Converting logic level sink to source for powering n-channel mosfet gate

Discussion in 'General Electronics Chat' started by MrSoftware, Nov 15, 2013.

  1. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    I need to power the gate of this n-channel MOSFET (provide +V), and I fear that my processor outputs won't give enough + current. The processor spec only says "weak pull up", and the MOSFET data sheet doesn't show the gate current required to maintain the on-state, so I am going on the assumption that my processor outputs won't provide enough source to activate the MOSFET.

    However, the processor outputs will sink up to 5mA, so somehow I need to turn this 5mA sink into a source to power the MOSFET gate. Normally I could just use a pull-up resistor to power the gate and use the processor output (sink) to keep it low, but power usage is critical. My whole control circuit must run 2 years on a pair of AAA batteries, about 1200mAh capacity, and a 10k pull-up resistor alone will eat the batteries in only 151 days so I need a different solution. Does anyone have any ideas? :confused:
     
  2. crutschow

    Expert

    Mar 14, 2008
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    How fast do you need the MOSFET to switch? A MOSFET gate is very high resistance but there is a significant capacitance that needs to be charged and discharged. If you aren't concerned about switching speed than you can use a high value resistor (say 1Mohm) for the pull-up. If you need fast switching speed then you can use a push-pull driver to increase the transient current available to drive the capacitance.

    Note that the MOSFET you show requires 5V to fully turn on. If you are operating from only 3V from the two AAA batteries you will need a logic-level MOSFET where Rds(on) is specified at 3V Vgs or less [not Vgs(th)].
     
  3. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    Thanks for the reply! Speed is not critical, it can take as long as a full second to turn on. Am I understanding your reply correctly that once enough charge has flowed into the gate to reach full voltage, there is essentially no more gate current? This is one area where I was confused; I assumed there would be some small continuous current into the gate during operation, but it sounds like the gate acts like a capacitor and there's essentially no more current once it has charged up?

    Thanks for pointing out the required Vgs, somehow I overlooked that and thought it would be full-on at 3V. If I can find a logic-level MOSFET, is it possible that the "weak pull up" from my processor output alone would be enough to activate it, without need for a pull-up resistor at all?

    Thanks for your help, I appreciate it!
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    The current required to turn on a MOSFET is so low you can't measure it.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Oh, you can measure it and it can be significant.

    In this device, for instance, the total gate charge is about 15nC. The turn-on rise time is about 300ns. So if you are pushing the device to achieve that performance your gate current during that time is about 50mA. If you are only switching it occasionally, the average current goes to negligible levels pretty quickly, but if you are switching it very often, the gate current can become a significant source of power dissipation.
     
  6. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    I will be switching it on once, and it will stay on for many seconds to several minutes. Then it will be switched off for many minutes or hours before being turned on again. So speed of on/off is relatively insignificant in this use case.

    I'm using it to turn on a piece of electrical equipment that will draw about .5A at 12V.
     
  7. BobTPH

    Active Member

    Jun 5, 2013
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    I have never heard of a weak pullup on an output, they are usually push-pull or open drain. What uP are you using? Cam you post a link to the datasheet?

    Bob
     
  8. MrChips

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    Oct 2, 2009
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    I was under the impression that this was a fairly static signal.
     
  9. t06afre

    AAC Fanatic!

    May 11, 2009
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    Would it not be better to use a large pull-down resistor. And then configure the pin used for controlling the FET as input then not used to turn the FET on
     
  10. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Perhaps a more important question is what is the min Vcc voltage of the processor? If it is only 2.2V as the batteries die, then you will need an "logic-level" NFet with very low threshold voltage (<1V).

    What is the off-state drain voltage of the NFet? I'm presuming that whatever load is being switched by the NFet is not powered by the two AAAs? :p
     
  11. crutschow

    Expert

    Mar 14, 2008
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    Yes, the gate does act as a capacitor and there is only a very small leakage current once the gate capacitance is charged.

    The "weak pull up" should be fine as long as the voltage is sufficient to fully turn on the MOSFET.
     
  12. inwo

    Well-Known Member

    Nov 7, 2013
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    Sounds like a job for bipolar man.
    Static switching an fet from a pnp should be able to use megohm resistances.
     
  13. crutschow

    Expert

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    Why add the pnp? It just takes extra current.
     
  14. inwo

    Well-Known Member

    Nov 7, 2013
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    No need if that's true.
    Any buffer will have some current penalty.
     
  15. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    Thanks to everyone for the great discussion, I am learning a lot. :) Here's a link to the data sheet for the processor I'm using (Maxim MAXQ610A):

    http://datasheets.maximintegrated.com/en/ds/MAXQ610.pdf
     
  16. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    It looks like I'll need to find a new FET since the one I had chosen won't be fully on even at max voltage of 3.3V. Do I need one with an integrated charge pump? Suggestions are welcome!


    The large load will be powered by a 12V battery, lead acid or lithium, probably around 9Ah capacity depending on how testing goes.
     
  17. crutschow

    Expert

    Mar 14, 2008
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    If you are operating with the source grounded and the load between the supply and the MOSFET drain, then you don't need a charge pump.

    Look for a logic level type MOSFET that will fully turn ON at at \leq2.5V. You can search Digikey or other suppliers for such devices.
     
  18. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    My intention was to place the FET between the (-) on the battery and the (-) on the load. The (-) from the load battery will also be connected to the (-) of the processor so they have a common ground. I'm thinking this FET looks good:

    http://www.aosmd.com/res/data_sheets/AO3414.pdf
     
  19. BobTPH

    Active Member

    Jun 5, 2013
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    From the datasheet:

    The MAXQ610 provides port pins for general-purpose
    I/Os that have the following features:
    • CMOS output drivers
    • Schmitt trigger inputs


    • Optional weak pullup to V
    DD when operating in input
    mode

    So it looks like the outputs are standard CMOS push-pull outputs, not a weak pullup.

    Bob

     
  20. MrSoftware

    Thread Starter Member

    Oct 29, 2013
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    Thanks for the clarification. So it sounds like the outputs should drive a logic level MOSFET then?

    I didn't understand what "CMOS output drivers" meant in terms of output power, so I mailed Maxim support for more detail. They wrote back indicating that the outputs had "weak pull-ups" that won't power the gate of a MOSFET so I just took their word for it. But it sounds like CMOS outputs should be plenty to power the gate of a logic level MOSFET? If yes then great, I think I found a good FET so I should be set. :)
     
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