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Finding the gain of a circuit (Part Three)

  1. Jul 13, 2011 #1
    As previously stated in other threads, I have started to work on a summer assignment for my Fall semester EE course. I was given a circuit and asked to derive an expression for the gain (Attachment 1). I have not yet started to write out the problem on paper, but have two thoughts as to where I need to begin based on referencing an EE book I've had from previous courses (Attachment 2). Right now all I'm asking for is some input on whether my thoughts are correct about where I should begin, or I'm just making things up. Any input would be much appreciated!!

    1. The problem statement, all variables and given/known data

    (Attachment 1)
    Derive an expression for the gain of the small-signal circuit shown.

    2. Relevant equations


    3. The attempt at a solution

    Ok so with looking at the book, it seems to be I need to begin on the right hand side of the circuit for this problem. Now my two thoughts are:
    1. ro and Rs are in parallel/series, which the result is then in parallel with RL
    2. ro and RL are in parallel
    My mind is 95% with my first thought, but the second thought is something that's making me second guess myself.. Any thoughts?
     

    Attached Files:

  2. jcsd
  3. Jul 13, 2011 #2

    gneill

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    Consider replacing the dependent current source and ro with their Thevenin equivalent. Then you should be able to write a loop equation and incorporate the appropriate dependencies. With the loop current you can then determine the output voltage across the load resistor.
     
  4. Jul 13, 2011 #3
    So what you're saying is to pretty much disregard the book example?
     
  5. Jul 13, 2011 #4

    gneill

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    The book example applies an analysis strategy that's appropriate for its given circuit. I'm just suggesting one possible strategy to solve for the output voltage for the different circuit you're given.
     
  6. Jul 13, 2011 #5
    Sounds good, just checking to see if there was any way I could incorporate the book example into my work. I will begin to work on this problem with the suggested route you gave me and show some updates/progress soon!
     
  7. Jul 13, 2011 #6
    Quick question, I asked one of my classmates as to how he went about solving this problem (I've never been the best with Thevenin), and he did it this way:
    1. He first came up with a KCL expression for the current (Equation 1 below)
    2. Then, solved for Vgs (Equation 2 below)
    3. Came up with a KCL expression at the ground (Equation 3 below)
    4. Plugged equation 2 into 1, getting rid of Vgs
    5. Plugged that into equation 3 and solve for Vo/Vsig

    Equation 1: [itex]\frac{V_o}{R_L}[/itex] = gmvgs + [itex]\frac{(Vo-Vs)}{R_o}[/itex]

    Equation 2: vgs = vg - vs = vsig - vs

    Equation 3: [itex]\frac{v_s}{r_s}[/itex] = -[itex]\frac{v_o}{R_L}[/itex]


    Any thoughts to as if this route is correct?
     
    Last edited: Jul 13, 2011
  8. Jul 13, 2011 #7

    gneill

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    Sure, you can use KCL to analyze the circuit. Equation 3 looks suspicious though. It's got a voltage over a resistance ( = current) set equal to a voltage over another voltage (no units).
     
  9. Jul 13, 2011 #8
    My fault, that was a typo on my end, I fixed it. It reads -[itex]\frac{v_o}{R_L}[/itex]
    If this seems fine I may go this route, as I said Thevenin hasn't been my strong point.
     
  10. Jul 13, 2011 #9

    gneill

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    It now looks fine.
     
  11. Jul 14, 2011 #10
    So I gave his route a try due to ease, and have scanned my progress (really do apoligize for the quality, if it's too difficult to read please let me know!).. My problem seems to be I have no idea how to bring Vsig over to make it [itex]\frac{V_o}{Vsig}[/itex]. I know I have to bring ro to the right hand side of the equation, I'll do that. But my thoughts towards the Vsig are adding it to the left side, then making it [itex]\frac{V_o}{Vsig}[/itex] + [itex]\frac{gmVsig}{Vsig}[/itex]... Is this a legal algebraic move? Or does anyone see something that I'm missing with a stupid mistake
     

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  12. Jul 14, 2011 #11

    gneill

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    I have to make a correction to my previous statement where I said that everything looked fine :frown:. Back in post #6 where the KCL current sum is being applied at the Vo node to arrive at equation #1, the terms on the either the LHS or RHS should be negated. In other words,
    [tex] \frac{V_o}{R_L} + \frac{V_o - V_s}{r_o} + g_m v_{gs} = 0 [/tex]

    In your derivation, why not use your equation (3) to find a substitution for vs, and plug THAT into your KCL equation? That would leave you with an equation involving Vo and Vsig only...
     
  13. Jul 14, 2011 #12
    Gotcha, I see what you're saying and took your advice, but now it's getting messy unless I made a mistake which I don't think I did :confused:
     

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  14. Jul 14, 2011 #13

    gneill

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    It might more obvious what you need to do if you leave all the terms in the KCL equation on the same side until later in the process; it'll make it easier to collect terms (vo and vsig terms). Here's what I get from your PDF derivations, with the alterations I've mentioned. It just remains to 'finish off' the algebra:

    attachment.php?attachmentid=37189&stc=1&d=1310663576.gif
     

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    • Fig1.gif
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  15. Jul 14, 2011 #14
    Ok I completely understand where you're coming from now, thanks for the visual :smile: But one thing I have a question on, your last step: Do you not need to have ...+RsRL rather than just Rs? (Within the parenthases)
     
  16. Jul 14, 2011 #15

    gneill

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    I don't believe so...

    attachment.php?attachmentid=37193&stc=1&d=1310671789.gif
     

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  17. Jul 14, 2011 #16
    Gotcha! I got mixed up with the [itex]\frac{R_L}{R_L}[/itex] step and didn't see where you came out with just the Rs :redface: I see now though, thanks for the clarification!
     
  18. Jul 14, 2011 #17
    So after all of the silly mistakes/errors I've made, I finally have found a solution for [itex]\frac{V_o}{Vsig}[/itex]!! Hope it's right....
     

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