Correct me if I am wrong

There are 3 modes for mosfet: 1, cut-off; 2, active (triode); 3, saturation; When VgsVt, it's mode 2; when Vgs>vt & Vds>=Vgs-vt, it's mode 3;

For N-mosfet, the drain is positive side, and source is the negative side, and the load should be connected to the drain side.

Here are my doubts

1, P0903BDG n-mos has 3 Vt values: min, typical, max (1, 1.6, 3V), which one should I use? Since this Vt is so critical here.

2, I setup a simple testing circuit with 4.07V Vdd -- red led -- 100 ohm -- drain, source -- ground and have a 50k pot connected to the gate.

According to Vgs>vt & Vds>=Vgs-vt and so when it's in saturation mode, Vds should be equal to 0v in my circuit. so Vds>=Vgs-vt == 0=Vgs-(1-3), and the result for Vt should be equal to (1-3) when it's in saturation.

Here are My measurements: when Vgs<1.2v, it's in cut off mode; When Vgs=1.3v, the led start to light up slightly; Vgs=1.3-2v = active mode; when Vgs=>2.1V, it's in saturation mode. When it's in sat mode, Vds=0, 0=vgs-vt, 0v=2.1v-2.1v. so Vt=2.1v for this mosfet? It can't be because when Vgs=1.3V, the led is already On. SO I am very confused here. What exactly is Vt? what is the exact problem here?

3, I want to operate this simple LED circuit, so when the pot is at 0 position, the led is off; when I turn the pot slightly, the led is slight on; and when I turn the pot to the other end, the LED is totally on with the mosfet in saturation mode. 1.2-2.1v is the range with my measurement, but how do I do it mathematically?

4, This should be a 50W mosfet. Yesterday, I tested this mos with a 12v vdd and a 12v power led. 12v-led-drain/ source-ground/ gate-10kohm-12v; and the current is just 550ma. It's not bright enough if I were connect the led directly to 12v power source. what is the problem here?

• if it's youe goal, TL431 makes a better voltage detector Nov 8, 2015 at 6:10
• Why the load should be connected to the drain side? Can be used ether in high side or low side configuration. And are you use this MOSFET for on-off application or to use its linear region (to dimm the LED)? In on-off application, a "logic level" MOSFET should be selected. As for the Vg value, see the "transfer characteristics". Nov 8, 2015 at 6:22
• The Vgs Threshold variations could make your circuit a production disaster .Do something to reduce this or use something more accurate. Nov 8, 2015 at 8:13

"What exactly is Vt? what is the exact problem here?"

The problem is that you're not paying enough attention to the data sheet.

First, when Vgs(th) is specified, note that the current level is .25 mA. Since I assume you want to drive more than .25 mA through your LED, you're going to need more gate drive. Second, for full brightness you'll want Vds to be as low as possible, so again you'll want more gate drive. Third, when the voltage is specified as min/typ/max, it means just that. The gate voltage required to drive 0.25 mA might be as low as the min, or it might be as high as the max, but typically it will be the typ voltage.

When you set up your circuit, let's assume your LED will have a nominal Vf of about 1.8 volts. Then a Vds of 0 volts means that the resistor will drop about (4.1 - 1.8 = 2.3) volts, and a 100 ohms means a current of 23 mA. However, remember that Vt is specified at .25 mA, which is probably just visible. You can check this by running your circuit with the LED just visible, measuring the voltage across your resistor, and finding the current. In this case 1.3 volts sounds perfectly reasonable.

If you want your pot to go from barely on to full on, you'll need to add two more resistors, as follows simulate this circuit – Schematic created using CircuitLab

Calculate as follows:

The current is the same through all 3 elements, but you know that the voltage across R3 is 0.9 volts. Since the voltage across R2 is 1.2, $$R2 = \frac{1.2}{0.9}\times 50k = 66.67k$$ and the voltage across R1 is (4.1 - 2.1 = 2.0) volts, so $$R1 = \frac{2.0}{0.9}\times 50k = 111.1k$$

As for your last question, I have no idea what the problem is. You should be aware that the power rating on the MOSFET has no connection to the power in the load - it simply deals with the power dissipated in the MOSFET. For instance, the maximum current is listed as 50 amps. A hard-driven MOSFET will have an Rds of .0095 ohms, and will dissipate $$P= i^2 R = ~25 \text{ watts}$$ while if the load voltage is 20 volts the load power will be $$P= i V = 50\times 20 = ~1000 \text{ watts}$$

• Some source resistance in M1 would help. Nov 8, 2015 at 8:14