P-Channel MOSFET Behaviour?

Discussion in 'General Electronics Chat' started by philipm, Jan 6, 2013.

  1. philipm

    Thread Starter Member

    Jun 27, 2012
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    My question boils down to - what is the behaviour of a P-Channel MOSFET when the gate is left floating? Does it conduct (Drain to Source) with minimal resistance in this case?

    I'm looking at this circuit:
    https://www.olimex.com/Products/Duino/AVR/OLIMEXINO-328/resources/OLIMEXINO-328-schematic.pdf

    Which uses this P-Channel MOSFET:
    http://www.mouser.com/ds/2/200/irlml6402-50379.pdf

    If you look at the middle left where it says "Battery Charger", I believe that the circuit is using the FET to switch the input to the MCP1700T step down regulator between the LIPO_BAT and +5V.
    I.e.:
    (1) when +5v is present (on the Gate), Gate to Source is Positive, so the FET turns off.
    (2) when the Gate is floating, Gate to Source is ... (negative?!?, so the FET is on?)

    It's the second scenario that I find hard to fathom, so wanted to check before I incorporate something similar in my own circuit. Many thanks.
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Because of the capacitance, turning on the gate then removing connection to the gate will leave a charge on the gate and the D-S will stay in a conducting state until the charge dissipates.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    When +5V is not present, the gate is not floating. It is pulled to zero volts through R3, R4, and R26 (near the bottom of the page), and possibly other miscellaneous loads on the board. This turns on FET2, which switches the LiPO battery to the input of VR1.
    Floating MOSFET gates are a no-no, because the gate is very high impedance, and has capacitance, and can hold a charge indefinitely, so the device could be on, or off, or anywhere in between.
     
  4. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Ah ha! Now I understand, thanks.

    So for my circuit design, I'd like to use similar logic, so that a regulator is supplied with Battery power when no usb is connected and disconnected from the battery when USB power is supplied. Hence I have two scenarios:
    (1) USB power is connected, turn off the MOSFET
    (2) no USB power, turn on the MOSFET

    To do this, I'll connect:
    (a) Drain to the Battery
    (b) Source to the Regulator
    (c) Gate to VUSB with a 10k pulldown resistor to ground

    I'm assuming that the Gate-Drain resistance is very large (say 10MOhm). Hence I believe a 10k pulldown should tie the Gate very close to 0v as well as discharge the potential 633pF@5V capacitance in well under a millisecond (if needed). Schematic attached. Does that sound correct?
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Sounds good.

    FYI "back in the day" when I would frequently have to test power MOSFETs in big TO-3 packages I found I could do a "dynamic" test with just my ohmmeter. Short gate to source gave a high resistance drain to source, then test gate to source (also high Z) would charge the gate so going back to test drain to source would now give a low resistance.

    Good FETs would fly thru this test, bad devices never did and got tossed.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Don't you want USB to power the output when it is present? Your circuit doesn't do that.
     
  7. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Good question, I do really want to just disable the regulator (and main circuit). The circuit is for a cycle light. Hence USB to charge and a battery when on the go. It also has a big power requirement and would try to draw >1amp from the USB (in addition to the 450mA charge current)! The regulator powers the Microcontroller and LED Driver, hence the desire to turn them off during charging, thus disabling everything but the charger. Thanks for checking.

    And thanks all for the kind replies. A floating gate could have been a disaster!
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    OK, I get it.
    What is the battery voltage?
    What kind of regulator is it going to?
     
  9. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    I'm planning to use this battery (probably a different LiPo after development/testing/refinement):
    https://www.olimex.com/Products/Power/BATTERY-LIPO1400mAh/

    The regulator will be an MCP1700 (3.0v version), rated to 250mA, but supplying less than 20mA (with only uA needed when the circuit is sleeping).
    http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en010642

    Hence I have a 4.2v->3v battery going to a 3v LDO regulator. The regulator will drop out a shade under 3.2v which is when the battery voltage goes "off a cliff" just before its own inbuilt cutoff at 3v.

    Edit - the big power requirement is to drive LED strings via a boost converter, and so doesn't need to go through the regulator.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    With the PMOS as you drew it, the battery will drive the regulator through the drain-source junction (shown in your schematic) when USB is present. If that is not what you intend, you should swap source and drain.
     
  11. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Are you sure? I've looked at this multiple times and still don't see the error. My reasoning is:
    1 the mosfet turn on when Vgs is negative
    2 the mosfet turns off when Vgs is positive
    3 the initial Source voltage is approx 3v (LiPo battery through the diode)
    4 the USB voltage is 5v

    Hence when vUSB is present on the Gate, Vgs = Vg - Vs = 5 - 3 = 2. So the mosfet is off.

    And when vUSB is not present, the Gate is pulled low to 0v, hence Vgs = Vg - Vs = 0 - 3 = -3 so the mosfet is on.

    It's my first mosfet circuit, so I'm keen to get it right! :)
     
  12. Ron H

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    Apr 14, 2005
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    See the attachment. The only thing wrong with you circuit is you need to swap source and drain.
     
    Last edited: Jan 9, 2013
  13. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    A second "Ah ha!" moment in one thread. I'm learning. Thank you for your patience.

    I actually have one of the olimex boards and so have done some testing just to get it really clear in my head. See attached pics. Text in blue are voltage readings with just the LiPo connected and text in blue are with vUSB and the LiPo connected.

    On reflection I think I've misunderstood what the circuit is trying to do. My initial belief is that it disconnects the LiPo from VR1 when 5V is available.

    Instead, I now believe that it allows the connection of LiPo to VR1 irrespective of +5V. At the same time, it protects LIPO from receiving 5V (which would exceed the charging limits). Clever.

    So the two scenarios are:
    (1) +5V present = LiPo is connected to VR1 via the diode in FET2 (FET is OFF). +5V is also connected through SD4 to VR1.
    (2) +5v is absent = FET2 Gate is pulled to 0V, FET2 is ON and LIPO is connected to VR1 with just a minimal voltage drop across FET2. So the circuit is efficient when battery powered.

    Is that correct?

    (I will also reverse the Source/Drain for my circuit. Thanks again!)
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You may have it right, depending on what you meant in (1). As you said, the FET is off. The battery is analog OR'ed with the +5V via the substrate diode and the Schottky diode. In an analog OR, the higher voltage wins. In this case, the battery would actually have to be above ≈5.3V before it would starrt to provide current to the regulator, because the Schottky has a forward voltage drop that is about 0.3v, while the substrate diode is about 0.6 - 0.7V.
    The FET could be replaced with a Schottky diode, but since the battery voltage is low to start with, the FET is used to minimize the voltage drop in that path when it is on.
     
  15. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Yes, that's my reading of the circuit too. I didn't know about the exact behaviour of the OR'd voltages, thanks for adding that too.

    Reversing the Source/Drain on my circuit also helped the board routing. =)

    Thanks again for all your kind assistance.
     
  16. philipm

    Thread Starter Member

    Jun 27, 2012
    47
    3
    Just a quick update - the circuit works perfectly. The MOSFET cuts the power to the regulator when charging the battery. Thanks again.
     
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