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Power Switching with mosfet

Discussion in 'Electronic Design' started by Jon Slaughter, Oct 23, 2007.

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  1. I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use it
    to switch power to an IC.

    I can only bring the gate voltage about 1.7V above the V_DS. I'm not
    switching the entire voltage to the IC.

    Some question I have, What is the difference between a high side mosfet
    driver and a low side? A low side brings the drain within a few mV above
    ground but the high side brings the source within only several hundred mV
    below the drain. Shouldn't it be symmetric?

    i.e., whats the difference between

    V
    |
    Mosfet
    |
    Load
    |
    G

    V
    |
    Load
    |
    Mosfet
    |
    G

    I'm calling the first one a high side and the second a low side, I think
    thats the correct terminology. Both mosfets are the same as is the load.

    The low side works fine but the high side doesn't. It seems that in either
    cause it should be the same. V_DS is the same in both cases? Obviously this
    isn't true but I can't seem to understand why?

    The way I'm looking at it is basically as two resistors in series so the
    voltage drop across each should be exactly the same.

    So whats wrong?

    Thanks,
    Jon
     
  2. D from BC

    D from BC Guest

    It's all about how Vgs is developed for saturation.

    High Side:
    (Vgs >Vdd )

    Low Side:
    (Vgs < Vdd )


    D from BC
     
  3. Winfield

    Winfield Guest

    It is completely symmetric, or shall we say identical.
    But this means you have to take the high-side gate as
    far above its source, e.g. 3.5 volts over the drain,
    as you took the low-side gate above its source=drain.
    This can mean taking the high-side gate higher than
    the supply voltage, which requires a special kind of
    driver circuit, called a high-side driver. These have
    a charged capacitor to act as the flying gate-driver
    power supply. That works for periodic switching, but
    for DC switching you need an isolated gate supply. I
    just finished a fast DC-capable 600V high-side pulser
    and powered the driver with its own 60Hz transformer.
     
  4. Winfield

    Winfield Guest

    It is completely symmetric, or shall we say identical.
    But this means you have to take the high-side gate as
    far above its source, e.g. 3.5 volts over the drain,
    as you took the low-side gate above its source=drain.
    This can mean taking the high-side gate higher than
    the supply voltage, which requires a special kind of
    driver circuit, called a high-side driver. These have
    a charged capacitor to act as the flying gate-driver
    power supply. That works for periodic switching, but
    for DC switching you need an isolated gate supply. I
    just finished a fast DC-capable 600V high-side pulser
    and powered the driver with its own 60Hz transformer.
     
  5. Tam/WB2TT

    Tam/WB2TT Guest

    You kind of answered your own question here. The 2N7002 won't even start to
    turn on before the Vgs reaches 2V. You are saying you can only supply 1.7.

    The recommended Gate to Source ON voltage is 5V. When the source is
    connected to ground that means a gate to ground voltage of 5V. For high side
    switching you will bring the source voltage up to near what you call V, and
    the gate to ground voltage now has to be V + 5. If you can't supply enough
    gate voltage, you need to use a P channel FET.

    Tam
     

  6. What sort of IC are you switching? In the case of low side switching,
    any input midway between V and G will end up below the load's ground
    rail. Likewise, for high side switching, an input may end up above the
    ICs supply rail. Some ICs don't like either (or both) of these
    situations.

    Others have pointed out the gate drive level issues.
     
  7. D from BC

    D from BC Guest

    Ooops ...let me fix that..

    High Side:
    (Vg > Vdd )

    Low Side:
    (Vg < Vdd)


    D from BC
     
  8. Ok guys, I think I got it. I was sorta thinking about it that way but
    couldn't put the pieces together cause I thought the gate voltage had only
    to be greater than V_DS to turn it on.

    So basically the gate voltage has to be so many volts above the source
    voltage? When the source is grounded, as in a low-side, then its very
    simple. If it's high side then you have increased the source voltage and
    have to increase the gate voltage to compensate.

