# working of a p channel mosfet

Discussion in 'General Electronics Chat' started by sharath_412, May 15, 2008.

1. ### sharath_412 Thread Starter Active Member

May 7, 2007
32
0
Hi,
Recently i was testing the functionality of a P channel mosfet.

1. I applied 3.3V to the source, varied the gate voltage from 0 to 5V. Till the gate voltage is 2.5 or 2.8 Mosfet was ON (I have put a resistor from drain to GND and checked the voltage across it)..

Inference:
Vgs<0 then it will be ON
Vgs>=0 then it will be OFF.

2. Now I applied 3.3V to Drain, varied gate voltage from 0 to 8V(I have put a resistor from source to GND and checked the voltage across it). The Mosfet was ON even if fthe gate voltage is more than 3.3V. How is this?

Did i do any mistake?

2. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
P-channel enh MOSFETS should be connected with the source to a positive potential voltage, load between the drain and a more negative potential.

When Vgs=0, the MOSFET will be off.
When Vgs is around -10, the MOSFET will be ON.
Some logic level MOSFETS need a much smaller Vgs.
If Vgs is somewhere between 0 and -10, the MOSFET will be in a partially conducting state. They are sometimes used this way in linear-type amplifiers. However, they dissipate a lot of power when used this way.

In your step 2, you connected the MOSFET backwards. They do not work correctly if you connect them up incorrectly.
With a P-channel MOSFET, you want the source to always be more positive than the drain.
With an N-channel MOSFET, you want the source to always be more negative than the drain.

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
171
In a pure MOSFET the source and drain are interchangeable entities are are defined only by the circuit configuration which dictates the source and drain. This is for both n- and p-channel MOSFETs.

You need to consider the threshold voltage of the transistor which describes the volatge at which the source-drain channel begins to conduct current. The MOSFET will turn ON when:

Vgs < Vth - both values will be negative

For pMOSFETs it is often better to consider the voltages as their magnitudes (i.e. ignore positive/negative), this way the pMOS equations are the same as for nMOS. Therefore, the pMOSFET is ON when:

|Vgs| > |Vth|

From your above measurements the FET turns off when the gate voltage gets to ~2.8V. Therefore the threshold voltage will be 3.3V (the magnitude of the voltage on the source) - 2.8V = 0.5V. The transistor will turn ON and OFF around this voltage (as an actual voltage it will turn off when Vgs = -0.5V).

Given what I said at the start about the interchangeability of the source and drain, this is exactly the same configuration as your first configuration.

Did you increase the gate voltage from 0 to 8V or 0 to -8V. If increased it from 0 to 8V then the magnitude of Vgs (3.3V to 11.3V) will always be greater than the threshold voltage Vth, therefore by virtue of the ON state being when |Vgs| > |Vth| the transistor won't turn off.

If you "increased" from 0 to -8V then there is an issue and we need to investigate further.

Dave