In looking at schematics using a MOSFET as a switch, I have noticed that BOTH the signal and load voltages use the same ground. What do I do if my signal source voltage and my load source voltage are from two entirely different circuits? For example, what if I wanted to use a battery for my 5v signal source and I wanted the MOSFET to switch on a 24 volt bulb? Thank you.

 

One thing that you may be able to do is to use the same common point, or ground, for both circuits, i.e. connect them together. The MOSFET will essentially keep the 24V away from the 5V. Be sure you use a MOSFET that can be 5V logic driven otherwise many require around 10V to turn them on. This is done quite often.

If you need electrical isolation for some specific reason, then think about using an opto-coupling device, where only light from an internal LED on the 5V side turns on a transistor on the 24V side.

 

When you say "logic driven," what does that mean exactly. I thought that an electron was an electron.

 

They are. However, most FET's require the gate being taken at least 10 volts above the source voltage to go into full conduction. A logic level FET will turn fully on with only 5 volts on the gate.

Look at the data sheets for a 2N7000 versus the one for a VN10LP.

 

A standard MOSFET can require a voltage of around 10V between the gate and source in order to turn the MOSFET completely ON, very low resistance from source to drain. A logic level MOSFET requires a lower level of voltage of 1.8-5V, digital logic levels, to turn ON.

Check the Vgs, voltage gate to source, ratings for the part you are using against the Rds(on), resistance drain to source while certain Vgs is applied, values.

 

Will a Panasonic AQV101 MOSFET Relay work for me in this application?

 

It could work, as long as the 24V dc bulb does not exceed the 700mA current rating of the opto device.

 

Thank you for your help. I'm all set.