberry Pi GPIO pin?

Answer 1

A fundamental misconception many have with MOSFETs is that the gate threshold is where the "magic" happens. The gate threshold is truly a threshold - it is the absolute threshold of conduction: where the device goes from totally of to jusssssssssssst a little bit on.

A meaningful diagram can be found a few pages deeper:

This shows various relationships between \$V_{GS}\$, \$I_D\$ and \$V_{DS}\$.

With a 12V \$V_{DS}\$ voltage, you can see that even with 4.5V on the gate, the MOSFET will conduct only a small fraction of the 110A that the device is rated for.

Yes, these curves are characterized for pulse behaviour but you get the idea - you need much more than the gate threshold voltage to really get the MOSFET "on" and working well.

If you want to drive a MOSFET from the GPIO line directly, you will need to find one which can sink sufficient current at that 3.3V drive level.

Consider the Fairchild FQP30N06L device:

You can see that even with 3V, the device will conduct a lot of current.

Answer 2

If you want to get rid of MOSFET threshold and turn-on voltage problems, completely, I suggest you try this:

R1 will dissipate about a watt while the LED's ON, so prudence would dictate using about a 2 watt resistor. I like these.

and here's the LTspice circuit list just in case you want to play with the circuit:

Version 4
SHEET 1 1140 708
WIRE 128 0 -16 0
WIRE 304 0 208 0
WIRE 304 32 304 0
WIRE 304 112 304 96
WIRE 128 160 80 160
WIRE 240 160 208 160
WIRE -16 224 -16 0
WIRE 80 224 80 160
WIRE -16 336 -16 304
WIRE 80 336 80 304
WIRE 80 336 -16 336
WIRE 304 336 304 208
WIRE 304 336 80 336
WIRE -16 384 -16 336
FLAG -16 384 0
SYMBOL res 224 -16 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 91
SYMBOL LED 288 32 R0
WINDOW 3 28 68 Left 2
SYMATTR InstName D1
SYMATTR Value QTLP690C
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL voltage 80 208 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName GPIO
SYMATTR Value PULSE(0 3.3 0 100n 100n 10m 20m)
SYMBOL voltage -16 208 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 12
SYMBOL npn 240 112 R0
SYMATTR InstName Q1
SYMATTR Value 2N2222
SYMBOL res 224 144 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 240
TEXT -8 360 Left 2 !.tran .5

Answer 3

I looked up the datasheet, and saw that the gate threshold voltage is 2-4v, meaning that this should work fine with the 3.3v Raspberry Pi GPIO.

That means 3.3 V could be below the gate threshold voltage.

Even if it were above, that's still not meaningful. You have to actually read the datasheet instead of skimming the highlights and then making (wrong) assumptions.

On page 2, V_GSth is clearly defined to be when the FET conducts 250 µA. It's hard to even imagine how you think that is relevant when you want 100 mA.

Just one line above that, the datasheet specifies R_DS(on) at 10 V, which is what this FET is intended to be operated at. Nothing says anything about what happens with only 3.3 V on the gate, other than sometimes maybe it will conduct 250 µA.

Again, this is all clearly spelled out. Sometimes you get ambiguous or poorly written datasheets, but this is not one of them. This FET is simply inappropriate for your application.

Answer 4

You should not need to burn 2W in the resistor. For driving 100mA into the diode you need to burn in a resistor : P = VI ; V = IR; P = I^2*R = .01*R W

Assuming the LED needs about 3-4V to be on, that leaves 8V leftover to burn across this resistor. V = IR; V = 8V, I = .1A; R = 80 Ohms

Going back to the power burned in the resistor, I^2*R = .1^2 * 80 = .01*80 = .8W You would also be burning 4V*.1A = .4W in the LED. This seems a little wasteful to me, and a better way might be to reduce the 12V to something more like 5V.

You need a more responsive FET (lower Vt). You could also just convert your output (0-4V) to (0-12V) using back to back BJT inverters, and use the current FET you have, along with 80Ohm resistor.