high side MOSFET switching

Discussion in 'General Electronics Chat' started by strantor, Sep 25, 2011.

  1. strantor

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    Why is it that when using a MOSFET for high side switching, the gate needs to be 10V higher than the drain to switch? and why is that a bootstrap capacitor is needed for high side switching but not low side?
     
  2. SgtWookie

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    That's not quite true.
    With a standard N-ch MOSFET, the gate needs to be 10v more positive than the source terminal. That's why Rds(on) is specified when Vgs = 10v.

    When used as a high side switch, when the MOSFET is fully turned on, the gate needs to be 10v higher than the source terminal, but when it's on, the source terminal and the drain terminal are at nearly the same potential, which is usually the highest voltage available from the power supply. So, you need some method of getting that gate ~10v higher than the power supply.
    For a low-side switch, you only need to get Vgs up to the point specified for Rds(on); your power supply is usually putting out that much. But with a high side switch, the MOSFET has to essentially "pack a lunch"; the low side of the boost cap is tied to the source terminal, so as the source terminal voltage rises, the voltage potential across the boost cap is carried along with it.

    [eta]
    If you want to experiment with a 555-powered high-side switch, have a look at the attached. You can get just about everything but the high-power LEDs and the 1N5817 Schottky diodes from a Radio Shack store (they don't carry any Schottky diodes, and they are needed here).
     
    Last edited: Sep 25, 2011
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  3. strantor

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    thank you sgt
     
  4. shortbus

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    @SgtWookie - But when a high side mosfet is 'ON' there is still only ~10V on the gate. it is just referenced to the source voltage, instead of ground. If the source has 60V on it, there isn't 70V on the gate. The mosfet is still limited to the Vgs limit, taking it beyond that (usually 20-30V limit) will blow the mosfet.
     
  5. SgtWookie

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    There is still only ~10v on the gate, as referenced to the source terminal - that's correct.

    If there were 6v, 60v or 600v on the source, you'd still have a required Vgs of 10v to turn it on.

    I see students and hobbyists constantly trying to reference the gate voltage to ground instead of the source terminal; that's just not the way to do it.

    Any time you start deviating from strictly Vgs, you start to get into trouble.
     
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  6. shortbus

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    Thanks for that. It took me awhile and a few mosfets to learn that you can't use just a diode from the source to the gate to keep the Vgs at 10V above source voltage.
     
  7. CraigHB

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    Aug 12, 2011
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    So, why would you use an N-channel as a high side switch instead of a P-channel which requires fewer parts? Not rhetorical, I'm asking.
     
  8. strantor

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    because I'm a NOOB and don't know any better. I have decided against making this BLDC controller that the question was based on. I have opted for making a simpler brushed DC motor controller to start off. Once I have success, I will be back to designing a BLDC controller and I will look into N-channel FETs at that time. Thanks for the suggestion, I won't forget it (hopefully)
     
  9. SgtWookie

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    Excellent question; I'm glad you asked that. ;)

    It has to do with the basic difference between P-channel and N-channel MOSFETs. It's easier for electrons to flow than holes.

    Between N-ch and P-ch MOSFETs, if all other parameters are complementary (identical, yet opposite in polarity) then the gate charge for the P-ch MOSFET will be about 2.5 times as high as that for the N-ch MOSFET. It's because you need a larger gate area to get the same Rds(on) for the same Vdss.

    The much larger gate charge means that it will require ~2.5 times more current flow to charge/discharge the gate in the same amount of time as the N-ch MOSFET. It requires less power to turn an N-ch MOSFET on and off.

    In a low-power application where the switch won't be changing frequently, it may not be much of an issue. However, if the item will be switching a lot, and will be in use for a considerable period of time, the cost of ownership will be higher over that period of time.

    Where the difference really comes into play is when you have both high and low side switches as in an H-bridge; it is far easier to get the switching times matched if you are using four identical switches.
     
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