I'm trying to understand a test result from a MOSFET-load switch circuit. The MOSFET tested after the load switch is DMN3404L, it's a N-channel MOSFET which has a typical Vgs around 1.5V. The load switch (MP5016H) has a Gate output which drives the Gate of the DMN3404L. In the test, the input of the load switch is 12V, the Gate output of the load switch is around 11.5V, since the source node of the DMN3404L is connected to the output of the MP5016H which is 12V as well, in this case, the Vgs is -0.5v, but the DMN3404L seems open which confused me. Is it supposed to be off because Vgs is <1.5V? The I did a test to change the input voltage from 5V to 16V, it seems that when the input is below 10V, the output of the MP5016H Gate pin is higher than the source. When the input is higher than 12V, the Gate output is always around 0.5V lower than the source, but the MOSFET still seems turned on, because the voltage measured at the Drain is the same as the voltage at the source. One thing I noticed is if there is no load at drain of DMN3404L, the voltage of the drain is around 0.7V less than the source, which seems that the MOSFET is off, but the diode is open, but as long as there is a load, the DMN3404L is turn on. Here is the test diagram and results are attached below, I appreciate if someone can give me an explain why the MOSFET is on when the Vgs is negative? Thank you,
I'm going out on a limb and say you have your circuit backwards. Typically for an N-channel mosfet you would put the voltage into load then connect the other side of the load to the the drain. The source would typically be connected to ground. That way your Vg is always referenced to ground so your Vgs is whatever your Vg is so to speak. Of course you may be doing something creative I don't understand.
As it is, any voltage you apply at the source will certainly show up at the drain minus the diode drop of 0.7V. This is when the MOSFET is off
I think if you just swap your drain and your source on the DMN3404L you should be good.
I think you need a pull down on the gate based on the data sheet. It seems that the load switch will pull down the gate under fault conditions but under normal conditions it does not say it will pull down the gate. I'm not immediately seeing how the gate will be drained otherwise
Also, with a load on the output the Drain voltage and thus the Source voltage could be lower. If the gate voltage is not changing that would explain why the load would cause the FET to turn on and no-load would cause it to remain off. It's changing the Vgs.