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.
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?