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Homework Help: Simple MOSFET Question

  1. Jun 21, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure

    2. Relevant equations

    3. The attempt at a solution

    I need to clear something up before I even attempt the question. In the question is states the the [tex]R_{on} = 100 ohm [/tex]

    is that the 100 ohm resistor I see in the circuit or is it literally inside the MOSFET?

    Any tips/suggestions on how to start this problem?

    EDIT: Disregard the messed up LaTeX there I can't seem to remove it.

    Attached Files:

  2. jcsd
  3. Jun 21, 2010 #2

    if the voltage across 19 ohm resistor(lower one) is greater than 2 v,
    then the mosfet will have gate voltage more than threshold voltage(i.e. 2 v).
    the conduction channel will be created or can be said induced in that nMOS.
    then the current will start to flow from drain to source, and the channel resistance will be, 100 ohm.
    so, the power dissipation needed here is across the nMOS.
    you calculate whether the lower 19 ohm resistor will have voltage drop of 2V or not.
    if it will able to posses 2V(using simple voltage divider rules etc),then the nMOS will be on and then you consider the nMOS as another 100 ohm resistor connected in series with 100 ohm(upper one) and in parallel to lower 19 ohm resistor.
    View attachment nMOS circuit.bmp

    ** the resistor values will be like previous ckt.
    and new one is Ron=100 ohm, obviously.
  4. Jun 21, 2010 #3
    I'm confused on how I would apply the voltage divider rule to a circuit like this. (I don't have any resistors in series?)

    If I apply KVL on the outer loop of the circuit I can solve for [tex]i_{2}[/tex]:

    [tex]KVLouterloop: 100i_{2} + 100i_{2} - 10 = 0 [/tex]

    [tex] i_{2} = \frac{1}{20} A[/tex]

    and applying KVL on the inner loop,

    [tex]KVLinnerloop: 19ki_{3} + 19ki_{3} - 10 = 0 [/tex]

    [tex] i_{3} = 2.63*10^{-4}[/tex]

    So the voltage across the bottom 19k resistor should be as follows:

    [tex] V_{2} = i_{3} * 19k = 5V [/tex]

    So the voltage across the gate is greater than the tolerance voltage so the MOSFET should be on and the voltage across Vo will be the voltage across a 100 ohm resistor.


    [tex] V_{o} = i_{2} * 100 = 5V [/tex]

    So then the power dissapated across should be as follows:

    [tex] P = i_{2} * V_{o} = \frac{1}{20} * 5 = 0.25W [/tex]

    Did I make any mistakes?
  5. Jun 21, 2010 #4


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    Gold Member

    These results look good.

    The voltage divider would have been faster, using the two 19k resistors in series. You could do the maths (or use experience to tell you that a voltage will be divided equally across two equal resistances) and see that the gate voltage is 5V.

    You could then see that since the MOSFET has been turned on, you again have 10V across two equal resistances (see Raj's circuit) and thus have 5V across each one, and so the power through the MOSFET could also be calculated using [tex]P=\frac{{V^2}}{R}[/tex] so you get [tex]P=\frac{{5^2}}{100}[/tex] and thus the same answer P = 0.25W.

    Your method is just as valid.
  6. Jun 21, 2010 #5
    Thanks for confirming.

    My method was somewhat systematic, your method is way more intuitive and simple.

    Thanks for sharing!
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