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Homework Help: Electrical Engineering - FET Common Source Configuration with Load

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data


    I'm given this configuration with a AC source at [itex]v_{i}[/itex] with a [itex]R_{sig}[/itex] connect to the source before the capacitor.

    [itex]R_{sig} = 0.6 KΩ[/itex]
    [itex]R_{G} = 1 MΩ[/itex]
    [itex]R_{D} = 2.7 KΩ[/itex]
    [itex]R_{L} = 4.7 KΩ[/itex]
    [itex]I_{DSS} = 10 mA[/itex]
    [itex]v_{p} = -6 V[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I was able to find [itex]z_{i}[/itex] and [itex]z_{o}[/itex] very easily. I'm trying to find [itex]A_{v}[/itex] and was able to find the formula [itex]A_{v} = g_{m}(R_{D}||R_{L})[/itex]. The only problem is that I don't know how to find [itex]g_{m}[/itex]. I know that [itex]g_{m} = \frac{2I_{DSS}}{|v_{P}|}(1 - \frac{v_{GS}}{v_{p}})[/itex]. I'm not so sure how to find [itex]v_{GS}[/itex]. I know that [itex]v_{GS} = v_{G} - v_{S}[/itex] and that [itex]v_{S} = 0[/itex] so this comes down to finding [itex]v_{G}[/itex]. I seem to be having some problems doing this. I now that [itex]v_{G} = \frac{R_{G}v_{I}}{R_{sig} + R_{G}}[/itex] but this doesn't seem to really help since I don't know the AC input [itex]v_{I}[/itex]. Thanks for any help.
  2. jcsd
  3. Nov 8, 2013 #2


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    What do you want to calculate, and what are all those variables?

    VG will depend on VI, and I think you will have to treat this as unknown input. Just calculate everything in terms of an unknown VI (if that is too hard in general, it might be interesting to consider a sine wave).
  4. Nov 8, 2013 #3
    Here's the actual picture from my book.

    http://imageshack.com/a/img600/8114/gz12.png [Broken]

    For part (a) I'm asked to determine [itex]A_{v_{NL}}[/itex], [itex]Z_{i}[/itex], and [itex]Z_{o}[/itex].

    Finding [itex]Z_{i}[/itex], and [itex]Z_{o}[/itex] isn't a problem I just have a problem trying to find [itex]A_{v_{NL}}[/itex]

    Below is my small signal equivalent circuit

    http://imageshack.com/a/img21/4747/y59u.png [Broken]

    What I get for the [itex]A_{v_{NL}}[/itex] is

    http://imageshack.com/a/img268/4091/f6ze.png [Broken]

    Which I don't see how it's wrong. My professor some how gets a value. He doesn't really show how he got his values but

    *edit* I realize that I forgot [itex]v_{p}[/itex] in my equation for [itex]g_{m}[/itex] and the negative sign in my equation for [itex]A_{v_{NL}}[/itex] which I'm adding now.

    http://imageshack.com/a/img713/7876/gi2i.png [Broken]

    I'm not sure how he got [itex]A_{v_{NL}}[/itex]

    I see how he gets [itex]g_{mo}[/itex]

    [itex]g_{mo} = \frac{2I_{DSS}}{|v_{p}|} = \frac{2(10 mA)}{6 V} ≈ 3.333 mS[/itex]

    I also know that

    [itex]g_{m} = g_{mo}(1 - \frac{v_{GS}}{v_{P}})[/itex]

    But I don't see how he got a value for [itex]g_{m}[/itex] because we don't know what [itex]v_{GS}[/itex] is other than [itex]v_{GS} = \frac{R_{G}v_{S}}{R_{sig} + R_{G}}[/itex] which doesn't really help because we don't know [itex]v_{s}[/itex] which leads me to believe that there must be some other way to find [itex]v_{GS}[/itex] that I'm not seeing.

    Oh apparently he's finding [itex]A_{v_{NL}} = \frac{v_{o}}{v_{s}}[/itex]
    Last edited by a moderator: May 6, 2017
  5. Nov 8, 2013 #4
    Oh apparently he's finding [itex]A_{v_{NL}} = \frac{v_{o}}{v_{s}}[/itex]

    which leaves me with this instead.

    http://imageshack.com/a/img811/3217/naxp.png [Broken]

    but still not able to estimate because we don't know [itex]v_{s}[/itex]

    Last edited by a moderator: May 6, 2017
  6. Nov 8, 2013 #5

    rude man

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    AVNL sounds like "no-load voltage gain". That would mean not including RL in your gain computation.
  7. Nov 8, 2013 #6
    I agree, but I am still left with the problem of not knowing how what [itex]v_{s}[/itex]. Taking what you say into consideration I get

    [itex]A_{v_{NL}} = -\frac{2I_{DSS}R_{D}}{|v_{P}|(R_{G} + R_{sig})}(1 - \frac{R_{G}v_{s}}{v_{p}(R_{G} + R_{sig})})[/itex]

    I do I go about getting around this?

