avrbasicmosfetrelay
I have 2 x 2N2222 BJTs powering 2 relays of 200mA each.
I am using 2 output pins with 500ohm resistors to power the transistors.
After powering it the circuit not working well, while replacing the relays with LEDs works well.
So I want to try and replace the 2 BJTs with 1 power FET. I have an IRFZ44N.
But after reading articles about the transistor, it seems like I cannot use it at all – looks like the gate needs +20V to work, while my AVR output 5V.
Is this true, and if so, what can I do?
Two things.
First, there is no reason for a 1 kΩ resistor between a 5V digital output and the FET gate, especially such a large resistor. There are certain cases when switching large loads where you do want to provide some finite impedance driving the gate, but that is not necessary here and the minimum impedance of the digital output is likely more than that anyway.
Second, what you are really seeing is not the FET turning off slowly (althouth the large gate resistor does cause some of that), but rather that there is nothing but the scope probe driving the source low after the FET has turned off. You are seeing the voltage on the inevitable capacitance on that line, which includes the scope probe capacitance, being discharged only by the high impedance of the scope probe. For a typical 1x probe, that is 1 MΩ. If it's a 10x probe, then there is probably only 10 MΩ discharging the aggregate capacitance. Note that even 1 MΩ and 100 pF is a time constant of 100 µs.
simulate this circuit – Schematic created using CircuitLab
The IRFZ44N would likely work, but it may be slower in switching and definitely overkill for your project. The 2n2222 transistor should work fine. I would take 200mA divided by the minimum gain of the transistor (hfe=35) and you'd get the current that you need to send into the base of your transistor. 200mA/35=5.714mA into the base of the transistor. The Vbe (voltage between the base and emitter) when on is generally around 0.7V. So you have 5V-0.7V = 4.3V. Then you need that 4.3V to push 5.7mA of current. Using ohm's law you get 4.3/0.0057 = 754 ohms of resistance.
In summary, you'll have about 6mA of current flowing through the base to the emitter which with minimal gain of 35 you'll end up with 200mA of current from collector to emitter. Attach the collector to the 5V relay coil, the other side of the relay coil to Vdd, and you should be set.
Best Answer
That part needs 10V to turn on to a guaranteed resistance. It's shown in this part of the datasheet:
20V is the absolute maximum (never exceed) value for the voltage.
It's also a beast of a MOSFET to replace a 2N2222A. To operate from a 5V input, you need a "Logic Level" MOSFET. A 2N7000 (TO-92) or 2N7002 (SOT-23) might work for you, but personally, I like some of the Alpha Omega parts such the AO3422 (SOT-23). It's not only rated for 55V and >2A, but it will also work well with 3V drive.
Edit: Vgs rating, as I said above, is an absolute maximum rating. You must never drive the gate beyond 12V in either direction with respect to the source. It has nothing directly to do with what it takes to turn the MOSFET on or ensure that it is off.
For that, you need to look at the Vgs for Rds(on) specification. The spec says it will have a resistance of < 160m\$\Omega\$ if you drive it with 4.5V (5V will be at least as good). So, at 200mA, you'll have a voltage drop of only 32mV, pretty close to a perfect switch. It's also pretty good with only 2.5V drive.
You also need to know what it takes to turn the MOSFET off, and the threshold voltage Vth gives you a clue:
250uA is not completely off, so you want to get well below 600mV on the gate to ensure it's fully off, although typically 1V would be okay. Any CMOS output will easily do this, unless you're sinking a lot of current through it for some other purpose.