understanding p-channel MOSFET unexpected behavior

Question

I'm fairly inexperienced and while I have an understanding of classic npn and pnp transistors, I'm shaky when it comes to MOSFETs.

I'm looking for an element to act as a switch. This element would be normally ON, when no stimulus is applied. The trigger needs to be a positive voltage, and is intended to turn OFF the element (open the circuit). After a bit of searching I figured that a P-channel JFET transistor might solve my problem. There would normally be continuity between Drain and Source, unless I applied a positive voltage to the Gate.

I'm talking about a low power application:

• Drain-source voltage: 5V or less
• Drain-source current: a few milliamps are enough
• Gate trigger voltage: 5V or less

Maybe hurriedly, I got a FQP27P06 MOSFET and set up a simple test circuit with a LED. I understand this is a MOSFET and not a JFET, but I thought it would work anyway.

I'm using a voltage divider to keep the gate voltage in the specified range (2 to 4V). TBH I don't understand why that's actually specified as a negative voltage: -2 to -4 V.

In my intentions, the LED should always be on unless the button is pressed.

^{simulate this circuit – Schematic created using CircuitLab}

What I'm observing is:

• the LED is normally ON
• when I press the button the LED goes OFF
• when I release the button the LED STAYS OFF. Or maybe it just glows very dimly and then turns OFF again. Sometimes it gets brighter and brighter but never reaches full brightness. This behavior seems to be randomic.
• in order to turn the LED back ON again, I need to short the Gate to ground!

I do not understand what's going on. Can someone please help me out? Thank you

Answer 1

You are disconnecting the gate of a mosfet by not pressing a button - this (due to high input impedance of the mosfet) leads to picking up charges from air (it acts like capacitor - try touching gate and ground/power), and partially turning mosfet on. To overcome this issue - add a high value resistor from Gate to ground - this will keep the mosfet on all the time, and when button is pressed - it will "overwrite" the resistor and turn transistor off as before. Value of added resistor should be in such range, that when button is pressed it will not drop the gate voltage to less than mosfet threshold voltage (or any voltage that is required for exact application)

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

The oxide layer in the transistor creates a capacitor with the gate as one plate and the rest of the device as the other. When you open the switch it's holding enough gate charge to deplete the channel. Try putting a pull-down resistor between the gate and ground.