If you look at the Vds vs Id vs Vgs characteristics curves, you will notice that the resistance of the IRF520 is about 0.4 ohm at Vgs = 5V.
Ok, so I take it that because the resistance is approx 0.4 ohm then I am effectively creating a voltage divider with the 2.2ohm resistor, so I will never be able to pull the voltage down to 0v, as some of it will always be across Vgs (approx 0.6v by my calcs). However, this obviously becomes a negligible voltage at higher resistances of R. Is that right?
You need a driver which creates more than 5V at the gate..
When the mosfet is not fully opened, it works in a linear region and heats up.
Ok, I have tested and you appear to be correct. When looking at the graphs on the datasheet (I hate logarithmic scales), it appears that at 5v Vgs & 5v Vds the drain current is circa 1.4-1.58A
However I cannot find anything in the datasheet that states in order to drive a Vds of 5v I need to have a Vgs geater than 5v...so what is happening here?? I am still a bit confused due to lack of information in the datasheet
Or use a logic level MOSFET. The spikes are generated by the power resistor since it's wirewound and therefore inductive.
I think you are correct, logic level MOSFET would be best. I did actually suspect inductance in the wirewound resistor causing the spikes, so I added a flyback diode to try and eliminate them, but it didn't actually do anything.
Adding a 1k resistor to the gate however, did remove the spikes when the gate powered off (voltage output went to 5v).
Any further explanations on the above?
It is clear you guys are correct, but due to lack of info on the datasheet I am still a bit perplexed as to why it reacts as it does.