integrated-circuitmosfetoperational-amplifierpower electronics
Hi,
I'm having some trouble understanding how the diode-connected device M3 maintains a fixed voltage difference between VDD (source) and node X (drain)? In other words, why is it called a diode?
I know that diode-connected devices are always in saturation if they are turned on. But who says that their Vgs will be equal exactly to their Vth. Can't Vgs be greater than Vth for a diode connected deivce?, so then the Vdd to Vx difference will no longer be a constant Vth.
A supply of 10 volts across 1k means 10mA is the maximum current you can get thru the MOSFET. That is the problem with your experiment - no matter how much more you turned the juice up on the gate, the aiming point for drain current is 10mA and can not be exceded no matter how hard you try.
Of course Vds decreased - the mosfet was turning really hard-on and acting as a very small value resistor. This small resistance forms a potential divider with the 1k to make a very small voltage that is nowhere near 10V.
"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 $$P= i V = 50\times 20 = ~1000 \text{ watts}$$
Best Answer
That's not the point here. The point is, for a given voltage difference between the gates of M1 and M2 there will be defined currents through M1 and M2 (sharing Iss). This current through M1 sets the \$V_{gs}\$ of M3 and thereby the voltage drop from \$V_{DD}\$ to point X. So for a given input the voltage the voltage on point X is always \$V_{DD} - V_{gs}\$.