    This is how I got that formula

    http://img89.imageshack.us/img89/8845/ls8c.png [Broken]

    Thanks for the help.
    Last edited by a moderator: May 6, 2017
  8. Nov 9, 2013 #7

    rude man

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    JFET circuit

    Haven't we discussed this before? You get VS by equating the expression for FET current to the current flowing thru RS. I thought you said you understood this.

    I also note that you are still using upper case vs. lower case in a confusing manner. For example, [itex]g_{m} = \frac{2I_{DSS}}{|v_{P}|}(1 - \frac{v_{GS}}{v_{p}})[/itex], VGS and VP are both dc quantities and so should be capitalized. Use vgs for ac signals. I found this problem in several of your equations.

    What BTW is gm0? How's it different from gm? There is only one value of gm unless you change the circuit.

    I think I'll let others take a shot at this since I feel you & I have already gone over this ground pretty thoroughly in another similar problem. Or am I mistaking you for someone else - that's happened too ...
    Last edited: Nov 9, 2013
  9. Nov 9, 2013 #8
    Well [itex]v_{s}[/itex] in this problem though is the input AC voltage. I know that [itex]i_{D} = i_{S}[/itex] and that I can find the voltage at the source of the transistor this way using the equation for [itex]i_{D}[/itex] for a FET, also symbolized by [itex]v_{S}[/itex] I do understand this process now. The only problem is that [itex]v_{s}[/itex] in my equation for [itex]V_{GS}[/itex] isn't the voltage at the source of the transistor but the input AC voltage to the circuit, which isn't given.

    In this problem I take [itex]v_{S}[/itex] the voltage at the source of the transistor to be zero since in the small signal equivalent it's connected to ground.

    I'll be sure to use capital values for [itex]V_{GS}[/itex] and [itex]V_{P}[/itex] since they aren't AC parameters.

    To me [itex]g_{mo}[/itex] is kind of silly. My book also defines
    [itex]g_{m} = g_{mo}(1 - \frac{V_{GS}}{V_{P}})[/itex]
    [itex]g_{mo} = \frac{2I_{DSS}}{|V_{P}|}[/itex]
    which would be the value of [itex]g_{m}[/itex] when [itex]V_{GS}[/itex] is zero
  10. Nov 9, 2013 #9

    rude man

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    That is correct. So that makes the problem extremely simple. So vgs in your equivalent circuit is just vg which is just a voltage divider away from vin. You should not have labeled the input Vs or vs. That's very confusing. Label it vin.

    So now surely you can solve for vo. What did you compute for Vs and gm?
  11. Nov 9, 2013 #10
    Ya I established that.

    so can I do this then.

    For FET
    [itex]i_{D} = i_{S}[/itex] [1]
    Here I use
    [itex]I_{D} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}[/itex]
    [itex]i_{S} = g_{m}V_{GS}[/itex] and plug these into [1]
    [itex]I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2} = g_{m}V_{GS}[/itex] [2]
    I use this formula for [itex]g_{m}[/itex] and plug into [2]
    [itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{v_{GS}}{V_{P}})[/itex]
    [itex]I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})V_{GS}[/itex][3]
    Here I try and find [itex]V_{GS}[/itex]
    [itex]V_{GS} = V_{G} - V_{S}[/itex] [4]
    I know that
    [itex]V_{S} = 0 V[/itex]
    so this simplifies [4] to this
    [itex]V_{GS} = V_{G}[/itex] [4]
    I apply a voltage divider to find [itex]v_{G}[/itex] and get this formula
    [itex]V_{G} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex]
    I plug this into [4]
    [itex]V_{GS} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex] [4]
    I plug this into [3]
    [itex]I_{DSS}(1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})})^{2} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})})\frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex] [3]
    [itex]1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})} = \frac{2}{|V_{P}|}\frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex]
    move second term on LHS to RHS
    [itex]1 = \frac{2}{|V_{P}|}\frac{v_{in}R_{G}}{R_{sig} + R_{G}} + \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})}[/itex]
    [itex]1 = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}(\frac{2}{|V_{P}|} + \frac{1}{V_{P}})[/itex]
    solve for [itex]v_{in}[/itex]
    [itex]\frac{R_{sig} + R_{G}}{R_{G}}\frac{1}{\frac{2}{|V_{P}|} + \frac{1}{V_{P}}} = v_{in}[/itex]
    [itex](\frac{R_{sig}}{R_{G}} + 1)\frac{1}{\frac{2}{|V_{P}|} + \frac{1}{V_{P}}} = v_{in}[/itex]
    plug in values
    [itex](\frac{0.6 KΩ}{1 MΩ} + 1)\frac{1}{\frac{2}{|-6 V|} - \frac{1}{6 V}} = v_{in}[/itex]
    [itex](6x10^{-4} + 1)\frac{1}{\frac{2}{6 V} - \frac{1}{6 V}} = v_{in}[/itex]
    [itex](6x10^{-4} + 1)\frac{1}{\frac{1}{6 V}} = v_{in}[/itex]
    [itex](6x10^{-4} + 1)6 V = v_{in}[/itex]
    [itex]6.0036 V = v_{in}[/itex]

    Does this look better for [itex]v_{in}[/itex]? If so from here it's easy.