    Ok, I think I got it for the most part(somewhat makes sense but not sure
    exactly why it works that way).

    I can't see what the difference between them are. Like why is one used over
    another. (I understand why one uses the mosfets but not when I should know
    to use one but not the other).

    This site talks about a high side using an n-channel,

    http://www.innovatia.com/Design_Center/High-Side Drivers.htm

    But it seems I can use a p-channel without all that mess?

    There are 4 combinations, n-channel high side, n-channel low side, p-channel
    high side, p-channel low side.

    The n-channel low side and p-channel high side seems to work the same? Is
    there a time when one is better to use than the other?

    What about a p-channel high side vs an n-channel high side? Why go through
    what that site says when a p-channel will simplify it?

    Thanks,
    Jon
     
  9. For a given price, you can get a N-channel with much better
    specifications than P-channel (on resistance, current and voltage
    handling). Especially at the high voltages and powers, where N-channel
    becomes the only option.
     
  10. http://en.wikipedia.org/wiki/Threshold_voltage

    I guess this might be where my confusion came from?

    "If the gate voltage is larger than the threshold voltage, the transistor is
    turned on"?

    But shouldn't that be the gate to source voltage?
     
  11. So its just a practical issue rather than a theoretical one...

    Thanks,
    Jon
     
  12. Its a Pic24.

    How can it end up above the rail? You mean input to the IC? Since the
    mosfet might not pull the down completely or pull up completely?

    If thats the case then how does one go about solving that problem? In my
    application my inputs the the IC will be at the rail since I'm using a
    pullup and very little current will flow. It will only be a few mV I imagine
    but maybe that will make a difference? R_DS is about 10 ohms which may or
    may not be significant? What kinda wiggle room do I have to work here?


    Oh, I think I can just use the mosfet to power up the pullups too though...
    so when the mosfet is triggered it will power all pullups used for
    inputs/outputs and so should reduce there voltages be the same amount... But
    assume I just didn't think about that ;)


    Thanks,
    Jon
     
  13. Tam/WB2TT

    Tam/WB2TT Guest

    Paul,
    You hit the nail on the head. From personal (unintentional) experience, at
    least for digital ICs, opening up either the ground or the Vdd lead can lead
    to unexpected and bizarre operation.

    Tam
     
  14. Tam/WB2TT

    Tam/WB2TT Guest

    By definition gate voltage means gate to source voltage when talking about
    the device characteristics. Also, see Paul's comments from a few posts back.
    It is not obvious that turning off power to the IC will do what you want it
    to. What do you expect the IC outputs to do when it is unpowered?

    Tam
     
  15. Sure. If you are only switching a small, low voltage load, by all
    means use a p-fet. (And it is usually a better idea to switch the +ve
    rail than the 0V).
     
  16. Why is that? I was thinking it might be better since it cuts the power(so
    the device isn't "floating" in some sense) but not sure...
     
  17. Huh? I want to cut the power to the IC because it requires a specific power
    on sequence to enter programming mode. Its not entirely clear that have to
    cut the power but I don't see the big deal in having that ability and I
    might need it so I can automatically reset the device if an error occurs.
     
  18. neon

    neon

    1,325
    0
    Oct 21, 2006
    it depends where the loading will be sourcing or outputs.
     
  19. Tam/WB2TT

    Tam/WB2TT Guest

    You are either going to leave VDD floating, or GND floating. Neither one is
    a good thing. Why don't you try it, open up one or the other manually, and
    restore with a clip lead. Why not just hold the chip in reset? It is
    improbable that the chip manufacturer intended for you to control power to
    the chip independent of other circuits.

    Tam

    Tam
     
  20. Well it depends on the circuit arrangement. But imagine you feed it
    from a power supply with an earthed 0V (A USB cable, perhaps). Your
    mosfet cuts the zero volts and the circuit turns off. But then you
    connect a RS232 cable, and the circuit powers up again, using the 0V
    connection in the RS232!
     
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