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Finding Thevenin Equivalent Circuit

  1. Jan 21, 2008 #1
    Hello,

    I am working on the following problem:

    [​IMG]

    I am stuck on this seemingly simple circuit problem. I've worked many Thevenin circuit problems in my introductory circuit analysis. This problem is from a new class and seems different. I realize that I must first find the open-circuit voltage at the terminals. After that I need to the Thevenin equivalent resistance by using

    [tex]\LARGE R_{Th}=V_{Th}/I_{sc}[/tex].

    First let me find the open-circuit voltage [tex]v_{oc}[/tex]. Is this correct?

    We know the circuit current is [tex]g_mv[/tex]. Thus the open circuit voltage which is the voltage across the voltage-controlled current source is

    Using KVL:

    [tex]\LARGE -v_s+v-v_{oc}=0 \implies v_{oc}=v-v_s[/tex]

    and

    [tex]\LARGE v=(g_mv)R_1[/tex].

    So we have

    [tex]
    \LARGE v_{oc} = (g_mv)R_1-v_s \implies
    \LARGE v_{oc} = (0.002v)(50k\Omega)-v_s \implies
    \LARGE v_{oc} = 100v-v_s
    [/tex]

    Does this look correct for the open circuit voltage?
     
    Last edited: Jan 21, 2008
  2. jcsd
  3. Jan 21, 2008 #2

    mjsd

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    V_oc = voltage across controlled current source, and assuming +polarity on the arrow-head end, you then have V_oc = Vs - V = Vs - IR where I is given by -gm V (passive ref scheme)
     
  4. Jan 22, 2008 #3
    Does this solution look correct? I see where I was wrong while finding the open-circuit voltage.

    [​IMG]

    [​IMG]

    [​IMG]
     
  5. Jan 22, 2008 #4

    mjsd

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    judging by the final diagram with + on top and - on bottom at output, your short circuit current may have been defined in the wrong direction, hence leading to an overall -ve sign error.
     
  6. Jan 23, 2008 #5

    CEL

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    The current [tex]g_mv[/tex] is entering the minus terminal of [tex]v[/tex], so you should have [tex]v = -g_mvR_1[/tex] or [tex]v\left[1+g_mR_1]=0[/tex] so, [tex]v = 0[/tex] and [tex]v_{oc}=v_s[/tex]
     
  7. Jan 23, 2008 #6
    How did you get this? Why isn't it

    [tex]v_{oc}=v_s-v[/tex] ?

    Thanks
     
  8. Jan 23, 2008 #7

    CEL

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    It is! But since [tex]v=0[/tex], [tex]v_{oc}=v_s-0=v_s[/tex].
     
  9. Jan 23, 2008 #8

    mjsd

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    your expression is pretty much correct, all you need is to look at the limit as v->0. basically R_th has to be infinite..... such equivalent circuit does not exist in real life, only in the realm of an analysis tool.
     
  10. Jan 23, 2008 #9
    I'm still confused. Why is [tex]v=0[/tex] and not left as [tex]v=-g_mvR_1[/tex]?

    Why do we need to look at the limit of [tex]v[/tex] as [tex]v \rightarrow 0[/tex] ?

    Correction: limit of [tex]R_{Th}[/tex] as [tex]v \rightarrow 0[/tex]
     
    Last edited: Jan 23, 2008
  11. Jan 24, 2008 #10

    CEL

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    You have [tex]v[/tex] in both members of the equation [tex]v=-g_mvR_1[/tex]. So, unless [tex]g_mR_1=1[/tex], which it is not, you must have [tex]v=0[/tex].
    Short circuiting the output you have [tex]i_{sc}=\frac{v_s}{R_1}+g_mv[/tex].
    Since [tex]v=v_s[/tex], [tex]i_{sc}=\frac{v_s}{R_1}+g_mv_s=v_s\left[\frac{1}{R_1}+g_m\right][/tex]
    So
    [tex]R_{th}=\frac{v_{oc}}{i_{sc}}=\frac{1}{\frac{1}{R_1}+g_m}=\frac{R_1}{1+g_mR_1}[/tex].
     
  12. Jan 24, 2008 #11
    Thanks CEL and mjsd for all of the help on this problem. It now makes sense. I found out the answer is R=495 ohms and vth=vs. This is definately the correct answer when I substitute the given values into the problem. Thanks again!
     
  13. Jan 24, 2008 #12

    mjsd

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    ok... i shouldn't have said R_th is infinite.. I evaluated the limit incorrectly.. since vs = v,
     
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