Why this circuit provides only 2V on the load (33 ohms resistor) when the switch to the gate of the MOSFET is closed ? I was expecting it to switch between 1.4V (caused by the 85 ohms resistor) and 5V (when the MOSFET gate is on). I'm using the MOSFET 2N7000. Is there any way to solve this to provide 1.4V and 5V when the gate of the MOSFET is on ?
Best Answer
An NMOS MOSFET makes a very poor high-side switch, which is what you are trying to use it for. The highest voltage that will appear at the source (which is connected to your 33 ohm resistor) is equal to the gate voltage (5V) minus the transistor threshold voltage (about 3V). You need to use a PMOS transistor instead, and look for one with a threshold voltage of 1.4V or less. The PMOS transistor will turn on (conduct well) when the gate is at ground, so the gate signal needs to be inverted with respect to the signal that controlled the NMOS gate.
And you are still going to have trouble meeting the 1% voltage tolerance required for the sensor, as we discussed in https://electronics.stackexchange.com/questions/76331/diode-causing-voltage-drop