# Power MOSFET internal diode: when does it conduct?

So I'm currently studying power MOSFETs and I'm getting a bit confused about when and how it conducts current.

Let's start with the normal MOSFET functioning. I'm considering a n-enhancement type MOSFET.

So, for $$\ v_{gs} <0 \$$ (or less than the threshold voltage), there is no inversion layer and therefor no channel. The MOSFET does not conduct. Now for $$\ v_{gs} >0 \$$ the channel is formed and the MOSFET can conduct through if a positive $$\ v_{ds} \$$ is applyed. Everything ok until here.

Now my doubt is how the internal diode takes place here. I know why the internal diode existes (shorting the p substract to the n+ terminal through the metalization) but what I don't know is when does it conduct.

For the situation we already checked: $$\ v_{gs} >0 \$$ and $$\ v_{ds} \ >0 \$$ it definitely does not conduct since it's a reversed junction (it has just a small parasitic current). The same can be said if $$\ v_{gs} <0 \$$ and $v_{ds} \ >0 . But now what if $$\ v_{gs} >0 \$$ and $$\ v_{ds} \ <0 \$$. I was told that until $$\ v_{ds} \$$ reaches a certain negative limit, the conduction still occurs essentially through the inversion layer. Then, after that limit, the conduction is taken by the diode. Is this correct? And what about when $$\ v_{gs} <0 \$$ and$ v_{ds} \ <0 \$. In this case there is no channel but the diode is forward biased therefore, will there be a current? Is this why the internal diode is useful? So that the MOSFET can also conduct when there is no channel and therefore, apply it to reversible converters.

I'm a bit confused if someone can clarify me all this situations it would be very helpful.

If $$\ V_{DS} \$$ is negative (more negative than forward diode drop 0.6 V to 0.9V typical) the diode will always conduct.
$$\v_{gs}>0\$$ and $$\v_{ds}<0\$$
In reverse conduction and with a positive gate bias, the channel conducts all the time. The current is split with the internal body diode according to its forward characteristics. Most of the current will go in the channel as long as the channel voltage drop is lower than the forward voltage of the diode, $$\ I_d \cdot R_{ds} < V_{f} \$$.
$$\ v_{gs}<0 \$$ and $$\ v_{ds} <0 \$$