The IRF520 FET switching module

Various webstores sell a FET module that is aimed at switching high current loads with an Arduino or even a Raspberry pi.

The module seems quite handy, especially if you do not like soldering. It has everything you need, including handy screw connectors for that heavy load you plan to switch. A quick glance on the specs tells you it can deliver 9.7 amps at a 100 Volts and it has a resistance (Rdson) of 0.2 ohm and it is “for the Arduino” so you should be fine right?

Well….not really. The FET used, the IRF520 is not really suitable for the logic levels the Arduino is working with, let alone the 3.3Volt levels of the Raspi. The IRF520 is more at ease with 10 Volt gate voltage. I wrote a post on this subject a while ago.

Using Mosfets with TTL levels.
Power LED Drivers.

So suppose you just bought a couple of those modules. Can’t you use them at all?

Well… maybe but let’s have a look at the datasheet first. In the ‘electric characteristics’ it says that the treshold Vgs, that is the voltage between gate and source (usually ground) should be 2-4 volts for the FET to start conducting. That sounds great -as the Arduino delivers 5 Volt- but it isn’t. At the threshold Vgs the current flowing is only minimal, in this case 250uAmp. That is not a current you bought a powerfet for.

But, i hear you say, the arduino delivers 5 volt, surely 5 Volt should be OK. Well if we again look in the datasheet (Fig 3), yes 5 Volt looks a bit better

Transfer characteristics

At 5 Volt the module should be able to deliver some 4 Amps and that ofcourse is not bad. But there is a caveat. Section 28.2 of the atmega328 datasheet says that the HIGH output voltage of the I/O pins is 0.7 Volt lower than the Vcc. So let’s take the scenario in which you are using a phone charger to power your arduino and it only provides 4.8 volt. The Arduino output then will be 4.1 volt. That brings the drain current in the 250uA range again. That’s not good.

So what to do? Well if you were smart you’d say ‘lesson learned’ throw them in a box and buy a decent logic level FET e.g. the IRL520. That one can deli er some 9 amps on a 4 volt gate to source voltage. Sadly however those do not come on such a handy little board as this IRF520 module. ( the IPP096N03L however is probably an even better choice than the IRL 520)

If you really want to use the IRF520 module, coz you hate to let it go to waste, there are some options, but these are only practical if you have say a 12 volt source availeble, for instance if you wanted to use the module to drive a 12 Volt load:

Use a transistor to drive the FET: usually this is done by pulling up the gate to 12 volt with a 10k resistor and switching the gate to LOW with a transistor (e.g. a BC547). (This will invert the signal though). Like so:

FET driver

In this case however that will not work because the module contains a 1k pulldown resistor. So when the transistor receives a LOW on its base to switch the FET on the 10k will form a voltage divider with the 1k, rendering only 1 volt on the gate. So we would need to remove the pull down resistor. It is the resistor directly next to the SIG label. Most likely you would also need to remove the LED (next to Vcc label) or its 1k seriesresistor (next to GND label).

IRF520 switch module

A more elegant method is using an optocoupler such as the ubiquitous PC817 to drive the FET. A look at the datasheet tells us it has a max collector current of 50mA and a Vce of 0.2Volt in saturation.

Connection with optocoupler

Thus when the optocoupler opens a current of 12/500=24mA will flow (500 because the module has 2x1k resistor on the input). The voltage on the gate will be 12-0.2Volt.

Should you still looking to buy a ready to use MOSFET module, consider the D4184 module

D4184 module