Is my mosfet IRF540N OK ? Help to understand MOSFET behaviour

Question

I am new to using MOSFETs. To help check the MOSFETs, I made a small circuit as in the schematic below on a perforated board. My test MOSFET is IRF540N. Supply source is a Lead Acid battery of 12V.

I am keeping \$V_{GS} = 0\$V (switch open). I expect when I apply the source, the MOSFET will hold and hold forever, as the supply voltage is well below the Breakdown Voltage.

But, I see the voltage across DS slowly falling and voltage across \$R_1\$ slowly rising till the voltage is about equal across \$R_1\$ and the MOSFET. Then it rises too. This keeps repeating.

If I short \$R_1\$, the MOSFET will blow in 1-2 min.

If anybody can help me understand where I am going wrong, that will be greatly appreciated.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Hook it up like this. R2 can be much higher than 10K, very high values will cause it to switch off slower.

The MOSFET is switched on by having a positive voltage of a few volts on the gate relative to the source. It is switched off by having 0V. The gate behaves electrically like a capacitor of some nF, so you have to charge and discharge it just like a capacitor. If you leave the gate open (floating) it will tend to maintain the same state, sometimes for a surprisingly long period of time, before eventually leakage causes it to change.

If you're using a heavy load instead of R1, you want to make sure that the MOSFET is always cleanly on or off, not somewhere in-between where it will get hot. For example, if your 12V was a car battery and R1 was a heater, if you allowed the car battery to discharge to flat, there might be a point where the MOSFET was destroyed because it got too hot. Same thing could happen a lot faster (with a heavy load) if you allow the gate to float.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

I am keeping VGS=0V (switch open)

No, you're not. When the switch is open \$V_{GS}\$ isn't 0V, it's undefined. The input is floating and has no set voltage.

Answer 3

Your test circuit won't work.

1. It is configured as a high side switch. \$V_{gs}\$ can never be high enough to turn it on.
2. \$V_{gs}\$ is not zero when the switch is open, it is floating.

Put it on the low side, add a \$100 \text k\Omega\$ pull-down to ground, and try again.

Answer 4

@gtn, with the same success you can control the MOSFET by swapping the switch SW1 and resistor R2... or replacing them with a 3-terminal switch. You can even connect a potentiometer to the input and explore (carefully in the middle) when the MOSFET switches.

Finally, if you want to impress others, you can control the MOSFET in a spectacular way, even without using the switch SW2 and resistor R2... but only using your body... and maybe even from some distance (a sensor switch). For this purpose, touch with a finger of one hand the input, and with a finger of the other hand - the positive rail or the ground (again, be careful and do not leave for a long time the input to "float"). The effect will be amazing:)