It's the first time I'm using a MOSFET. It's called IR540 and is apparently an Enhancement-mode N-channel MOSFET.

I made a simple a circuit to test its switching operation. I am using a 9V battery. I connect a 10k resistor between Vdd and the drain of the FET and ground the source. It starts conducting after I connect Vdd to the gate but it continues conducting even after I remove Vdd from it.

I tried using another FET of the same model, but that one behaves even more strange. It works without any gate voltage at all from the get go.

I have rechecked the which terminals are which several times and I don't think I'm making a mistake there.
I'm working with the assumption that with the front of FET facing me, the leftmost pin is G, middle D and rightmost S. I don't think I'm wrong there.

Any ideas?

 

MOSFETs are controlled by voltage. There is considerable capacitance between the gate and the other electrodes, and there is some leakage. When you touched the gate to the drain voltage you charged up the gate capacitance, and unless you dump the charge, such as by connecting the gate to the source directly or through a resistor or other component, the gate will stay charged for a long time.

Depending upon the leakage current, if you don't discharge the gate capacitance it might never discharge or do so slowly.

Strange, the IR datasheet does not give the pin assignments. I think your assignment is correct, but I like to see it in the datasheet just in case.

 

Add a resistor from gate to source. About 10k will work.
The gate is very high impedance and will hold a charge. You have to let it drain out to ground.

 

It's the first time I'm using a MOSFET. It's called IR540 and is apparently an Enhancement-mode N-channel MOSFET.

I made a simple a circuit to test its switching operation. I am using a 9V battery. I connect a 10k resistor between Vdd and the drain of the FET and ground the source. It starts conducting after I connect Vdd to the gate but it continues conducting even after I remove Vdd from it...

Congratulations! you have a working MOSFET!

 

I am using a 9V battery. I connect a 10k resistor between Vdd and the drain of the FET and ground the source. It starts conducting after I connect Vdd to the gate but it continues conducting even after I remove Vdd from it.

That is perfectly normal. Once you placed a positive charge on the gate by connecting it to +9V, disconnecting the +9V simply left the charge on there. With nothing to drain off the charge (like a resistor between gate and source), the gate stayed positively biased and kept the MOSFET on.

 

You made the common newbie mistake of thinking an open input is at zero potential.
For high impedance devices, such as CMOS logic circuits and other MOSFETs, an open gate input is floating, and can be at any potential.

 

You made the common newbie mistake of thinking an open input is at zero potential.
For high impedance devices, such as CMOS logic circuits and other MOSFETs, an open gate input is floating, and can be at any potential.

Hehe. It is funny because I have previously corrected my friends for making the same mistake that I have made here.

Thanks a lot everyone! Makes perfect sense now.

Though, just to be sure, I have to ask: I want to use MOSFET in a switching circuit triggered by pulses. Would the MOSFET turn off when the gate pulse reaches 0V without there being a resistor between gate and source? I'm pretty sure it would, but just want to be sure.

Final question: What happens if Vgs goes negative in an n-channel E-MOSFET? Does it remain off?

Thanks again, people!

Congratulations! you have a working MOSFET!

LOL!

 

I want to use MOSFET in a switching circuit triggered by pulses. Would the MOSFET turn off when the gate pulse reaches 0V without there being a resistor between gate and source? I'm pretty sure it would, but just want to be sure.

You know a mosfet is a complete cripple when it comes to controlling its own gate. The answer depends entirely on the driver. An open collector comparator? Must have a pull-up resistor. A Class A amplifier? Must have a load. An op-amp? Usually has a bipolar driver that sources and sinks current.

 

Though, just to be sure, I have to ask: I want to use MOSFET in a switching circuit triggered by pulses. Would the MOSFET turn off when the gate pulse reaches 0V without there being a resistor between gate and source?

Sure. The important thing is that whatever drives the MOSFET gate must drive it to the desired voltage; simply making it open-circuit is a no-no.

Final question: What happens if Vgs goes negative in an n-channel E-MOSFET? Does it remain off?

Yes, it remains off. What else it does depends on the MOSFET, and what kind of gate protection is built into it: some have a gate-to-source Zener diode to protect the gate from electrostatic discharge, and will not tolerate a negative drive voltage on the gate. Always consult the MOSFET data sheet to see what the gate drive requirements and restrictions are.

 

Hehe. It is funny because I have previously corrected my friends for making the same mistake that I have made here.
.........................

They say experience is what allows you to recognize a mistake when you make it again.

 

What happens if Vgs goes negative in an n-channel E-MOSFET? Does it remain off?

With sufficiently high Vds, there will be some drain current flowing with Vgs=0V. Making Vgs negative will increase channel resistance and decrease leakage current; but you can't exceed the max specified in the datasheet.

 

They say experience is what allows you to recognize a mistake when you make it again.

Well-said. Point taken!

I thank you all once again.