Resistance Troubles (n-Channel MOSFET)

Discussion in 'General Electronics Chat' started by Guest3123, Aug 2, 2016.

  1. Guest3123

    Thread Starter Member

    Oct 28, 2014
    312
    18
    I don't understand how something with less resistance conducts better than something that has a higher resistance.

    To Me : Duh.. it has more resistance, so it doesn't conduct as well.

    Copper Wire
    AWG : 12
    Length : 2106 Inches
    Voltage : 12vdc
    Resistance : 28.8 Ohm's
    It heats up, and dissipates 5 watts of power, and 0.416A of current passes threw it, to the ground.

    Copper Wire
    AWG : 12
    Length : 210.6 Inches
    Voltage : 12vdc
    Resistance : 2.88 Ohm's
    It heats up, and dissipates 50 watts of power, and 4.16 Amps of current passes threw it, to the ground.

    Great. So wouldn't an n-Channel Mosfet with an Rds(on) of 1.2mOhm's heat up quicker, than let's say one that was rated at 5 Ohm's?

    The loads current & voltage goes from the Drain to the Source. The Mosfet get's turned on if their is a positive voltage on the gate, and it's greater than the Vgs(th). Got it.

    All Ground connections are connected to the Source. The Drain is connected to the negative of the load. Understood.

    A 15,000Ω resistor on the gate and source, as a pulldown resistor with a positive 5 volts on the gate. The gate draws 0.000333 Amps (0.33uA), and dissipates 0.0016 Watts (1.6mW) of power. I tested, and understand it.

    Why is an n-Channel MOSFET with a (R) Resistance of the (ds) Drain & Source, when the MOSFET is in (on) complete saturation mode, better to have a lower Rds(on) value, than one that has a higher Rds(on) resistance?

    It shouldn't heat up or anything if it's higher. According to Ohm's law.


    Here's some MOSFETs I pulled from Mouser Electronics.

    2.5 Vgs(th), 3.6mOhm's Rds(on)
    Mouser Electronics Part # : 78-SUP85N03-3M6P-GE3

    2.5Vgs(th), 2.5MOhm's
    Mouser Electronics Part # : 595-CSD19536KCS

    So what's the difference? Is it a matter of how well one conducts better or worse than the other?


    Thanks so much, this has been bothering me for quite some time. I've tried for a couple hours to read and watch videos about it, and I just don't understand. Could someone please help?
     
    Last edited: Aug 2, 2016
  2. SLK001

    Well-Known Member

    Nov 29, 2011
    825
    229
    Just remember that P = I^2 x R is also part of Ohm's law.

    Something may have more resistance and so it doesn't conduct as well, but that doesn't mean that it won't conduct! The current thru a wire/device depends only on the voltage across it. So, in your N-Chan MOSFET question, a lower Rds(on) will generate less heat (power) by reason of the equation for power above, given that both MOSFETs have the same current passing thru them.
     
    Guest3123 likes this.
  3. #12

    Expert

    Nov 30, 2010
    16,355
    6,852
    That seems to be the problem, they don't have equal currents. Our guest seems to be trying to short out a massive 12 volt battery, applying constant voltage to all devices. It makes as much sense as connecting a switch to a car battery and then wondering why the switch melted.

    A mosfet can be used as a switch, but it can't be used to short out a car battery. It's supposed to switch the current on and off for a load resistance of some sort, not BE the load.
     
    Guest3123 likes this.
  4. wayneh

    Expert

    Sep 9, 2010
    12,156
    3,063
    The part that's missing is the voltage drop across the load. The two hypothetical MOSFETs are not seeing the same voltage drop of 12V. Suppose the higher Rds MOSFET has the same impedance as the load. You'd have 6V across the load and 6V across the MOSFET. They's both dissipate the same power.

    Now cut Rds in half. Now you have 3/4 of the total resistance you had before, so the amperage goes up to 4/3 (for purely passive resistive loads). Two thirds of the total voltage or 8V are dropped across the load while only 4V are dropped by the MOSFET. The power dissipated by the MOSFET is I^2•R which is now (4/3)^2•0.5 = 0.89. That's 89% of the first scenario.

    The Rds of the MOSFET is usually quite a bit less than the load, but the same factors are relevant. Less Rds means lower ∆V and less power dissipation by the MOSFET.
     
    Guest3123 likes this.
  5. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    Your confusion comes from not understanding that whether a change in resistance generates more or less dissipation depends upon what is driving the resistance.
    If you have a fixed voltage across the resistance, than a lower resistance will have more dissipation since power equals V² / R.
    That's the case with your example of the two different wire lengths.

    If you have a fixed current through the resistance, than a lower resistance will have less dissipation since power also equals I²R.
    That's basically the case when you have a MOSFET switching a load.
    In that case the resistance of the MOSFET is normally much smaller than the load resistance so a change in the MOSFET resistance causes only a very minor change in the load current (and thus the current through the MOSFET).

    So you can't mix the two situations when determining the dissipation in a resistance.

    Make more sense now?
     
    Guest3123 likes this.
Loading...