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,
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\$\begingroup\$ What makes it "seem open"? Don't forget about the body diode. \$\endgroup\$– HearthCommented Mar 22, 2020 at 16:26
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\$\begingroup\$ I thought if the body diode would have some voltage drop between the source and drain, that happens when there is no load at the drain side. There is about 0.7v drop between source and drain. I thought that's the body diode. \$\endgroup\$– JasonCommented Mar 22, 2020 at 17:06
2 Answers
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.
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1\$\begingroup\$ The DMN3404L is for reverse polarity protection - there is already one within the load switch. It is wired correctly. \$\endgroup\$ Commented Mar 22, 2020 at 18:14
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\$\begingroup\$ Hi Carl, thanks for your comments, it's not an usual application to me as well. It's a suggested reverse current block protection circuit on the MP5016H datasheet. If I flip the Drain and source, the body diode will conduct the reverse current back to the load switch which will damage the part. \$\endgroup\$– JasonCommented Mar 22, 2020 at 18:20
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\$\begingroup\$ I see. Perhaps you need a pull down on your gate. \$\endgroup\$ Commented Mar 22, 2020 at 18:33
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\$\begingroup\$ @Jason Don't flip drain and source. Follow the datasheet. The datasheet is correct like also Kevin White said. This answer is not applicable to MP5016 and neither to your question \$\endgroup\$– HuismanCommented Mar 22, 2020 at 18:48
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\$\begingroup\$ Hi, @Huisman, the external MOSFET connections follow the MP5016H datasheet, the external MOSFET is used to block the reverse current. The source of the external MOSFET is connected to the MP5016 output(Source), the drain of the external MOSFET is connected to the load. The Gate output make sense when the input of the MP5016H is below 10V, the Vgate is always higher which turns on the MOSFET, just don't understand when the input is higher, the Vgate is lower than the Source, how it can turn on the MOSFET. I can try to put pull down resistor at Gate pin, see how it works. Thanks \$\endgroup\$– JasonCommented Mar 22, 2020 at 22:59
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.
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\$\begingroup\$ Hi Carl, it looks like if I add a pull down resistor at the Gate output of MP5016H, it will always pull down the gate and shut off the load switch. \$\endgroup\$– JasonCommented Mar 23, 2020 at 15:02
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\$\begingroup\$ That's strange. Your pulldown was at least 10k ohms? \$\endgroup\$ Commented Mar 24, 2020 at 2:56
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\$\begingroup\$ It looks like it's the load switch function, my application has only about 200mA current in which MP5016 is still in the sleep mode, when the load current goes higher than 500mA, it should be fine. \$\endgroup\$– JasonCommented Mar 26, 2020 at 13:21
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