Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Another Simple MOSFET Question

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

    2. Relevant equations
    V = IR

    3. The attempt at a solution

    I have a few questions about notation before I begin with an attempt at the problem.

    Are there 2 voltage sources in this circuit diagram? One on the left, one ontop?

    Also it says that Vo = 5v, does that mean they want to me to create this "buffer" so that the voltage measured at the port Vo is 5V?

    Also what do they mean by a buffer?

    I'm pretty confused as to how I should be manipulating the resistance RL, what kind of effect am I looking to create to make a "buffer"?

    Thanks again guys!

    Attached Files:

  2. jcsd
  3. Jun 21, 2010 #2


    User Avatar
    Gold Member

    Yes* (Though a down pointing arrow is sometimes ground, I'm assuming this symbol which I have never seen before is a 5V source, since I get an answer that is one of the five given if I do)


    A buffer is some sort of separation device usually. It has the same value input and output and provides a barrier against an undesired situation occuring.

    You know that the MOSFETS's are turned on by 2V, the MOSFET's can be replaced with a 100Ohm resister if they are turned on, and that the first MOSFET has 5V going into its gate.

    This implies the first MOSFET can be replaced with a 100Ohm resistor. To get 5V at the output the second MOSFET needs to be turned on though. How will you do that?

    * (Given my assumption that the up pointing arrow is a 5V source, leaving the second MOSFET off would result in no voltage drop across the 1kOhm resister and therefore 5V at the output ... can anyone comment on that?)
    Last edited: Jun 21, 2010
  4. Jun 21, 2010 #3

    No, by buffer they are referring to the whole circuit. When the circuit is operating properly the output voltage will be 5 V.

    All they want you to do is find the minimum value of resistance such that the voltage at the gate of the last transistor will result in an output of 5 V at the drain. What are the conditions that need to be satisfied to get 5V on the output?
  5. Jun 21, 2010 #4
    Does my circuit look like what I've drawn? (see figure)

    If I want to get 5V on the output I need the second MOSFET turned off correct? Otherwise Vo would read 0V, no?

    I need RL to have the voltage drop of atleast 3.1V so that then second MOSFET won't turn on, correct?

    EDIT: Disregard the KVL equation in the picture.

    Attached Files:

  6. Jun 21, 2010 #5


    User Avatar
    Gold Member

    If the second MOSFET is off, you get no drop across the 1k resistance and Vo = 5V from the source, but that begs the question of why have the circuit at all if youre just having a source provide 5V?

    The idea of a buffer is to have an input and an output, so if you turn both MOSFET's on, you get this input and output, however the voltage divider on the output has 10/11*5V dropped across the 1k resistance so Vo = 0.45V?

    What does an up pointing arrow mean, and why is it different from the voltage source picture on the input (standard circle with polarity)?
  7. Jun 22, 2010 #6
    Anyone else care to shed further insight?

    Why is there another voltage source simply feeding the output 5V?

    How does manipulating the resistor RL effect the 2nd voltage source?
  8. Jun 22, 2010 #7
    Zryn, the arrow is simply a voltage source just like the circle, its just another notation.
    Its polarity is given by either positive or negative, in this case positive 5V.

    jegues, you seem to be right the MOSFET on the right must be off, therefore the voltage of the gate of the second MOSFET must be less than 2V. Now remember Von is Vgs, and the source is at 0V therefore the gate must be less than 2V.

    Doing a simple voltage divider.
    2=5*(100/(RL+100)), we find that RL=150 ohms.
    I hope this helps.
  9. Jun 22, 2010 #8
    This all makes perfect sense, the only trouble I'm having is seeing how the 100ohm resistor from the first MOSFET is in parallel with RL.

    Does the circuit look like the figure I've drawn? Does current run through the red wire even though it's a short? Wouldn't all the current just run down the red wire and completely skip over the other black wire containing RL?

    Hopefully you can help me clear up any misconceptions I still have.

    Thanks again!

    Attached Files:

  10. Jun 22, 2010 #9
    Ok jegues, let’s take it one step at a time. Remember that a MOSFET is basically a switch is going to be either ON or OFF. From previous analysis we have concluded that the MOSFET on the left will be ON while the one on the right will be OFF. Taking this into account, I have provided an equivalent circuit which is attached. The mistake you made is that you included the 5 volt source for the first MOSFET. This is not correct because this voltage simply determines whether the MOSFET is ON or OFF. Once the MOSFET is ON then the drain and source become “one wire”. Looking at the equivalent circuit we see that RL and the 100 ohm resistance are in series. We know we want less than 2V at the top of the 100 ohm resistor because this is where the gate of the MOSFET is being fed. Now we simply perform the voltage divider 2=5*(100/(RL+100)). Also, the right MOSFET can be modeled as an open, since we are designing it to be OFF.

    The current would not run down the red wire you draw simply because this is NOT a short. As a matter of fact the key characteristics of MOSFET’s is that gate current is very near 0 amps.

    Attached Files:

  11. Jun 22, 2010 #10
    Thank you your post has addressed my confusion. I thought in order to create a voltage divider it required you to have one resistor in series, and the other in parallel?

    This isn't the case right? You seemed to have created a voltage divider with two resistors that are in series!
  12. Jun 22, 2010 #11


    User Avatar
    Gold Member

    Given that RL = 150 ohms, the gate of the second MOSFET will have 2V going into it and thus turn on ... so if you want it to be turned off the minimum from your options would be the next thing larger than 150 ohms, i.e. 500 ohms.

    But whats the point of that circuit, an academic theory tester? It's been labeled as a buffer.

    I think this question is dodgy.

    *Resistors in parallel have the same voltage across them. Resistors in series have the voltage divided between them by the ratio of their resistances. Can you draw a picture of one resistor in series and the other in parallel out of curiosity?
    Last edited: Jun 22, 2010
  13. Jun 22, 2010 #12
    You're correct, the answer must be 500 ohms.

    I was confusing myself earlier, voltage dividers are indeed made with two resistors in series. (Not one in parallel and one in series)

    If I have 5 resistors in series and I want to apply voltage divider across the first two, can I do so without any discrepencies? Or do I need to reduce the circuit to a maximum of 2 resistors in series to use the voltage divider technique?
  14. Jun 22, 2010 #13
    You will need to reduce the circuit to 2 resistors before doing the voltage divider.
  15. Jun 22, 2010 #14


    User Avatar
    Gold Member

    The standard voltage divider equation is: V=[tex]\frac{V_{s}*R_{1}}{R_{1}+R_{2}}[/tex]

    What it actually represents is: V=[tex]\frac{V_{s}*R_{desired}}{R_{desired}+R_{remaining}}[/tex]

    So if you have 5 resistors in series, you can find the voltage drop across the 3rd one by using the value of the 3rd resistor as [tex]R_{desired}[/tex] and the sum of all the remaining resistors as [tex]R_{remaining}[/tex].

    Alternatively, yes, you can reduce the first two resistors down to one value by summing them (call this value [tex]R_{1}[/tex]), and then reduce the remaining three down to another value by summing them (call this value [tex]R_{2}[/tex]) then use the initial formula to find the voltage dissipated across each resistor section.
  16. Jun 22, 2010 #15
    Thank you both for clearing things up with me.

    It all makes sense now!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook