Electrical – Why is the drain circuit always drawing current, regardless of MOSFET gate state

ledmicrocontrollermosfet

I'm building a circuit that will drive various 12V solenoids and a pump. I use an MDF1903 MOSFET to switch the pump and a ULN2003A to drive the various low-power solenoids. A micro-controller signals the MOSFET and transistors accordingly.

In order to simulate the 12V components, I've replaced them with 2V LEDs and appropriate resistors to limit current at 12V. This works fine for the LEDs switched under the transistors, however I've noticed that the LED switched by the MOSFET is always on (albeit at low intensity). Why is this? Shouldn't this only occur when the gate voltage is high? The program on the micro-controller initializes the output pin to the low state and the output pin is grounded with a 10K pull-down resistor.

Here is a simplified circuit to illustrate my dilemma.

schematic

simulate this circuit – Schematic created using CircuitLab

For what its worth, the ground inputs and outputs on my Buck converter are common. The LED draws the full 20mA in its "on" state, and only 0.9mA in its "off" state. I also measured 10V across drain and source/ground, which is the 12V supply less the 2V dropped from the active LED. Regardless, I'm a bit confused as to why there is any current draw at all. Have I done something wrong, or am I thinking about this too hard?

Best Answer

Thank you to those that provided their feedback in the comments to my question above. I've learned a lot more about how MOSFETs work, and also got some hands-on testing experience.

I built a simple test circuit (below) to demonstrate any failure of the component. In doing so, I removed the micro-controller and Buck converter to eliminate any bias that these components may introduce.

schematic

simulate this circuit – Schematic created using CircuitLab

Clearly, there should be no signal to the gate channel when the circuit is energized. However the LED illuminates in this configuration, and becomes only slightly brighter when the push-button is closed. In the open-switch state, Vgs is ~2.5V and I don't think that can come from anywhere other than an internal short/failure.

Here's a photo of my breadboard showing an implementation of the circuit.

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