P-Channel Pickle

Discussion in 'General Electronics Chat' started by jbriaris, Apr 16, 2013.

  1. jbriaris

    Thread Starter New Member

    Apr 11, 2013
    Hi All

    I'm looking to use a p-channel MOSFET as a logic level switch. I'm working between 0 and 5V, and want the FET to switch on (i.e. saturated) at 0V and to switch off at a positive voltage less than 5V.

    I've been playing with different spice models (e.g. IRLML6302, BSS84, etc.) but whilst they all have positive gate threshold voltages Vgs, I'm still operating in the linear region at 0V (i.e. Vgs is the 'turn off' voltage, not the 'turn on' voltage). I want the FET to be saturated at 0V.

    Does anyone know of any logic level p-channel MOSFETs that can do this? Or, perhaps, I'm missing something and can obtain the desired result using an N-channel MOSFET?

    Thanx! :)
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
    Where exactly is the threshold? 0V and lower means on, anything above 0V means off?
    It seems you would want a comparator driving the mosfet.
  3. jbriaris

    Thread Starter New Member

    Apr 11, 2013
    Thanks kubeek.

    Yes, 0V should be off and anything greater than 5V should be on. A comparator would be an option, but I was trying to keep the number of components to a minimum.

    I ask because I'm designing an LED driver circuit and already using some n-channel MOSFETS (2N7007) to do switching using 0 and 5V as logic levels, i.e., 5V on, and 0V off. I was hoping that I could 'mirror' the 2N7007 FET using a similar p-channel equivalent to provide 5V off and 0V on. Perhaps some SPSTs/SPDTs would be better? :confused:
  4. #12


    Nov 30, 2010
    You seem to be looking for some logic level, p-channel, depletion type mosfets. I don't think they exist.

    How much current?
  5. jbriaris

    Thread Starter New Member

    Apr 11, 2013
    It wouldn't surprise me if I was looking for something that's not available :p.

    The role of the FET isn't to draw current through a load, but to tie another MOSFET that forms part of a constant current circuit to ground (see circ1.png attached). I might not need a p-channel FET, but I was trying to understand what was going on by trying one out - let me explain more:

    The circuit attached is a simplified part of a larger circuit that drives two LEDs back-to-back. It is grossly simplified but exhibits the same phenomenon that I don't quite understand. The FET U3 is switched off asynchronously with FET U2, i.e., the output from the opamp is 'grounded' via R3 when V2 is low (0V), but delivered to the gate of U2 when V2 is high (again, this isn't the complete circuit, but does demonstrate what I'm getting at).

    plot1.png and plot2.png shows what I'm trying to understand. plot1.png shows pulse trains V2 and V3, and the current through V5 (null voltage source just to measure current). The current is pretty much as expected at 20mA, but with peaking either side. First few questions:
    1. Is this peaking the direct result of the small rise- and fall-times of the pulses?
    2. There will always be some rise-/fall-time; is there any way to minimize the peaking effect on the current?
    3. More of a general point: I set the rise/fall times to zero in the SPICE PULSE parameters for V2 and V3 - why are they there?
    plot2.png shows the output voltage from the 'ideal' opamp U1. When V3 is low and U2 is tied to ground, the gain becomes massive before slowly falling away. Again,
    1. Is this the result of the rise/fall-times of the pulse trains?
    2. Do I need to worry about this? (I guess in a real opamp the gain will be limited by Vcc)
    Would really appreciate some pointers. Thanks. :)
  6. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    What happens is when V3 is high U2 is held off, but the op amp U1 wants to see current so it's output goes to it's positive saturation. When you release the drive on U2 a HUGE drive voltage is there so it turns on into an overshoot.

    You would get a better response by moving R3/U3 from the output of U1 to the input, using it to "short out" the V2 signal from U1.
    jbriaris likes this.
  7. takao21203

    Distinguished Member

    Apr 28, 2012
    They do exist. All P-CH MOSFETs have negative gate voltages.

    I've abused some regular p-CH MOSFETs for a 3V circuit, it is enough for small currents, right on the threshold margin.
    They are not neccessarily expensive.

    But normally even for 5V, use a logic level MOSFET.