Krupski:
I know I wasn't going to post in this thread anymore, but what you mentioned is worth a comment.
The two "exponentials" that you see have different causes.
The first one is due to the Miller effect combined with the inherent gate capacitance of the MOSFET. The "plateau" you see is when the drain is pulled as low as it's going to go, then finally the second exponential is the gate capacitance alone charging to Vgs equilibrium.
I don't think most people really know what the Miller effect is. Anywhere you look it up online, there are all kinds of formulas and hokey-pokey, but no clear explanation of the actual mechanism.
Imagine any 3 terminal (triode) device such as a BJT, a MOSFET and even a vacuum tube triode.
There is inherent capacitance that exists between each terminal and every other terminal, but the one that causes the Miller effect is the drain to gate capacitance (or collector to base, or plate to grid).
Imagine a MOSFET with the drain is being used to switch an LED on through the appropriate resistor.
Initially, the drain is at Vdd (minus the drop across the LED) and the gate is at 0.
Therefore, the inherent capacitor between the drain and gate is charged to Vdd - V-LED.
Now, begin to raise the gate voltage. As the MOSFET begins to turn on, the drain begins to lower, beginning to turn on the LED and also lowering the drain side of the inherent capacitor.
Of course, the gate side of the capacitor will also start to lower, which is FIGHTING the thing trying to drive the gate high.
The gate ends up looking like a LARGER capacitor which takes a lot of current to charge.
Once the MOSFET is saturated (that is, Vgs is at it's minimum), the "capacitance amplifying effect" stops and you get the plateau. Finally, you get another, less "severe" exponential as the gate (and it's inherent capacitance alone) is charged past the Vgs threshold and ends when Vgs is equal to the voltage of the gate driver.
As you can see, if the MOSFET were not connected to a load and the drain instead tied to ground, there would be virtually no Miller effect since the drain can't "push against" the gate. It's already at minimum and won't move any further.
Make sense?
Hi,
Well that's an interesting read.
I do have to disagree slightly though, and i think it is only because you probably mis-spoke that's all. The part is here:
StartQuote
Once the MOSFET is saturated (that is, Vgs is at it's minimum), the "capacitance amplifying effect" stops and you get the plateau. Finally, you get another, less "severe" exponential as the gate (and it's inherent capacitance alone) is charged past the Vgs threshold and ends when Vgs is equal to the voltage of the gate driver.
EndQuote
I agree that the Miller effect stops when Vds (not Vgs but that's probably a typo) is at it's minimum, but that is also when the plateau should end. That's because the reason for the plateau in the first place is because the the Vds is falling and that provides feedback to the gate which ends up being held somewhat constant.
The way i like to explain the plateau is like a DC voltage regulator. Imagine you want to hold the gate voltage perfectly constant for a short time, and you had to use the drain of the mosfet to do that. The mosfet drain and associated capacitance acts like negative feedback, so when the drain voltage is falling you get feedback to the gate which holds it constant. Of course once Vds falls completely you loose the negative feedback and so the gate is no longer regulated, so the plateau time is over and the gate voltage can start to rise again.
However, after taking a second look at the equivalent circuit, i see we mainly have to understand what happens at the drain rather than the gate, because the drain is what we see at the Arduino pin. That's the main power consumer so we should turn our attention to that instead i think, and see what we can find.
I see some other posts now with some waveforms so i am going to take a look at that next. If we get an Arduino scope pic we should be able to tell a lot about what is happening.