    Now that I have [itex]v_{in}[/itex] I solve for [itex]V_{GS} = V_{G}[/itex]
    [itex]v_{G} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}} = \frac{(6.0036 V)(1 MΩ)}{0.6 KΩ + 1 MΩ} = 6 V[/itex]
    I plug this into the equation for [itex]g_{m}[/itex]
    [itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})^{2} = \frac{2(10 MΩ)}{6 V}(1 + \frac{6 V}{6 V})^{2} = \frac{1}{3000}(1 + 1)^{2} = \frac{1}{3000}2^{2} = \frac{4}{3000} ≈ 1.333 mS[/itex]

    This isn't what my professor got for the supposed solution.

    *edit* I added in some explanations.
    Last edited: Nov 9, 2013
  12. Nov 9, 2013 #11

    rude man

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    vin is a given. It can be anything you like - within limits.

    You are supposed to solve for vo, given vin. AvNL = vo/vin with RL removed.

    I don't understand your equations.

    I didn't see numbers for gm and Vs.
  13. Nov 9, 2013 #12
    Wait really I can make up [itex]v_{in}[/itex] to be anything I want? So I can just go with one volt? That seems weird, I was able to solve for it some how.

    By [itex]v_{s}[/itex] you mean the the AC voltage? It's just zero because it's connected to ground in the small signal equivalent circuit.
  14. Nov 9, 2013 #13

    rude man

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    You need to understand what is meant by "gain". Gain = v_out/v_in. That ratio is independent of v_in.

    I did not say v_s. I said V_s.
  15. Nov 9, 2013 #14
    Gain is the output voltage divided by the AC input? I solved for the AC input voltage and got 6.0036 volts. Oh I need the DC voltage V_S and use that value for V_GS


    But wait,

    In order to find [itex]V_{G}[/itex] I just use the DC schematic diagram and get

    [itex]V_{G} = \frac{R_{G}v_{in}}{R_{sig} + R_{G}}[/itex]

    So I don't know [itex]v_{in}[/itex] but I should be able to solve without hm.
  16. Nov 9, 2013 #15

    rude man

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    Vin = 0.
    You need VGS to solve for gm. You know VG = 0.

    Equate the expression for the dc FET current to the dc drain current to get VS.

    Then use your formula for gm.
  17. Nov 9, 2013 #16
    This is starting to make sense, but why do you take Vin = 0?
  18. Nov 9, 2013 #17
    Why can you let [itex]v_{in}[/itex] be anything you want?
  19. Nov 9, 2013 #18
    [itex]A_{V_{NL}} = \frac{v_{O}}{v_{in}}[/itex]
    [itex]v_{O} = -g_{m}V_{GS}R_{D}[/itex]
    [itex]A_{V_{NL}} = -\frac{g_{m}V_{GS}R_{D}}{v_{in}}[/itex]
    [itex]V_{GS} = V_{G} - V_{S}[/itex]
    [itex]V_{G} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}}[/itex]
    [itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - V_{S}[/itex]
    [itex]V_{S} = I_{S}R_{S}[/itex]
    [itex]I_{S} = I_{D} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}[/itex]
    [itex]V_{S} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}[/itex]
    [itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}[/itex]
    [itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{2V_{GS}}{V_{P}} + \frac{V_{GS}^{2}}{V_{P}^{2}})R_{S}[/itex]
    [itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}}[/itex]
    [itex]0 = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}} - V_{GS}[/itex]
    [itex]0 = -\frac{R_{S}I_{DSS}}{V_{P}^{2}}V_{GS}^{2} + (\frac{2I_{DSS}R_{S}}{V_{P}} - 1)V_{GS} + \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S}[/itex]
    [itex]V_{GS} = \frac{-(\frac{2I_{DSS}R_{S}}{V_{P}} - 1) \underline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{2(-\frac{R_{S}I_{DSS}}{V_{P}^{2}})}[/itex]
    [itex]V_{GS} = \frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}}[/itex]
    [itex]A_{V_{NL}} = -\frac{g_{m}(\frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}})R_{D}}{v_{in}}[/itex]
    [itex]A_{V_{NL}} = -\frac{g_{m}(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}}[/itex]
    [itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})[/itex]
    [itex]A_{V_{NL}} = -\frac{2I_{DSS}(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}|V_{P}|}[/itex]
    [itex]A_{V_{NL}} = -\frac{(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}}{v_{in}R_{S}}[/itex]
    etc... this problem sucks. I'm not sure how to solve this problem when I don't know [itex]v_{in}[/itex]. The previous problem, this information was given.
  20. Nov 10, 2013 #19

    rude man

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    Because that is the dc operating point of your circuit.
  21. Nov 10, 2013 #20

    rude man

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    You have an amplifier. It has a fixed gain vout/vin. And you can make vin anything you want. Otherwise it's not an amplifier.

    I am signing off this thread, sorry.
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