Switching high side of LED with N channel

Discussion in 'General Electronics Chat' started by stoopkid, Feb 18, 2013.

  1. stoopkid

    Thread Starter Member

    Mar 3, 2011
    136
    1
    I'm trying to throw something together and I need to use small parts. But I realized I have no SMD PNP transistors. I have an LED that I need to switch on the high side (it is part of an LCD with only one common ground). The best I can find in my inventory is a BSS138N N Channel Mosfet. Now, I'm only just getting the hang of BTJs so I'm hoping I can get some help making sure this thing will work for me since I can't breadboard it first.

    The gate will be controlled from a 3v source and will be switching a 3v source and only need to pass less than 75mA. Will that be sufficient gate voltage for this device? Should I use a gate resistor, and of what value? Is a pulldown resistor of approx 10x the gate resistor appropriate as with a BTJ? I believe table 5 of that sheet is what I need to be looking at and that it will pass more than 0.1A at 3v. Is that correct?

    Should the Gate be pulled down to the source pin/LED anode or just down to ground?

    Thanks.
     
    Last edited: Feb 18, 2013
  2. #12

    Expert

    Nov 30, 2010
    16,261
    6,770
    Yeah. About .24 amp (max) with 3.2V gate to source.
    In a DC condition, mosfets don't have gate current. Forget the 10 to 1 ratio.
    The problem I am imagining is that you're going to use an n-channel MOSFET like an emitter follower and try to drive an LED that needs at least 2.2 volts from a 3.3 volt source. Imagine it as an npn transistor with a high base voltage and no base current.

    You wanna give a drawing? I am not getting you a positive answer from what I have so far.

    Edit: Volts gate to source is what matters. I can't tell if "ground" will satisfy the MOSFET without a drawing of some sort. And, when volts gate to source is zero, the mosfet will be "off".
     
    Last edited: Feb 18, 2013
  3. stoopkid

    Thread Starter Member

    Mar 3, 2011
    136
    1
    This is basically what I mean. Somehow I've got to switch the high side, I can't switch that ground without switching the entire LCD. NPN BTJ's and this N Channel mosfet are all I've got to do it.

    I was under the impression that it was ok to switch the high side of a load with a mosfet because it had no voltage drop, just some amount of resistance. Is that not the case?

    In the schematic, I didn't include a gate resistor or a pulldown resistor. Is a gate resistor necessary? I figure a pulldown is necessary but I wasn't sure if I should pull down to ground or to the source pin of the mosfet.

    Edit: ah, I see what you're saying. Do I simply need a PNP or P Channel to get this done?

    I just thought I wouldn't have to worry about that with a mosfet but I guess the easiest thing to do would be to find a way to squeeze a 3906 through hole onto the board. Thanks.
     
    Last edited: Feb 18, 2013
  4. stoopkid

    Thread Starter Member

    Mar 3, 2011
    136
    1
    Will this work? Sometimes I see PNP transistors being pulled down by an NPN transistor. Do I need that here or am I ok?
     
  5. #12

    Expert

    Nov 30, 2010
    16,261
    6,770
    Yes, your pnp transistor will work.

    Irrelevant now, but the mosfet would be safe with the gate resistor to ground or the source because the mosfet can survive +/- 20 volts on its gate.

    The problem with your first approach is the emitter follower idea. The LED and resistor will have voltage across them and the gate of the mosfet has to be higher than its source to pass current. You have very little voltage left to drive the gate with after the LED and resistor use up their share.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    Others have already said this, but I'll add a little more info.
    You need a PNP, or a logic level p-channel MOSFET which is specified for Rds(on) with |Vgs|<3V.
    You will not be able to drive your LED with an NMOS unless the gate voltage is at least a couple of volts higher than the drain supply voltage, and then you have to carefully choose one which has Rds(on) specified with low Vgs.
     
  7. tinamishra

    New Member

    Dec 1, 2012
    39
    1
    Note that When using a N-Channel MOSFETs as a a high-side power switch, the FET will not turn completely ‘on’ by simply bringing the gate voltage up to V+. You will need to bring the gate voltage ABOVE the rail voltage to get the FET to turn completely on.
     
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