From what I understand, MOSFETs inherently have a diode between the source and drain known as a body diode. If I were to use an nMOSFET as a low side switch, would it be bad to have a diode between the load and the drain of the MOSFET? What effect would it have?
Schematic is a representation only and is not intended to produce accurate simulation results
Edit to clarify comment:
The goal is to enable 2 low side switches (not at the same time) to switch between 2 rgb controllers. Originally I wanted to multiplex between the 2 controllers, but the couldn't find any with a large current rating. Without D1 and D2, the switching of controller 1 would affect current through controller 2 and vice versa. If there's a better way to switch between low side switches I am open to suggestions. (RGB Led should be LED Strip. Resistors are embedded)
Best Answer
It would decrease the max voltage across the load by a voltage equal to the forward voltage (Vf) of D1. One reason I can think of to do this is if whatever the load is cannot handle the full Vdd across it.
I'm not sure what you mean by "cancel the body diode". When the FET is fully enhanced, it will effectively short out the diode and current can move through the FET in both directions. When the FET is off, the body diode is just like a power rectifier; it will pass current in one direction. In your circuit it will be reverse-biased by Vdd, so there will be no electron flow with or without D1.
If you have a situation where it is possible for both positive and negative current to appear across the FET, and you want to block both directions when the FET is off, such as reverse-polarity protection of a power supply input, then D1 will do this in the circuit as shown. There are two better ways to do this.